This document discusses Z-transforms and their properties. It begins by defining a Z-transform as a mathematical operation that transforms a discrete function into another function using a Z-transform operator and parameter z. It then outlines several key properties of Z-transforms, including linearity, damping, left and right shifting properties. The document also provides the Z-transform definitions and formulas for standard functions like 1.
The document discusses the one-sided z-transform, which is used to solve difference equations with initial conditions. It defines the one-sided z-transform and explains that it contains information only about causal signals that are zero for negative time values. Some key properties are described, including shifting properties for time delay and advance, and the final value theorem. An example shows how to use the one-sided z-transform to solve a simple difference equation.
This document provides an overview of the z-transform and its properties. It defines the z-transform as a power series representation of a discrete-time signal and explains that it exists only when the series converges within a region of convergence (ROC). Several examples of calculating the z-transform of finite and infinite duration signals are shown along with the corresponding ROCs. Common methods for taking the inverse z-transform like synthetic division, partial fraction expansion, and Cauchy's integration method are also briefly mentioned.
The z-transform provides a third domain (besides time and frequency domains) to represent discrete-time signals and systems. It transforms a sequence into a complex function of z by taking the sum of the sequence multiplied by powers of z. The z-transform has useful properties like superposition and time-delay that allow convolutions to be represented as multiplications in the z-domain. Cascading systems in the time-domain corresponds to multiplying their z-transforms. Factoring z-transforms splits systems into subsystems, related to the roots of the z-transform polynomial.
This document provides information about the Digital Signal Processing course EC362/CE362 at Pharos University in Alexandria, Egypt. It includes the contact details for the course instructor Dr. Heba Raafat and course assistant. It outlines the schedule, topics to be covered in the course, assessment criteria, textbooks and references. The course aims to provide students with fundamental knowledge of digital signal processing and explore applications. Key topics include Z-transforms, discrete Fourier transforms, digital filters and their design. Assessment is based on a final exam, midterm, lab works, presentations and quizzes.
The document provides solutions to problems from a discrete-time signal processing textbook. It includes:
1) Solutions to convolution problems graphically representing signals and their convolution.
2) Derivations of impulse responses from difference equations and system functions using the z-transform.
3) Analyses of signals as eigenfunctions of linear time-invariant systems.
The document provides solutions to problems from a discrete-time signal processing textbook. It includes:
1) Solutions to convolution problems graphically representing signals and their convolution.
2) Derivations of impulse responses from system functions using the z-transform.
3) Analyses of signals as eigenfunctions and determining if systems are linear and time-invariant.
4) Solutions involving filtering, modulation, and determining system properties from inputs and outputs.
The document provides solutions to problems from a discrete-time signal processing textbook. It includes:
1) Solutions to convolution problems graphically representing signals and their convolution.
2) Derivations of impulse responses from system functions using the z-transform.
3) Analyses of signals as eigenfunctions and determining if systems are linear and time-invariant.
4) Solutions involving filtering, modulation, and determining system properties from input-output relations.
The document discusses the one-sided z-transform, which is used to solve difference equations with initial conditions. It defines the one-sided z-transform and explains that it contains information only about causal signals that are zero for negative time values. Some key properties are described, including shifting properties for time delay and advance, and the final value theorem. An example shows how to use the one-sided z-transform to solve a simple difference equation.
This document provides an overview of the z-transform and its properties. It defines the z-transform as a power series representation of a discrete-time signal and explains that it exists only when the series converges within a region of convergence (ROC). Several examples of calculating the z-transform of finite and infinite duration signals are shown along with the corresponding ROCs. Common methods for taking the inverse z-transform like synthetic division, partial fraction expansion, and Cauchy's integration method are also briefly mentioned.
The z-transform provides a third domain (besides time and frequency domains) to represent discrete-time signals and systems. It transforms a sequence into a complex function of z by taking the sum of the sequence multiplied by powers of z. The z-transform has useful properties like superposition and time-delay that allow convolutions to be represented as multiplications in the z-domain. Cascading systems in the time-domain corresponds to multiplying their z-transforms. Factoring z-transforms splits systems into subsystems, related to the roots of the z-transform polynomial.
This document provides information about the Digital Signal Processing course EC362/CE362 at Pharos University in Alexandria, Egypt. It includes the contact details for the course instructor Dr. Heba Raafat and course assistant. It outlines the schedule, topics to be covered in the course, assessment criteria, textbooks and references. The course aims to provide students with fundamental knowledge of digital signal processing and explore applications. Key topics include Z-transforms, discrete Fourier transforms, digital filters and their design. Assessment is based on a final exam, midterm, lab works, presentations and quizzes.
The document provides solutions to problems from a discrete-time signal processing textbook. It includes:
1) Solutions to convolution problems graphically representing signals and their convolution.
2) Derivations of impulse responses from difference equations and system functions using the z-transform.
3) Analyses of signals as eigenfunctions of linear time-invariant systems.
The document provides solutions to problems from a discrete-time signal processing textbook. It includes:
1) Solutions to convolution problems graphically representing signals and their convolution.
2) Derivations of impulse responses from system functions using the z-transform.
3) Analyses of signals as eigenfunctions and determining if systems are linear and time-invariant.
4) Solutions involving filtering, modulation, and determining system properties from inputs and outputs.
The document provides solutions to problems from a discrete-time signal processing textbook. It includes:
1) Solutions to convolution problems graphically representing signals and their convolution.
2) Derivations of impulse responses from system functions using the z-transform.
3) Analyses of signals as eigenfunctions and determining if systems are linear and time-invariant.
4) Solutions involving filtering, modulation, and determining system properties from input-output relations.
This document contains lecture notes for Engineering Mathematics - I. It covers topics in differential calculus including differential calculus I, II, and III. For each unit, it provides introductions to concepts and worked examples of finding derivatives of various functions up to the nth order. These include standard functions like exponential, trigonometric, logarithmic, and algebraic functions. Formulas for derivative rules like the product rule, quotient rule, and chain rule are also stated.
This document provides an introduction to the quantum theory of angular momentum. It begins by discussing how angular momentum arises in quantum mechanics when moving from one-dimensional to three-dimensional systems. It then defines the components of angular momentum as linear operators and establishes their commutation relations. Next, it introduces raising and lowering operators and shows how they relate to the components. It also proves several theorems, such as that the square of the total angular momentum commutes with its individual components. Finally, it discusses how angular momentum eigenstates can be rotated using rotation operators.
This document contains a chapter on complex numbers from an Oxford textbook. It includes examples of adding, multiplying, dividing and simplifying complex numbers. It also covers topics like modulus, argument and solving complex number equations. Several worked examples are provided with step-by-step solutions.
The document discusses various 2D and 3D transformations including translation, scaling, rotation, reflection, shearing, and homogeneous coordinates. It provides the mathematical definitions and matrix representations for each transformation type in 2D and 3D. It also covers topics like composition and inverse of transformations, classification of transformations, and properties of rigid body transformations.
The document provides solutions to questions from an IIT-JEE mathematics exam. It includes 8 questions worth 2 marks each, 8 questions worth 4 marks each, and 2 questions worth 6 marks each. The solutions solve problems related to probability, trigonometry, geometry, calculus, and loci. The summary focuses on the high-level structure and content of the document.
Higher order derivatives for N -body simulationsKeigo Nitadori
This document discusses higher order derivatives that are useful for N-body simulations. It presents formulas for calculating higher order derivatives of power functions like y=xn, and applies this to derivatives of gravitational force f=mr-3. Specifically:
1) It derives recursive formulas for calculating higher order derivatives of power functions y=xn in terms of previous derivatives.
2) It applies these formulas to calculate derivatives of the gravitational force f=mr-3 in terms of derivatives of r and q=r-3/2.
3) It also describes an alternative approach by Le Guyader (1993) for calculating derivatives of r and q in terms of dot products of r with itself.
The document is an index for a course on complex functions and conformal mapping. It outlines 31 methods that will be covered across 5 units. Unit 1 focuses on complex functions and conformal mapping, exploring methods for basic examples, square roots of complex numbers, nth roots, trigonometric functions of complex numbers, logarithms, differentiability, analyticity, and more. Subsequent units cover complex integrals, sequences and series, Laurent series and residues, and first and higher order partial differential equations. Examples are provided for many of the methods throughout the document.
The document discusses ordinary differential equations (ODEs) and methods for solving separable differential equations. It defines ODEs, order, and degree of differential equations. It then introduces the method of separating variables for solving separable differential equations. This method involves rearranging the differential equation so that the variables are separated on each side. The document provides examples of using this method to solve three separable differential equations, finding the integral relations between the variables in each case.
The document provides an overview of the theory of complex numbers. It defines key concepts such as the Argand diagram, polar and rectangular forms, and operations involving complex numbers like addition, subtraction, multiplication, and division. It also discusses applications of complex numbers to electrical circuits and provides sample exercises for working with complex numbers in different forms and operations.
This document provides an overview of the algebra of complex numbers. It begins with an introduction to complex numbers, defining them as numbers of the form a + bi, where a is the real part and b is the imaginary part. It then covers operations with complex numbers like addition, multiplication, and division. Next, it discusses the Argand diagram, which represents complex numbers graphically as points in a plane. It also defines the argument and modulus of a complex number. The document aims to outline the key concepts and operations involving complex numbers.
Engineering Maths N4
Rectangular Form
Polar Form
Relationship between Polar and Rectangular Complex NUmbers
FET College Registrations in Johannesburg Cbd
November 2015 Exams Registrations
Enrolment for Fet College Exams
Distance Learning
Correspondence
Unisa Tutorials
Extra Lessons
E Learning
Engineering Certificates
Engineering Diplomas
Business Certificates
Phone 073 090 2954
Fax 086 244 2355
Email topstudentz017@gmail.com
This document contains notes from a signals and systems chapter about the z-transform and representing discrete-time systems using difference equations. It defines the z-transform, provides examples of elementary z-transforms, and discusses using the z-transform to analyze discrete-time linear systems. It also describes representing the input-output relationship of a discrete-time system using a linear constant-coefficient difference equation and gives examples of problems involving determining z-transforms and deriving difference equations from transfer functions.
- The z-transform is a mathematical tool that converts discrete-time sequences into complex functions, analogous to how the Laplace transform handles continuous-time signals.
- Key properties and sequences that are transformed include the unit impulse δn, unit step un, and geometric sequences an.
- The z-transform is computed by taking the z-transform definition, which is an infinite summation, and obtaining closed-form expressions using properties like linearity and geometric series sums.
- Common transforms include U(z) for the unit step, 1/1-az^-1 for geometric sequences an, and expressions involving z, sinh/cosh, and sin/cos for exponential and trigonometric sequences.
Important formulas for JEE Main 2020 - Subject-wise ListKanhaMalik
This document provides important formulas for the JEE Main 2020 exam across various subjects - mathematics, physics, and chemistry. In mathematics, formulas are provided for topics like roots of unity, complex numbers, trigonometry, quadratic equations, and vectors. In physics, formulas cover topics such as electricity, magnetism, motion, gravitation, work, and friction. Formulas in chemistry include those for temperature conversion, molarity, molecular mass, atomic structure, gas laws, thermodynamics, and electrochemistry. Mastering these essential formulas can help JEE aspirants save time during the exam, better prepare, and reduce mistakes.
Z TRANSFORM PROPERTIES AND INVERSE Z TRANSFORMTowfeeq Umar
The document discusses various methods for computing the inverse z-transform including inspection, partial fraction expansion, and power series expansion. It provides examples to illustrate each method. The inverse z-transform finds the original time domain sequence from its z-transform. Key properties like linearity, time shifting, and convolution are also covered.
Here are the key steps to solve this problem:
1. Write the rotation matrix for 45 degree rotation about the x-axis:
R = [1 0 0]
[0 cos(45) -sin(45)]
[0 sin(45) cos(45)]
2. Write the scaling matrix with factors of 4 in each direction:
S = [4 0 0]
[0 4 0]
[0 0 4]
3. Pre-multiply the original coordinates by the rotation matrix, then scale the results by pre-multiplying the rotation matrix by the scaling matrix:
A' = SR(0,1,0)
= S(0,0.
This document discusses the discrete Fourier transform (DFT) and its inverse, the inverse discrete Fourier transform (IDFT). It defines the DFT and IDFT formulas, introduces twiddle factors, and shows how to represent DFTs and IDFTs using matrices. Examples are provided to calculate DFTs and IDFTs both directly from the definition and using matrix representations.
1. The document discusses using Z-transforms to solve linear difference equations with constant coefficients.
2. It provides the working procedure which involves taking the Z-transform of both sides of the difference equation, rearranging to isolate the Z-transform of the unknown function U(z), and then taking the inverse Z-transform to find the solution in terms of n.
3. As an example, it shows the step-by-step solution of the difference equation un+2 - 2un+1 + un = 3n + 5 using this method. The solution is found to be 1/2n(n-1)(n+3) + c0 + (c1-c0)n
Fast and efficient exact synthesis of single qubit unitaries generated by cli...JamesMa54
The document describes a presentation on an algorithm for exact synthesis of single qubit unitaries generated by Clifford and T gates. The algorithm reduces the problem of implementing a unitary to the problem of state preparation. It then uses a series of HT gates to iteratively decrease the smallest denominator exponent of the state entries until it reaches a base case that can be looked up. The algorithm runs in time linear in the initial smallest denominator exponent and provides an optimal sequence of H and T gates for implementing the input unitary exactly.
This document contains lecture notes for Engineering Mathematics - I. It covers topics in differential calculus including differential calculus I, II, and III. For each unit, it provides introductions to concepts and worked examples of finding derivatives of various functions up to the nth order. These include standard functions like exponential, trigonometric, logarithmic, and algebraic functions. Formulas for derivative rules like the product rule, quotient rule, and chain rule are also stated.
This document provides an introduction to the quantum theory of angular momentum. It begins by discussing how angular momentum arises in quantum mechanics when moving from one-dimensional to three-dimensional systems. It then defines the components of angular momentum as linear operators and establishes their commutation relations. Next, it introduces raising and lowering operators and shows how they relate to the components. It also proves several theorems, such as that the square of the total angular momentum commutes with its individual components. Finally, it discusses how angular momentum eigenstates can be rotated using rotation operators.
