Signals and Systems Jayapradha.pdf

SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDYALAYA
(University established under section 3of UGC Act 1956)
(Accredited with ‘A’ Grade by NAAC)
Enathur, Kanchipuram – 631 561
DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Prepared by:
Dr. V.Jayapradha, Assistant Professor
Course Material
Signals and Systems
FULL TIME B.E II YEAR, IIIrd
SEMESTER

Sri ChandrasekharendraSaraswathiViswaMahavidyalaya
Department of Electronics and Communication Engineering
SIGNALS AND SYSTEMS
Prerequisite: Basic knowledge on Mathematics which includes Fourier Series and Laplace
Transform
OBJECTIVES:
To understand the properties and representation of discrete and continuous signals
To understand the sampling process and analysis of discrete systems using z-transforms
To study the analysis and synthesis of discrete time systems
UNIT-I CLASSIFICATION OF SIGNALS AND SYSTEMS
Continuous Time Signals(CT signals) , Discrete time signals (DT signals) step, ramp, pulse,
impulse, exponential, Classification of CT and DT signals - periodic, aperiodic, random signals
- CT systems and DT systems, Basic properties of systems - Linear Time invariant systems and
properties
UNIT-II ANALYSIS OF CONTINUOUS TIME SIGNALS
Fourier Series Analysis- Representation of periodic signals in trigonometric and exponential
form, Spectrum of CT signals-Fourier Transform and Laplace Transform in signal analysis
UNIT-III LINEAR TIME INVARIANT – CONTINUOUS TIME SYSTEMS
Differential Equation - Block diagram Representation, Impulse response, Convolution Integral-
Frequency response, Fourier and Laplace Transforms in analysis, State variable equations and
Matrix representation of systems
UNIT-IV ANALYSIS OF DISCRETE TIME SYSTEMS
Sampling of CT signals and aliasing, DTFT and properties, Z-transform and properties of Z-
transform
UNIT-V LINEAR TIME INVARIANT – DISCRETE TIME SYSTEMS
Difference equations, Block Diagram representation, Impulse response, Convolution sum, LTI
systems analysis using DTFT and Z-transforms, State variable equations and matrix
representation of systems
OUTCOMES:
Students will be able to:
Understand the properties and representation of continuous and discrete time signals.
Analyze the discrete time systems using z-transforms.
TEXTBOOKS:
1. P.Ramesh Babu & R.Anandanatarajan, signals and systems, 4th edition, scitech publication
private limited, 2009.
2. Allam V. Oppenheim, S.Wilsky and S.H.Nawab, Signals and systems, Pearson
Education,2007
REFERENCES:
1. Robert A.Gabel and Richard A.Roberts, Signals & Linear Systems, John Wiley & Sons 2004
2. Simon Haykins and Barry Van Veen, Signals and Systems, John Wiley & Sons,2004

SIGNALS AND SYSTEMS
Prerequisite: Basic knowledge on Mathematics
Objectives:
To understand the properties and representation of discrete and continuous signals
Unit-I Classification of Signals and Systems
Continuous Time Signals(CT signals) , Discrete time signals (DT signals) step, ramp,
pulse, impulse, exponential, Classification of CT and DT signals - periodic, aperiodic,
random signals - CT systems and DT systems, Basic properties of systems - Linear
Time invariant systems and properties
Outcome:
On completion of the course, student will be able to
Understand mathematical description and representation of continuous and
discrete time signals and systems.

UNIT- I – Pre – Test - MCQ
1. _________ is defined as any physical quantity that varies with time, space or any other
independent variable
A. Signal
B. system
C. signals & system
D. none of the above
2. Which of the following time system operates with a continuous time signal & produces a
continuous time output signal
A. Continuous time system
B. Discrete time system
C. Time invariant system
D. Time variant system
3. Frequency and Time period are ______
A. Proportional to each other
B. Inverse of each other
C. Same
D. None of the above
4. Which of the following can weaken a signal
A. Attenuation
B. Distortion
C. Noise
D. All of the above
5. Which is a physical device that performs an operation on the signal
Signal
Response
System
All the above

SUBJECT: EC8352- Signals and Systems
AUTHOR: M.SHAKUNTHALA, A.P/ECE/RMDEC
Dr.M.N.VIMAL KUMAR, AP/ECE/RMDEC
1
0 1 0 1 0 0 1 0
Unit 1: Classification of signals and systems
1.1 Signal
Signal is one that carries information and is defined as a physical quantity that varies
with one or more independent variable.
Example: Music, speech
1.2 Classification of signals
1.2.1 Analog and Digital signal
Analog signal:
A signal that is defined for every instants of time is known as analog signal. Analog
signals are continuous in amplitude and continuous in time. It is denoted by x(t). It is also called
as Continuous time signal. Example for Continuous time signal is shown in Fig 1.1
Fig 1.1 Continuous time signal
Fig 1.2 Digital Signal
Digital signal:
The signals that are discrete in time and quantized in amplitude is called digital signal
(Fig 1.2)
1.2.2 Continuous time and discrete time signal
Continuous time signal:
A signal that is defined for every instants of time is known as continuous time signal.
Continuous time signals are continuous in amplitude and continuous in time. It is denoted by
x(t) and shown in Fig 1.1
Discrete time signal:
A signal that is defined for discrete instants of time is known as discrete time signal.
Discrete time signals are continuous in amplitude and discrete in time. It is also obtained by
sampling a continuous time signal. It is denoted by x(n) and shown in Fig 1.3
Amplitude Sine Wave
time
www.studentsfocus.com

2
Fig 1.3 Discrete time signal
1.2.3 Even (symmetric) and Odd (Anti-symmetric) signal
Continuous domain:
Even signal:
A signal that exhibits symmetry with respect to t=0 is called even signal
Even signal satisfies the condition 𝑥(𝑡) = 𝑥(−𝑡)
Odd signal:
A signal that exhibits anti-symmetry with respect to t=0 is called odd signal
Odd signal satisfies the condition 𝑥(𝑡) = −𝑥(−𝑡)
Even part 𝒙𝒆(𝒕) and Odd part 𝒙𝟎(𝒕) of continuous time signal 𝒙 𝒕 :
Even part 𝑥𝑒 𝑡 =
1
2
[𝑥 𝑡 + 𝑥 −𝑡 ]
Odd part 𝑥𝑜 𝑡 =
1
2
[𝑥 𝑡 − 𝑥 −𝑡 ]
Discrete domain:
Even signal:
A signal that exhibits symmetry with respect to n=0 is called even signal
Even signal satisfies the condition 𝑥(𝑛) = 𝑥(−𝑛).
Odd signal:
A signal that exhibits anti-symmetry with respect to n=0 is called odd signal Odd signal
satisfies the condition 𝑥(𝑛) = −𝑥(−𝑛).
Even part 𝒙𝒆(𝒏) and Odd part 𝒙𝟎(𝒏) of discrete time signal 𝒙 𝒏 :
Even part 𝑥𝑒 𝑛 =
1
2
[𝑥 𝑛 + 𝑥 −𝑛 ]
Odd part 𝑥𝑜 𝑛 =
1
2
[𝑥 𝑛 − 𝑥 −𝑛 ]

3
1.2.4 Periodic and Aperiodic signal
Periodic signal:
A signal is said to periodic if it repeats again and again over a certain period of time.
Aperiodic signal:
A signal that does not repeat at a definite interval of time is called aperiodic signal.
Continuous domain:
A Continuous time signal is said to periodic if it satisfies the condition
𝑥 𝑡 = 𝑥 𝑡 + 𝑇 𝑤𝑕𝑒𝑟𝑒 𝑇 𝑖𝑠 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑
If the above condition is not satisfied then the signal is said to be aperiodic
Fundamental time period 𝐓 =
𝟐𝛑
Ω
, where Ω is fundamental angular frequency in rad/sec
Discrete domain:
A Discrete time signal is said to periodic if it satisfies the condition
𝑥 𝑛 = 𝑥 𝑛 + 𝑁 𝑤𝑕𝑒𝑟𝑒 𝑁 𝑖𝑠 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑
If the above condition is not satisfied then the signal is said to be aperiodic
Fundamental time period 𝐍 =
𝟐𝛑𝐦
𝛚
, where ω is fundamental angular frequency in rad/sec, 𝑚 is
smallest positive integer that makes N as positive integer
1.2.5 Energy and Power signal
Energy signal:
The signal which has finite energy and zero average power is called energy signal. The
non periodic signals like exponential signals will have constant energy and so non periodic
signals are energy signals.
i.e., For energy signal, 0 < 𝐸 < ∞ 𝑎𝑛𝑑 𝑃 = 0
For Continuous time signals,
𝐸𝑛𝑒𝑟𝑔𝑦 𝐸 = lim
𝑇→∞
|𝑥 𝑡 |2
𝑑𝑡
𝑇
−𝑇
For Discrete time signals,
𝐸𝑛𝑒𝑟𝑔𝑦 𝐸 = lim
𝑁→∞
𝑥(𝑛) 2
𝑁
𝑛=−𝑁
Power signal:
The signal which has finite average power and infinite energy is called power signal. The
periodic signals like sinusoidal complex exponential signals will have constant power and so
periodic signals are power signals.
i.e., For power signal, 0 < 𝑃 < ∞ 𝑎𝑛𝑑 𝐸 = ∞
For Continuous time signals,

4
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 𝑃 = lim
𝑇→∞
1
2𝑇
|𝑥 𝑡 |2
𝑑𝑡
𝑇
−𝑇
For Discrete time signals,
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 𝑃 = lim
𝑁→∞
1
2𝑁 + 1
𝑥(𝑛) 2
𝑁
𝑛=−𝑁
1.2.6 Deterministic and Random signals
Deterministic signal:
A signal is said to be deterministic if there is no uncertainity over the signal at any
instant of time i.e., its instantaneous value can be predicted. It can be represented by
mathematical equation.
Example: sinusoidal signal
Random signal (Non-Deterministic signal):
A signal is said to be random if there is uncertainity over the signal at any instant of time
i.e., its instantaneous value cannot be predicted. It cannot be represented by mathematical
equation.
Example: noise signal
Deterministic signal
Random signal
1.2.7 Causal and Non-causal signal
Continuous domain:
Causal signal:
A signal is said to be causal if it is defined for t≥0.
𝑖. 𝑒., 𝑥 𝑡 = 0 𝑓𝑜𝑟 𝑡 < 0
Non-causal signal:
A signal is said to be non-causal, if it is defined for t< 0 or for both
𝑡 < 0 and 𝑡 ≥ 0
𝑖. 𝑒., 𝑥 𝑡 ≠ 0 𝑓𝑜𝑟 𝑡 < 0
When a non-causal signal is defined only for t<0, it is called as anti-causal signal
Random Signal
Sinusoidal Signal

5
Discrete domain:
Causal signal:
A signal is said to be causal, if it is defined for n≥0.
𝑖. 𝑒., 𝑥 𝑛 = 0 𝑓𝑜𝑟 𝑛 < 0
Non-causal signal:
A signal is said to be non-causal, if it is defined for n< 0
or for both n < 0 𝑎𝑛𝑑 𝑛 ≥ 0
𝑖. 𝑒., 𝑥 𝑛 ≠ 0 𝑓𝑜𝑟 𝑛 < 0
When a non-causal signal is defined only for n<0, it is called as anti-causal signal
1.3 Basic(Elementary or Standard) continuous time signals
1.3.1 Step signal
Unit Step signal is defined as
𝑢 𝑡 = 1 𝑓𝑜𝑟 𝑡 ≥ 0
= 0 𝑓𝑜𝑟 𝑡 < 0
Unit step signal
1.3.2 Ramp signal
Unit ramp signal is defined as
𝑟 𝑡 = 𝑡 𝑓𝑜𝑟 𝑡 ≥ 0
= 0 𝑓𝑜𝑟 𝑡 < 0
Unit ramp signal
1.3.3 Parabolic signal
Unit Parabolic signal is defined as
r (t)
t
0 1 2 3 . .
.
.
3
2
1
u (t)
t
0
1

6
𝑥 𝑡 =
𝑡2
2
𝑓𝑜𝑟 𝑡 ≥ 0
= 0 𝑓𝑜𝑟 𝑡 < 0
Unit Parabolic signal
Relation between Unit Step signal, Unit ramp signal and Unit Parabolic signal:
x Unit ramp signal is obtained by integrating unit step signal
𝑖. 𝑒. , 𝑢 𝑡 𝑑𝑡 = 1𝑑𝑡 = 𝑡 = 𝑟(𝑡)
x Unti Parabolic signal is obtained by integrating unit ramp signal
𝑖. 𝑒. , 𝑟 𝑡 𝑑𝑡 = 𝑡𝑑𝑡 =
𝑡2
2
= 𝑝(𝑡)
x Unit step signal is obtained by differentiating unit ramp signal
𝑖. 𝑒. ,
𝑑
𝑑𝑡
𝑟 𝑡 =
𝑑
𝑑𝑡
𝑡 = 1 = 𝑢(𝑡)
x Unit ramp signal is obtained by differentiating unit Parabolic signal
𝑖. 𝑒. ,
𝑑
𝑑𝑡
𝑝 𝑡 =
𝑑
𝑑𝑡
𝑡2
2
=
1
2
2𝑡 = 𝑡 = 𝑟(𝑡)
1.3.4 Unit Pulse signal is defined as
∏ 𝑡 = 1 𝑓𝑜𝑟 𝑡 ≤
1
2
= 0 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒
Unit Pulse signal
1.3.5 Impulse signal
Unit Impulse signal is defined as
𝛿 𝑡 = 0 𝑓𝑜𝑟 𝑡 ≠ 0
𝛿 𝑡 𝑑𝑡 = 1
∞
−∞
Unit Impulse signal
-1/2 1/2 t
1
∏ (t)
:
4.5
2
0.5
p(t)
0 1 2 3 . . . t

7
Properties of Impulse signal:
Property 1:
𝒙(𝒕)𝜹 𝒕
∞
−∞
𝒅𝒕 = 𝒙(𝟎)
Proof:
𝑥(𝑡)𝛿 𝑡
∞
−∞
𝑑𝑡 = 𝑥 0 𝛿 0 = 𝑥 0 [∵ 𝛿 𝑡 𝑒𝑥𝑖𝑠𝑡𝑠 𝑜𝑛𝑙𝑦 𝑎𝑡 𝑡 = 0 𝑎𝑛𝑑 𝛿 0 = 1]
Thus proved
Property 2:
𝒙(𝒕)𝜹 𝒕 − 𝒕𝟎
∞
−∞
𝒅𝒕 = 𝒙(𝒕𝟎)
Proof:
𝑥(𝑡)𝛿 𝑡 − 𝑡0
∞
−∞
𝑑𝑡 = 𝑥 𝑡0 𝛿 𝑡0 − 𝑡0 = 𝑥 𝑡0 𝛿 0 = 𝑥 𝑡0
∵ 𝛿 𝑡 − 𝑡0 𝑒𝑥𝑖𝑠𝑡𝑠 𝑜𝑛𝑙𝑦 𝑎𝑡 𝑡 = 𝑡0 𝑎𝑛𝑑 𝛿 0 = 1
Thus proved
1.3.6 Sinusoidal signal
Cosinusoidal signal is defined as
𝑥 𝑡 = 𝐴𝑐𝑜𝑠 𝛺𝑡 + 𝛷
Sinusoidal signal is defined as
𝑥 𝑡 = 𝐴𝑠𝑖𝑛 𝛺𝑡 + 𝛷
where Ω = 2πf =
2π
T
and Ω is angular frequency in rad/sec
f is frequency in cycles/sec or Hertz and
A is amplitude
T is time period in seconds
𝛷 is phase angle in radians
Cosinusoidal signal
𝑤𝑕𝑒𝑛 𝜙 = 0, 𝑥 𝑡 = 𝐴𝑐𝑜𝑠 𝛺𝑡
Sinusoidal signal
𝑤𝑕𝑒𝑛 𝜙 = 0, 𝑥 𝑡 = 𝐴𝑠𝑖𝑛 𝛺𝑡
x(t)
A
-A
t
φ=0
x(t)
A
-A
t
φ=0

8
Cosinusoidal signal
Sinusoidal signal
1.3.7 Exponential signal
Real Exponential signal is defined as 𝑥 𝑡 = 𝐴𝑒𝑎𝑡
where A is amplitude
Depending on the value of ‘a’ we get dc signal or growing exponential signal or decaying
exponential signal
DC signal Exponentially
growing signal
Exponentially
decaying signal
Complex exponential signal is defined as 𝑥 𝑡 = 𝐴𝑒𝑠𝑡
where 𝐴 is amplitude, s is complex variable and 𝑠 = 𝜎 + 𝑗Ω
𝑥 𝑡 = 𝐴𝑒𝑠𝑡
= 𝐴𝑒 𝜎+𝑗Ω 𝑡
= 𝐴𝑒𝜎𝑡
𝑒𝑗Ω𝑡
= 𝐴𝑒𝜎𝑡
(𝑐𝑜𝑠Ω𝑡 + 𝑗𝑠𝑖𝑛Ω𝑡)
𝑤𝑕𝑒𝑛 𝜎 = +𝑣𝑒, 𝑡𝑕𝑒𝑛 𝑥 𝑡 = 𝐴𝑒𝜎𝑡
(𝑐𝑜𝑠Ω𝑡 + 𝑗𝑠𝑖𝑛Ω𝑡),
𝑤𝑕𝑒𝑟𝑒 𝑥𝑟 𝑡 = 𝐴𝑒𝜎𝑡
𝑐𝑜𝑠Ω𝑡 𝑎𝑛𝑑 𝑥𝑖 𝑡 = 𝐴𝑒𝜎𝑡
𝑠𝑖𝑛Ω𝑡
Exponentially growing Cosinusoidal signal
Exponentially growing sinusoidal signal
𝑤𝑕𝑒𝑛 𝜎 = −𝑣𝑒, 𝑡𝑕𝑒𝑛 𝑥 𝑡 = 𝐴𝑒−𝜎𝑡
(𝑐𝑜𝑠Ω𝑡 + 𝑗𝑠𝑖𝑛Ω𝑡),
𝑤𝑕𝑒𝑟𝑒 𝑥𝑟 𝑡 = 𝐴𝑒−𝜎𝑡
𝑐𝑜𝑠Ω𝑡 𝑎𝑛𝑑 𝑥𝑖 𝑡 = 𝐴𝑒−𝜎𝑡
𝑠𝑖𝑛Ω𝑡
xi(t)
A
t
xr(t)
A
t
xi(t)
A
t
xr(t)
A
t
x(t)
A t
a<0
x(t)
A t
a>0
x(t)
A t
a=0

9
Exponentially decaying Cosinusoidal signal Exponentially decaying sinusoidal signal
1.4 Basic(Elementary or Standard) Discrete time signals
1.4.1 Step signal
Unit Step signal is defined as
𝑢 𝑛 = 1 𝑓𝑜𝑟 𝑛 ≥ 0
= 0 𝑓𝑜𝑟 𝑛 < 0
Unit step signal
1.4.2 Unit Ramp signal
Unit Ramp signal is defined as
𝑟 𝑛 = 𝑛 𝑓𝑜𝑟 𝑛 ≥ 0
= 0 𝑓𝑜𝑟 𝑛 < 0
1.4.3 Pulse signal (Rectangular pulse function)
Pulse signal is defined as
𝑥 𝑛 = 𝐴 𝑓𝑜𝑟 𝑛1 ≤ 𝑛 ≤ 𝑛2
= 0 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒
Pulse signal
1.4.4 Unit Impulse signal
Unit Impulse signal is defined as
A
n1 -1 0 1 n2 n
x(n)
. . . . . .
r(n)
:
4
3
2
1
Unit Ramp signal
. . .
n
0 1 2 3 4
1
0 1 2 n
u(n)
3 4
. . .

10
𝛿 𝑛 = 1 𝑓𝑜𝑟 𝑛 = 0
𝛿 𝑛 = 0 𝑓𝑜𝑟 𝑛 ≠ 0
Unit Impulse signal
1.4.5 Sinusoidal signal
Cosinusoidal signal is defined as
𝑥 𝑛 = 𝐴𝑐𝑜𝑠(𝜔𝑛)
Sinusoidal signal is defined as
𝑥 𝑛 = 𝐴𝑠𝑖𝑛(𝜔𝑛)
where 𝜔 = 2πf =
2π
N
m and 𝜔 is frequency in radians/sample
m is smallest integer
f is frequency in cycles/sample, A is amplitude
Cosinusoidal signal
Cosinusoidal signal
Sinusoidal signal
Sinusoidal signal
1.4.6 Exponential signal
Real Exponential signal is defined as 𝑥 𝑛 = 𝑎𝑛
𝑓𝑜𝑟 𝑛 ≥ 0
Decreasing exponential signal Increasing exponential signal
Complex Exponential signal is defined as 𝑥 𝑛 = 𝑎𝑛
𝑒𝑗 (𝜔0𝑛)
= 𝑎𝑛
[𝑐𝑜𝑠𝜔0𝑛 + 𝑗𝑠𝑖𝑛𝜔0𝑛]
𝑤𝑕𝑒𝑟𝑒 𝑥𝑟 𝑛 = 𝑎𝑛
𝑐𝑜𝑠𝜔0𝑛 𝑎𝑛𝑑 𝑥𝑖 𝑛 = 𝑎𝑛
𝑠𝑖𝑛𝜔0𝑛

11
Exponentially decreasing Cosinusoidal signal
Exponentially growing Cosinusoidal signal
Exponentially decreasing sinusoidal signal
Exponentially growing sinusoidal signal
1.5 Classification of System
x Continuous time and Discrete time system
x Linear and Non-Linear system
x Static and Dynamic system
x Time invariant and Time variant system
x Causal and Non-Causal system
x Stable and Unstable system
1.5.1 Continuous time and Discrete time system
Continuous time system:
Continuous time system operates on a continuous time signal (input or excitation) and
produces another continuous time signal (output or response) which is shown in Fig 1.84. The
signal 𝑥(𝑡) is transformed by the system into signal 𝑦(𝑡), this transformation can be expressed
as,

12
Response 𝑦 𝑡 = 𝑇 𝑥 𝑡
where 𝑥(𝑡) is input signal, 𝑦(𝑡) is output signal, and T denotes transformation
Fig 1.84 Representation of continuous time system
Discrete time system:
Discrete time system operates on a discrete time signal (input or excitation) and
produces another discrete time signal (output or response) which is shown in Fig 1.85.
The signal 𝑥(𝑛) is transformed by the system into signal 𝑦(𝑛), this transformation can be
expressed as,
Response 𝑦 𝑛 = 𝑇 𝑥 𝑛
where x(n) is input signal, y(n) is output signal, and T denotes transformation
Fig 1.85 Representation of discrete time system
1.5.2 Linear system and Non Linear system
Continuous time domain:
Linear system:
A system is said to be linear if it obeys superposition theorem. Superposition theorem
states that the response of a system to a weighted sum of the signals is equal to the
corresponding weighted sum of responses to each of the individual input signals.
Condition for Linearity:
𝑇 𝑎𝑥1 𝑡 + 𝑏𝑥2 𝑡 = 𝑎𝑦1 𝑡 + 𝑏𝑦2(𝑡)
𝑤𝑕𝑒𝑟𝑒 𝑦1 𝑡 𝑎𝑛𝑑 𝑦2 𝑡 𝑎𝑟𝑒 𝑡𝑕𝑒 𝑟𝑒𝑠𝑝𝑜𝑛𝑠𝑒𝑠 𝑜𝑓 𝑥1 𝑡 𝑎𝑛𝑑 𝑥2 𝑡 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦
Non Linear system:
A system is said to be Non linear if it does not obeys superposition theorem.
𝑖. 𝑒. , 𝑇 𝑎𝑥1 𝑡 + 𝑏𝑥2 𝑡 ≠ 𝑎𝑦1 𝑡 + 𝑏𝑦2(𝑡)
where y1 t and y2 t are the responses of x1 t and x2 t respectively
Discrete time domain:
Linear system:
A system is said to be linear if it obeys superposition theorem. Superposition theorem
states that the response of a system to a weighted sum of the signals is equal to the
corresponding weighted sum of responses to each of the individual input signals.
Condition for Linearity:
𝑇 𝑎𝑥1 𝑛 + 𝑏𝑥2 𝑛 = 𝑎𝑦1 𝑛 + 𝑏𝑦2(𝑛)
where y1 n and y2 n are the responses of x1 n and x2 n respectively
Non Linear system:
A system is said to be Non linear if it does not obeys superposition theorem.
𝑖. 𝑒. , 𝑇 𝑎𝑥1 𝑛 + 𝑏𝑥2 𝑛 ≠ 𝑎𝑦1 𝑛 + 𝑏𝑦2(𝑛)
T
𝑥(𝑛) 𝑦(𝑛)
T
𝑥(𝑡) 𝑦(𝑡)

13
where y1 n and y2 n are the responses of x1 n and x2 n respectively
1.5.3 Static (Memoryless) and Dynamic (Memory) system
Static system:
A system is said to be memoryless or static if the response of the system is due to
present input alone.
Example: 𝑦(𝑡) = 2𝑥(𝑡)
𝑦(𝑡) = 𝑥2
(𝑡) + 𝑥(𝑡)
Dynamic system:
A system is said to be memory or dynamic if the response of the system depends on
factors other than present input also.
Example: 𝑦(𝑡) = 2𝑥(𝑡) + 𝑥(−𝑡)
𝑦(𝑡) = 𝑥2
(𝑡) + 𝑥(2𝑡)
Static system:
A system is said to be memoryless or static if the response of the system is due to
present input alone.
Example: 𝑦(𝑛) = 𝑥(𝑛)
𝑦(𝑛) = 𝑥2
(𝑛) + 3𝑥(𝑛)
Dynamic system:
A system is said to be memory or dynamic if the response of the system depends on
factors other than present input also.
Example: 𝑦(𝑛) = 2𝑥(𝑛) + 𝑥(−𝑛)
𝑦(𝑛) = 𝑥2
(1 − 𝑛) + 𝑥(2𝑛)
1.5.4 Time invariant (Shift invariant) and Time variant (Shift variant) system
Time invariant system:
A system is said to time invariant if the relationship between the input and output does
not change with time.
If 𝑦 𝑡 = 𝑇 𝑥 𝑡
Then 𝑇 𝑥 𝑡 − 𝑡0 = 𝑦(𝑡 − 𝑡0) should be satisfied for the system to be time invariant
Time variant system:
A system is said to time variant if the relationship between the input and output changes
with time.
If 𝑦 𝑡 = 𝑇 𝑥 𝑡
Then 𝑇 𝑥 𝑡 − 𝑡0 ≠ 𝑦(𝑡 − 𝑡0) should be satisfied for the system to be time variant

14
Time invariant system:
A system is said to time invariant if the relationship between the input and output does
not change with time.
If 𝑦 𝑛 = 𝑇 𝑥 𝑛
Then 𝑇 𝑥 𝑛 − 𝑛0 = 𝑦(𝑛 − 𝑛0) should be satisfied for the system to be time invariant
Time variant system:
A system is said to time variant if the relationship between the input and output changes
with time.
If 𝑦 𝑛 = 𝑇 𝑥 𝑛
Then 𝑇 𝑥 𝑛 − 𝑛0 ≠ 𝑦(𝑛 − 𝑛0) should be satisfied for the system to be time variant
1.5.5 Causal and Non-Causal system
Causal system:
A system is said to be causal if the response of a system at any instant of time depends
only on the present input, past input and past output but does not depends upon the future
input and future output.
Example: 𝑦(𝑡) = 3𝑥(𝑡) + 𝑥(𝑡 − 1)
A system is said to be causal if impulse response 𝑕(𝑡) is zero for negative values of 𝑡
i.e., 𝑕(𝑡) = 0 𝑓𝑜𝑟 𝑡 < 0
Non-Causal system:
A system is said to be Non-causal if the response of a system at any instant of time
depends on the future input and also on the present input, past input, past output.
Example: 𝑦(𝑡) = 𝑥(𝑡 + 2) + 𝑥(𝑡 − 1)
𝑦(𝑡) = 𝑥(−𝑡) + 𝑥(𝑡 + 4)
A system is said to be non-causal if impulse response 𝑕(𝑡) is non-zero for negative values of 𝑡
i.e., 𝑕(𝑡) ≠ 0 𝑓𝑜𝑟 𝑡 < 0
Causal system:
A system is said to be causal if the response of a system at any instant of time depends
only on the present input, past input and past output but does not depends upon the future
input.
Example: 𝑦(𝑛) = 3𝑥(𝑛) + 𝑥(𝑛 − 1)
A system is said to be causal if impulse response h(n) is zero for negative values of n
i.e., 𝑕(𝑛) = 0 𝑓𝑜𝑟 𝑛 < 0
Non-Causal system:
A system is said to be Non-causal if the response of a system at any instant of time
depends on the future input and also on the present input, past input, past output.
Example: 𝑦(𝑛) = 𝑥(𝑛 + 2) + 𝑥(𝑛 − 1)

15
𝑦(𝑛) = 𝑥(−𝑛) + 𝑥(𝑛 + 4)
A system is said to be non-causal if impulse response 𝑕(𝑛) is non-zero for negative values of n
i.e., 𝑕(𝑛) ≠ 0 for 𝑛 < 0
1.5.6 Stable and Unstable system
A system is said to be stable if and only if it satisfies the BIBO stability criterion.
BIBO stable condition:
x Every bounded input yields bounded output.
𝑖. 𝑒. , if 0 < 𝑥 t < ∞ 𝑡𝑕𝑒𝑛 0 < 𝑦 t < ∞ should be satisfied for the system to be stable
x Impulse response should be absolutely integrable
𝑖. 𝑒. , 0 < 𝑕(𝜏) 𝑑𝜏 < ∞
∞
−∞
If the BIBO stable condition is not satisfied, then the system is said to be unstable system
A system is said to be stable if and only if it satisfies the BIBO stability criterion.
BIBO stable condition:
x Every bounded input yields bounded output.
x Impulse response should be absolutely summable
𝑖. 𝑒. , 0 < 𝑕(𝑘) < ∞
∞
𝑘=−∞
If the BIBO stable condition is not satisfied, then the system is said to be unstable system
1.6 Solved Problems
1. Draw 𝑟(𝑡 + 3), where r(t) is ramp signal
Solution:
𝑟 𝑡 = 𝑡; 𝑡 ≥ 0
r(t+3)
:
4
3
2
1
-3 -2 -1 0 1 . . . t
r(t)
:
:
3
2
1
0 1 2 3 . . . t

16
2. Sketch 𝑥(𝑡) = 3𝑟(𝑡 − 1) + 𝑟(−𝑡 + 2)
𝑥 𝑡 = 3𝑟 𝑡 − 1 + 𝑟(−𝑡 + 2)
= 0 + 4 − 𝑡 𝑓𝑜𝑟 − 2 ≤ 𝑡 ≤ −1
= 0 + 3 − 𝑡 𝑓𝑜𝑟 − 1 ≤ 𝑡 ≤ 0
= 0 + 2 − 𝑡 𝑓𝑜𝑟 0 ≤ 𝑡 ≤ 1
= 3𝑡 + 1 − 𝑡 𝑓𝑜𝑟 1 ≤ 𝑡 ≤ 2
= 3 + 3𝑡 + 0 𝑓𝑜𝑟 2 ≤ 𝑡 ≤ 3
= 6 + 3𝑡 + 0 𝑓𝑜𝑟 3 ≤ 𝑡 ≤ 4
and so on
Fig 1.163
3. Draw time reversal signal of unit step signal
Solution:
𝑢 𝑛 = 1; 𝑛 ≥ 0