This document contains a chapter on complex numbers from an Oxford textbook. It includes examples of adding, multiplying, dividing and simplifying complex numbers. It also covers topics like modulus, argument and solving complex number equations. Several worked examples are provided with step-by-step solutions.
The document discusses various 2D and 3D transformations including translation, scaling, rotation, reflection, shearing, and homogeneous coordinates. It provides the mathematical definitions and matrix representations for each transformation type in 2D and 3D. It also covers topics like composition and inverse of transformations, classification of transformations, and properties of rigid body transformations.
The document provides solutions to questions from an IIT-JEE mathematics exam. It includes 8 questions worth 2 marks each, 8 questions worth 4 marks each, and 2 questions worth 6 marks each. The solutions solve problems related to probability, trigonometry, geometry, calculus, and loci. The summary focuses on the high-level structure and content of the document.
Higher order derivatives for N -body simulationsKeigo Nitadori
This document discusses higher order derivatives that are useful for N-body simulations. It presents formulas for calculating higher order derivatives of power functions like y=xn, and applies this to derivatives of gravitational force f=mr-3. Specifically:
1) It derives recursive formulas for calculating higher order derivatives of power functions y=xn in terms of previous derivatives.
2) It applies these formulas to calculate derivatives of the gravitational force f=mr-3 in terms of derivatives of r and q=r-3/2.
3) It also describes an alternative approach by Le Guyader (1993) for calculating derivatives of r and q in terms of dot products of r with itself.
The document is an index for a course on complex functions and conformal mapping. It outlines 31 methods that will be covered across 5 units. Unit 1 focuses on complex functions and conformal mapping, exploring methods for basic examples, square roots of complex numbers, nth roots, trigonometric functions of complex numbers, logarithms, differentiability, analyticity, and more. Subsequent units cover complex integrals, sequences and series, Laurent series and residues, and first and higher order partial differential equations. Examples are provided for many of the methods throughout the document.
The document discusses ordinary differential equations (ODEs) and methods for solving separable differential equations. It defines ODEs, order, and degree of differential equations. It then introduces the method of separating variables for solving separable differential equations. This method involves rearranging the differential equation so that the variables are separated on each side. The document provides examples of using this method to solve three separable differential equations, finding the integral relations between the variables in each case.
The document provides an overview of the theory of complex numbers. It defines key concepts such as the Argand diagram, polar and rectangular forms, and operations involving complex numbers like addition, subtraction, multiplication, and division. It also discusses applications of complex numbers to electrical circuits and provides sample exercises for working with complex numbers in different forms and operations.
This document provides an overview of the algebra of complex numbers. It begins with an introduction to complex numbers, defining them as numbers of the form a + bi, where a is the real part and b is the imaginary part. It then covers operations with complex numbers like addition, multiplication, and division. Next, it discusses the Argand diagram, which represents complex numbers graphically as points in a plane. It also defines the argument and modulus of a complex number. The document aims to outline the key concepts and operations involving complex numbers.
Engineering Maths N4
Rectangular Form
Polar Form
Relationship between Polar and Rectangular Complex NUmbers
FET College Registrations in Johannesburg Cbd
November 2015 Exams Registrations
Enrolment for Fet College Exams
Distance Learning
Correspondence
Unisa Tutorials
Extra Lessons
E Learning
Engineering Certificates
Engineering Diplomas
Business Certificates
Phone 073 090 2954
Fax 086 244 2355
Email topstudentz017@gmail.com
This document contains notes from a signals and systems chapter about the z-transform and representing discrete-time systems using difference equations. It defines the z-transform, provides examples of elementary z-transforms, and discusses using the z-transform to analyze discrete-time linear systems. It also describes representing the input-output relationship of a discrete-time system using a linear constant-coefficient difference equation and gives examples of problems involving determining z-transforms and deriving difference equations from transfer functions.
- The z-transform is a mathematical tool that converts discrete-time sequences into complex functions, analogous to how the Laplace transform handles continuous-time signals.
- Key properties and sequences that are transformed include the unit impulse δn, unit step un, and geometric sequences an.
- The z-transform is computed by taking the z-transform definition, which is an infinite summation, and obtaining closed-form expressions using properties like linearity and geometric series sums.
- Common transforms include U(z) for the unit step, 1/1-az^-1 for geometric sequences an, and expressions involving z, sinh/cosh, and sin/cos for exponential and trigonometric sequences.
Important formulas for JEE Main 2020 - Subject-wise ListKanhaMalik
This document provides important formulas for the JEE Main 2020 exam across various subjects - mathematics, physics, and chemistry. In mathematics, formulas are provided for topics like roots of unity, complex numbers, trigonometry, quadratic equations, and vectors. In physics, formulas cover topics such as electricity, magnetism, motion, gravitation, work, and friction. Formulas in chemistry include those for temperature conversion, molarity, molecular mass, atomic structure, gas laws, thermodynamics, and electrochemistry. Mastering these essential formulas can help JEE aspirants save time during the exam, better prepare, and reduce mistakes.
Z TRANSFORM PROPERTIES AND INVERSE Z TRANSFORMTowfeeq Umar
The document discusses various methods for computing the inverse z-transform including inspection, partial fraction expansion, and power series expansion. It provides examples to illustrate each method. The inverse z-transform finds the original time domain sequence from its z-transform. Key properties like linearity, time shifting, and convolution are also covered.
Here are the key steps to solve this problem:
1. Write the rotation matrix for 45 degree rotation about the x-axis:
R = [1 0 0]
[0 cos(45) -sin(45)]
[0 sin(45) cos(45)]
2. Write the scaling matrix with factors of 4 in each direction:
S = [4 0 0]
[0 4 0]
[0 0 4]
3. Pre-multiply the original coordinates by the rotation matrix, then scale the results by pre-multiplying the rotation matrix by the scaling matrix:
A' = SR(0,1,0)
= S(0,0.
This document discusses the discrete Fourier transform (DFT) and its inverse, the inverse discrete Fourier transform (IDFT). It defines the DFT and IDFT formulas, introduces twiddle factors, and shows how to represent DFTs and IDFTs using matrices. Examples are provided to calculate DFTs and IDFTs both directly from the definition and using matrix representations.
1. The document discusses using Z-transforms to solve linear difference equations with constant coefficients.
2. It provides the working procedure which involves taking the Z-transform of both sides of the difference equation, rearranging to isolate the Z-transform of the unknown function U(z), and then taking the inverse Z-transform to find the solution in terms of n.
3. As an example, it shows the step-by-step solution of the difference equation un+2 - 2un+1 + un = 3n + 5 using this method. The solution is found to be 1/2n(n-1)(n+3) + c0 + (c1-c0)n
Fast and efficient exact synthesis of single qubit unitaries generated by cli...JamesMa54
The document describes a presentation on an algorithm for exact synthesis of single qubit unitaries generated by Clifford and T gates. The algorithm reduces the problem of implementing a unitary to the problem of state preparation. It then uses a series of HT gates to iteratively decrease the smallest denominator exponent of the state entries until it reaches a base case that can be looked up. The algorithm runs in time linear in the initial smallest denominator exponent and provides an optimal sequence of H and T gates for implementing the input unitary exactly.
End-to-end pipeline agility - Berlin Buzzwords 2024Lars Albertsson
We describe how we achieve high change agility in data engineering by eliminating the fear of breaking downstream data pipelines through end-to-end pipeline testing, and by using schema metaprogramming to safely eliminate boilerplate involved in changes that affect whole pipelines.
A quick poll on agility in changing pipelines from end to end indicated a huge span in capabilities. For the question "How long time does it take for all downstream pipelines to be adapted to an upstream change," the median response was 6 months, but some respondents could do it in less than a day. When quantitative data engineering differences between the best and worst are measured, the span is often 100x-1000x, sometimes even more.
A long time ago, we suffered at Spotify from fear of changing pipelines due to not knowing what the impact might be downstream. We made plans for a technical solution to test pipelines end-to-end to mitigate that fear, but the effort failed for cultural reasons. We eventually solved this challenge, but in a different context. In this presentation we will describe how we test full pipelines effectively by manipulating workflow orchestration, which enables us to make changes in pipelines without fear of breaking downstream.
Making schema changes that affect many jobs also involves a lot of toil and boilerplate. Using schema-on-read mitigates some of it, but has drawbacks since it makes it more difficult to detect errors early. We will describe how we have rejected this tradeoff by applying schema metaprogramming, eliminating boilerplate but keeping the protection of static typing, thereby further improving agility to quickly modify data pipelines without fear.
We are pleased to share with you the latest VCOSA statistical report on the cotton and yarn industry for the month of March 2024.
Starting from January 2024, the full weekly and monthly reports will only be available for free to VCOSA members. To access the complete weekly report with figures, charts, and detailed analysis of the cotton fiber market in the past week, interested parties are kindly requested to contact VCOSA to subscribe to the newsletter.
Codeless Generative AI Pipelines
(GenAI with Milvus)
https://ml.dssconf.pl/user.html#!/lecture/DSSML24-041a/rate
Discover the potential of real-time streaming in the context of GenAI as we delve into the intricacies of Apache NiFi and its capabilities. Learn how this tool can significantly simplify the data engineering workflow for GenAI applications, allowing you to focus on the creative aspects rather than the technical complexities. I will guide you through practical examples and use cases, showing the impact of automation on prompt building. From data ingestion to transformation and delivery, witness how Apache NiFi streamlines the entire pipeline, ensuring a smooth and hassle-free experience.
Timothy Spann
https://www.youtube.com/@FLaNK-Stack
https://medium.com/@tspann
https://www.datainmotion.dev/
milvus, unstructured data, vector database, zilliz, cloud, vectors, python, deep learning, generative ai, genai, nifi, kafka, flink, streaming, iot, edge
Open Source Contributions to Postgres: The Basics POSETTE 2024ElizabethGarrettChri
Postgres is the most advanced open-source database in the world and it's supported by a community, not a single company. So how does this work? How does code actually get into Postgres? I recently had a patch submitted and committed and I want to share what I learned in that process. I’ll give you an overview of Postgres versions and how the underlying project codebase functions. I’ll also show you the process for submitting a patch and getting that tested and committed.
Orchestrating the Future: Navigating Today's Data Workflow Challenges with Ai...Kaxil Naik
Navigating today's data landscape isn't just about managing workflows; it's about strategically propelling your business forward. Apache Airflow has stood out as the benchmark in this arena, driving data orchestration forward since its early days. As we dive into the complexities of our current data-rich environment, where the sheer volume of information and its timely, accurate processing are crucial for AI and ML applications, the role of Airflow has never been more critical.
In my journey as the Senior Engineering Director and a pivotal member of Apache Airflow's Project Management Committee (PMC), I've witnessed Airflow transform data handling, making agility and insight the norm in an ever-evolving digital space. At Astronomer, our collaboration with leading AI & ML teams worldwide has not only tested but also proven Airflow's mettle in delivering data reliably and efficiently—data that now powers not just insights but core business functions.
This session is a deep dive into the essence of Airflow's success. We'll trace its evolution from a budding project to the backbone of data orchestration it is today, constantly adapting to meet the next wave of data challenges, including those brought on by Generative AI. It's this forward-thinking adaptability that keeps Airflow at the forefront of innovation, ready for whatever comes next.
The ever-growing demands of AI and ML applications have ushered in an era where sophisticated data management isn't a luxury—it's a necessity. Airflow's innate flexibility and scalability are what makes it indispensable in managing the intricate workflows of today, especially those involving Large Language Models (LLMs).
This talk isn't just a rundown of Airflow's features; it's about harnessing these capabilities to turn your data workflows into a strategic asset. Together, we'll explore how Airflow remains at the cutting edge of data orchestration, ensuring your organization is not just keeping pace but setting the pace in a data-driven future.
Session in https://budapestdata.hu/2024/04/kaxil-naik-astronomer-io/ | https://dataml24.sessionize.com/session/667627
Introduction to Jio Cinema**:
- Brief overview of Jio Cinema as a streaming platform.
- Its significance in the Indian market.
- Introduction to retention and engagement strategies in the streaming industry.
2. **Understanding Retention and Engagement**:
- Define retention and engagement in the context of streaming platforms.
- Importance of retaining users in a competitive market.
- Key metrics used to measure retention and engagement.
3. **Jio Cinema's Content Strategy**:
- Analysis of the content library offered by Jio Cinema.
- Focus on exclusive content, originals, and partnerships.
- Catering to diverse audience preferences (regional, genre-specific, etc.).
- User-generated content and interactive features.
4. **Personalization and Recommendation Algorithms**:
- How Jio Cinema leverages user data for personalized recommendations.
- Algorithmic strategies for suggesting content based on user preferences, viewing history, and behavior.
- Dynamic content curation to keep users engaged.
5. **User Experience and Interface Design**:
- Evaluation of Jio Cinema's user interface (UI) and user experience (UX).
- Accessibility features and device compatibility.
- Seamless navigation and search functionality.
- Integration with other Jio services.
6. **Community Building and Social Features**:
- Strategies for fostering a sense of community among users.
- User reviews, ratings, and comments.
- Social sharing and engagement features.
- Interactive events and campaigns.
7. **Retention through Loyalty Programs and Incentives**:
- Overview of loyalty programs and rewards offered by Jio Cinema.
- Subscription plans and benefits.
- Promotional offers, discounts, and partnerships.
- Gamification elements to encourage continued usage.
8. **Customer Support and Feedback Mechanisms**:
- Analysis of Jio Cinema's customer support infrastructure.
- Channels for user feedback and suggestions.
- Handling of user complaints and queries.
- Continuous improvement based on user feedback.
9. **Multichannel Engagement Strategies**:
- Utilization of multiple channels for user engagement (email, push notifications, SMS, etc.).
- Targeted marketing campaigns and promotions.
- Cross-promotion with other Jio services and partnerships.
- Integration with social media platforms.
10. **Data Analytics and Iterative Improvement**:
- Role of data analytics in understanding user behavior and preferences.
- A/B testing and experimentation to optimize engagement strategies.
- Iterative improvement based on data-driven insights.