17
4. Check whether the following is periodic or not. If periodic, determine fundamental time
period
a. 𝒙 𝒕 = 𝟐 𝐜𝐨𝐬 𝟓𝒕 + 𝟏 − 𝐬𝐢𝐧 𝟒𝒕
Here Ω1 = 5, Ω2 = 4
𝑇1 =
2π
Ω1
=
2π
5
=
2π
5
𝑇2 =
2π
Ω2
=
2π
4
=
π
2
𝑇1
𝑇2
=
2π
5
π
2
=
4
5
(It is rational number)
Hence 𝑥(𝑡) is periodic
𝑇 = 5𝑇1 = 4𝑇2 = 2𝜋
∴ 𝑥(𝑡) is periodic with period 𝟐𝝅
b. 𝑥 𝑛 = 3 cos 4𝜋𝑛 + 2 sin𝜋𝑛
Here ω1 = 4π, ω2 = π
𝑁1 =
2πm
ω1
=
2πm
4π
=
m
2
𝑁1 = 1 (taking m = 2)
𝑁2 =
2πm
ω2
=
2πm
π
= 2m
𝑁2 = 2 (taking m = 1)
𝑁 = 𝐿𝐶𝑀 1,2 = 2
Hence x(n) ∴ 𝑥(𝑛) is periodic with period 2
5. Determine whether the signals are energy or power signal
𝒙 𝒕 = 𝒆−𝟑𝒕
𝒖(𝒕)
𝑬𝒏𝒆𝒓𝒈𝒚 𝑬∞ = lim
𝑇→∞
|𝑥 𝑡 |2
𝑑𝑡 =
𝑇
−𝑇
lim
𝑇→∞
|𝑒−3𝑡
|2
𝑑𝑡 = lim
𝑇→∞
𝑒−6𝑡
𝑇
0
𝑇
0
𝑑𝑡 = lim
𝑇→∞
𝑒−6𝑡
−6 0
𝑇
= lim
𝑇→∞
𝑒−6𝑇
−6
−
𝑒−0
−6
=
𝟏
𝟔
< ∞ ∵ 𝑒−∞
= 0, 𝑒−0
= 1

18
𝑷𝒐𝒘𝒆𝒓 𝑷∞ = lim
𝑇→∞
1
2𝑇
𝑥 𝑡 2
𝑑𝑡 =
𝑇
−𝑇
lim
𝑇→∞
1
2𝑇
𝑒−3𝑡 2
𝑑𝑡 =
𝑇
0
lim
𝑇→∞
1
2𝑇
𝑒−6𝑡
𝑇
0
𝑑𝑡
= lim
𝑇→∞
1
2𝑇
𝑒−6𝑡
−6 0
𝑇
= lim
𝑇→∞
1
2𝑇
𝑒−6𝑇
−6
−
𝑒−0
−6
= lim
𝑇→∞
1
2𝑇
1
6
= 𝟎 ∵ 𝑒−∞
= 0, 𝑒−0
= 1,
1
∞
= 0
Since energy value is finite and average power is zero, the given signal is an energy signal.
6. Determine whether the signals are energy or power signal
𝒙 𝒏 = 𝒆𝒋
𝝅𝒏
𝟒
+
𝝅
𝟐
𝑬𝒏𝒆𝒓𝒈𝒚 𝑬∞ = lim
𝑁→∞
𝑥(𝑛) 2
𝑁
𝑛=−𝑁
= lim
𝑁→∞
𝒆𝒋(
𝝅𝒏
𝟒
+
𝝅
𝟐
)
2
=
𝑁
𝑛=−𝑁
lim
𝑁→∞
12
= lim
𝑁→∞
2𝑁 + 1 = ∞
𝑁
𝑛=−𝑁
∵ 𝑒𝑗(𝜔𝑛 +𝜃)
= 1 𝑎𝑛𝑑 1 = 2𝑁 + 1
𝑁
𝑛=−𝑁
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝒑𝒐𝒘𝒆𝒓 𝑷∞ = lim
𝑁→∞
1
2𝑁 + 1
𝑥(𝑛) 2
𝑁
𝑛=−𝑁
= lim
𝑁→∞
1
2𝑁 + 1
𝒆𝒋(
𝝅𝒏
𝟒
+
𝝅
𝟐
)
2
𝑁
𝑛=−𝑁
lim
=𝑁→∞
1
2𝑁 + 1
12
= lim
𝑁→∞
1
2𝑁 + 1
𝑁
𝑛=−𝑁
2𝑁 + 1 = 𝟏
Since energy value is infinite and average power is finite, the given signal is power signal
7. Determine whether the following systems are linear or not
𝒅𝒚(𝒕)
𝒅𝒕
+ 𝒕𝒚 𝒕 = 𝒙𝟐
(𝒕)
Output due to weighted sum of inputs:
𝑑[𝑎𝑦1 𝑡 + 𝑏𝑦2 𝑡 ]
𝑑𝑡
+ 𝑡 𝑎𝑦1 𝑡 + 𝑏𝑦2 𝑡 = 𝑎𝑥1 𝑡 + 𝑏𝑥2 𝑡 2
… (1)
Weighted sum of outputs:
For input 𝑥1 𝑡 :
𝑑𝑦1 𝑡
𝑑𝑡
+ 𝑡𝑦1 𝑡 = 𝑥1
2
𝑡 … (2)
For input 𝑥2 𝑡 :
𝑑𝑦2 𝑡
𝑑𝑡
+ 𝑡𝑦2 𝑡 = 𝑥2
2
𝑡 … (3)
2 × 𝑎 + 3 × 𝑏 ⇒ 𝑎
𝑑𝑦1 𝑡
𝑑𝑡
+ 𝑎𝑡𝑦1 𝑡 + 𝑏
𝑑𝑦2 𝑡
𝑑𝑡
+ 𝑏𝑡𝑦2 𝑡 = 𝑎𝑥1
2
𝑡 + 𝑏𝑥2
2
𝑡 … (4)

19
1 ≠ (4)
The given system is Non-Linear
8. Determine whether the following systems are linear or not
𝒚 𝒏 = 𝒙 𝒏 − 𝟐 + 𝒙(𝒏𝟐
)
Output due to weighted sum of inputs:
𝑦3 𝑛 = 𝑎𝑥1 𝑛 − 2 + 𝑏𝑥2 𝑛 − 2 + 𝑎𝑥1 𝑛2
+ 𝑏𝑥2(𝑛2
)
Weighted sum of outputs:
For input 𝑥1 𝑛 :
𝑦1 𝑛 = 𝑥1 𝑛 − 2 + 𝑥1 𝑛2
For input 𝑥2 𝑛 :
𝑦2 𝑛 = 𝑥2 𝑛 − 2 + 𝑥2(𝑛2
)
𝑎𝑦1 𝑛 + 𝑏𝑦2 𝑛 = 𝑎𝑥1 𝑛 − 2 + 𝑎𝑥1 𝑛2
+ 𝑏𝑥2 𝑛 − 2 + 𝑏𝑥2(𝑛2
)
∵ 𝑦3 𝑛 = 𝑎𝑦1 𝑛 + 𝑏𝑦2 𝑛
9. Determine whether the following systems are static or dynamic
𝒚 𝒕 = 𝒙 𝟐𝒕 + 𝟐𝒙 𝒕
𝑦 0 = 𝑥 0 + 2𝑥 0 ⇒ present inputs
𝑦 −1 = 𝑥 −2 + 2𝑥 −1 ⇒ past and present inputs
𝑦 1 = 𝑥 2 + 2𝑥 1 ⇒ future and present inputs
Since output depends on past and future inputs the given system is dynamic system
10. Determine whether the following systems are static or dynamic
𝒚(𝒏) = 𝒔𝒊𝒏𝒙(𝒏)
𝑦 0 = 𝑠𝑖𝑛𝑥(0) ⇒ present input
𝑦 −1 = 𝑠𝑖𝑛𝑥(−1) ⇒ present input
Since output depends on present input the given system is Static system
11. Determine whether the following systems are time invariant or not
𝒚(𝒕) = 𝒙(𝒕)𝒔𝒊𝒏𝒘𝒕
Output due to input delayed by T seconds
𝑦(𝑡, 𝑇) = 𝑥(𝑡 − 𝑇)𝑠𝑖𝑛𝑤𝑡
Output delayed by T seconds
𝑦(𝑡 − 𝑇) = 𝑥(𝑡 − 𝑇)𝑠𝑖𝑛𝑤(𝑡 − 𝑇)
∵ 𝑦 𝑡, 𝑇 ≠ 𝑦 𝑡 − 𝑇
The given system is time variant
12. Determine whether the following systems are time invariant or not
𝒚 𝒏 = 𝒙(−𝒏 + 𝟐)
Output due to input delayed by k seconds
𝑦 𝑛, 𝑘 = 𝑥(−𝑛 + 2 − 𝑘)
Output delayed by k seconds
𝑦 𝑛 − 𝑘 = 𝑥 −(𝑛 − 𝑘 + 2) = 𝑥(−𝑛 + 𝑘 + 2)
∵ 𝒚 𝒏,𝒌 ≠ 𝒚 𝒏 − 𝒌
The given system is time variant

20
13. Determine whether the following systems are causal or not
𝒚 𝒕 =
𝒅𝒙(𝒕)
𝒅𝒕
+ 𝟐𝒙(𝒕)
The given equation is differential equation and the output depends on past input. Hence the
given system is 𝐂𝐚𝐮𝐬𝐚𝐥
14. Determine whether the following systems are causal or not
𝒚(𝒏) = 𝒔𝒊𝒏𝒙(𝒏)
𝑦 −1 = 𝑠𝑖𝑛𝑥(−1) ⇒ present input
Since output depends on present input the given system is Causal system
15. Determine whether the following systems are stable or not
𝒉 𝒕 = 𝒆−𝟒𝒕
𝒖(𝒕)
Condition for stability 𝑕(𝜏) 𝑑𝜏 < ∞
∞
−∞
𝑕(𝜏) 𝑑𝜏 = 𝑒−4𝜏
𝑢(𝜏)
∞
−∞
∞
−∞
𝑑𝜏 = 𝑒−4𝜏
𝑑𝜏
∞
0
=
𝑒−4𝜏
−4 0
∞
=
1
4
∵ 𝑕(𝜏) 𝑑𝜏 < ∞
∞
−∞
the given system is 𝐬𝐭𝐚𝐛𝐥𝐞
16. Determine whether the following systems are stable or not
𝒚 𝒏 = 𝟑𝒙(𝒏)
Let 𝑥 𝑛 = 𝛿 𝑛 , 𝑦 𝑛 = 𝑕(𝑛)
⇒ 𝑕 𝑛 = 3𝛿(𝑛)
Condition for stability 𝑕(𝑘) < ∞
∞
𝑘=−∞
𝑕(𝑘) =
∞
𝑘=−∞
3𝛿(𝑘)
∞
𝑘=0
= 3𝛿(𝑘)
∞
𝑘=0
= 3
∵ 𝛿 𝑘 = 0 𝑓𝑜𝑟 𝑘 ≠ 0 𝑎𝑛𝑑 𝛿 𝑘 = 1 𝑓𝑜𝑟 𝑘 = 0
∵ 𝑕(𝑘) <
∞
𝑘=−∞
∞ the given system is 𝐬𝐭𝐚𝐛𝐥𝐞

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GENERATION OF CONTINUOUS TIME SIGNALS
AIM:
To generate a functional sequence of a signal (Sine, Cosine, triangular, Square, Saw
tooth and sinc ) using MATLAB function.
APPARATUS REQUIRED:
HARDWARE : Personal Computer
SOFTWARE : MATLABR2014a
PROCEDURE:
1. Start the MATLAB program.
2. Open new M-file
3. Type the program
4. Save in current directory
5. Compile and Run the program
6. If any error occurs in the program correct the error and run it again
7. For the output see command window Figure window
8. Stop the program.
PROGRAM: (Generation of Continuous Time Signals)
%Program for sine wave
t=0:0.1:10;
y=sin(2*pi*t);
subplot(3,3,1);
plot(t,y,'k');
xlabel('Time');
ylabel('Amplitude');
title('Sine wave');
%Program for cosine wave
t=0:0.1:10;
y=cos(2*pi*t);
subplot(3,3,2);
plot(t,y,'k');
xlabel('Time');
title('Cosine wave');
%Program for square wave
t=0:0.001:10;
y=square(t);
subplot(3,3,3);

plot(t,y,'k');
xlabel('Time');
title('Square wave');
%Program for sawtooth wave
t=0:0.1:10;
y=sawtooth(t);
subplot(3,3,4);
plot(t,y,'k');
xlabel('Time');
title('Sawtooth wave');
%Program for Triangular wave
t=0:.0001:20;
y=sawtooth(t,.5); % sawtooth with 50% duty cycle
(triangular)
subplot(3,3,5);
plot(t,y);
ylabel ('Amplitude');
xlabel ('Time Index');
title('Triangular waveform');
%Program for Sinc Pulse
t=-10:.01:10;
y=sinc(t);
axis([-10 10 -2 2]);
subplot(3,3,6)
plot(t,y)
title('Sinc Pulse');
% Program for Exponential Growing signal
t=0:.1:8;
a=2;
y=exp(a*t);
subplot(3,3,7);
plot(t,y);
title('Exponential growing Signal');
% Program for Exponential Growing signal
t=0:.1:8;
a=2;
y=exp(-a*t);
subplot(3,3,8);
plot(t,y);
title('Exponential decaying Signal');

OUTPUT: (Generation of Continuous Time Signals)
RESULT:
Thus the MATLAB programs for functional sequence of a signal (Sine, Cosine,
triangular, Square, Saw tooth and sinc ) using MATLAB function written and the results were
plotted.

Ex. No: 1b
Date:
AIM:
GENERATION OF DISCRETE TIME SIGNALS
To generate a discrete time signal sequence (Unit step, Unit ramp, Sine, Cosine,
Exponential, Unit impulse) using MATLAB function.
APPARATUS REQUIRED:
HARDWARE : Personal Computer
SOFTWARE : MATLABR2014a
PROCEDURE:
1. Start the MATLAB program.
2. Open new M-file
3. Type the program
4. Save in current directory
5. Compile and Run the program
6. If any error occurs in the program correct the error and run it again
7. For the output see command window Figure window
8. Stop the program.

PROGRAM: (Generation of Discrete Time Signals)
%Program for unit step sequence
clc;
N=input('Enter the length of unit step sequence(N)= ');
n=0:1:N-1;
y=ones(1,N);
subplot(3,2,1);
stem(n,y,'k');
xlabel('Time')
ylabel('Amplitude')
title('Unit step sequence');
%Program for unit ramp sequence
N1=input('Enter the length of unit ramp sequence(N1)= ');
n1=0:1:N1-1;
y1=n1;
subplot(3,2,2);
stem(n1,y1,'k');
xlabel('Time');
title('Unit ramp sequence');
%Program for sinusoidal sequence
N2=input('Enter the length of sinusoidal sequence(N2)=
');
n2=0:0.1:N2-1;
y2=sin(2*pi*n2);
subplot(3,2,3);
stem(n2,y2,'k');
xlabel('Time');
title('Sinusoidal sequence');
%Program for cosine sequence
N3=input('Enter the length of the cosine sequence(N3)=');
n3=0:0.1:N3-1;
y3=cos(2*pi*n3);
subplot(3,2,4);
stem(n3,y3,'k');
xlabel('Time');
title('Cosine sequence');
%Program for exponential sequence
N4=input('Enter the length of the exponential
sequence(N4)= ');
n4=0:1:N4-1;
a=input('Enter the value of the exponential sequence(a)=
');
y4=exp(a*n4);
subplot(3,2,5);
stem(n4,y4,'k');
xlabel('Time');
title('Exponential sequence');
%Program for unit impulse
n=-3:1:3;
y=[zeros(1,3),ones(1,1),zeros(1,3)];

subplot(3,2,6);
stem(n,y,'k');
xlabel('Time');
title('Unit impulse');
OUTPUT: (Generation of Discrete Time Signals)
RESULT:
Thus the MATLAB programs for discrete time signal sequence (Unit step, Unit ramp,
Sine, Cosine, Exponential, Unit impulse) using MATLAB function written and the results were
plotted.