Build applications with generative AI on Google CloudMárton Kodok
We will explore Vertex AI - Model Garden powered experiences, we are going to learn more about the integration of these generative AI APIs. We are going to see in action what the Gemini family of generative models are for developers to build and deploy AI-driven applications. Vertex AI includes a suite of foundation models, these are referred to as the PaLM and Gemini family of generative ai models, and they come in different versions. We are going to cover how to use via API to: - execute prompts in text and chat - cover multimodal use cases with image prompts. - finetune and distill to improve knowledge domains - run function calls with foundation models to optimize them for specific tasks. At the end of the session, developers will understand how to innovate with generative AI and develop apps using the generative ai industry trends.
1. Unit - II: Z - TRANSFORMS ...
Transformation OR Transform: A transform is a mathematical devices which
transforms one function to the another function. Most of the science and engineering
problems are ending with the mathematical equations like equations, ordinary
differential equations, partial differential equations and difference equations.
In the present topic, we discuss the basic definition of Z - transform, properties of Z -
transforms, Z - transforms of the standard functions, problems on Z - transforms,
initial value and final value theorem of the Z - transforms and problems, inverse Z -
transforms, definition of the inverse Z - transforms, formulas of the inverse Z
-transforms of standard functions and problems. Application of Z -transforms to
solve difference equations.
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 1 / 99
2. Continued Unit - II on Z - Transforms ...
Definition of the Z - transforms: The Z - transforms of the discrete function un,
defined for all n ≥ 0 and un = 0, for n < 0, is denoted by ZT [un] OR Z[un] OR U(z)
OR Ū(z) and is defined as,
ZT [un] = Z[un] =
∞
X
n=0
unz−n
= U(z) = Ū(z)
Where Z OR ZT is the Z - transforms operator and z is the Z - transforms parameter.
Definition of the Inverse Z - Transforms: The function U(z) is said to be
inverse Z - transforms of the function un. The inverse Z - transforms of the function
U(z) is denoted by Z−1
[U(z)] and is given by
un = Z−1
[U(z)]
where, Z−1
is the inverse Z - transforms operator.
Properties of the Z - transforms:
1. Linearity Property: If c1, c2, ..., cp all are constants then
Z[c1u1(n) + c2u2(n) + ... + cpup(n)] = c1Z[u1(n)] + c2Z[f2(x)] + ... + cpZ[up(x)]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 2 / 99
3. Continued Unit - II on Properties of the Z - Transforms ...
Poof: We have, by definition of Z - transforms
Z[c1u1(n) + c2u2(n) + ... + cpup(n)] =
∞
X
n=0
(c1u1(n) + c2u2(n) + ... + cpup(n))z−n
Z[c1u1(n)+c2u2(n)+...+cpup(n)] = c1
∞
X
n=0
u1(n)z−n
+c2
∞
X
n=0
u2(n)z−n
+...+cp
∞
X
n=0
up(n)z−
Z[c1u1(n) + c2u2(n) + ... + cpup(n)] = c1Z[u1(n)] + c2Z[u2(n)] + ... + cpZ[up(n)]
2. Damping Property OR Damping rule: If Z[un = U(z)], then prove that (i)
Z[an
un] = U(z
a
), (ii) Z[a−n
un] = U(az).
Poof: (i) We have, by definition of Z - transforms
Z[un] =
∞
X
n=0
unz−n
= U(z)
Therefore,
Z[an
un] =
∞
X
n=0
(an
un)z−n
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 3 / 99
4. Continued Unit - II on Properties of the Z - Transforms ...
Z[an
un] =
∞
X
n=0
un
z−n
a−n
Z[an
un] =
∞
X
n=0
un(
z
a
)−n
Comparing this with the definition here instead of z is z
a
Z[an
un] = U(
z
a
)
(ii) We have, by definition of Z - transforms
Z[un] =
∞
X
n=0
unz−n
= U(z)
Therefore,
Z[a−n
un] =
∞
X
n=0
(a−n
un)z−n
Z[a−n
un] =
∞
X
n=0
un(az)−n
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 4 / 99
5. Continued Unit - II on Properties of the Z - Transforms ...
Comparing this with the definition here instead of z is az
Z[a−n
un] = U(az)
3. Left Shifting Property: If Z[un = U(z)], then prove that (i)
Z[un+1] = z[U(z) − u0],
(ii) Z[un+2] = z2
[U(z) − u0 − u1
z
],
(iii)Z[un+3] = z3
[U(z) − u0 − u1
z
− u2
z2 ] and
(iv) Z[un+k] = zk
[U(z) −
Pk−1
r=0 urzr
].
Poof: (i) We have, by definition of Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
∴
Z[un+1] =
∞
X
n=0
un+1z−n
Multiply and dividing by z in the RHS
Z[un+1] = (z)(
1
z
)
∞
X
n=0
un+1z−n
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 5 / 99
6. Continued Unit - II on Properties of the Z - Transforms ...
Z[un+1] = z
∞
X
n=0
un+1
1
z
z−n
Z[un+1] = z
∞
X
n=0
un+1z−1
z−n
Z[un+1] = z
∞
X
n=0
un+1z−n−1
Z[un+1] = z
∞
X
n=0
un+1z−(n+1)
Expanding the summation in the RHS
Z[un+1] = z[u1z−1
+ u2z−2
+ u3z−3
+ ...]
Now, adding and subtracting the missing term u0 in the RHS
Z[un+1] = z[u0 + u1z−1
+ u2z−2
+ u3z−3
+ ... − u0]
Z[un+1] = z[(u0 + u1z−1
+ u2z−2
+ u3z−3
+ ...) − u0]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 6 / 99
7. Continued Unit - II on Properties of the Z - Transforms ...
Z[un+1] = z[
∞
X
n=0
unz−n
− u0]
Z[un+1] = z[U(z) − u0]
(ii) We have, by definition of Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
∴
Z[un+2] =
∞
X
n=0
un+2z−n
Multiply and dividing by z2
in the RHS
Z[un+2] = (z2
)(
1
z2
)
∞
X
n=0
un+2z−n
Z[un+2] = z2
∞
X
n=0
un+2
1
z2
z−n
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 7 / 99
8. Continued Unit - II on Properties of the Z - Transforms ...
Z[un+2] = z2
∞
X
n=0
un+2z−2
z−n
Z[un+2] = z2
∞
X
n=0
un+2z−n−2
Z[un+2] = z2
∞
X
n=0
un+2z−(n+2)
Expanding the summation in the RHS
Z[un+2] = z2
[u2z−2
+ u3z−3
+ u4z−4
+ ...]
Now, adding and subtracting the missing terms u0 and u1z−1
in the RHS
Z[un+2] = z2
[u0 + u1z−1
+ u2z−2
+ u3z−3
+ ... − u0 − u1z−1
]
Z[un+2] = z2
[(u0 + u1z−1
+ u2z−2
+ u3z−3
+ ...) − u0 − u1z−1
]
Z[un+2] = z2
[
∞
X
n=0
unz−n
− u0 −
u1
z
]
Z[un+2] = z2
[U(z) − u0 −
u1
z
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 8 / 99
9. Continued Unit - II on Properties of the Z - Transforms ...
(iii) We have, by definition of Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
∴
Z[un+3] =
∞
X
n=0
un+3z−n
Multiply and dividing by z3
in the RHS
Z[un+3] = (z3
)(
1
z3
)
∞
X
n=0
un+3z−n
Z[un+3] = z3
∞
X
n=0
un+3
1
z3
z−n
Z[un+3] = z3
∞
X
n=0
un+3z−3
z−n
Z[un+3] = z3
∞
X
n=0
un+3z−n−3
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 9 / 99
10. Continued Unit - II on Properties of the Z - Transforms ...
Z[un+3] = z3
∞
X
n=0
un+3z−(n+3)
Expanding the summation in the RHS
Z[un+3] = z3
[u3z−3
+ u4z−4
+ u5z−5
+ ...]
Now, adding and subtracting the missing terms u0, u1z−1
and u2z−2
in the RHS
Z[un+3] = z3
[u0 + u1z−1
+ u2z−2
+ u3z−3
+ ... − u0 − u1z−1
− u2z−2
]
Z[un+3] = z3
[(u0 + u1z−1
+ u2z−2
+ u3z−3
+ ...) − u0 − u1z−1
− u2z−2
]
Z[un+3] = z3
[
∞
X
n=0
unz−n
− u0 −
u1
z
−
u2
z2
]
Z[un+3] = z3
[U(z) − u0 −
u1
z
−
u2
z2
]
Similarly, we can prove (iv) Z[un+k] = zk
[U(z) −
Pk−1
r=0 urzr
].
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 10 / 99
11. Continued Unit - II on Properties of the Z - Transforms ...
4. Right Shifting Property: If Z[un = U(z)], then prove that
Z[un−k] = z−k
U(z), k > 0.
Property - 05: If Z[un = U(z)], then prove that Z[np
] = −z d
dz
[Z(np−1
)], where p is
any positive integer.
Poof: We have, by definition of Z - transforms
Z[np
] =
∞
X
n=0
np
z−n
Multiply and dividing by n and z in the RHS
Z[np
] = (z)
∞
X
n=0
(n)(
1
n
)np
(
1
z
)z−n
Z[np
] = z
∞
X
n=0
nn−1
np
z−1
z−n
Z[np
] = z
∞
X
n=0
nnp−1
z−n−1
(1)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 11 / 99
12. Continued Unit - II on Properties of the Z - Transforms ...
Again, by definition of Z - transforms
Z[np−1
] =
∞
X
n=0
np−1
z−n
Differentiating w.r.t. z on both sides
d
dz
Z[np−1
] =
d
dz
[
∞
X
n=0
np−1
z−n
]
d
dz
[Z(np−1
)] =
∞
X
n=0
d
dz
[np−1
z−n
]
d
dz
[Z(np−1
)] =
∞
X
n=0
np−1
(−n)z−n−1
d
dz
[Z(np−1
)] = −
∞
X
n=0
nnp−1
z−n−1
∞
X
n=0
nnp−1
z−n−1
= −
d
dz
[Z(np−1
)] (2)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 12 / 99
13. Continued Unit - II on Properties of the Z - Transforms ...
Substituting equation (2) in equation (1), we get
Z[np
] = z[−
d
dz
[Z(np−1
)]]
Z[np
] = −z
d
dz
[Z(np−1
)]
Hence, the property is proved.
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 13 / 99
14. Continued Unit - II on Z - Transforms of the Standard Functions ...
1. Find the Z - transforms of 1.
Solution: We know that, by definition of the Z - transforms
Z[un] =
∞
X
n=0
unz−n
= U(z)
but, here un = 1
Z[1] =
∞
X
n=0
(1)z−n
Z[1] =
∞
X
n=0
1
zn
Z[1] =
∞
X
n=0
1
z
!n
We have, by Binomial expansion
(1 − x)−1
= 1 + x + x2
+ x3
+ ... =
∞
X
n=0
xn
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 14 / 99
15. Continued Unit - II on Z - Transforms of the Standard Functions ...
here, x = 1
z
in the above equation
Z[1] = 1 −
1
z
!−1
Z[1] =
z − 1
z
!−1
Z[1] =
z
z − 1
Replace 1 by −1 , we get
Z[−1] =
z
z + 1
2. Find the Z - transforms of an
.
Solution: We know that, by definition of the Z - transforms
Z[un] =
∞
X
n=0
unz−n
= U(z)
but, here un = an
Z[an
] =
∞
X
n=0
(an
)z−n
n
∞
X an
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 15 / 99
16. Continued Unit - II on Z - Transforms of the Standard Functions ...
Z[an
] =
∞
X
n=0
(
a
z
)n
We have, by Binomial expansion
(1 − x)−1
= 1 + x + x2
+ x3
+ ... =
∞
X
n=0
xn
here, x = a
z
in the above equation
Z[an
] = 1 −
a
z
!−1
Z[an
] =
z − a
z
!−1
Z[an
] =
z
z − a
Replace a by −a , we get
Z[(−a)n
] =
z
z + a
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 16 / 99
17. Continued Unit - II on Z - Transforms of the Standard Functions ...
Similarly, the Z - transforms
(a)
Z[enθ
] = Z[(eθ
)n
] =
z
z − eθ
(b)
Z[e−nθ
] = Z[(e−θ
)n
] =
z
z − e−θ
(c)
Z[einθ
] = Z[(eiθ
)n
] =
z
z − eiθ
(d)
Z[e−inθ
] = Z[(e−iθ
)n
] =
z
z − e−iθ
(e) Z - transforms of any constant k
Z[k] =
kz
z − 1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 17 / 99
18. Continued Unit - II on Z - Transforms of the Standard Functions ...
We know that,
I.
einθ
= cos(nθ) + i sin(nθ)
II.
e−inθ
= cos(nθ) − i sin(nθ)
III.
cos(nθ) =
einθ
+ e−inθ
2
IV.
einθ
+ e−inθ
= 2 cos(nθ)
V.
sin(nθ) =
einθ
− e−inθ
2i
VI.
einθ
− e−inθ
= 2i sin(nθ)
VII.
sinh(nθ) =
enθ
− e−nθ
2
VIII.
cosh(nθ) =
enθ
+ e−nθ
2
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 18 / 99
19. Continued Unit - II on Z - Transforms of the Standard Functions ...
3. Obtain the Z - transforms of sinh(nθ) and cosh(nθ).
Solution: (i) To obtain the Z - transforms of sinh(nθ)
We know that,
Z[enθ
] =
z
z − eθ
and
Z[e−nθ
] =
z
z − e−θ
Also, we have
sinh(nθ) =
enθ
− e−nθ
2
Applying the Z - transforms on the both sides
Z[sinh(nθ)] = Z[
enθ
− e−nθ
2
]
Z[sinh(nθ)] =
1
2
"
Z[enθ
] − Z[e−nθ
]
#
Substituting Z[enθ
] = z
z−eθ and Z[e−nθ
] = z
z−e−θ
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 19 / 99
20. Continued Unit - II on Z - Transforms of the Standard Functions ...
Z[sinh(nθ)] =
1
2
"
z
z − eθ
−
z
z − e−θ
#
Z[sinh(nθ)] =
1
2
"
z(z − e−θ
) − z(z − eθ
)
(z − eθ)(z − e−θ)
#
Z[sinh(nθ)] =
1
2
[
z2
− ze−θ
− z2
+ zeθ
z2 − ze−θ − zeθ + eθe−θ
]
Z[sinh(nθ)] =
1
2
"
z(eθ
− e−θ
)
z2 − z(eθ + e−θ) + e0
#
Z[sinh(nθ)] =
1
2
"
z(2 sinh(θ))
z2 − z(2 cosh(θ)) + 1
#
Z[sinh(nθ)] =
1
1
"
z(sinh(θ))
z2 − 2z cosh(θ) + 1
#
Z[sinh(nθ)] =
z sinh(θ)
z2 − 2z cosh(θ) + 1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 20 / 99
21. Continued Unit - II on Z - Transforms of the Standard Functions ...
(ii) To obtain the Z - transforms of cosh(nθ)
We know that,
Z[enθ
] =
z
z − eθ
and
Z[e−nθ
] =
z
z − e−θ
Also, we have
cosh(nθ) =
enθ
+ e−nθ
2
Applying the Z - transforms on the both sides
Z[cosh(nθ)] = Z[
enθ
+ e−nθ
2
]
Z[cosh(nθ)] =
1
2
"
Z[enθ
] + Z[e−nθ
]
#
Substituting Z[enθ
] = z
z−eθ and Z[e−nθ
] = z
z−e−θ
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 21 / 99
22. Continued Unit - II on Z - Transforms of the Standard Functions ...
Z[cosh(nθ)] =
1
2
"
z
z − eθ
+
z
z − e−θ
#
Z[cosh(nθ)] =
1
2
"
z(z − e−θ
) + z(z − eθ
)
(z − eθ)(z − e−θ)
#
Z[cosh(nθ)] =
1
2
"
z2
− ze−θ
+ z2
− zeθ
z2 − ze−θ − zeθ + eθe−θ
#
Z[cosh θ)] =
1
2
"
2z2
− z(eθ
+ e−θ
)
z2 − z(eθ + e−θ) + e0
#
Z[cosh(nθ)] =
1
2
"
2z2
− z(2 cosh(θ))
z2 − z(2 cosh(θ)) + 1
#
Z[cosh(nθ)] =
1
1
"
z2
− z(cosh(θ))
z2 − 2z cosh(θ) + 1
#
Z[cosh(nθ)] =
z2
− z cosh(θ)
z2 − 2z cosh(θ) + 1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 22 / 99
23. Continued Unit - II on Z - Transforms of the Standard Functions ...
4. Obtain the Z - transforms of sin(nθ) and cos(nθ).
Solution: We know that,
einθ
= cos(nθ) + i sin(nθ)
Applying Z - transforms on both sides
Z[einθ
] = Z[cos(nθ) + i sin(nθ)]
Z[cos(nθ)] + iZ[sin(nθ)] = Z[einθ
]
Substituting Z[einθ
] = z
z−eiθ in the RHS
Z[cos(nθ)] + iZ[sin(nθ)] =
z
z − eiθ
Z[cos(nθ)] + iZ[sin(nθ)] =
z
z − (cos(θ) + i sin(θ))
Z[cos(nθ)] + iZ[sin(nθ)] =
z
(z − cos(θ)) − i sin(θ)
Now, rationalize RHS
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 23 / 99
24. Continued Unit - II on Z - Transforms of the Standard Functions ...
Z[cos(nθ)] + iZ[sin(nθ)] =
z((z − cos(θ)) + i sin(θ))
((z − cos(θ)) − i sin(θ))((z − cos(θ)) + i sin(θ))
Z[cos(nθ)] + iZ[sin(nθ)] =
(z2
− z cos(θ)) + iz sin(θ)
(z − cos(θ))2 − (i sin(θ))2
Z[cos(nθ)] + iZ[sin(nθ)] =
(z2
− z cos(θ)) + iz sin(θ)
z2 − 2z cos(θ) + cos2(θ) − i2 sin2
(θ)
Z[cos(nθ)] + iZ[sin(nθ)] =
(z2
− z cos(θ)) + iz sin(θ)
z2 − 2z cos(θ) + cos2(θ) + sin2
(θ)
Z[cos(nθ)] + iZ[sin(nθ)] =
(z2
− z cos(θ)) + iz sin(θ)
z2 − 2z cos(θ) + 1
Separate the real and imaginary parts in the RHS
Z[cos(nθ)] + iZ[sin(nθ)] =
z2
− z cos(θ)
z2 − 2z cos(θ) + 1
+ i
z sin(θ)
z2 − 2z cos(θ) + 1
Comparing the real and imaginary parts on both sides, we get
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 24 / 99
25. Continued Unit - II on Z - Transforms of the Standard Functions ...
Z[sin(nθ)] =
z sin(θ)
z2 − 2z cos(θ) + 1
and
Z[cos(nθ)] =
z2
− z cos(θ)
z2 − 2z cos(θ) + 1
5. Find the Z - transforms of n.
Solution: We know that, by property
Z[np
] = −z
d
dz
[Z(np−1
)]
Substitute p = 1
Z[n1
] = −z
d
dz
[Z(n1−1
)]
Z[n] = −z
d
dz
[Z(n0
)]
Z[n] = −z
d
dz
[Z(1)]
substituting Z[1] = z
z−1
in the RHS
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 25 / 99
26. Continued Unit - II on Z - Transforms of the Standard Functions ...
Z[n] = −z
d
dz
[
z
z − 1
]
Differentiate using the quotient rule
Z[n] = −z[
(z − 1)(1) − (z)(1)
(z − 1)2
]
Z[n] = −z[
z − 1 − z
(z − 1)2
]
Z[n] = −z[
−1
(z − 1)2
]
Z[n] =
z
(z − 1)2
6. Find the Z - transforms of n2
.
Solution: We know that, by property
Z[np
] = −z
d
dz
[Z(np−1
)]
Substitute p = 2
Z[n2
] = −z
d
dz
[Z(n2−1
)]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 26 / 99
27. Continued Unit - II on Z - Transforms of the Standard Functions ...
Z[n2
] = −z
d
dz
[Z(n1
)]
Z[n2
] = −z
d
dz
[Z(n)]
substituting Z[n] = z
(z−1)2 in the RHS
Z[n2
] = −z
d
dz
[
z
(z − 1)2
]
Differentiate using the quotient rule
Z[n2
] = −z[
(z − 1)2
(1) − (z)2(z − 1)(1)
(z − 1)4
]
Z[n2
] = −z(z − 1)[
z − 1 − 2z
(z − 1)4
]
Z[n2
] = −z[
−z − 1
(z − 1)3
]
Z[n2
] =
z2
+ z
(z − 1)3
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 27 / 99
28. Continued Unit - II on Z - Transforms of the Standard Functions ...
7. Find the Z - transforms of n3
.
Solution: We know that, by property
Z[np
] = −z
d
dz
[Z(np−1
)]
Substitute p = 3
Z[n3
] = −z
d
dz
[Z(n3−1
)]
Z[n3
] = −z
d
dz
[Z(n2
)]
substituting Z[n2
] = z2
+z
(z−1)3 in the RHS
Z[n3
] = −z
d
dz
[fracz2
+ z(z − 1)3
]
Differentiate using the quotient rule
Z[n3
] = −z[
(z − 1)3
(2z + 1) − (z2
+ z)3(z − 1)2
(1)
(z − 1)6
]
Z[n3
] = −z(z − 1)2
[
(z − 1)(2z + 1) − 3(z2
+ z)
(z − 1)6
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 28 / 99
29. Continued Unit - II on Z - Transforms of the Standard Functions ...
Z[n3
] = −z[
2z2
+ z − 2z − 1 − 3z2
− 3z
(z − 1)4
]
Z[n3
] = −z[
−z2
− 4z − 1
(z − 1)4
]
Z[n3
] =
z3
+ 4z2
+ z
(z − 1)4
8. Find the Z - transforms of n4
.
Solution: We know that, by property
Z[np
] = −z
d
dz
[Z(np−1
)]
Substitute p = 4
Z[n4
] = −z
d
dz
[Z(n4−1
)]
Z[n4
] = −z
d
dz
[Z(n3
)]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 29 / 99
30. Continued Unit - II on Z - Transforms of the Standard Functions ...
substituting Z[n3
] = z3
+4z2
+z
(z−1)4 in the RHS
Z[n4
] = −z
d
dz
[
z3
+ 4z2
+ z
(z − 1)4
]
Differentiate using the quotient rule
Z[n4
] = −z[
(z − 1)4
(3z2
+ 8z + 1) − (z3
+ 4z2
+ z)4(z − 1)3
(1)
(z − 1)8
]
Z[n4
] = −z(z − 1)3
[
(z − 1)(3z2
+ 8z + 1) − 4(z3
+ 4z2
+ z)
(z − 1)8
]
Z[n4
] = −z[
(z − 1)(3z2
+ 8z + 1) − 4z3
− 16z2
− 4z
(z − 1)5
]
Z[n4
] = −z[
3z3
+ 8z2
+ z − 3z2
− 8z − 1 − 4z3
− 16z2
− 4z
(z − 1)5
]
Z[n4
] = −z[
−z3
− 11z2
− 11z − 1
(z − 1)5
]
Z[n4
] =
z4
+ 11z3
+ 11z2
+ z
(z − 1)5
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 30 / 99
31. Continued Unit - II on Z - Transforms of the Standard Functions ...
9. Find the Z - transforms of 1
n
, for n ≥ 1.
Solution: We know that, by definition of the Z - transforms
Z
"
1
n
#
=
∞
X
n=1
1
n
!
z−n
Expanding the summation in the RHS
Z
"
1
n
#
=
1
1
!
z−1
+
1
2
!
z−2
+
1
3
!
z−3
+ ...
Z
"
1
n
#
=
1
z
+
1
2z2
+
1
3z3
+ ...
Z
"
1
n
#
=
(1
z
)
1
+
(1
z
)2
2
+
(1
z
)3
3
+ ...
Z[
1
n
] = −[−
1
z
!
1
−
1
z
!2
2
−
1
z
!3
3
− ...]
We know that, by logarithmic series
log(1 − x) = −x −
x2
−
x3
− ...
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 31 / 99
32. Continued Unit - II on Z - Transforms of the Standard Functions ...
Here, x = 1
z
Z
"
1
n
#
= −
"
log 1 −
1
z
!#
Z
"
1
n
#
= − log
z − 1
z
!
Z
"
1
n
#
= log
z − 1
z
!−1
Z
"
1
n
#
= log
z
z − 1
!
10. Find the Z - transforms of 1
n+1
, for n ≥ 0.
Solution: We know that, by definition of the Z - transforms
Z
"
1
n + 1
#
=
∞
X
n=0
1
n + 1
!
z−n
Expanding the summation in the RHS
Z
"
1
n + 1
#
=
1
1
!
z0
+
1
2
!
z−1
+
1
3
!
z−2
+
1
4
!
z−3
+ ...
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 32 / 99
33. Continued Unit - II on Z - Transforms of the Standard Functions ...
Multiply and dividing by −z in the RHS
Z
"
1
n + 1
#
= (−z)[−
(1
z
)
1
−
(1
z
)2
2
−
(1
z
)3
3
− ...]
We know that, by logarithmic series
log(1 − x) = −x −
x2
2
−
x3
3
− ...
Z
"
1
n + 1
#
= (−z)[log(1 −
1
z
)]
Z
"
1
n + 1
#
= −z log
z − 1
z
!
Z
"
1
n + 1
#
= z log
z − 1
z
!−1
Z
"
1
n + 1
#
= z log
z
z − 1
!
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 33 / 99
34. Continued Unit - II on Z - Transforms of the Standard Functions ...
11. Define unit step sequence for discrete and hence find its Z - transforms.
Solution:
Definition of the unit step sequence for discrete:
The unit step sequence for discrete is denoted by u(n) or H(n) and is defined as
u(n) = H(n) =
(
1, for n ≥ 0
0, for n < 0
To find the Z - transforms of the unit step sequence for discrete:
We have, by definition of Z - transforms
Z[H(n)] =
∞
X
n=0
(H(n))z−n
Expanding the summation in the RHS
Z[H(n)] = H(0)z0
+ H(1)z−1
+ H(2)z−2
+ H(3)z−3
+ ...
Z[H(n)] = 1 +
1
z
!
+
1
z
!2
+
1
z
!3
+ ...
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 34 / 99
35. Continued Unit - II on Z - Transforms of the Standard Functions ...
We have, by Binomial expansion
(1 − x)−1
= 1 + x + x2
+ x3
+ ...
Z[H(n)] = 1 −
1
z
!−1
Z[H(n)] =
z − 1
z
!−1
Z[H(n)] =
z
z − 1
12. Define unit impulse sequence for discrete and hence find its Z - transforms.
Solution:
Definition of the unit impulse sequence for discrete:
The unit impulse sequence for discrete is denoted by δ(n) and is defined as
δ(n) =
(
1, for n = 0
0, for n 6= 0
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 35 / 99
36. Continued Unit - II on Z - Transforms of the Standard Functions ...
To find the Z - transforms of the unit impulse sequence for discrete:
We have, by definition of Z - transforms
Z[δ(n)] =
∞
X
n=0
(δ(n))z−n
Expanding the summation in the RHS
Z[δ(n)] = δ(0)z0
+ δ(1)z−1
+ δ(2)z−2
+ δ(3)z−3
+ ...
Z[δ(n)] = 1 + (0) + (0) + (0) + ...
Z[δ(n)] = 1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 36 / 99
37. Continued Unit - II on Formulas of the Z - Transforms of the Standard
Functions ...
1.
Z[1] =
z
z − 1
2.
Z[−1] =
z
z + 1
3.
Z[an
] =
z
z − a
4.
Z[(−a)n
] =
z
z + a
5.
Z[sinh(nθ)] =
z sinh(θ)
z2 − 2z cosh(θ) + 1
6.
Z[cosh(nθ)] =
z2
− z cosh(θ)
z2 − 2z cosh(θ) + 1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 37 / 99
38. Continued Unit - II on Formulas of the Z - Transforms of the Standard
Functions ...
7.
Z[sin(nθ)] =
z sin(θ)
z2 − 2z cos(θ) + 1
8.