UNIT- I – Post – Test – MCQ
1. Which one of the following is correct? Energy of a power signal is
a) finite
b) zero
c) infinite
d) between 1 and 2
2. If x (-t) = -x (t) then the signal is said to be _____________
a. Even signal
b. Odd signal
c. Periodic signal
d. Non periodic signal
3. Determine the nature of the system y(t) = 10 x(t) +5 is
a. causal, linear and time invarient
b. causal, non-linear and time invarient
c. non causal, non non-linear and time varient
d. non causal, linear and time varient
4. Which of the following is an example for non- causal system?
a) y[n] = 1
⁄3 {x[n-1] + x[n] + x[n-2]}
b) y[n] = 1
⁄3 {x[n-1] + x[n] + x[n+1]}
c) y[n] = 1
⁄2 {x[n-1] + x[n]}
d) y[n] = 1
⁄2 {x[n] + x[n-2]}
5. Which of the following systems is memoryless?
a) y(t) = x(2t) + x(t)
b) y(t) = -x(t) + x(1-t)
c) y(t) = x(t) + 2x(t)
d) y(t) = x(t) + 2x(t+2)

SIGNALS AND SYSTEMS
Prerequisite: Basic operations of signals and Basic knowledge on Mathematics
which includes Fourier series and Laplace Transform
Objectives:
To understand the properties and representation of Fourier series, Fourier
Transform and Laplace transform
Unit-II Analysis of Continuous Time Signals
Fourier Series Analysis- Representation of periodic signals in trigonometric and
exponential form, Spectrum of CT signals-Fourier Transform and Laplace Transform
in signal analysis
Outcome:
 Understand and resolve the signals in frequency domain using Fourier series
and Fourier transforms.
 Understand the limitations of Fourier transform and need for Laplace
transform and develop the ability to analyze the system in s- domain

UNIT- II – Pre – Test – MCQ
1. Which mathematical notation species the condition of periodicity for a continuous time
signal?
a. x(t) = x(t +T0)
b. x(n) = x(n+ N)
c. x(t) = 𝑒−𝑎𝑡
d. None of the above
2. The inverse Laplace transform of
1
𝑠+𝑎
is
a) 𝑒𝑎𝑡
b) t 𝑒𝑎𝑡
c) t2
𝑒𝑎𝑡
d) t
3. Double-sided phase & amplitude spectra _____
a. Possess an odd & even symmetry respectively
b. Possess an even & odd symmetry respectively
c. Both possess an odd symmetry
d. Both possess an even symmetry
4. The final value theorem is used to find the
a) Steady state value of the system
b) Initial value of the system output
c) Transient behavior of the system output
d) None of these
5. What are the two types of Fourier series?
a) Trigonometric and exponential
b) Trigonometric and logarithmic
c) Exponential and logarithmic
d) Trigonometric only

Dr.M.N.VIMAL KUMAR , AP/ECE/ RMDEC
1
Unit 2: Analysis of continuous time signals
2.1 Fourier series analysis
The Fourier representation of signals can be used to perform frequency domain analysis of
signals in which we can study the various frequency components present in the signal, magnitude
and phase of various frequency components.
Conditions for existence of Fourier series:
The Fourier series exist only if the following Dirichlet’s conditions are satisfied.
 The signal 𝑥(𝑡) must be single valued function.
 The signal 𝑥(𝑡) must possess only a finite number of discontinuous in the period T.
 The signal must have a finite number of maxima and minima in the period T.
 𝑥(𝑡) must be absolutely integrable. i.e., 𝑥(𝑡) 𝑑𝑡
𝑇
0
< ∞
Types of Fourier series:
 Trigonometric Fourier series
 Exponential Fourier series
 Cosine Fourier series
2.2 Trigonometric Fourier series
The trigonometric form of Fourier series of a periodic signal, 𝑥(𝑡) with period 𝑇 is defined
as
𝑥 𝑡 = 𝑎0 + 𝑎𝑛 cos 𝑛Ω0𝑡
∞
𝑛=1
+ 𝑏𝑛 sin 𝑛Ω0𝑡
∞
𝑛=1
… … … 1
𝑎0, 𝑎𝑛 ,𝑏𝑛 → Fourier coefficients of trigonometric form of Fourier series
𝑎0 =
1
𝑇
𝑥 𝑡 𝑑𝑡
𝑡0+𝑇
𝑡0
𝑎𝑛 =
2
𝑇
𝑥 𝑡 cos 𝑛Ω0𝑡 𝑑𝑡
𝑡0+𝑇
𝑡0
𝑏𝑛 =
2
𝑇
𝑥 𝑡 sin 𝑛Ω0𝑡 𝑑𝑡
𝑡0+𝑇
𝑡0

2
Example 1 Find the trigonometric Fourier series for the periodic signal x(t) as shown in Figure
Solution:
𝑇 = 3 − −1 = 4 𝑎𝑛𝑑 Ω0 =
2𝜋
𝑇
=
𝜋
2
Evaluation of 𝒂𝟎
𝒂𝟎 =
1
𝑇
𝑥 𝑡 𝑑𝑡
𝑡0+𝑇
𝑡0
=
1
4
1
1
−1
𝑑𝑡 + −1
3
1
𝑑𝑡 =
1
4
𝑡 −1
1
− 1 𝑡 1
3
=
1
4
(1 − (−1 ) − (3 − 1)]
=
1
4
2 − 2 = 𝟎
Evaluation of 𝒂𝒏
𝒂𝒏 =
2
𝑇
𝑥 𝑡 cos 𝑛𝛺0𝑡
𝑡0+𝑇
𝑡0
𝑑𝑡 =
2
4
cos 𝑛𝛺0𝑡 𝑑𝑡 + (−1)cos 𝑛𝛺0𝑡 𝑑𝑡
3
1
1
−1
=
1
2
sin𝑛𝛺0𝑡
𝑛𝛺0 −1
1
−
sin𝑛𝛺0𝑡
𝑛𝛺0 1
3
=
1
2
sin 𝑛
𝜋
2
𝑡
𝑛
𝜋
2 −1
1
−
sin𝑛
𝜋
2
𝑡
𝑛
𝜋
2 1
3
=
1
2
2
𝑛𝜋
sin𝑛
𝜋
2
− sin 𝑛
𝜋
2
−1 − sin𝑛
𝜋
2
3 − sin 𝑛
𝜋
2
=
1
𝑛𝜋
sin𝑛
𝜋
2
+ sin 𝑛
𝜋
2
−sin 3𝑛
𝜋
2
+sin 𝑛
𝜋
2
=
1
𝑛𝜋
3sin
𝑛𝜋
2
− sin(2n𝜋 −
𝑛𝜋
2
)
=
1
𝑛𝜋
3sin
𝑛𝜋
2
− −sin𝑛
𝜋
2
=
𝟒
𝒏𝝅
𝐬𝐢𝐧 𝒏
𝝅
𝟐
Evaluation of 𝒃𝒏
𝒃𝒏 =
2
𝑇
𝑥
𝑡0+𝑇
𝑡0
𝑡 sin 𝑛𝛺0𝑡 𝑑𝑡 =
2
4
sin 𝑛𝛺0𝑡 𝑑𝑡 + − sin𝑛𝛺0𝑡 𝑑𝑡
3
1
1
−1
=
1
2
−cos 𝑛𝛺0𝑡
𝑛𝛺0 −1
1
−
−cos 𝑛𝛺0𝑡
𝑛𝛺0 1
3
=
1
2
−cos 𝑛
𝜋
2
𝑡
𝑛
𝜋
2 −1
1
+
cos 𝑛
𝜋
2
𝑡
𝑛
𝜋
2 1
3
=
1
2
−2
𝑛𝜋
cos 𝑛
𝜋
2
− cos 𝑛
𝜋
2
−1 +
2
𝑛𝜋
cos 𝑛
𝜋
2
3 − cos 𝑛
𝜋
2
=
1
2
0 +
2
𝑛𝜋
cos 2𝑛𝜋 −
𝑛𝜋
2
− cos 𝑛
𝜋
2
=
1
𝑛𝜋
cos 𝑛
𝜋
2
− cos 𝑛
𝜋
2
= 𝟎
Trigonometric Fourier series
𝑥 𝑡 = 𝑎0 + 𝑎𝑛
∞
𝑛=1
cos 𝑛𝛺0𝑡 + 𝑏𝑛
∞
𝑛=1
sin𝑛𝛺0𝑡
=
4
𝑛𝜋
sin
𝑛𝜋
2
cos 𝑛𝛺0𝑡 =
4
𝑛𝜋
∞
𝑛=1
∞
𝑛=1
sin
𝑛𝜋
2
cos 𝑛
𝜋
2
𝑡
Example 2 Obtain Fourier series of the following full wave rectified sine wave shown in figure
--1 0 1 3 t
x(t)
1
-1
-7 -5 -3 -1 0 1 3 5 7 9 t
T
x(t)
1
-1

3
Solution:
𝑥 𝑡 = 𝑥 −𝑡 ; ∴ 𝐺𝑖𝑣𝑒𝑛 𝑠𝑖𝑔𝑛𝑎𝑙 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑠𝑖𝑔𝑛𝑎𝑙, 𝑠𝑜 𝒃𝒏 = 𝟎
𝑇 = 1 𝑎𝑛𝑑 Ω0 =
2𝜋
1
= 2𝜋
The given signal is sinusoidal signal, ∴ 𝑥 𝑡 = 𝐴 sin Ω𝑡
Here Ω =
2π
T
=
2π
2
= π and A = 1
∴ 𝒙(𝒕) = 𝐬𝐢𝐧𝝅𝒕
𝒂𝟎 =
2
𝑇
𝑥 𝑡 𝑑𝑡
𝑇
2
0
=
2
1
𝑥 𝑡 𝑑𝑡
1
2
0
= 2 sin𝜋𝑡 𝑑𝑡
1
2
0
= 2 −
cos 𝜋𝑡
𝜋 0
1
2
= −
2
𝜋
cos
𝜋
2
− cos 0 =
𝟐
𝝅
𝒂𝒏 =
4
𝑇
𝑥 𝑡 cos 𝑛Ω0𝑡 𝑑𝑡
𝑇
2
0
=
4
1
sin𝜋𝑡 cos 𝑛2𝜋𝑡 𝑑𝑡
1
2
0
= 2 sin (1 + 2𝑛)𝜋𝑡 + sin (1 − 2𝑛)𝜋𝑡 𝑑𝑡
1
2
0
= 2 −
cos (1 + 2𝑛)𝜋𝑡
1 + 2𝑛 𝜋
−
cos (1 − 2𝑛)𝜋𝑡
1 − 2𝑛 𝜋 0
1
2
=
2
𝜋
−
cos (1 + 2𝑛)
𝜋
2
1 + 2𝑛
−
cos (1 − 2𝑛)
𝜋
2
1 − 2𝑛
+
1
1 + 2𝑛
+
1
1 − 2𝑛
=
2
𝜋
1
1 + 2𝑛
+
1
1 − 2𝑛
=
2
𝜋
1 − 2𝑛 + 1 + 2𝑛
1 − 4𝑛2
=
𝟒
𝝅 𝟏 − 𝟒𝒏𝟐
Trigonometric Fourier series
𝑥 𝑡 = 𝑎0 + 𝑎𝑛
∞
𝑛=1
cos 𝑛𝛺0𝑡 + 𝑏𝑛
∞
𝑛=1
sin 𝑛𝛺0𝑡
∴ 𝑥 𝑡 =
2
𝜋
+
4
𝜋 1 − 4𝑛2
∞
𝑛=1
cos 𝑛2𝜋𝑡
2.3 Exponential Fourier series
The exponential form of Fourier series of a periodic signal 𝑥(𝑡) with period 𝑇 is defined as,
𝑥 𝑡 = 𝑐𝑛 𝑒𝑗𝑛 Ω0𝑡
∞
𝑛=−∞
The Fourier coefficient 𝑐𝑛 can be evaluated using the following formulae
-2 -1 0 1 2 t
x(t)
1

4
𝑐𝑛 =
1
𝑇
𝑥 𝑡 𝑒−𝑗𝑛 Ω0𝑡
𝑑𝑡
𝑇
2
−𝑇
2
Example 3 Find exponential series for the signal shown in figure
Fig 2.26
Solution:
𝑇 = 1, Ω0 =
2π
T
=
2π
1
= 2π
Consider the equation of a straight line
𝑦 − 𝑦1
𝑦2 − 𝑦1
=
𝑥 − 𝑥1
𝑥2 − 𝑥1
… … … 9
Consider one period of the given signal Fig 2.26 as shown in Fig 2.27
Consider points P,Q as shown in fig 2.27
Coordinates of point 𝑃 = 0,0
Coordinates of point 𝑄 = 1,1
On substituting the coordinates of points 𝑃 and 𝑄 in eq 9
𝑥 𝑡 − 0
1 − 0
=
𝑡 − 0
1 − 0
⇒ 𝑥 𝑡 = 𝑡
∵ 𝑥 = 𝑡, 𝑦 = 𝑥(𝑡)
Fig 2.27
Evaluation of 𝒄𝟎
𝑐0 =
1
𝑇
𝑥(𝑡)𝑑𝑡
𝑇
0
=
1
1
(𝑡)𝑑𝑡
1
0
=
𝑡2
2 0
1
=
1
2
Evaluation of 𝒄𝒏
𝑐𝑛 =
1
𝑇
𝑥(𝑡)𝑒−𝑗𝑛 Ω0𝑡
𝑑𝑡
𝑇
0
=
1
1
𝑡𝑒−𝑗𝑛 2𝜋𝑡
𝑑𝑡
1
0
= 𝑡
𝑒−𝑗𝑛 2𝜋𝑡
−𝑗𝑛2𝜋 0
1
−
−𝑗𝑛2𝜋
𝑑𝑡
1
0
=
𝑒−𝑗𝑛 2𝜋
−𝑗𝑛2𝜋
+ 0 +
−𝑗2(𝑛2𝜋)2
0
1
= 𝑗
𝑛2𝜋
+
𝑛24𝜋2
−
1
𝑛24𝜋2
=
𝑗
𝑛2𝜋
+
1
𝑛24𝜋2
−
1
𝑛24𝜋2
=
𝑗
𝑛2𝜋
𝑐𝑛 =
𝑗
𝑛2𝜋
𝑐1 =
𝑗
2𝜋
, 𝑐2 =
𝑗
4𝜋
, 𝑐3 =
𝑗
6𝜋
, 𝑐−1 =
𝑗
−2𝜋
, 𝑐−2 =
𝑗
−4𝜋
, 𝑐−3 =
𝑗
−6𝜋
Exponential Fourier series
𝑥 𝑡 = 𝑐𝑛 𝑒𝑗𝑛 𝛺0𝑡
∞
𝑛=−∞
0 1 t
x(t)
1 Q[1,1]
P[0,0]
0 1 2 3
t
x(t)
1