Z[cos(nθ)] =
z2
− z cos(θ)
z2 − 2z cos(θ) + 1
9.
Z[n] =
z
(z − 1)2
10.
Z[n2
] =
z2
+ z
(z − 1)3
11.
Z[n3
] =
z3
+ 4z2
+ z
(z − 1)4
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 38 / 99
39. Continued Unit - II on Formulas of the Z - Transforms of the Standard
Functions ...
12.
Z[n4
] =
z4
+ 11z3
+ 11z2
+ z
(z − 1)5
13.
Z[
1
n
] = log(
z
z − 1
)
14.
Z[
1
n + 1
] = z log(
z
z − 1
)
15.
Z[H(n)] =
z
z − 1
16.
Z[δ(n)] = 1
17.
Z[k] =
kz
z − 1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 39 / 99
40. Continued Unit - II Problems on Z - Transforms ...
Problem - 01 : Find the Z - transforms of the 3n + sin(5n) − cosh(4n) + 5.
Solution: By data 3n + sin(5n) − cosh(4n) + 5
Applying the Z - transforms for the problem
Z[3n + sin(5n) − cosh(4n) + 5] = 3Z[n] + Z[sin(5n)] − Z[cosh(4n)] + Z[5]
Z[3n+sin(5n)−cosh(4n)+5] = 3(
z
(z − 1)2
)+(
z sin(5)
z2 − 2z cos(5) + 1
)−(
z2
− z cosh(4)
z2 − 2z cosh(4) + 1
)+
Z[3n+sin(5n)−cosh(4n)+5] =
3z
(z − 1)2
+
z sin(5)
z2 − 2z cos(5) + 1
−
z2
− z cosh(4)
z2 − 2z cosh(4) + 1
+
5z
z − 1
Problem - 02 : Find the Z - transforms of the cos(3n + 2).
Solution: By data cos(3n + 2)
cos(3n + 2) = cos(3n) cos(2) − sin(3n) sin(2)
Applying the Z - transforms on both sides
Z[cos(3n + 2)] = Z[cos(3n) cos(2) − sin(3n) sin(2)]
Z[cos(3n + 2)] = Z[cos(3n) cos(2)] − Z[sin(3n) sin(2)]
Z[cos(3n + 2)] = (cos(2))Z[cos(3n)] − (sin(2))Z[sin(3n)]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 40 / 99
41. Continued Unit - II Problems on Z - Transforms ...
Z[cos(3n + 2)] = (cos(2))(
z2
− z cos(3)
z2 − 2z cos(3) + 1
) − (sin(2))(
z sin(3)
z2 − 2z cos(3) + 1
)
Z[cos(3n + 2)] =
z2
cos(2) − z cos(3) cos(2) − z sin(3) sin(2)
z2 − 2z cos(3) + 1
Z[cos(3n + 2)] =
z2
cos(2) − z(cos(3) cos(2) + sin(3) sin(2))
z2 − 2z cos(3) + 1
Z[cos(3n + 2)] =
z2
cos(2) − z(cos(3 − 2))
z2 − 2z cos(3) + 1
Z[cos(3n + 2)] =
z2
cos(2) − z cos(1)
z2 − 2z cos(3) + 1
Problem - 03 : Find the Z - transforms of the sin(π
4
− nπ
2
).
Solution: Given sin(π
4
− nπ
2
)
sin(
π
4
−
nπ
2
) = sin(
π
4
) cos(
nπ
2
) − cos(
π
4
) sin(
nπ
2
)
But, sin(π
4
) = 1
√
2
and cos(π
4
) = 1
√
2
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 41 / 99
42. Continued Unit - II Problems on Z - Transforms ...
sin(
π
4
−
nπ
2
) = (
1
√
2
) cos(
nπ
2
) − (
1
√
2
) sin(
nπ
2
)
Applying the Z - transforms on both sides
Z[sin(
π
4
−
nπ
2
)] = (
1
√
2
)Z[cos(
nπ
2
) − sin(
nπ
2
)]
Z[sin(
π
4
−
nπ
2
)] =
1
√
2
[(
z2
− z cos(π
2
)
z2 − 2z cos(π
2
) + 1
) − (
z sin(π
2
)
z2 − 2z cos(π
2
) + 1
)]
sin(π
2
) = 1 and cos(π
2
) = 0
Z[sin(
π
4
−
nπ
2
)] =
1
√
2
[(
z2
− z(0)
z2 − 2z(0) + 1
) − (
z(1)
z2 − 2z(0) + 1
)]
Z[sin(
π
4
−
nπ
2
)] =
z2
− z
√
2(z2 + 1)
Problem - 04 : Find the Z - transforms of the 4n2
+ cos(nπ
4
) + 10a3
.
Solution: Given 4n2
+ cos(nπ
4
) + 10a3
Applying the z - transforms for the given
problem
Z[4n2
+ cos(
nπ
4
) + 10a3
] = 4Z[n2
] + Z[cos(
nπ
4
)] + Z[10a3
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 42 / 99
43. Continued Unit - II Problems on Z - Transforms ...
Z[4n2
+ cos(
nπ
4
) + 10a3
] = 4(
z2
+ z
(z − 1)3
) + (
z2
− z cos(π
4
)
z2 − 2z cos(π
4
) + 1
) + (
10a3
z
z − 1
)
but, cos(π
4
= 1
√
2
Z[4n2
+ cos(
nπ
4
) + 10a3
] =
4(z2
+ z)
(z − 1)3
+
z2
− z( 1
√
2
)
z2 − 2z( 1
√
2
) + 1
+
10a3
z
z − 1
Z[4n2
+ cos(
nπ
4
) + 10a3
] =
4(z2
+ z)
(z − 1)3
+
√
2z2
− z
√
2z2 − 2z +
√
2
+
10a3
z
z − 1
Problem - 05 : Find the Z - transforms of the (3n + 2)2
.
Solution: By data (3n + 2)2
(3n + 2)2
= (3n)2
+ (2)2
+ 2(3n)(2)
(3n + 2)2
= 9n2
+ 4 + 12n
Applying the z - transforms on both sides
Z[(3n + 2)2
] = Z[9n2
+ 4 + 12n]
Z[(3n + 2)2
] = 9Z[n2
] + 4Z[1] + 12Z[n]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 43 / 99
44. Continued Unit - II Problems on Z - Transforms ...
Z[(3n + 2)2
] = 9(
z2
+ z
(z − 1)3
) + 4(
z
z − 1
) + 12(
z
(z − 1)2
)
Z[(3n + 2)2
] =
9(z2
+ z)
(z − 1)3
+
4z
z − 1
+
12z
(z − 1)2
Problem - 06 : Find the Z - transforms of the (2n − 1)3
.
Solution: By data (2n − 1)3
(2n − 1)3
= (2n)3
− 3(2n)2
(1) + 3(2n)(1)2
− (1)3
(2n − 1)3
= 8n3
− 12n2
+ 6n − 1
Applying the z - transforms on both sides
Z[(2n − 1)3
] = Z[8n3
− 12n2
+ 6n − 1]
Z[(2n − 1)3
] = 8Z[n3
] − 12Z[n2
] + 6Z[n] − Z[1]
Z[(2n − 1)3
] = 8(
z3
+ 4z2
+ z
(z − 1)4
) − 12(
z2
+ z
(z − 1)3
) + 6(
z
(z − 1)2
) − (
z
z − 1
)
Z[(2n − 1)3
] =
8(z3
+ 4z2
+ z)
(z − 1)4
−
12(z2
+ z)
(z − 1)3
+
6z
(z − 1)2
−
z
z − 1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 44 / 99
45. Continued Unit - II Problems on Z - Transforms ...
Problem - 07 : Find the Z - transforms of the an
n.
Solution: By data an
n
We know that, by damping property
Z[an
un] = U(
z
a
)
Here, un = n and U(z) = Z[n] = z
(z−1)2 , Replace z by z
a
Z[an
n] = [
z
(z − 1)2
]z→ z
a
Z[an
n] = [
(z
a
)
(z
a
− 1)2
]
Z[an
n] =
az
(z − a)2
Problem - 08 : Find the Z - transforms of the a−n
n2
.
Solution: By data a−n
n2
We know that, by damping property
Z[a−n
un] = U(az)
Here, un = n2
and U(z) = Z[n2
] = z2
+z
(z−1)3 , Replace z by az
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 45 / 99
46. Continued Unit - II Problems on Z - Transforms ...
Z[a−n
n2
] = [
z2
+ z
(z − 1)3
]z→az
Z[a−n
n2
] = [
(az)2
+ az
(az − 1)3
]
Z[a−n
n2
] =
a2
z2
+ az
(az − 1)3
Problem - 09 : Find the Z - transforms of the an
n2
.
Solution: By data an
n2
We know that, by damping property
Z[an
un] = U(
z
a
)
Here, un = n2
and U(z) = Z[n2
] = z2
+z
(z−1)3 , Replace z by z
a
Z[an
n2
] = [
z2
+ z
(z − 1)3
]z→ z
a
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 46 / 99
47. Continued Unit - II Problems on Z - Transforms ...
Z[an
n2
] = [
(z
a
)2
+ (z
a
)
(z
a
− 1)3
]
Z[an
n2
] =
az2
+ a2
z
(z − a)3
Problem - 10 : Find the Z - transforms of the a−n
ebn
.
Solution: By data a−n
ebn
We know that, by damping property
Z[a−n
un] = U(az)
Here, un = ebn
and U(z) = Z[ebn
] = z
z−eb , Replace z by az
Z[a−n
ebn
] = [
z
z − eb
]z→az
Z[a−n
ebn
] = [
az
az − eb
]
Z[a−n
ebn
] =
az
az − eb
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 47 / 99
48. Continued Unit - II Problems on Z - Transforms ...
Problem - 11 : Find the Z - transforms of the an
cos(nθ).
Solution: By data an
cos(nθ)
We know that, by damping property
Z[an
un] = U(
z
a
)
Here, un = cos(nθ) and U(z) = Z[cos(nθ)] = z2
−z cos(θ)
z2−2z cos(θ)+1
, Replace z by z
a
Z[an
cos(nθ)] = [
z2
− z cos(θ)
z2 − 2z cos(θ) + 1
]z→ z
a
Z[an
cos(nθ)] =
(z
a
)2
− (z
a
) cos(θ)
(z
a
)2 − 2(z
a
) cos(θ) + 1
Z[an
cos(nθ)] =
z2
− az cos(θ)
z2 − 2az cos(θ) + a2
Problem - 12 : Find the Z - transforms of the a−n
sin(nθ).
Solution: By data a−n
sin(nθ)
We know that, by damping property
Z[a−n
un] = U(az)
Here, un = sin(nθ) and U(z) = Z[sin(nθ)] = z sin(θ)
z2−2z cos(θ)+1
, Replace z by az
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 48 / 99
49. Continued Unit - II Problems on Z - Transforms ...
Z[a−n
sin(nθ)] = [
z sin(θ)
z2 − 2z cos(θ) + 1
]z→az
Z[a−n
sin(nθ)] =
az sin(θ)
(az)2 − 2az cos(θ) + 1
Z[a−n
sin(nθ)] =
az sin(θ)
a2z2 − 2az cos(θ) + 1
Problem - 13 : Find the Z - transforms of the an
sinh(nθ).
Solution: By data an
sinh(nθ)
We know that, by damping property
Z[an
un] = U(
z
a
)
Here, un = sinh(nθ) and U(z) = Z[sinh(nθ)] = z sinh(θ)
z2−2z cosh(θ)+1
, Replace z by z
a
Z[an
sinh(nθ)] = [
z sinh(θ)
z2 − 2z cosh(θ) + 1
]z→ z
a
Z[an
sinh(nθ)] =
(z
a
) sinh(θ)
(z
a
)2 − 2(z
a
) cosh(θ) + 1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 49 / 99
50. Continued Unit - II Problems on Z - Transforms ...
Z[an
sinh(nθ)] =
az sinh(θ)
z2 − 2az cosh(θ) + a2
Problem - 14 : Find the Z - transforms of the a−n
cosh(nθ).
Solution: By data a−n
cosh(nθ)
We know that, by damping property
Z[a−n
un] = U(az)
Here, un = cosh(nθ) and U(z) = Z[cosh(nθ)] = z2
−z cosh(θ)
z2−2z cosh(θ)+1
, Replace z by az
Z[a−n
cosh(nθ)] = [
z2
− z cosh(θ)
z2 − 2z cosh(θ) + 1
]z→az
Z[a−n
cosh(nθ)] =
(az)2
− (az) cosh(θ)
(az)2 − 2(az) cosh(θ) + 1
Z[a−n
cosh(nθ)] =
a2
z2
− az cosh(θ)
a2z2 − 2az cos(θ) + 1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 50 / 99
51. Continued Unit - II Problems on Z - Transforms ...
Problem - 15 : Find the Z - transforms of the 1
n(n+1)
.
Solution: By data 1
n(n+1)
, adding and subtracting n in the numerator
1
n(n + 1)
=
n + 1 − n
n(n + 1)
1
n(n + 1)
=
(n + 1) − n
n(n + 1)
1
n(n + 1)
=
(n + 1)
n(n + 1)
−
n
n(n + 1)
1
n(n + 1)
=
1
n
−
1
n + 1
Applying the Z - transforms on both sides
Z[
1
n(n + 1)
] = Z[
1
n
] − Z[
1
n + 1
]
We know that
Z[
1
n
] = log(
z
z − 1
)
and
Z[
1
n + 1
] = z log(
z
z − 1
)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 51 / 99
52. Continued Unit - II Problems on Z - Transforms ...
Using these results in the above equation
Z[
1
n(n + 1)
] = (log(
z
z − 1
)) − (z log(
z
z − 1
))
Z[
1
n(n + 1)
] = (1 − z) log(
z
z − 1
)
Problem - 16 : Find the Z - transforms of the n
n+1
.
Solution: By data n
n+1
, adding and subtracting 1 in the numerator
n
n + 1
=
n + 1 − 1
n + 1
n
n + 1
=
(n + 1) − 1
n + 1
n
n + 1
=
(n + 1)
n + 1
−
1
n + 1
n
n + 1
= 1 −
1
n + 1
Applying the Z - transforms on both sides
Z[
n
n + 1
] = Z[1] − Z[
1
n + 1
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 52 / 99
53. Continued Unit - II Problems on Z - Transforms ...
We know that
Z[1] =
z
z − 1
and
Z[
1
n + 1
] = z log(
z
z − 1
)
Using these results in the above equation
Z[
n
n + 1
] =
z
z − 1
− z log(
z
z − 1
)
Problem - 17 : Show that ZT [ 1
n!
] = e
1
z , and hence evaluate (i) ZT [ 1
(n+1)!
],
(ii) ZT [ 1
(n+2)!
].
Solution: We have, by definition of Z - transform
ZT [
1
n!
] =
∞
X
n=0
(
1
n!
)z−n
Expanding the summation in the RHS
ZT [
1
n!
] = [(
1
0!
)z0
+ (
1
1!
)z−1
+ (
1
2!
)z−2
+ (
1
3!
)z−3
+ ...]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 53 / 99
54. Continued Unit - II Problems on Z - Transforms ...
ZT [
1
n!
] = 1 +
(1
z
)
1!
+
(1
z
)2
2!
+
(1
z
)3
3!
+ ...
We know that, by exponential series
ex
= 1 +
x
1!
+
x2
2!
+
x3
3!
+ ...
Here x = 1
z
ZT [
1
n!
] = e
1
z
(i) To evaluate ZT [ 1
(n+1)!
]
Taking un = 1
n!
, un+1 = 1
(n+1)!
, u0 = 1
0!
= 1, and U(z) = ZT [un] = ZT [ 1
n!
] = e
1
z
We have, by shifting property
ZT [un+1] = z[U(z) − u0]
Now, substituting all results
ZT [
1
(n + 1)!
] = z[e
1
z − 1]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 54 / 99
55. Continued Unit - II Problems on Z - Transforms ...
(ii) To evaluate ZT [ 1
(n+2)!
]
Taking un = 1
n!
, un+2 = 1
(n+2)!
, u0 = 1
0!
= 1, u1 = 1
1!
= 1, and
U(z) = ZT [un] = ZT [ 1
n!
] = e
1
z
We have, by shifting property
ZT [un+2] = z2
[U(z) − u0 −
u1
z
]
Now, substituting all results
ZT [
1
(n + 2)!
] = z2
[e
1
z − 1 −
1
z
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 55 / 99
56. Continued Unit - II on Initial Value of the Z - Transforms ...
Initial Value Theorem of the Z - transforms:
Statement: If Z[un = U(z)], then prove that u0 = lim
z→
U(z). Hence prove that
(i) u1 = lim
z→
z[U(z) − u0],
(ii) u2 = lim
z→
z2
[U(z) − u0 − u1
z
],
(iii) u3 = lim
z→
z3
[U(z) − u0 − u1
z
− u2
z2 ].
Proof: We know that, by definition of the Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
Expanding the summation in the RHS
U(z) = u0z0
+ u1z−1
+ u2z−2
+ ...
U(z) = u0(1) +
u1
z
+
u2
z2
+ ...
Applying the limit z → ∞ on both sides
lim
z→∞
U(z) = lim
z→∞
[u0 +
u1
z
+
u2
z2
+ ...]
lim
z→∞
U(z) = [u0 + (0) + (0) + ...]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 56 / 99
57. Continued Unit - II on Initial Value of the Z - Transforms ...
u0 = lim
z→∞
U(z)
Hence initial value theorem is proved (i) To prove this u1 = lim
z→
z[U(z) − u0],
We know that, by definition of the Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
Expanding the summation in the RHS
U(z) = u0z0
+ u1z−1
+ u2z−2
+ u3z−3
+ ...
U(z) − u0 =
u1
z
+
u2
z2
+
u3
z3
+ ...
Multiplying throughout by z
z[U(z) − u0] = u1 +
u2
z
+
u3
z2
+ ...
Applying the limit z → ∞ on both sides
lim
z→∞
z[U(z) − u0] = lim
z→∞
[u1 +
u2
z
+
u3
z2
+ ...]
lim
z→∞
z[U(z) − u0] = [u1 + (0) + (0) + ...]
u1 = lim
z→∞
z[U(z) − u0]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 57 / 99
58. Continued Unit - II on Initial Value of the Z - Transforms ...
(ii) To prove this result
u2 = lim
z→
z2
[U(z) − u0 − u1
z
],
We know that, by definition of the Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
Expanding the summation in the RHS
U(z) = u0z0
+ u1z−1
+ u2z−2
+ u3z−3
+ u4z−4
+ ...
U(z) − u0 −
u1
z
=
u2
z2
+
u3
z3
+
u4
z4
+ ...
Multiplying throughout by z2
z2
[U(z) − u0 −
u1
z
] = u2 +
u3
z
+
u4
z2
+ ...
Applying the limit z → ∞ on both sides
lim
z→∞
z2
[U(z) − u0 −
u1
z
] = lim
z→∞
[u2 +
u3
z
+
u4
z2
+ ...]
lim
z→∞
z2
[U(z) − u0 −
u1
z
] = [u2 + (0) + (0) + ...]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 58 / 99
59. Continued Unit - II on Initial Value of the Z - Transforms ...
u2 = lim
z→∞
z[U(z) − u0 −
u1
z
]
(iii) To prove this result
u3 = lim
z→
z3
[U(z) − u0 − u1
z
− u2
z2 ],
We know that, by definition of the Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
Expanding the summation in the RHS
U(z) = u0z0
+ u1z−1
+ u2z−2
+ u3z−3
+ u4z−4
+ ...
U(z) − u0 −
u1
z
−
u2
z2
=
u3
z3
+
u4
z4
+ ...
Multiplying throughout by z3
z3
[U(z) − u0 −
u1
z
−
u2
z2
] = u3 +
u4
z
+
u5
z2
+ ...
Applying the limit z → ∞ on both sides
lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
] = lim
z→∞
[u3 +
u4
z
+
u5
z2
+ ...]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 59 / 99
60. Continued Unit - II on Initial Value and Final Value Theorem of the Z -
Transforms ...
lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
] = [u3 + (0) + (0) + ...]
u3 = lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
]
Final Value Theorem of the Z - Transforms:
Statement: If Z[un] = U(z), then prove that lim
n→∞
un = lim
z→1
(z − 1)U(z).
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 60 / 99
61. Continued Unit - II Problems on Initial Value of the Z - Transforms ...
Problem - 01: Find the values of the u0, u1, u2 and u3, given that
U(z) = 5z2
+3z+12
(z−1)4 .
Solution: By data U(z) = 5z2
+3z+12
(z−1)4
(i) To find the value of u0
We know that, by initial value of Z - transforms
u0 = lim
z→∞
U(z)
Substituting U(z) in the RHS
u0 = lim
z→∞
[
5z2
+ 3z + 12
(z − 1)4
] = lim
z→∞
z2
[
5 + 3
z
+ 12
z2
z4(1 − 1
z
)4
] = lim
z→∞
[
5 + 3
z
+ 12
z2
z2(1 − 1
z
)4
]
u0 = lim
z→∞
1
z2
[
5 + 3
z
+ 12
z2
(1 − 1
z
)4
] = (0)[
5 + (0) + (0)
(1 − 0)4
] = (0)(5)
u0 = 0
(ii) To find the value of u1
We know that, by initial value of Z - transforms
u1 = lim
z→∞
z[U(z) − u0]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 61 / 99
62. Continued Unit - II Problems on Initial Value of the Z - Transforms ...
Substituting U(z) and u0 in the RHS
u1 = lim
z→∞
z[
5z2
+ 3z + 12
(z − 1)4
− 0] = lim
z→∞
z[
5z2
+ 3z + 12
(z − 1)4
] = lim
z→∞
z3
[
5 + 3
z
+ 12
z2
z4(1 − 1
z
)4
]
u1 = lim
z→∞
[
5 + 3
z
+ 12
z2
z(1 − 1
z
)4
] = lim
z→∞
1
z
[
5 + 3
z
+ 12
z2
(1 − 1
z
)4
] = (0)[
5 + (0) + (0)
(1 − 0)4
] = (0)(5)
u1 = 0
(iii) To find the value of u2
We know that, by initial value of Z - transforms
u2 = lim
z→∞
z2
[U(z) − u0 −
u1
z
]
Substituting U(z), u0 and u1 in the RHS
u2 = lim
z→∞
z2
[
5z2
+ 3z + 12
(z − 1)4
− 0 − 0] = lim
z→∞
z2
[
5z2
+ 3z + 12
(z − 1)4
] = lim
z→∞
z4
[
5 + 3
z
+ 12
z2
z4(1 − 1
z
)4
]
u2 = lim
z→∞
[
5 + 3
z
+ 12
z2
(1 − 1
z
)4
] = [
5 + (0) + (0)
(1 − 0)4
] =
5
1
u1 = 5
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 62 / 99
63. Continued Unit - II Problems on Initial Value of the Z - Transforms ...
(iv) To find the value of u3
We know that, by initial value of Z - transforms
u3 = lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
]
Substituting U(z), u0, u1 and u2 in the RHS
u3 = lim
z→∞
z3
[
5z2
+ 3z + 12
(z − 1)4
− 0 − 0 −
5
z2
] = lim
z→∞
z3
[
5z2
+ 3z + 12
(z − 1)4
−
5
z2
]
u3 = lim
z→∞
z3
[
5z4
+ 3z3
+ 12z2
− 5(z − 1)4
z2(z − 1)4
]
But (z − 1)4
= (z − 1)2
(z − 1)2
(z − 1)4
= (z2
− 2z + 1)(z2
− 2z + 1)
(z − 1)4
= (z4
− 2z3
+ z2
− 2z3
4z2
− 2z + z2
− 2z + 1)
(z − 1)4
= (z4
− 4z3
+ 6z2
− 4z + 1)
u3 = lim
z→∞
z3
[
5z4
+ 3z3
+ 12z2
− 5(z4
− 4z3
+ 6z2
− 4z + 1)
z2(z − 1)4
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 63 / 99
64. Continued Unit - II Problems on Initial Value of the Z - Transforms ...
u3 = lim
z→∞
z3
[
5z4
+ 3z3
+ 12z2
− 5z4
+ 20z3
− 30z2
+ 20z − 5
z2(z − 1)4
]
u3 = lim
z→∞
z3
[
23z3
− 18z2
+ 20z − 5
z2(z − 1)4
] = lim
z→∞
z6
[
23 − 18
z
+ 20
z2 − 5
z3
z6(1 − 1
z
)4
]
u3 = lim
z→∞
[
23 − 18
z
+ 20
z2 − 5
z3
(1 − 1
z
)4
] = [
23 − (0) + (0) − (0)
(1 − 0)4
] =
23
1
u3 = 23
Problem - 02: Find the values of the u0, u1, u2 and u3, when U(z) = 2z2
+4z+12
(z−1)4 .
Solution: By data U(z) = 2z2
+4z+12
(z−1)4
(i) To find the value of u0
We know that, by initial value of Z - transforms
u0 = lim
z→∞
U(z)
Substituting U(z) in the RHS
u0 = lim
z→∞
[
2z2
+ 4z + 12
(z − 1)4
] = lim
z→∞
z2
[
2 + 4
z
+ 12
z2
z4(1 − 1
z
)4
] = lim
z→∞
[
2 + 4
z
+ 12
z2
z2(1 − 1
z
)4
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 64 / 99
65. Continued Unit - II Problems on Initial Value of the Z - Transforms ...
u0 = lim
z→∞
1
z2
[
2 + 4
z
+ 12
z2
(1 − 1
z
)4
] = (0)[
2 + (0) + (0)
(1 − 0)4
] = (0)(2)
u0 = 0
(ii) To find the value of u1
We know that, by initial value of Z - transforms
u1 = lim
z→∞
z[U(z) − u0]
Substituting U(z) and u0 in the RHS
u1 = lim
z→∞
z[
2z2
+ 4z + 12
(z − 1)4
− 0] = lim
z→∞
z[
2z2
+ 4z + 12
(z − 1)4
] = lim
z→∞
z3
[
2 + 4
z
+ 12
z2
z4(1 − 1
z
)4
]
u1 = lim
z→∞
[
2 + 4
z
+ 12
z2
z(1 − 1
z
)4
] = lim
z→∞
1
z
[
2 + 4
z
+ 12
z2
(1 − 1
z
)4
] = (0)[
2 + (0) + (0)
(1 − 0)4
] = (0)(2)
u1 = 0
(iii) To find the value of u2
We know that, by initial value of Z - transforms
u2 = lim
z→∞
z2
[U(z) − u0 −
u1
z
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 65 / 99
66. Continued Unit - II Problems on Initial Value of the Z - Transforms ...
Substituting U(z), u0 and u1 in the RHS
u2 = lim
z→∞
z2
[
2z2
+ 4z + 12
(z − 1)4
− 0 − 0] = lim
z→∞
z2
[
2z2
+ 4z + 12
(z − 1)4
] = lim
z→∞
z4
[
2 + 4
z
+ 12
z2
z4(1 − 1
z
)4
]
u2 = lim
z→∞
[
2 + 4
z
+ 12
z2
(1 − 1
z
)4
] = [
2 + (0) + (0)
(1 − 0)4
] =
2
1
u1 = 2
(iv) To find the value of u3
We know that, by initial value of Z - transforms
u3 = lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
]
Substituting U(z), u0, u1 and u2 in the RHS
u3 = lim
z→∞
z3
[
2z2
+ 4z + 12
(z − 1)4
− 0 − 0 −
2
z2
] = lim
z→∞
z3
[
2z2
+ 4z + 12
(z − 1)4
−
2
z2
]
u3 = lim
z→∞
z3
[
2z4
+ 4z3
+ 12z2
− 2(z − 1)4
z2(z − 1)4
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 66 / 99
68. Continued Unit - II Problems on Initial Value of the Z - Transforms ...
Problem - 03: If ZT (un) = z
z−1
− z
z2+1
, then find ZT (un+1).