5
∴ 𝑥 𝑡 = + ⋯ −
𝑗
6𝜋
𝑒−𝑗6𝜋𝑡
−
𝑗
4𝜋
−
𝑗
2𝜋
+
1
2
+
𝑗
2𝜋
𝑒𝑗2𝜋𝑡
+
𝑗
4𝜋
𝑒𝑗4𝜋𝑡
+
𝑗
6𝜋
𝑒𝑗6𝜋𝑡
+ ⋯
=
1
2
+
𝑗
2𝜋
𝑒𝑗2𝜋𝑡
− 𝑒−𝑗2𝜋𝑡
+
𝑗
4𝜋
𝑒𝑗4𝜋𝑡
+
𝑗
6𝜋
𝑒𝑗6𝜋𝑡
+ ⋯
=
1
2
+
1
𝜋
𝑒𝑗2𝜋𝑡
−1 2𝑗
+
1
2𝜋
𝑒𝑗4𝜋𝑡
−1 2𝑗
+
1
3𝜋
𝑒𝑗6𝜋𝑡
−1 2𝑗
=
1
2
+
−1
𝜋
𝑠𝑖𝑛 2𝜋𝑡 −
1
2𝜋
𝑠𝑖𝑛 4𝜋𝑡 −
1
3𝜋
sin 6𝜋𝑡
=
1
2
−
1
𝜋
𝑠𝑖𝑛 2𝜋𝑡 +
1
2
𝑠𝑖𝑛 4𝜋𝑡 +
1
3
𝑠𝑖𝑛 6𝜋𝑡 + ⋯
2.4 Cosine Fourier series
Cosine representation of 𝑥 𝑡 is
𝑥(𝑡) = 𝐴0 + 𝐴𝑛 𝑐𝑜𝑠 (𝑛𝛺0𝑡 + 𝜃𝑛 )
∞
𝑛=1
Where 𝐴0 is dc component, 𝐴𝑛 is harmonic amplitude or spectral amplitude and 𝜃𝑛 is phase
coefficient or phase angle 𝑜𝑟 𝑠𝑝𝑒𝑐𝑡𝑟𝑎𝑙 𝑎𝑛𝑔𝑙𝑒
Example 4 Determine the cosine Fourier series of the signal shown in Figure
Solution:
The signal shown in is periodic with period 𝑇 = 2𝜋 𝑎𝑛𝑑 Ω0 =
2𝜋
2𝜋
= 1
The given signal is sinusoidal signal, ∴ 𝑥 𝑡 = 𝐴 sin Ω𝑡
Here Ω =
2π
T
=
2π
2π
= 1, A = 1
∴ 𝒙(𝒕) = 𝒔𝒊𝒏 𝒕
𝑎0 =
1
𝑇
𝑥(𝑡)𝑑𝑡
𝑇
0
=
1
2𝜋
𝑠𝑖𝑛 𝑡 𝑑𝑡
𝜋
0
=
1
2𝜋
− 𝑐𝑜𝑠 𝑡 0
𝜋
=
1
2𝜋
[− 𝑐𝑜𝑠 𝜋 + 𝑐𝑜𝑠 0] =
1
2𝜋
[2] =
1
𝜋
x(t)
1
-2π -π 0 π 2π 3π t

6
𝑎𝑛 =
2
𝑇
𝑥 𝑡 𝑐𝑜𝑠 𝑛𝛺0𝑡 𝑑𝑡
𝑇
0
=
2
2𝜋
𝑠𝑖𝑛 𝑡 𝑐𝑜𝑠 𝑛𝑡
𝜋
0
𝑑𝑡 =
1
2𝜋
𝑠𝑖𝑛 1 + 𝑛 𝑡 + 𝑠𝑖𝑛 1 − 𝑛 𝑡
𝜋
0
𝑑𝑡
=
1
2𝜋
−
𝑐𝑜𝑠 1 + 𝑛 𝑡
1 + 𝑛
−
𝑐𝑜𝑠 1 − 𝑛 𝑡
1 − 𝑛 0
𝜋
=
1
2𝜋
−
𝑐𝑜𝑠 1 + 𝑛 𝜋
1 + 𝑛
−
𝑐𝑜𝑠 1 − 𝑛 𝜋
1 − 𝑛
+
1
1 + 𝑛
+
1
1 − 𝑛
𝑓𝑜𝑟 𝑛 = 𝑜𝑑𝑑 ∶ 𝑎𝑛 =
1
2𝜋
−
1
1 + 𝑛
−
1
1 − 𝑛
+
1
1 + 𝑛
+
1
1 − 𝑛
= 0
𝑓𝑜𝑟 𝑛 = 𝑒𝑣𝑒𝑛 ∶ 𝑎𝑛 =
1
2𝜋
1
1 + 𝑛
+
1
1 − 𝑛
+
1
1 + 𝑛
+
1
1 − 𝑛
=
1
2𝜋
2
1 + 𝑛
+
2
1 − 𝑛
=
1
𝜋
1 − 𝑛 + 1 + 𝑛
1 − 𝑛2
=
2
𝜋(1 − 𝑛2)
∴ 𝒂𝒏 =
𝟎 𝒇𝒐𝒓 𝒏 = 𝒐𝒅𝒅
𝟐
𝝅(𝟏 − 𝒏𝟐)
𝒇𝒐𝒓 𝒏 = 𝒆𝒗𝒆𝒏
Evaluation of 𝒃𝒏
𝒃𝒏 =
2
𝑇
𝑥(𝑡) 𝑠𝑖𝑛 𝑛𝛺0𝑡 𝑑𝑡
𝑇
0
=
2
2𝜋
𝑠𝑖𝑛 𝑡 𝑠𝑖𝑛 𝑛𝑡
𝜋
0
𝑑𝑡 =
1
2𝜋
𝑐𝑜𝑠 1 − 𝑛 𝑡 − 𝑐𝑜𝑠 1 + 𝑛 𝑡
𝜋
0
𝑑𝑡
=
1
2𝜋
𝑠𝑖𝑛 1 − 𝑛 𝑡
(1 − 𝑛)
−
𝑠𝑖𝑛 1 + 𝑛 𝑡
(1 + 𝑛) 0
𝜋
=
1
2𝜋
𝑠𝑖𝑛 1 − 𝑛 𝜋
(1 − 𝑛)
−
𝑠𝑖𝑛 1 + 𝑛 𝜋
(1 + 𝑛)
− 0 = 𝟎
Evaluation of Fourier coefficients of Cosine Fourier series from Trigonometric Fourier series:
𝐴0 = 𝑎0 =
1
𝜋
𝐴𝑛 = 𝑎𝑛
2 + 𝑏𝑛
2
=
2
𝜋(1 − 𝑛2)
, 𝑓𝑜𝑟 𝑛 = 𝑒𝑣𝑒𝑛
𝜃𝑛 = −𝑡𝑎𝑛−1
𝑏𝑛
𝑎𝑛
= 0
Cosine Fourier series
x t = A0 + An cos nΩ0t + θn
∞
n=1
∴ 𝑥(𝑡) =
1
𝜋
+
2
𝜋(1 − 𝑛2)
𝑐𝑜𝑠 𝑛𝑡
∞
𝑛=1
(𝑛=𝑒𝑣𝑒𝑛 )
=
1
𝜋
+
2
𝜋(1 − 4)
𝑐𝑜𝑠 2𝑡 +
2
𝜋(1 − 16)
𝑐𝑜𝑠 4𝑡 + ⋯
=
1
𝜋
−
2
3𝜋
𝑐𝑜𝑠 2𝑡 −
2
15𝜋
𝑐𝑜𝑠 4𝑡 + ⋯ =
1
𝜋
−
2
𝜋
[
1
3
𝑐𝑜𝑠 2𝑡 +
1
15
𝑐𝑜𝑠 4𝑡 + ⋯ ]
2.5 Fourier transform
The Fourier representation of periodic signals has been extended to non-periodic signals by
letting the fundamental period T tend to infinity and this Fourier method of representing non-
periodic signals as a function of frequency is called Fourier transform.
Definition of Continuous time Fourier Transform

7
The Fourier transform (FT) of Continuous time signals is called Continuous Time Fourier
Transform
𝐿𝑒𝑡 𝑥 𝑡 = 𝐶𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑡𝑖𝑚𝑒 𝑠𝑖𝑔𝑛𝑎𝑙
𝑋 𝑗Ω = 𝐹 𝑥(𝑡)
The Fourier transform of continuous time signal, x(t) is defined as,
𝑋 𝑗Ω = 𝐹 𝑥(𝑡) = 𝑥 𝑡 𝑒−𝑗Ω𝑡
∞
−∞
𝑑𝑡
Conditions for existence of Fourier transform
The Fourier transform 𝑥(𝑡) exist if it satisfies the following Dirichlet condition
1. 𝑥(𝑡) should be absolutely integrable
𝑖𝑒 , 𝑥 𝑡 𝑑𝑡 < ∞
∞
−∞
2. 𝑥(𝑡) should have a finite number of maxima and minima with in any finite interval.
3. 𝑥 𝑡 should have a finite number of discontinuities with in any interval.
Definition of Inverse Fourier Transform
The inverse Fourier Transform of 𝑋 𝑗Ω is defined as,
𝑥 𝑡 = 𝐹−1
𝑋 𝑗Ω =
1
2𝜋
𝑋 𝑗Ω 𝑒𝑗Ω𝑡
𝑑Ω
∞
−∞
Example 5 Find Fourier transform of impulse signal
Solution:
By definition of Fourier transform
𝐹 𝑥 𝑡 = 𝑋 𝑗Ω = 𝑥 𝑡 𝑒−𝑗Ω𝑡
∞
−∞
𝑑𝑡
∴ 𝐹[𝛿(𝑡)] = 𝛿(𝑡)𝑒−𝑗 𝛺𝑡
𝑑𝑡
∞
−∞
𝐹 𝛿 𝑡 = 𝛿 0 𝑒−𝑗 𝛺 0
= 1 ∵ 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 𝑠𝑖𝑔𝑛𝑎𝑙 𝛿 𝑡 =
1 𝑓𝑜𝑟 𝑡 = 0
0 𝑓𝑜𝑟 𝑡 ≠ 0
Example 6 Find Fourier transform of double sided exponential signal
Solution:
Double sided exponential signal is given by
𝐹[𝑒−𝑎|𝑡|
] = 𝑒−𝑎𝑡
∶ 𝑡 ≥ 0
𝑒𝑎𝑡
∶ 𝑡 ≤ 0
𝐹[𝑒−𝑎|𝑡|
] = 𝑒𝑎𝑡
𝑒−𝑗 𝛺𝑡
𝑑𝑡
0
−∞
+ 𝑒−𝑎𝑡
.𝑒−𝑗 𝛺𝑡
𝑑𝑡
∞
0
= 𝑒(𝑎−𝑗 𝛺)𝑡
𝑑𝑡
0
−∞
+ 𝑒−(𝑎+𝑗 𝛺)𝑡
𝑑𝑡
∞
0
=
𝑒(𝑎−𝑗 𝛺)𝑡
𝑎 − 𝑗 𝛺 −∞
0
+
𝑒−(𝑎+𝑗 𝛺)𝑡
−(𝑎 + 𝑗 𝛺) 0
∞
=
1
𝑎 − 𝑗 𝛺
+
1
𝑎 + 𝑗 𝛺
=
𝑎 + 𝑗 𝛺 + 𝑎 − 𝑗 𝛺
𝑎2 + 𝛺2
=
2𝑎
𝑎2 + 𝛺2

8
Example 7 Find Fourier transform of rectangular pulse function shown in figure
Solution:
𝑥 𝑡 = 𝜋(𝑡) = 𝐴 ;
−𝑇
2
≤ 𝑡 ≤
𝑇
2
𝐹[𝜋(𝑡)] = 𝐴𝑒−𝑗𝛺𝑡
𝑑𝑡
𝑇
2
−𝑇
2
= 𝐴
𝑒−𝑗𝛺𝑡
−𝑗𝛺 −
𝑇
2
𝑇
2
=
𝐴
−𝑗𝛺
[𝑒−𝑗𝛺
𝑇
2 − 𝑒𝑗𝛺
𝑇
2] =
2𝐴
𝑗𝛺
[
𝑒𝑗𝛺
𝑇
2 − 𝑒−𝑗𝛺
𝑇
2
2
] =
2𝐴
𝛺
𝑠𝑖𝑛 𝛺
𝑇
2
=
2𝐴
𝛺𝑇
𝑇 𝑠𝑖𝑛 𝛺
𝑇
2
= 𝐴𝑇
𝑠𝑖𝑛 𝛺
𝑇
2
𝛺
𝑇
2
𝐴𝑇 𝑠𝑖𝑛 𝑐 𝛺
𝑇
2
Example 8 Find inverse Fourier transform 𝑋(𝑗 𝛺) = 𝛿(𝛺)
Solution:
∴ 𝐹−1
[𝑋(𝑗 𝛺)] = 𝐹−1
[𝛿(𝛺)]
𝑥 𝑡 =
1
2𝜋
𝑋 𝑗 𝛺 𝑒𝑗 𝛺𝑡
𝑑𝛺
∞
−∞
=
1
2𝜋
𝛿 𝛺 𝑒𝑗 𝛺𝑡
𝑑𝛺
∞
−∞
=
1
2𝜋
1 ∵ 𝛿 𝛺 =
1 𝑓𝑜𝑟 𝛺 = 0
0 𝑓𝑜𝑟 𝛺 ≠ 0
𝐹−1
[𝛿(𝛺)] =
1
2𝜋
2.6 Laplace transform
It is used to transform a time domain to complex frequency domain signal(s-domain)
Two Sided Laplace transform (or) Bilateral Laplace transform
Let 𝑥(𝑡) be a continuous time signal defined for all values of 𝑡. Let 𝑋(𝑆) be Laplace transform of
𝑥(𝑡).
𝐿 𝑥(𝑡) = 𝑋 𝑆 = 𝑥(𝑡)𝑒−𝑆𝑡
𝑑𝑡
∞
−∞
One sided Laplace transform (or) Unilateral Laplace transform
Let 𝑥(𝑡) be a continuous time signal defined for 𝑡 ≥ 0 (ie If 𝑥(𝑡) is causal) then,
𝐿 𝑥(𝑡) = 𝑋 𝑆 = 𝑥(𝑡)𝑒−𝑆𝑡
𝑑𝑡
∞
0
Inverse Laplace transform
The S-domain signal 𝑋(𝑆) can be transformed to time domain signal x(t) by using inverse Laplace
transform.
-T/2 0 T/2 t
x(t)
A

9
The inverse Laplace transform of X(S) is defined as,
𝐿−1
𝑋(𝑠) = 𝑥 𝑡 =
1
2𝜋𝑗
𝑋 𝑆 𝑒𝑠𝑡
𝑑𝑠
𝑠=𝜎+𝑗Ω
𝑠=𝜎−𝑗Ω
Existence of Laplace transform
The necessary and sufficient conditions for the existence of Laplace transform are
 𝑥(𝑡) should be continuous in the given closed interval
 𝑥 𝑡 𝑒−𝜎𝑡
must be absolutely intergrable
𝑖. 𝑒. , 𝑋 𝑆 𝑒𝑥𝑖𝑠𝑡𝑠 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥(𝑡)𝑒−𝜎𝑡
𝑑𝑡
∞
−∞
< ∞
Example 9 Find unilateral Laplace transform for the following signals
𝒊) 𝒙(𝒕) = 𝜹(𝒕)
X S = x t e−st
dt
∞
0
= δ t e−st
dt
∞
0
= e−s 0
= 1 ∵ δ t =
1 for t = 0
0 for t ≠ 0
𝒊𝒊) 𝒙(𝒕) = 𝒖(𝒕)
𝑋 𝑆 = x t e−st
dt
∞
0
= u t e−st
dt
∞
0
= 1𝑒−𝑠𝑡
𝑑𝑡
∞
0
=
𝑒−𝑠𝑡
−𝑠 0
∞
=
1
𝑠
∵ u t =
1 for t ≥ 0
0 for t < 0
Example 10 Find Laplace transform of 𝑥 𝑡 = 𝑒𝑎𝑡
𝑢 𝑡
Solution:
𝑋 𝑆 = 𝐿 𝑒𝑎𝑡
𝑢 𝑡 = 𝑒𝑎𝑡
𝑒−𝑠𝑡
∞
0
𝑑𝑡 = 𝑒− 𝑠−𝑎 𝑡
∞
0
𝑑𝑡 =
𝑒− 𝑠−𝑎 𝑡
− 𝑠 − 𝑎 0
∞
=
1
𝑠 − 𝑎
Example 11 Determine initial value and final value of the following signal 𝑋(𝑆) =
1
𝑠 𝑠+2
Solution:
Initial value
𝑥 0 = Lt
𝑠→∞
𝑆𝑋 𝑆 = Lt
𝑠→∞
𝑠
1
𝑠 𝑠 + 2
=
1
∞
= 0
Final value
𝑥 ∞ = Lt
𝑠→0
𝑆𝑋 𝑆 = Lt
𝑠→0
𝑠
1
𝑠 𝑠 + 2
=
1
2
Example 12 Find inverse Laplace Transform of 𝑋 𝑆 =
𝑆2+9𝑆+1
𝑆[𝑆2+6𝑆+8]
. Find ROC for 𝑖) 𝑅𝑒 𝑠 > 0
𝑖𝑖) 𝑅𝑒 𝑠 < −4 𝑖𝑖𝑖) − 2 > 𝑅𝑒 𝑠 > −4
Solution:

10
X 𝑆 =
𝑆2
+ 9𝑆 + 1
𝑆[𝑆2 + 6𝑆 + 8]
=
𝑆2
+ 9𝑆 + 1
𝑆 𝑆 + 4 (𝑆 + 2)
=
𝐴
𝑆
+
𝐵
𝑆 + 4
+
𝐶
(𝑆 + 2)
𝑆2
+ 9𝑆 + 1 = 𝐴 𝑆 + 4 𝑆 + 2 + B𝑆 𝑆 + 2 + C𝑆 𝑆 + 4
at 𝑆 = 0
𝐴 =
1
8
𝑆 = −4
𝐵 = −
19
8
𝑆 = −2
𝐶 =
13
4
∴ 𝑋 𝑆 =
1
8
𝑆
+
−
19
8
𝑆 + 4
+
13
4
(𝑆 + 2)
Applying inverse Laplace transform
𝑥 𝑡 =
1
8
𝑢 𝑡 −
19
8
𝑒−4𝑡
𝑢(𝑡) +
13
4
𝑒−2𝑡
𝑢(𝑡)
ROC
𝒊) 𝑹𝒆 𝒔 > 𝟎
ROC lies right side of all poles
∴ 𝑥 𝑡 =
1
8
𝑢 𝑡 −
19
8
𝑒−4𝑡
𝑢(𝑡) +
13
4
𝑒−2𝑡
𝑢(𝑡)
𝒊𝒊) 𝑹𝒆 𝒔 < −𝟒
ROC lies left side of all poles
∴ 𝑥 𝑡 = −
1
8
𝑢 −𝑡 +
19
8
𝑒−4𝑡
𝑢 −𝑡 −
13
4
𝑒−2𝑡
𝑢(−𝑡)
𝒊𝒊𝒊) − 𝟐 > 𝑅𝑒 𝒔 > −𝟒
ROC lies left side of poles s=-2, s=0 and right side of
pole s=-4
∴ 𝑥 𝑡 = −
1
8
𝑢 −𝑡 −
19
8
𝑒−4𝑡
𝑢 𝑡 −
13
4
𝑒−2𝑡
𝑢(−𝑡)
jΩ
-4 -2 0 σ
jΩ
-4 -2 0 σ
jΩ
-4 -2 0 σ

Cosine Fourier Ses tes
xCt)= A+ +2 AnCas(nagt+0,)
n-
Ao De o m p o o n t
Hormonic Componen
PhaPhove an
Detumun Cosinafs dhe& pe
E
&n
2
So T E)A Siodbe
lt) uot
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dk

o Ca
Co i-Caso]
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Sin t (as nt ot
Stn ACasg Sin(1)+io (4-e) 7
Costt+). + Cas (1-n)E
tn
Do
1-Y
Cos (Itr)i Cas(i-n)
AT tn I-n + -n
edo -
Cos A = v e
-n -
on o:@xen
7
Oun +
L

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A(n) 2(Hn)
2
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1t)C-m T(-n)
toc no del
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b xt) So na,t dt
Siot otSin n dt
2TT
Sin Cl+n)t4So- k
s Cos (Hn)t
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d

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Cos~s s
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Find tho Cnponent F3 and Plok e,
ans bosu pecbru
Mgwinelo
x C )
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n Do
C
n z - P
C
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2h
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appl UV wehe
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e 21nE
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an- 1-0
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een
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Ar
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an
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eve n

Jio
Cn
Cn
Cn
Hante Co xCE) dt
C
x() 2 C
Jaiot
CE)- + e

Mesgnifelo Col Te
-tan
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PC)
-qo don n>d
90 dor ncd
1Cnl
sfo -2foH

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•,· X(f) : A(f ) --f- j J
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eC+)
?<(1) z
-=> J6 -&pL1
b rJnb tP1
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CE) jw)
E) T B2(t) «,iu)+ ex,Ow)
xu) SxE)e
jwt
t ) e dt +8 x,t)e
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Ct) x(w)
J(t)edt
u

tea Under zlw)
zt) = Jwt
dw
Sto
z(o) x(w) do
Jx(hu dw
2Ce) J xow) do > Avea undes (iw)
Scaleng
lat)
xCw
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d
atC
t

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Cal-) e
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xCus)
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ztato) C)
t t x{u)
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olt
(3u)J (ito)e
t(t-t tT+ E
dt dt
xCw)
ft-to)
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Fourter Jrencfosm Leaplace Transfor m
fropesty
lin2ily FLa,xt)+
+ak2(t)J=
a, X,Cw+aX2(iw)
La,,(e)+a2X2()
wto
shisin x[E-t) e
xCyw) (tt*) e XC
me
Tima
x(at) x
Scalin
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t Cut
e x(E) xCpa)
x(e) SXa)-x(0)
x) oxClw)
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Jw
Tme
rteaTakon Jxt) dt|ts
S
txt) -otxa)
s
reyuasN tzt x )
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-volubtun x,(E) ,[t)= *,(z) X, (t-2)|
SA) x l t ) A
x s ) X,)
C'w) Y, ()w)

le-A4.1: PropertiesofLaplaceTransform
Note:x(1)}=X(s); x,0X(8) L,(0 X
(8)
Property Timedomain signal S-domain signal
Amplitude sealing A x(t) A X(S)
Linearity 1x(t) a,x.(0) a,X,(9)a, X,()
Time differentiation x(t)
dt
s X() x(0)
d"
Kd xt)
d
s X(s)
d X ( t )
wheren1,2,3
X(s) x0 d
inmentegrati0n
x() dt
..x()(dt)"
Jora
X(s)
s
where n 1,2,3.. (0
Frequency shifting e x(t) X(s+a)
Timeshifting x(tt a) c Xs)
dX(s)
Frequency differentiation 1 x(1)
ds
(-1" d X(s)
t x()
wheren 1,2,3..
ds"
Frequency ntegration x(t) X(s) ds
Time scaling x(at)
Periodicity x(t + nT) x(1) e d
Initial value theorem Lt x()x(0) Lt SX
Final value theorem LI X(t) x()
x(t)x,(1)
onyolution theorein
0)st-2)dh

Table-A4.3: Properties ofContinuousTimeFourierTransform
Note:x(O}=Xj)Fx(0=X,G2)Fx0}-X0)
Property Time domain function Frequeney domain funetion
Linearity 2, X,(1) a, x,(t) 2, X,G2)-a, X.(Q
Time shiling x(t-) i Xi2)
x)
Time scaling x(at)
Tme reversal X-1) Xj2)
Conjugation x(1) X-iQ)
Frequency shifting eox(t) XG(2-2))
Time diffe entiation x(t)
XG2) TX(0) 02)
x(T) dr
Time integration
j2
Frequency differentiation t x(t) JXj2)
x,(t)#x (t)-X,(T) x,(t-T)dr XG2) X.(j2)
Time convolution
X()X.(i-2)di
Frequency convolution
(or Multiplication)
x(t)x.(t)
XGQ)X G2)
XGO-X(-i0
Symmetry of real signals x(1) 1s real
X(GQ)-Xj0)
Re XGO)) ReX-10
Real and Even x(1) 1s real and even
XG2)are realandeven
Real and Odd x(t) 18 Teal and od XGQhare imaginaryand odd
Duality If x.(t) X(2) 1.e.,x(nandX,(G)are sinilar fiunctons
then X(2)20x, j2) ie.X02)and 2rx12ure sinilar functions
Area under frequency
doman Signal KG d s0
Arca mder me
domangnal wdt

Finding the Fourier Transform of a given signal and plotting its magnitude and phase
Spectrum
AIM: To find the Fourier Transform of a given signal and plotting its magnitude and phase
spectrum.
Software Required:
Matlab software
Theory:
Fourier Transform:
The Fourier transform as follows. Suppose that ƒ is a function which is zero outside of some
interval [−L/2, L/2]. Then for any T ≥ L we may expand ƒ in a Fourier series on the interval
[−T/2,T/2], where the "amount" of the wave e2πinx/T in the Fourier series of ƒ is given by
By definition Fourier Transform of signal f(t) is defined as
Program:
clc;
clear all;
close all;
fs=1000;
N=1024; % length of fft sequence
t=[0:N-1]*(1/fs);
% input signal
x=0.8*cos(2*pi*100*t);
subplot(3,1,1);
plot(t,x);
axis([0 0.05 -1 1]);
grid;
xlabel('t');
ylabel('amplitude');
title('input signal');
% Fourier transformof given signal
x1=fft(x);
% magnitude spectrum

k=0:N-1;
Xmag=abs(x1);
subplot(3,1,2);
plot(k,Xmag);
grid;
xlabel('t');
ylabel('amplitude');
title('magnitude of fft signal')
%phase spectrum
Xphase=angle(x1);
subplot(3,1,3);
plot(k,Xphase);
grid;
xlabel('t');
ylabel('angle');
title('phase of fft signal');
Result: Magnitude and phase spectrum of FFT of a given signal is plotted.
Output:

UNIT- II – Post – Test – MCQ
1. What are the conditions called which are required for a signal to fulfil to be represented
as Fourier series?
a) Dirichlet’s conditions
b) Gibbs phenomenon
c) Fourier conditions
d) Fourier phenomenon
2. How is a trigonometric Fourier series represented?
a) A0 +∑ [an cos(ω0t)+ bn sin(ω0t)]
b) ∑ [an cos(ω0t)+ an sin(ω0t)]
c) A0 *∑ [an cos(ω0t )+ an sin(ω0t )]
d) A0 +∑ [an cos(ω0t)+ an sin(ω0t)] + sin ω0t
3. Which of the following is a Dirichlet condition with respect to the signal x(t)? a) x(t) has a
finite number of discontinuities in any period
b) x(t) has finite number of maxima and minima during any period
c) x(t) is absolutely integrable in any period
d) all of the mentioned
4. The Fourier transform of a function x(t) is X(f). The Fourier transform of 𝑑𝑥(𝑡)/ 𝑑𝑡 will be
(a) (𝑓) 𝑑𝑡
(b)𝑗2𝜋𝑓 (𝑓)
(c)𝑗𝑓 𝑋(𝑓)
(d)𝑋(𝑓) 𝑗𝑓
5. What should be the value of laplace transform for the time domain signal equation
𝑒−𝑎𝑡
cos ωt.u(t)?
a)
1
𝑠+𝑎
with ROC σ > – a
b)
𝜔
(𝑠+𝑎)2+ 𝜔2 with ROC σ > – a
c)
𝑠+𝑎
(𝑠+𝑎)2+ 𝜔2 with ROC σ > – a
d)
𝜔
𝑠2+ 𝜔2 with ROC σ > 0

SIGNALS AND SYSTEMS
Prerequisite:
Knowledge about properties of Fourier and Laplace transform
Objectives:
To understand the linear time invariant systems and its properties
Unit-III Linear Time Invariant – Continuous Time Systems
Differential Equation - Block diagram Representation, Impulse response,
Convolution Integral- Frequency response, Fourier and Laplace Transforms in
analysis, State variable equations and Matrix representation of systems.
Outcome:
 understand linear time-invariant systems theory and applications

UNIT- III – Pre – Test – MCQ
1. What should be location of poles corresponding to ROC for bilateral Inverse Laplace
Transform especially for determining the nature of time domain signal?
a. On L.H.S of ROC
b. On R.H.S of ROC
c. On both sides of ROC
d. None of the above
2. Define transfer function in CT system using Fourier transform.
a. 𝐻(𝑠) =
𝑌(𝑠)
𝑋(𝑠)
b. 𝐻(𝑓) =
𝑌(𝑓)
𝑋(𝑓)
c. 𝐻(𝑠) =
𝑤(𝑠)
𝑋(𝑠)
𝐻(𝑗𝜔) =
𝑌(jω)
𝑤(jω)
3. Let h(t) be the impulse response of a linear time invariant system. Then the response of
the system for any input u(t) is
(a) ∫ h(τ) u(t − τ)dτ
𝑡
0
(b)
𝑑
𝑑𝑡
)
𝑑
𝑑𝑡
∫ h(τ) u(t − τ)dτ
𝑡
0
(c)∫ | ∫ h(τ) u(t − τ)dτ
𝑡
0
|dt
𝑡
0
(d)∫ ℎ2
(τ) u(t − τ)dτ
𝑡
0
4. Let x(𝑡) ↔X (𝑗𝜔) be Fourier Transform pair. The Fourier Transform of the signal 𝑥(5𝑡 − 3)
in terms of 𝑋(𝑗𝜔) is given as
(a)
1
5
𝑒
−𝑗3𝜔
5 𝑋(
𝑗𝜔
5
)
(b)
1
5
𝑒
𝑗3𝜔
5 𝑋(
𝑗𝜔
5
)
(c) )
1
5
𝑒−𝑗3𝜔
𝑋(
𝑗𝜔
5
)
(d) )
1
5
𝑒𝑗3𝜔
𝑋(
𝑗𝜔
5
)
5. When is the system said to be causal as well as stable in accordance to pole/zero of ROC
specfied by system transfer function?
a. Only if all the poles of system transfer function lie in left-half of S-plane
b. Only if all the poles of system transfer function lie in right-half of S-plane c. Only if all
the poles of system transfer function lie at the centre of S-plane d. None of the above

Faculty: Mrs.M.Shakunthala & Dr.M.N.Vimal Kumar
202
Impulse input
𝛿(𝑡)
Continuous time system
h(t)
H
Unit 3: Linear Time Invariant-Continuous Time
Systems
3.1 LTI-CT (Linear Time Invariant-Continuous Time) Systems
When continuous time system satisfies the properties of linearity and time invariant
then it is called an LTI-CT (Linear Time Invariant-Continuous Time) System.
3.2 Impulse Response
When the input to a continuous time system is an unit impulse signal δ(t) then the
output is called an impulse response of the system and it is denoted by h(t)
Impulse response, h(t) = H{𝛿(t)}
Fig 3.1
3.3 Convolution Integral
𝑦(𝑡 = 𝑥(𝜏 𝑕(𝑡 − 𝜏 𝑑𝜏
∞
−∞
This is called convolution integral or simply convolution. The convolution of two signal x(t) and
h(t) can be represented as
𝑦(𝑡 = 𝑥(𝑡 ∗ 𝑕(𝑡)
3.4 Systems connected in series/parallel(Block diagram representation)
3.4.1 System Realization
There are four types of system realization in continuous time linear time invariant
systems.
They are
x Direct form I realization
x Direct form II realization
x Cascade form realization
x Parallel form realization
3.4.2 Direct form I realization
It is the direct implementation of differential equation or transfer function describing the
system. It uses separate integrators for input and output variables. It provides direct relation
between time domain and s-domain equations. In general, this form requires 2N delay elements
(for both input and output signals) for a filter of order N. This form is practical for small filters.
Advantages:
x Simplicity
x Most straight forward realization

Linear Time Invariant-Continuous Time systems
203
Disadvantages:
x More number of integrators are used
x Inefficient and impractical (numerically unstable) for complex design
3.4.3 Direct form II realization
It is the direct implementation of differential equation or transfer function describing the
system. Instead of using separate integrators for integrating input and output variables
separately, an intermediate variable is integrated. It provides direct relation between time
domain and s-domain equations.
Advantages:
x It uses minimum number of integrators
x Straight forward realization
Disadvantages:
x It increases the possibility of arithmetic overflow for filters of high Q or resonance
3.4.4 Cascade form
In cascade form realization the given transfer function is expressed as a product of several
transfer function and each of these transfer function is realized in direct form II and then all those
realized structures are cascaded i.e., is connected in series.
3.4.5 Parallel form realization
The given transfer function is expressed into its partial fractions and each factor is
realized in direct form II and all those realized structures are connected in parallel.
3.5 Solved Problems
Example 3.1: Find the convolution by graphical method
𝑥(𝑡 =
1 𝑓𝑜𝑟 0 ≤ 𝑡 ≤ 2
0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒
; 𝑕(𝑡 =
1 𝑓𝑜𝑟 0 ≤ 𝑡 ≤ 3
Solution:
𝐼𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑥1(𝑡 ∗ 𝑥2(𝑡 = 𝑥1(𝜏 𝑥2(𝑡 − 𝜏 𝑑𝜏
∞
−∞
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 𝑕(𝑡 ∗ 𝑥(𝑡 = 𝑕(𝜏 𝑥(𝑡 − 𝜏 𝑑𝜏
∞
−∞
Replacing 𝑡 by 𝜏 in 𝑥(𝑡 and 𝑕(𝑡)
𝑥(𝜏 =
1 𝑓𝑜𝑟 0 ≤ 𝜏 ≤ 2
; 𝑕(𝜏 =
1 𝑓𝑜𝑟 0 ≤ 𝜏 ≤ 3