Solution: By data ZT (un) = U(z) = z
z−1
− z
z2+1
We have, by shifting property
ZT (un+1) = z2
[U(z) − u0 −
u1
z
]
To find the value of u0, by initial value theorem
u0 = lim
z→∞
U(z)
u0 = lim
z→∞
[
z
z − 1
−
z
z2 + 1
] = lim
z→∞
[
z
z(1 − 1
z
)
−
z
z2(1 + 1
z2 )
]
u0 = lim
z→∞
[
1
(1 − 1
z
)
− (
1
z
)
1
(1 + 1
z2 )
] = [
1
(1 − 0)
− (0)
1
(1 + 0)
] = [
1
(1)
− (0)(
1
(1)
)]
u0 = 1
To find the value of u1, by initial value theorem
u1 = lim
z→∞
z[U(z) − u0]
u1 = lim
z→∞
z[
z
z − 1
−
z
z2 + 1
− 1] = lim
z→∞
z[
z(z2
+ 1) − z(z − 1) − (z − 1)(z2
+ 1)
(z − 1)(z2 + 1)
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 68 / 99
69. Continued Unit - II Problems on Initial Value of the Z - Transforms ...
u1 = lim
z→∞
z[
z3
+ z − z2
+ z − z3
− z + z2
+ 1
(z − 1)(z2 + 1)
] = lim
z→∞
z[
z + 1
(z − 1)(z2 + 1)
]
u1 = lim
z→∞
z2
[
(1 + 1
z
)
z3(1 − 1
z
)(1 + 1
z2 )
] = lim
z→∞
[
(1 + 1
z
)
z(1 − 1
z
)(1 + 1
z2 )
] = lim
z→∞
(
1
z
)[
(1 + 1
z
)
(1 − 1
z
)(1 + 1
z2 )
]
u0 = (0)[
(1 + 0)
(1 − 0)(1 + 0)
] = (0)(1) = 0
Substituting U(z), u0 and u1 in the above equation
ZT (un+1) = z2
[
z
z − 1
−
z
z2 + 1
− 1 −
(0)
z
]
ZT (un+1) = z2
[
z
z − 1
−
z
z2 + 1
− 1]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 69 / 99
70. Continued Unit - II on Inverse Z - Transforms and Formulas ...
Definition of the Inverse Z - Transforms: The function U(z) is said to be
inverse Z - transforms of the function un. The inverse Z - transforms of the function
U(z) is denoted by Z−1
[U(z)] and is given by
un = Z−1
[U(z)]
where, Z−1
is the inverse Z - transforms operator.
Formulas of the Inverse Z - Transforms of standard functions:
1.
Z−1
[
z
z − 1
] = 1
2.
Z−1
[
z
z + 1
] = −1
3.
Z−1
[
z
z − a
] = an
4.
Z−1
[
z
z + a
] = (−a)n
5.
Z−1
[
1
z − a
] = an−1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 70 / 99
71. Continued Unit - II on Formulas of the Inverse Z - Transforms ...
6.
Z−1
[
1
z + a
] = (−a)n−1
7.
Z−1
[
1
(z − a)2
] = (n − 1)an−2
8.
Z−1
[
1
(z + a)2
] = (n − 1)(−a)n−2
9.
Z−1
[
1
(z − a)3
] =
1
2
(n − 1)(n − 2)an−3
10.
Z−1
[
z2
(z − a)2
] = (n + 1)an
11.
Z−1
[
z3
(z − a)3
] =
1
2!
(n + 1)(n + 2)an
un
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 71 / 99
72. Continued Unit - II on Formulas of the Inverse Z - Transforms ...
12.
Z−1
[
z
(z − a)2
] = nan−1
13.
Z−1
[
az
(z − a)2
] = nan
14.
Z−1
[
az2
+ a2
z
(z − a)3
] = n2
an
15.
Z−1
[
z
z − 1
] = H(n)
16.
Z−1
[1] = δ(n)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 72 / 99
73. Continued Unit - II Problems on Inverse Z - Transforms ...
Problem - 01 : Find the inverse Z - transforms of z
(z−1)(z+2)
.
Solution: by data z
(z−1)(z+2)
, Take
U(z) =
z
(z − 1)(z + 2)
U(z)
z
=
1
(z − 1)(z + 2)
(3)
Applying the partial fraction for the RHS expression
1
(z − 1)(z + 2)
=
A
(z − 1)
+
B
(z + 2)
1 = A(z + 2) + B(z − 1)
put, z = 1, we get 1 = A(3) + B(0), A = 1
3
put, z = −2, we get 1 = A(0) + B(−3), B = −1
3
. Substituting the value of A and B
in the above equation
1
(z − 1)(z + 2)
=
1
3(z − 1)
−
1
3(z + 2)
(4)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 73 / 99
74. Continued Unit - II Problems on Inverse Z - Transforms ...
Using Eqn. (4) in Eqn. (3)
U(z)
z
=
1
3(z − 1)
−
1
3(z + 2)
Multiply throughout by z
U(z) =
1
3
(
z
z − 1
) −
1
3
(
z
z + 2
)
Applying inverse Z - transforms on both sides
Z−1
[U(z)] =
1
3
Z−1
[
z
z − 1
] −
1
3
Z−1
[
z
z + 2
]
Z−1
[
z
(z − 1)(z + 2)
] =
1
3
(1) −
1
3
(−2)n
Z−1
[
z
(z − 1)(z + 2)
] =
1
3
[1 − (−2)n
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 74 / 99
75. Continued Unit - II Problems on Inverse Z - Transforms ...
Problem - 02 : Find the inverse Z - transforms of 5z
(2−z)(3z−1)
.
Solution: by data 5z
(2−z)(3z−1)
, Take
U(z) =
5z
(2 − z)(3z − 1)
U(z)
z
=
5
(2 − z)(3z − 1)
(5)
Applying the partial fraction for the RHS expression
5
(2 − z)(3z − 1)
=
A
(2 − z)
+
B
(3z − 1)
5 = A(3z − 1) + B(2 − z)
put, z = 2, we get 5 = A(5) + B(0), A = 1
put, z = 1
3
, we get 5 = A(0) + B(5
3
), B = 3. Substituting the value of A and B in the
above equation
5
(2 − z)(3z − 1)
=
1
(2 − z)
+
3
(3z − 1)
(6)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 75 / 99
76. Continued Unit - II Problems on Inverse Z - Transforms ...
Using Eqn. (6) in Eqn. (5)
U(z)
z
=
1
(2 − z)
+
3
(3z − 1)
Multiply throughout by z
U(z) =
z
(2 − z)
+ 3
z
(3z − 1)
Applying inverse Z - transforms on both sides
Z−1
[U(z)] = Z−1
[
z
−(z − 2)
] + 3Z−1
[
z
3(z − 1
3
)
]
Z−1
[
5z
(2 − z)(3z − 1)
] = (−1)(2)2
+
3
3
(
1
3
)n
Z−1
[
5z
(2 − z)(3z − 1)
] = −(2)2
+ (
1
3
)n
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 76 / 99
77. Continued Unit - II Problems on Inverse Z - Transforms ...
Problem - 03 : Find the inverse Z - transforms of 2z2
+3z
(z+2)(z−4)
.
Solution: by data 2z2
+3z
(z+2)(z−4)
, Take
U(z) =
2z2
+ 3z
(z + 2)(z − 4)
U(z)
z
=
2z + 3
(z + 2)(z − 4)
(7)
Applying the partial fraction for the RHS expression
2z + 3
(z + 2)(z − 4)
=
A
(z + 2)
+
B
(z − 4)
2z + 3 = A(z − 4) + B(z + 2)
put, z = −2, we get −1 = A(−6) + B(0), A = 1
6
put, z = 4, we get 11 = A(0) + B(6), B = 11
6
. Substituting the value of A and B in
the above equation
2z + 3
(z + 2)(z − 4)
=
1
6(z + 2)
+
11
6(z − 4)
(8)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 77 / 99
78. Continued Unit - II Problems on Inverse Z - Transforms ...
Using Eqn. (8) in Eqn. (7)
U(z)
z
=
1
6(z + 2)
+
11
6(z − 4)
Multiply throughout by z
U(z) =
1
6
[
z
(z + 2)
] +
11
6
[
z
(z − 4)
]
Applying inverse Z - transforms on both sides
Z−1
[U(z)] =
1
6
Z−1
[
z
(z + 2)
] +
11
6
Z−1
[
z
(z − 4)
]
Z−1
[
2z2
+ 3z
(z + 2)(z − 4)
] =
1
6
(−2)n
+
11
6
(4)n
Z−1
[
2z2
+ 3z
(z + 2)(z − 4)
] =
1
6
[(−2)n
+ 11(4)n
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 78 / 99
79. Continued Unit - II Problems on Inverse Z - Transforms ...
Problem - 04 : Find the inverse Z - transforms of 4z2
−2z
z3−5z2+8z−4
.
Solution: by data 4z2
−2z
z3−5z2+8z−4
, Take
U(z) =
4z2
− 2z
z3 − 5z2 + 8z − 4
Factorize the denominator expression z3
− 5z2
+ 8z − 4 = (z − 1)(z − 2)2
U(z) =
4z2
− 2z
(z − 1)(z − 2)2
U(z)
z
=
4z − 2
(z − 1)(z − 2)2
(9)
Applying the partial fraction for the RHS expression
4z − 2
(z − 1)(z − 2)2
=
A
(z − 1)
+
B
(z − 2)
+
C
(z − 2)2
4z − 2 = A(z − 2)2
+ B(z − 1)(z − 2) + C(z − 1)
put, z = 1, we get 2 = A(1) + B(0) + C(0), A = 2,
put, z = 2, we get 6 = A(0) + B(0) + C(1), C = 6,
put, z = 0, we get −2 = A(4) + B(2) + C(−1), −2 = 8 + 2B − 6, B = 0. Substituting
the value of A, B and C in the above equation
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 79 / 99
80. Continued Unit - II Problems on Inverse Z - Transforms ...
4z − 2
(z − 1)(z − 2)2
=
2
(z − 1)
+
0
(z − 2)
+
6
(z − 2)2
4z − 2
(z − 1)(z − 2)2
=
2
(z − 1)
+
6
(z − 2)2
(10)
Using Eqn. (10) in Eqn. (9)
U(z)
z
=
2
(z − 1)
+
6
(z − 2)2
Multiply throughout by z
U(z) = 2
z
(z − 1)
+ 6
z
(z − 2)2
Applying inverse Z - transforms on both sides
Z−1
[U(z)] = 2Z−1
[
z
(z − 1)
] + 6Z−1
[
z
(z − 2)2
]
Z−1
[
4z2
− 2z
z3 − 5z2 + 8z − 4
] = 2(1) + 6n(2)n−1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 80 / 99
81. Continued Unit - II Problems on Inverse Z - Transforms ...
Z−1
[
4z2
− 2z
z3 − 5z2 + 8z − 4
] = 2 + 6n(2)n−1
Problem - 05 : Find the inverse Z - transforms of 8z−z3
(4−z)3 .
Solution: by data 8z−z3
(4−z)3 , Take
U(z) =
8z − z3
(−1)3(z − 4)3
=
8z − z3
−(z − 4)3
U(z) =
z3
− 8z
(z − 4)3
We can observe in the denominator expression has repeated linear factors and we
have Z−1
[ z
(z−4)
] = 4n
, Z−1
[ 4z
(z−4)2 ] = n4n
and Z−1
[4z2
+16z
(z−4)3 ] = n2
4n
We have, apply
the partial fractions for the RHS expression for the above equation
z3
− 8z
(z − 4)3
= A
z
z − 4
+ B
4z
(z − 4)2
+ C
4z2
+ 16z
(z − 4)3
z3
− 8z = Az(z − 4)2
+ 4Bz(z − 4) + 4Cz(z + 4)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 81 / 99
82. Continued Unit - II Problems on Inverse Z - Transforms ...
Canceling z on both sides
z2
− 8 = A(z − 4)2
+ 4B(z − 4) + 4C(z + 4)
put, z = 4, we get −8 = A(0) + 4B(0) + 4C(8), C = 1
4
.
Equating the coefficient of z2
on both sides, we get, A = 1.
Again by equating the coefficient of z on both sides, we have, 0 = −8A + 4B + 4C,
we get, B = 7
4
. Substituting the value of A, B and C in the above equation
z3
− 8z
(z − 4)3
= (1)
z
z − 4
+ (
7
4
)
4z
(z − 4)2
+ (
1
4
)
4z2
+ 16z
(z − 4)3
Applying the inverse Z - transforms on both sides
Z−1
[
z3
− 8z
(z − 4)3
] = Z−1
[
z
z − 4
] +
7
4
Z−1
[
4z
(z − 4)2
] +
1
4
Z−1
[
4z2
+ 16z
(z − 4)3
]
Using the above results, we get
Z−1
[
z3
− 8z
(z − 4)3
] = 4n
+
7
4
n4n
+
1
4
n2
4n
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 82 / 99
83. Continued Unit - II Problems on Inverse Z - Transforms ...
Problem - 06 : Find the inverse Z - transforms of z3
−20z
(z−2)3(z−4)
.
Solution: by data z3
−20z
(z−2)3(z−4)
, Take
U(z) =
z3
− 20z
(z − 2)3(z − 4)
We can observe in the denominator expression has repeated linear factors in first
factor and we have Z−1
[ z
(z−2)
] = 2n
, Z−1
[ 2z
(z−2)2 ] = n2n
, Z−1
[2z2
+4z
(z−2)3 ] = n2
4n
and
Z−1
[ z
(z−4)
] = 4n
We have, apply the partial fractions for the RHS expression for the
above equation
z3
− 20z
(z − 2)3(z − 4)
= A
z
z − 2
+ B
2z
(z − 2)2
+ C
2z2
+ 4z
(z − 2)3
+ D
z
z − 4
z3
− 20z = Az(z − 2)2
(z − 4) + 2Bz(z − 2)(z − 4) + C(2z2
+ 4z)(z − 4) + Dz(z − 2)3
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 83 / 99
84. Continued Unit - II Problems on Inverse Z - Transforms ...
Canceling z on both sides
z2
− 20 = A(z − 2)2
(z − 4) + 2B(z − 2)(z − 4) + C(2z + 4)(z − 4) + D(z − 2)3
put, z = 2, we get −16 = A(0) + 2B(0) + C(−16) + D(0), C = 1.
put, z = 4, we get −4 = A(0) + 2B(0) + C(0) + D(8), D = −1
2
.