204
Fig 3.21 Fig 3.22 Fig 3.23
Case (i) 𝒕 < 0
Fig 3.24
Since overlap is absent between 𝑕(𝜏) and 𝑥(−𝜏 + 𝑡)
∴ 𝑦(𝑡 = 𝑕(𝑡 ∗ 𝑥(𝑡 = 0
Case (ii) 𝟎 ≤ 𝒕 < 2
Fig 3.25
Since overlap is present
∴ 𝑦(𝑡 = 𝑕(𝑡 ∗ 𝑥(𝑡 = 𝑕(𝜏 𝑥(𝑡 − 𝜏 𝑑𝜏
∞
−∞
= (1 (1) 𝑑𝜏
𝑡
0
= 𝜏 0
𝑡
= 𝑡
Case (iii) 𝟐 ≤ 𝒕 < 3
Fig 3.26
∴ 𝑦(𝑡 = 𝑕(𝑡 ∗ 𝑥(𝑡 = 𝑕(𝜏 𝑥(𝑡 − 𝜏 𝑑𝜏
∞
−∞
= (1 (1) 𝑑𝜏
𝑡
𝑡−2
= 𝜏 𝑡−2
𝑡
= 2
Case (iv) 𝟑 ≤ 𝒕 < 5
Fig 3.27
∴ 𝑦(𝑡 = 𝑕(𝑡 ∗ 𝑥(𝑡 = 𝑕(𝜏 𝑥(𝑡 − 𝜏 𝑑𝜏
∞
−∞
= (1 (1) 𝑑𝜏
3
𝑡−2
= 𝜏 𝑡−2
3
= 5 − 𝑡
Case (v) 𝒕 > 5
Fig 3.28
Since overlap is absent
∴ 𝑦(𝑡 = 𝑕(𝑡 ∗ 𝑥(𝑡 = 0
0 3 t-2 t τ
h(τ)x(t-τ)
1
0 t-2 3 t τ
h(τ)x(t-τ)
1
0 t-2 t 3 τ
h(τ)x(t-τ)
1
t-2 0 t 3 τ
h(τ)x(t-τ)
1
t-2 t 0 3 τ
h(τ)x(t-τ)
1
-2 0 τ
x(-τ)
1
0 2 τ
x(τ)
1
0 3 τ
h(τ)
1

205
∴ 𝑦(𝑡 = 𝑕(𝑡 ∗ 𝑥(𝑡 =
0 𝑓𝑜𝑟 𝑡 < 0
𝑡 𝑓𝑜𝑟 0 ≤ 𝑡 < 2
2 𝑓𝑜𝑟 2 ≤ 𝑡 < 3
5 − 𝑡 𝑓𝑜𝑟 3 ≤ 𝑡 < 5
0 𝑓𝑜𝑟 𝑡 ≥ 5
Example 3.2: Find impulse response of the following equation
𝑑2
𝑦(𝑡)
𝑑𝑡2
+ 5
𝑑𝑦(𝑡)
𝑑𝑡
+ 6𝑦(𝑡) = 𝑥(𝑡)
Solution:
𝑑2
𝑦(𝑡)
𝑑𝑡2
+ 5
𝑑𝑦(𝑡)
𝑑𝑡
+ 6𝑦(𝑡) = 𝑥(𝑡)
Assume all the initial conditions are zero
Applying Laplace transform of the given equation
𝑆2
𝑌(𝑆 + 5𝑆𝑌(𝑆 + 6𝑌(𝑆 = 𝑋(𝑆)
𝑌(𝑆 (𝑆2
+ 5𝑆 + 6) = 𝑋(𝑆)
Transfer function 𝐻(𝑆 =
𝑌(𝑆
𝑋(𝑆)
=
1
(𝑆2+5𝑆+6)
𝐻(𝑆 = 𝑌(𝑆 =
1
𝑆2 + 5𝑆 + 6
(∵ 𝐹𝑜𝑟 𝑖𝑚𝑝𝑢𝑙𝑠𝑒 𝑖𝑛𝑝𝑢𝑡 𝑥(𝑡) = 𝛿(𝑡) => 𝑋(𝑆) = 1)
𝐻(𝑆 =
1
(𝑆 + 3 (𝑆 + 2)
=
𝐴
𝑆 + 3
+
𝐵
𝑆 + 2
1 = 𝐴(𝑆 + 2 + 𝐵(𝑆 + 3)
at 𝑆 = −3
𝐴 = −1
𝑆 = −2
𝐵 = 1
∴ 𝐻(𝑆 = −
1
𝑆 + 3
+
1
𝑆 + 2
Applying Inverse Laplace transform
𝒉(𝒕 = −𝒆−𝟑𝒕
𝒖(𝒕 + 𝒆−𝟐𝒕
𝒖(𝒕)
Example 3.3: Using Laplace transform solve differential equation
𝑑2
𝑦(𝑡)
𝑑𝑡2
+ 𝑦(𝑡) = 𝑥(𝑡)
Where y′
(0) = 2; 𝑦(0 = 1; 𝑖𝑛𝑝𝑢𝑡 𝑥(𝑡 = 𝐶𝑜𝑠2𝑡
Solution:
𝑑2
𝑦(𝑡)
𝑑𝑡2
+ 𝑦(𝑡) = 𝑥(𝑡)
Applying Laplace transform
𝑆2
𝑌(𝑆 − 𝑆𝑦(0 − 𝑦′ (0 + 𝑌(𝑆 = 𝑋(𝑆)
𝑆2
𝑌(𝑆 − 𝑆 − 2 + 𝑌(𝑆 =
𝑆
𝑆2 + 4
∴ 𝑌(𝑆 (𝑆2
+ 1 =
𝑆
𝑆2 + 4
+ 𝑆 + 2

206
𝑌(𝑆 =
𝑆
(𝑆2 + 4)(𝑆2 + 1
+
𝑆
(𝑆2 + 1
+
2
(𝑆2 + 1
𝐿𝑒𝑡
𝑆
(𝑆2 + 4)(𝑆2 + 1
=
𝐴𝑆 + 𝐵
(𝑆2 + 4)
+
𝐶𝑆 + 𝐷
(𝑆2 + 1
𝑆 = (𝐴𝑆 + 𝐵 (𝑆2
+ 1 + (𝐶𝑆 + 𝐷)(𝑆2
+ 4
𝑆 = 𝐴𝑆3
+ 𝐵𝑆2
+ 𝐴𝑆 + 𝐵 + 𝐶𝑆3
+ 𝐷𝑆2
+ 4𝐶𝑆 + 4𝐷
Comparing constant term
0 = 𝐵 + 4𝐷
𝐵 = −4𝐷 … (7)
Comparing coeff of 𝑆3
0 = 𝐴 + 𝐶
𝐴 = −𝐶 … (8)
Comparing coeff of 𝑆2
0 = 𝐵 + 𝐷 … (9)
Comparing coeff of S
1 = 𝐴 + 4𝐶 … (10)
Substitute 𝑒𝑞 8 in 𝑒𝑞 10 and 𝑒𝑞 7 in 𝑒𝑞 9
𝑪 =
𝟏
𝟑
, 𝑫 = 𝟎
Substitute value of C and D in 𝑒𝑞 8 and 𝑒𝑞 7
𝑨 = −
𝟏
𝟑
, 𝑩 = 𝟎
∴
𝑆
(𝑆2 + 4)(𝑆2 + 1
=
−
1
3
𝑆
(𝑆2 + 4)
+
1
3
𝑆
(𝑆2 + 1
𝑌(𝑆 =
−
1
3 𝑆
(𝑆2 + 4)
+
1
3 𝑆
(𝑆2 + 1
+
𝑆
(𝑆2 + 1
+
2
(𝑆2 + 1
𝑌(𝑆 =
−
1
3 𝑆
(𝑆2 + 4)
+
4
3 𝑆
(𝑆2 + 1
+
2
(𝑆2 + 1
Taking Inverse Laplace transform
𝒚(𝒕 = −
𝟏
𝟑
𝒄𝒐𝒔𝟐𝒕 𝒖(𝒕 +
𝟒
𝟑
𝒄𝒐𝒔𝒕 𝒖(𝒕 + 𝟐𝒔𝒊𝒏𝒕 𝒖(𝒕)
Example 3.4: Find step response of the circuit shown in Fig 3.30
Fig 3.30
Solution:
Applying KVL to the circuit shown in Fig 3.30
𝑥(𝑡 = 𝑅𝑖(𝑡 + 𝐿
𝑑𝑖(𝑡)
𝑑𝑡
Applying Laplace transform
𝑋(𝑆 = 𝑅𝐼(𝑆 + 𝐿𝑆𝐼(𝑆)
𝑋(𝑆 = [𝑅 + 𝐿𝑆]𝐼(𝑆)
𝑦(𝑡 = 𝐿
𝑑𝑖(𝑡)
𝑑𝑡
𝑌(𝑆 = 𝐿𝑆𝐼(𝑆)
R
x(t) L
i(t)
y(t)

207
𝐼(𝑆 =
𝑋(𝑆
[𝑅 + 𝐿𝑆]
𝑌(𝑆 = 𝐿𝑆
𝑋(𝑆
[𝑅 + 𝐿𝑆]
For Step response 𝑥(𝑡 = 𝑢(𝑡 => 𝑋(𝑆 =
1
𝑆
𝑌(𝑆 = 𝐿𝑆
1
𝑆
𝑅 + 𝐿𝑆
=
𝐿
𝐿𝑆 + 𝑅
=
1
𝑆 +
𝑅
𝐿
Applying Inverse Laplace transform
𝒚(𝒕 = 𝒆−
𝑹
𝑳
𝒕
𝒖(𝒕)
Example 3.5: Solve the differential equation using Fourier transform
𝑑2
𝑦(𝑡)
𝑑𝑡2
+ 6
𝑑𝑦(𝑡)
𝑑𝑡
+ 8𝑦(𝑡) = 2𝑥(𝑡)
(i) Find the impulse response of the system
(ii) What is the response of the system if 𝑥(𝑡) = 𝑡𝑒−2𝑡
𝑢(𝑡)
Solution:
𝑑2
𝑦(𝑡)
𝑑𝑡2
+ 6
𝑑𝑦(𝑡)
𝑑𝑡
+ 8𝑦(𝑡) = 2𝑥(𝑡)
Applying Fourier transform
(𝑗𝛺 2
Y(𝑗Ω + 6𝑗Ω𝑌(𝑗Ω + 8𝑌(𝑗Ω = 2X(𝑗Ω)
Y(𝑗Ω [(𝑗𝛺 2
+ 6𝑗Ω + 8] = 2X(𝑗Ω)
H(𝑗Ω =
Y(𝑗Ω
X(𝑗Ω)
=
2
[(𝑗𝛺 2 + 6𝑗Ω + 8]
(i) Impulse response 𝑥(𝑡 = 𝛿(𝑡 => 𝑋(𝑗Ω = 1
∴ H(𝑗Ω = Y(𝑗Ω =
2
[(𝑗𝛺 2 + 6𝑗Ω + 8]
=
𝐴
𝑗Ω + 4
+
𝐵
𝑗Ω + 2
2 = 𝐴(𝑗Ω + 2 + 𝐵(𝑗Ω + 4
at 𝑗Ω = −4
𝐴 = −1
𝑗𝛺 = −2
𝐵 = 1
H(𝑗Ω =
−1
𝑗Ω + 4
+
1
𝑗Ω + 2
Applying Inverse Fourier Transform
𝒉(𝒕 = −𝒆−𝟒𝒕
𝒖(𝒕 + 𝒆−𝟐𝒕
𝒖(𝒕
(ii) 𝑥(𝑡 = 𝑡𝑒−2𝑡
𝑢(𝑡
𝑋(𝑗Ω =
1
(𝑗Ω + 2 2
Y(𝑗Ω
X(𝑗Ω)
=
2
[(𝑗𝛺 2 + 6𝑗Ω + 8]
∴ Y(𝑗Ω =
2
[(𝑗𝛺 2 + 6𝑗Ω + 8]
.
1
(𝑗Ω + 2 2
=
2
(𝑗Ω + 4 (𝑗Ω + 2 3
=
𝐴
𝑗Ω + 4
+
𝐵
𝑗Ω + 2
+
𝐶
(𝑗Ω + 2 2
+
𝐷
(𝑗Ω + 2 3

208
at 𝑗Ω = −4
𝐴 =
2
(𝑗Ω + 4 (𝑗Ω + 2 3
(𝑗Ω + 4
𝑗Ω=−4
𝐴 = −
1
4
𝑗𝛺 = −2
𝐵 =
1
2!
𝑑2 2
(𝑗Ω + 4 (𝑗Ω + 2 3 (𝑗Ω + 2 3
𝑑(𝑗Ω 2
𝑗𝛺 =−2
𝐵 =
1
4
𝑗𝛺 = −2
𝐶 =
𝑑
2
(𝑗Ω + 4 (𝑗Ω + 2 3 (𝑗Ω + 2 3
𝑑(𝑗Ω
𝑗𝛺 =−2
𝐶 = −
1
2
𝑗𝛺 = −2
𝐷 =
2
(𝑗Ω + 4 (𝑗Ω + 2 3
(𝑗Ω + 2 3
𝑗𝛺 =−2
𝐷 = 1
∴ Y(𝑗Ω =
−
1
4
𝑗Ω + 4
+
1
4
𝑗Ω + 2
+
−
1
2
(𝑗Ω + 2 2
+
1
(𝑗Ω + 2 3
Applying Inverse Fourier Transform
𝒚(𝒕 = −
𝟏
𝟒
𝒆−𝟒𝒕
𝒖(𝒕 +
𝟏
𝟒
𝒆−𝟐𝒕
𝒖(𝒕 −
𝟏
𝟐
𝒆−𝟐𝒕
𝒕𝒖(𝒕 +
𝟏
𝟐
𝒆−𝟐𝒕
𝒕𝟐
𝒖(𝒕
Example 3.6: Find the direct form II structure of
𝐻(𝑆 =
5𝑆3
− 4𝑆2
+ 11𝑆 − 2
𝑆 −
1
4
𝑆2 − 𝑆 +
1
2
Solution:
𝐻(𝑆 =
5𝑆3
− 4𝑆2
+ 11𝑆 − 2
𝑆 −
1
4 𝑆2 − 𝑆 +
1
2
=
5𝑆3
− 4𝑆2
+ 11𝑆 − 2
𝑆3 −
𝑆2
4 − 𝑆2 +
𝑆
4 +
𝑆
2 −
1
8
𝐻(𝑆 =
5𝑆3
− 4𝑆2
+ 11𝑆 − 2
𝑆3 −
5𝑆2
4 +
3𝑆
4 −
1
8
=
5 −
4
𝑆
+
11
𝑆2 −
2
𝑆3
1 −
5
4𝑆 +
3
4𝑆2 −
1
8𝑆3
Direct form II structure
Fig 3.39
Example 3.7: Realize the system with following differential equation in direct form I
𝑑3𝑦(𝑡)
𝑑𝑡 3 + 3
𝑑2𝑦(𝑡)
𝑑𝑡 2 + 5
𝑑𝑦(𝑡)
𝑑𝑡
+ 7𝑦(𝑡) = 2
𝑑2𝑥(𝑡)
𝑑𝑡 2 + 0.4
𝑑𝑥(𝑡)
𝑑𝑡
+ 0.5𝑥(𝑡)
1/S
1/S
X(S) W(S) Y(S)
5/4
-3/4
-4
11
1/8 -2
5