Equating the coefficient of z3
on both sides, we get, A + D = 0, A = 1
2
.
put, z = 0, we get −20 = A(4)(−4) + 2B(8) + C(−16) + D(−8), B = 0. Substituting
the value of A, B, C and D in the above equation
z3
− 20z
(z − 2)3(z − 4)
= (
1
2
)
z
z − 2
+ (0)
2z
(z − 2)2
+ (1)
2z2
+ 4z
(z − 2)3
+ (
−1
2
)
z
z − 4
Applying the inverse Z - transforms on both sides
Z−1
[
z3
− 20z
(z − 2)3(z − 4)
] =
1
2
Z−1
[
z
z − 2
] + Z−1
[
2z2
+ 4z
(z − 2)3
] −
1
2
Z−1
[
z
z − 4
]
Using the above results, we get
Z−1
[
z3
− 20z
(z − 2)3(z − 4)
] =
1
2
2n
+ n2
2n
−
1
2
4n
= 2n−1
+ n2
2n
− 22n−1
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 84 / 99
85. Continued Unit - II on Application of the Z - Transforms to Solve the
Difference Equations ...
Introduction:
Definition of Difference equation: An equation which involving the set of values
of the dependent variable is known as a difference equation.
Let us consider the second order forward difference of y0 is equal to zero, that is
∆2
y0 = 0
∆(∆y0) = 0
we have, ∆y0 = y1 − y0
∆(y1 − y0) = 0
∆y1 − ∆y0 = 0
(y2 − y1) − (y1 − y0) = 0
y2 − 2y1 + y0 = 0
This kind of equation is called as a difference equation Examples:
1. yn+2 + 5yn+1 + 6yn = 3n
.
2. un+2 − 3yn+1 + 2yn = n2
.
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 85 / 99
86. Continued Unit - II on Application of the Z - Transforms to Solve the
Difference Equations ...
Working procedure to solve the difference equations:
Step - 01: Applying the Z - transforms on both sides.
Step - 02: Substituting the shifting property of Z - transforms, that is
Z[un+1] = z[U(z) − u0]
Z[un+2] = z2
[U(z) − u0 − u1
z
]
Z[un+3] = z3
[U(z) − u0 − u1
z
− u2
z2 ]
Step - 03: Substituting the given initial conditions and simplify U(z) in terms of z.
Step - 04: Applying the inverse Z - transforms on both sides, we get the solution of
the given difference equation.
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 86 / 99
87. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
Problem - 01: Solve the difference equation un+2 − 5un+1 + 6un = 2, given that
u0 = 3, u1 = 7, by using Z - transforms.
Solution: By data un+2 − 5un+1 + 6un = 2, and u0 = 3, u1 = 7
Applying the Z - transforms on both sides
Z[un+2] − 5Z[un+1] + 6Z[un] = Z[2]
Substituting the shifting property of the Z - transforms
z2
[U(z) − u0 −
u1
z
] − 5(z[U(z) − u0]) + 6(U(z)) = (
2z
z − 1
)
Substituting the given initial conditions u0 = 3 and u1 = 7
z2
[U(z) − (3) −
7
z
] − 5(z[U(z) − (3)]) + 6(U(z)) = (
2z
z − 1
)
Simplify
z2
U(z) − 3z2
− 7z − 5zU(z) + 15z + 6U(z) =
2z
z − 1
z2
U(z) − 5zU(z) + 6U(z) =
2z
z − 1
+ 3z2
+ 7z − 15z
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 87 / 99
88. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
(z2
− 5z + 6)U(z) =
2z
z − 1
+ 3z2
− 8z
(z − 2)(z − 3)U(z) =
2z + 3z2
(z − 1) − 8z(z − 1)
z − 1
U(z) =
3z3
− 11z2
+ 10z
(z − 1)(z − 2)(z − 3)
U(z)
z
=
3z2
− 11z + 10
(z − 1)(z − 2)(z − 3)
(11)
Now, applying the partial fractions for the RHS expression
3z2
− 11z + 10
(z − 1)(z − 2)(z − 3)
=
A
z − 1
+
B
z − 2
+
C
z − 3
3z2
− 11z + 10 = A(z − 2)(z − 3) + B(z − 1)(z − 3) + C(z − 1)(z − 2)
put, z = 1, we get, 2 = A(−1)(−2) + B(0) + C(0), A = 1.
put, z = 2, we get, 0 = A(0) + B(1)(−1) + C(0), B = 0.
put, z = 3, we get, 4 = A(0) + B(0) + C(2)(1), C = 2. Substituting the values of the
A, B and C in the above equation
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 88 / 99
89. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
3z2
− 11z + 10
(z − 1)(z − 2)(z − 3)
=
1
z − 1
+
0
z − 2
+
2
z − 3
3z2
− 11z + 10
(z − 1)(z − 2)(z − 3)
=
1
z − 1
+
2
z − 3
Using Eqn. (12) in Eqn. (11)
U(z)
z
=
1
z − 1
+
2
z − 3
(12)
Multiply throughout by z
U(z) =
z
z − 1
+ 2
z
z − 3
Applying inverse Z - transforms on both sides
Z−1
[U(z)] = Z−1
[
z
z − 1
] + 2Z−1
[
z
z − 3
]
but, Z−1
[U(z)] = un
un = (1) + 2(3n
) = 1 + 23n
This is required solution for the given difference equation.
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 89 / 99
90. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
Problem - 02: Solve yn+2 + 2yn+1 + yn = n, given that y0 = 0, y1 = 0, by using Z -
transforms.
Solution: Given yn+2 + 2yn+1 + yn = n and y0 = 0, y1 = 0
Applying the Z - transforms on both sides
Z[yn+2] + 2Z[yn+1] + Z[yn] = Z[n]
Substituting the shifting property of the Z - transforms
z2
[Y (z) − y0 −
y1
z
] + 2(z[Y (z) − y0]) + (Y (z)) = (
z
(z − 1)2
)
Substituting the given initial conditions y0 = 0 and y1 = 0
z2
[Y (z) − (0) −
0
z
] + 2(z[Y (z) − (0)]) + (Y (z)) =
z
(z − 1)2
Simplify
z2
Y (z) − (0) − (0) + 2zY (z) − (0) + Y (z) =
z
(z − 1)2
(z2
+ 2z + 1)Y (z) =
z
(z − 1)2
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 90 / 99
91. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
(z + 1)2
Y (z) =
z
(z − 1)2
Y (z) =
z
(z − 1)2(z + 1)2
Y (z)
z
=
1
(z − 1)2(z + 1)2
(13)
Applying the partial fractions for the RHS expression
1
(z − 1)2(z + 1)2
=
A
(z − 1)
+
B
(z − 1)2
+
C
(z + 1)
+
D
(z + 1)2
1 = A(z − 1)(z + 1)2
+ B(z + 1)2
+ C(z + 1)(z − 1)2
+ D(z − 1)2
put, z = 1, we get, 1 = A(0) + B(4) + C(0) + D(0), B = 1
4
.
put, z = −1, we get, 1 = A(0) + B(0) + C(0) + D(4), D = 1
4
.
Rewrite the above equation
1 = A(z3
+ z2
− z − 1) + B(z2
+ 2z + 1) + C(z3
− z2
− z + 1) + D(z2
− 2z + 1)
Equating the coefficient of z3
on both sides A + C = 0 Equating the coefficient of z2
on both sides A + B − C + D = 0, A − C = −1
2
,
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 91 / 99
92. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
solving these two equations, we get A = −1
4
and C = 1
4
. Substituting the values of A,
B, C and D in the above equation.
1
(z − 1)2(z + 1)2
=
−1
4(z − 1)
+
1
4(z − 1)2
+
1
4(z + 1)
+
1
4(z + 1)2
(14)
Using Eqn. (14) in (13)
Y (z)
z
=
−1
4(z − 1)
+
1
4(z − 1)2
+
1
4(z + 1)
+
1
4(z + 1)2
Multiplying throughout by z
Y (z) = −
1
4
z
z − 1
+
1
4
z
(z − 1)2
+
1
4
z
z + 1
+
1
4
z
(z + 1)2
Applying the inverse Z - transforms on both sides
Z−1
[Y (z)] = −
1
4
[Z−1
[
z
z − 1
] + Z−1
[
z
(z − 1)2
] + Z−1
[
z
z + 1
] + Z−1
[
z
(z + 1)2
]]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 92 / 99
93. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
but, Z−1
[Y (z)] = yn
yn =
1
4
[−(1) + n(1)n−1
+ (−1)n
+ n(−1)n−1
] =
1
4
[−1 + n + (−1)n
+ n(−1)n−1
]
This is required solution for the given difference equation.
Problem - 03: Solve yn+2 − 4yn+1 + 3yn = H(n), given that y0 = 0, y1 = 0, by
using Z - transforms.
Solution: Given yn+2 − 4yn+1 + 3yn = H(n) and y0 = 0, y1 = 0
Applying the Z - transforms on both sides
Z[yn+2] − 4Z[yn+1] + 3Z[yn] = Z[H(n)]
Substituting the shifting property of the Z - transforms
z2
[Y (z) − y0 −
y1
z
] − 4(z[Y (z) − y0]) + 3(Y (z)) = (
z
(z − 1)
)
Substituting the given initial conditions y0 = 0 and y1 = 0
z2
[Y (z) − (0) −
0
z
] − 4(z[Y (z) − (0)]) + 3(Y (z)) =
z
(z − 1)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 93 / 99
94. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
Simplify
z2
Y (z) − (0) − (0) − 4zY (z) − (0) + 3Y (z) =
z
(z − 1)
(z2
− 4z + 3)Y (z) =
z
(z − 1)
(z − 1)(z − 3)Y (z) =
z
(z − 1)
Y (z) =
z
(z − 1)2(z − 3)
Y (z)
z
=
1
(z − 1)2(z − 3)
(15)
Applying the partial fraction for RHS expression in the above equation
1
(z − 1)2(z − 3)
=
A
(z − 1)
+
B
(z − 1)2
+
C
(z − 3)
1 = A(z − 1)(z − 3) + B(z − 3) + C(z − 1)2
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 94 / 99
95. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
put, z = 1, we get, 1 = A(0) + B(−2) + C(0), B = −1
2
.
put, z = 3, we get, 1 = A(0) + B(0) + C(4), C = 1
4
.
put, z = 0, we get, 1 = A(3) + B(−3) + C(1), A = −1
4
. Substituting the values of the
A, B and C in the above equation
1
(z − 1)2(z − 3)
=
−1
4(z − 1)
+
−1
2(z − 1)2
+
1
4(z − 3)
(16)
Using Eqn. (16) IN Eqn.. (15)
Y (z)
z
=
−1
4(z − 1)
+
−1
2(z − 1)2
+
1
4(z − 3)
Multiplying throughout by z
Y (z) = −
1
4
z
(z − 1)
−
1
2
z
2(z − 1)2
+
1
4
z
(z − 3)
Applying the inverse Z - transforms on both sides
Z−1
[Y (z)] = −
1
4
Z−1
[
z
(z − 1)
] −
1
2
Z−1
[
z
(z − 1)2
] +
1
4
Z−1
[
z
(z − 3)
]
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 95 / 99
96. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
But, Z−1
[Y (z)] = yn
yn = −
1
4
(1) −
1
2
(n) +
1
4
(3)n
=
−1
4
−
n
2
+
(3)n
4
This is required solution for the given difference equation.
Problem - 04: Solve the difference equation un+2 − 5un+1 + 6un = Hn, given that
u0 = 0, u1 = 1, where Hn is a unit step sequence, by using Z - transforms.
Solution: Given un+2 − 5un+1 + 6un = Hn and u0 = 0, u1 = 1
Applying the Z - transforms on both sides
Z[un+2] − 5Z[un+1] + 6Z[un] = Z[Hn]
Substituting the shifting property of the Z - transforms
z2
[U(z) − u0 −
u1
z
] − 5(z[U(z) − u0]) + 6(U(z)) = (
z
(z − 1)
)
Substituting the given initial conditions u0 = 0 and u1 = 1
z2
[U(z) − (0) −
1
z
] − 5(z[U(z) − (0)]) + 6(U(z)) =
z
(z − 1)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 96 / 99
97. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
z2
U(z) − 0 − z − 5zU(z) − 0 + 6U(z) =
z
(z − 1)
(z2
− 5z + 6)U(z) =
z
(z − 1)
+ z =
z + z2
− z
(z − 1)
(z − 2)(z − 3)U(z) =
z2
(z − 1)
U(z) =
z2
(z − 1)(z − 2)(z − 3)
U(z)
z
=
z
(z − 1)(z − 2)(z − 3)
(17)
Applying the partial fractions for the RHS expression
z
(z − 1)(z − 2)(z − 3)
=
A
(z − 1)
+
B
(z − 2)
+
C
(z − 3)
z = A(z − 2)(z − 3) + B(z − 1)(z − 3) + C(z − 1)(z − 2)
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 97 / 99
98. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
put, z = 1, we get, 1 = A(2) + B(0) + C(0), A = 1
2
.
put, z = 2, we get, 2 = A(0) + B(−2) + C(0), B = −1
2
.
put, z = 3, we get, 3 = A(0) + B(0) + C(2), C = 3
2
. Substituting the values of the A,
B and C in the above equation
z
(z − 1)(z − 2)(z − 3)
=
1
2(z − 1)
+
−1
2(z − 2)
+
3
2(z − 3)
Using Eqn. (18) in Eqn. (17)
U(z)
z
=
1
2(z − 1)
+
−1
2(z − 2)
+
3
2(z − 3)
Multiplying throughout by z
U(z) =
1
2
z
z − 1
−
1
2
z
z − 2
+
3
2
z
z − 3
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 98 / 99
99. Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
Applying the inverse Z - transforms on both sides
Z−1
[U(z)] =
1
2
Z−1
[
z
z − 1
] −
1
2
Z−1
[
z
z − 2
] +
3
2
Z−1
[
z
z − 3
]
But, Z−1
[U(z)] = un
un =
1
2
(1) −
1
2
(2)n
+
3
2
(3)n
=
1
2
− (2)n−1
+
(3)n+1
2
This is required solution for the given difference equation.
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 99 / 99