209
Solution:
𝑑3𝑦(𝑡)
𝑑𝑡 3 + 3
𝑑2𝑦(𝑡)
𝑑𝑡 2 + 5
𝑑𝑦(𝑡)
𝑑𝑡
+ 7𝑦(𝑡 = 2
𝑑2𝑥(𝑡)
𝑑𝑡 2 + 0.4
𝑑𝑥(𝑡)
𝑑𝑡
+ 0.5𝑥(𝑡)
Taking Laplace transform
𝑆3
𝑌(𝑆 + 3𝑆2
𝑌(𝑆 + 5𝑆𝑌(𝑆 + 7𝑌(𝑆 = 2𝑆2
𝑋(𝑆 + 0.4𝑆𝑋(𝑆 + 0.5𝑋(𝑆
Dividing both the side by 𝑆3
𝑌(𝑆 +
3
𝑆
𝑌(𝑆 +
5
𝑆2
𝑌(𝑆 +
7
𝑆3
𝑌(𝑆 =
2
𝑆
𝑋(𝑆 +
0.4
𝑆2
𝑋(𝑆 +
0.5
𝑆3
𝑋(𝑆
𝑌(𝑆 =
2
𝑆
𝑋(𝑆 +
0.4
𝑆2
𝑋(𝑆 +
0.5
𝑆3
𝑋(𝑆 −
3
𝑆
𝑌(𝑆 −
5
𝑆2
𝑌(𝑆 −
7
𝑆3
𝑌(𝑆
Direct form I structure
Fig 3.42
Example 3.8: Realize the system with transfer function in cascade form
𝐻(𝑆 =
4(𝑆2
+ 4𝑆 + 3
𝑆3 + 6.5𝑆2 + 11𝑆 + 4
Solution:
𝐻(𝑆 =
4(𝑆2
+ 4𝑆 + 3
𝑆3 + 6.5𝑆2 + 11𝑆 + 4
=
4(𝑆 + 1 (𝑆 + 3
(𝑆 + 0.5 (𝑆 + 2 (𝑆 + 4
=
4
𝑆 + 0.5
.
𝑆 + 1
𝑆 + 2
.
𝑆 + 3
𝑆 + 4
𝐻1(𝑆 𝐻2(𝑆 𝐻3(𝑆 =
4
𝑆 + 0.5
.
𝑆 + 1
𝑆 + 2
.
𝑆 + 3
𝑆 + 4
𝐻1(𝑆 =
4
𝑆 + 0.5
=
4
𝑆
1 + 0.5
𝑆
𝑌1(𝑆)
𝑊1(𝑆)
=
4
𝑆
𝑊1(𝑆)
𝑋1(𝑆)
=
1
1 +
0.5
𝑆
Fig 3.54
𝐻2(𝑆 =
𝑆 + 1
𝑆 + 2
=
1 + 1
𝑆
1 + 2
𝑆
𝑌2(𝑆)
𝑊2(𝑆)
= 1 + 1
𝑆
𝑊2(𝑆)
𝑋2(𝑆)
=
1
1 + 2
𝑆
Fig 3.55
𝐻3(𝑆 =
𝑆 + 3
𝑆 + 4
=
1 + 3
𝑆
1 + 4
𝑆
𝑌3(𝑆)
𝑊3(𝑆)
= 1 + 3
𝑆
𝑊3(𝑆)
𝑋3(𝑆)
=
1
1 + 4
𝑆
Fig 3.56
1/S
X3(S) W3(S) Y3(S)
-4 3
1/S
X2(S) W2(S) Y2(S)
-2 1
Y1(S)
X1(S) W1(S)
-0.5 4
1/S
1/S 1/S
1/S
1/S
X(S) W(S) Y(S)
2
0.4
-3
-5
1/S 1/S
0.5 -7

210
Cascade form:
Fig 3.57
Example 3.9: Realize the following system in parallel form
𝐻(𝑆 =
𝑆(𝑆 + 2
(𝑆 + 1 (𝑆 + 3 (𝑆 + 4
Solution:
𝐻(𝑆 =
𝑆(𝑆 + 2
(𝑆 + 1 (𝑆 + 3 (𝑆 + 4
=
𝐴
𝑆 + 1
+
𝐵
𝑆 + 3
+
𝐶
𝑆 + 4
𝑆(𝑆 + 2 = 𝐴(𝑆 + 3 (𝑆 + 4 + 𝐵(𝑆 + 1 (𝑆 + 4 + 𝐶(𝑆 + 1 (𝑆 + 3)
Let S= −1
−1(1 = 𝐴(2 (3)
𝑨 = −
𝟏
𝟔
Let S= −3
−3(−1 = 𝐵(−2 (1)
𝑩 = −
𝟑
𝟐
Let S= −4
−4(−2 = 𝐶(−3 (−1)
𝑪 =
𝟖
𝟑
∴ 𝐻(𝑆 =
−
1
6
𝑆 + 1
+
−
3
2
𝑆 + 3
+
8
3
𝑆 + 4
Parallel form structure
X (S)
-1 -1/6
-3
-3/2
-4
8/3 Y (S)
1/S
1/S
1/S
X(S)
-0.5 4
-2 1
Y(S)
-4 3
1/S
1/S
1/S

2.57. Determine the output of the systems described by the following differential equations with input
andinitial conditions as specified:
(a) d
dt y(t) + 10y(t) = 2x(t), y(0−
) = 1, x(t) = u(t)
t ≥ 0 natural: characteristic equation
r + 10 = 0
r = −10
y(n)
(t) = ce−10t
particular
y(p)
(t) = ku(t) =
1
5
u(t)
y(t) =
1
5
+ ce−10t
y(0−
) = 1 =
1
5
+ c
c =
4
5
y(t) =
1
5

1 + 4e−10t

u(t)
(b) d2
dt2 y(t) + 5 d
dt y(t) + 4y(t) = d
dt x(t), y(0−
) = 0, d
dt y(t)

t=0− = 1, x(t) = sin(t)u(t)
r2
+ 5r + 4 = 0
r = −4, − 1
y(n)
(t) = c1e−4t
+ c2e−t
particular
y(p)
(t) = A sin(t) + B cos(t)
=
5
34
sin(t) +
3
34
cos(t)
y(t) =
5
34
sin(t) +
3
34
cos(t) + c1e−4t
+ c2e−t
y(0−
) = 0 =
3
34
+ c1 + c2
d
dt
y(0)

t=0−
= 1 =
5
34
− 4c1 − c2
c1 = −
13
51
c2 =
1
6
y(t) =
5
34
sin(t) +
3
34
cos(t) −
13
51
e−4t
+
1
6
e−t
(c) d2
dt2 y(t) + 6 d
dt y(t) + 8y(t) = 2x(t), y(0−
) = −1, d
dt y(t)

t=0− = 1, x(t) = e−t
u(t)
61

r2
+ 6r + 8 = 0
r = −4, − 2
y(n)
(t) = c1e−2t
+ c2e−4t
particular
y(p)
(t) = ke−t
u(t)
=
2
3
e−t
u(t)
y(t) =
2
3
e−t
u(t) + c1e−2t
+ c2e−4t
y(0−
) = −1 =
2
3
+ c1 + c2
d
dt
y(0)

t=0−
= 1 = −
2
3
− 2c1 − 4c2
c1 = −
5
2
c2 =
5
6
y(t) =
2
3
e−t
u(t) −
5
2
e−2t
+
5
6
e−4t
(d) d2
dt2 y(t) + y(t) = 3 d
dt x(t), y(0−
) = −1, d
dt y(t)

t=0− = 1, x(t) = 2te−t
u(t)
r2
+ 1 = 0
r = ±j
y(n)
(t) = A cos(t) + B sin(t)
particular
y(p)
(t) = kte−t
u(t)
d2
dt2
y(p)
(t) = −2ke−t
+ kte−t
−2ke−t
+ kte−t
+ kte−t
= 3[2e−t
− 2te−t
]
k = −3
y(p)
(t) = −3te−t
u(t)
y(t) = −3te−t
u(t) + A cos(t) + B sin(t)
y(0−
) = −1 = 0 + A + 0
d
dt
y(t)

t=0−
= 1 = −3 + 0 + B
y(t) = −3te−t
u(t) − cos(t) + 4 sin(t)
62

3.88. Use the result of Problem 3.87 to determine the frequency response, impulse response, and dif-
ferential equation descriptions for the continuous-time systems described by the following state variable
descriptions.
(a) A =

−2 0
0 −1

, b =

0
2

, c = 1 1 , D = [0]
H(jω) = c(jωI − A)−1
b + D
=
2
jω + 1
h(t) = 2e−t
u(t)
=
Y (jω)
X(jω)
Y (jω)(jω + 1) = 2X(jω)
d
dt
y(t) + y(t) = 2x(t)
(b) A =

1 2
−3 −4

, b =

1
2

, c = 0 1 , D = [0]
H(jω) = c(jωI − A)−1
b + D
=
2jω − 5
(jω)2 + 3jω + 2
=
Y (jω)
X(jω)
Y (jω)((jω)2
+ 3jω + 2) = X(jω)(2jω − 5)
d2
dt2
y(t) + 3
d
dt
y(t) + 2y(t) = 2
d
dt
x(t) − 5x(t)
H(jω) =
A
2 + jω
+
B
1 + jω
2 = A + B
−5 = A + 2B
=
9
2 + jω
−
7
1 + jω
h(t) = (9e−2t
− 7e−t
)u(t)
3.89. Use the result of Problem 3.87 to determine the frequency response, impulse response, and
diﬀerence equation descriptions for the discrete-time systems described by the following state variable
descriptions.
(a) A =

−1
2 1
0 1
4

, b =

0
1

, c = 1 0 , D = [1]
H(ejΩ
) = c(ejΩ
I − A)−1
b + D
= 1 +
1
(ejΩ + 1
2 )(ejΩ − 1
4 )
58

ONTT-3
TT- CT
Theso a ems Oua cheracteri zud
J mpulce res poN Sa
iderentel ecuotton
3 Block diaare ms
A Stle Vesiele dseription
vounufer duncton
Difexntal Fauabo
M
N
a dAk CE) b (6)
dt
d t K
DE velatos the put anod op 0f the
Natusrel Kesponse
The
The output Procucael oy te Sustawo only deu
Es otbia Condibions TP u zero dov natureul
Teaponso CE
FoYCac Kasponse S yC
The Output Proclucaol by to Sysbn only du
2/P. La:ial Conclstiona ave CeonsMelereel a koro.
Th

Sr. No. Input Particular solution
(t)
1 k
keat
e a t
3 cos(of+ 0) k COs Of + ko sin ot
ea cos(of+o) e kicosot+k,sinot]
5 ko + kt
6 ko +kt+k,t2+.
terat
ko+ kt)

Complela Raxpense to ha
4 s Capol
aanclferCac Ras ponL
etermi ne Cemplala e a pons o
dg) ds 4gt
d
coth ()=o duCE) I 4 x(t)-e
+
olt
-26
dlet-o N
Selo
k=o
Complela Soln
(P)
CE)
Notsed apone
stup 1 Reeb e Chavecturiehe Caahe
sdyt) y (t) dE)
N-2 Bn sY+yYao
6ru C)-o
Rs

Stap
4, C, e 2 e 2 t
Ce
A C e
shp3
Particular ubo
Paut o
orCad osponse
P E) eul
From ble
-2
e) ke
Sap h
To ind k, .2t
Pat ut)=4EN H E, xlt)= e ut)
Pot
d 2
( k ) + 5 ( k e 4 -2
ax-2ke2t) )
Ak e - 10 ke 4k e -2e
-2t
-Zke
-2t
-2e
k
- 2 E
e

Ce4Ce -2t
e
Stp S:
To dmd C C2
T]e Reso Use Tnitved Conelhun
Ste)= Ce c,e xe 2
2 e 6
D.r.to t.
cdylt) 4 ce-jc, 2 e 2 6
-4t
dt
Lwk yCo)= o dy(o) 1 t= o
de
-4 C-C2
-4C-Ca 3
4C+C2 -3
- 2
Lub tzo n 8CE
9(o) C+Czt
Ct
C= -)-
- S
4C-C
C,--C)

2
C
Complett Ksponse C
ty
C)- e e-y,e 2
+e
'
te deteantal aquobon of t gen
d() +5 agt)+6%(e) = - d)
Te
2
de de
seterm)ne o dreaony esponte
mpule o ta un
sesponAe
dxt)
d t d cul)LqCt)= -
Taka FGuter renaferm
Siako
U) Yw)+sbw) Y(w)+6 Y(w)-Jw X(a)

ylwou)+sjw+
-
Juo Yleo)
Trouder dunc bo n
YCeo)
HC)= xlw) u)sjw+b
-J A
+
Cu)sjwt b Jw +2 Ju +3
jw
ACw+:)+ e(juwf2) - Jw
Jw-3 o+ BC-) =-3)
3 B-3
A-2
jw-2 A+o -C-2)
2 3
H( wt2 wt3
FT E
OL+jw
-2t
ht) 2e -3e tCt)

to Ceonvolubon of two dunebu
obteuin
(E) 2 - 2 t 2
else
ht)- ost2
eLye
C
7
2
1h-)
SCE)x(z)h
(t-z)dt
Case E-2
h)
( e )
t
-5) xt)-f
h t ) =
«(t)
A[-TDele
CE)
4h
8(E)o No oveslap
he z
4h

Cosp
e- 2 E
St J 2C) h[t-d
oxhlt-)dz+ C) h[-)d
-2
CE) O dr
E
CT)hCt-colz
2 dz 8z
-2 -2
- 2 )
T+ 2 or-2 stco

lap In) ocE2
2
- 2 0
SC-J(t)h{t-)d
( ) h l t - e l
t-2 E
st-(e- -t+2] -1b
Cae v) 2te4
2
(3)

9l) xC hE-c)d
J -() hCt- ) d
E-2
d z
E-2 -
8 [ , 2--2)
E-2
[2-E +2
-E3 dor 2ete
CepC) E4
-2 2
Ovexlap
dov E-2
RE) 8E2)or 2t20
fov b t 2
C4-) tor 2 t 4

Block apom kopresetetoo
SCculcr Mbelication
xC
CE)
8CE cxCE)
Acdelttion
,LE)
3 Intearatio n
() (E)dz
Caslado
H)- (w)-Rsl)
s -
Paclal
H(w)- H)4Ha(u)
EO-yCu
CoH

DrOLd dtvect dorm I dree o
3trwctuses de TT-CT
dly(t) 12Ct) dxt) x(t)
dt
6reweal Cquahoo
(N-)
CE)
M
Ct)
ko
N-2
2
C2-k)
CE) b tE)
(2-k)
(2)
C
ag (E) a,u ) + yCt) =
boYCe)+ 6,0) b,2(t)
k- 4e dorn E)
8e)-, sre)-yte)b,x(o + b,x

DxecEtorm T
S-
b
a
CED
o
PCE
bo
derm
D r e e t
z ( t )
b
Oo bo

7 g (E)- dx[t)
[E)
7 dgCe)
dt t
NE) S ahndd be 1,o divido b
C 1 dx(e) x ()
bp
de
12
a b
7
Drect Jorm T
3CE)
Drect dovm
by
bo1

)
dyc
+4 dat) +7dyCe)
de(E) +4 d)7lt)
deT de
dt
N:3
24 Ol 8B
b
bgO b,4 bo7
b3 glt
- Np.h
b 4
-A
bo1
b3 )
b S
- c - 4
-1
bo
o

atrik
Equeuon
dyl
dE
(e)
day le b
d
VC)
a y, a lt) + x(t)
olt
d qy Ct)
-V,C
dt
ye) b,,CE) +bo7,(t)+ bzlt)
at
da,le) 0
9,0t)
dE

dey AT ) +fs]« (E)
de
9( b b]lt+b zlt)
Cbb
a
A
D [b]
Matsices
A D
equata of the CT-LTT
ootaiin
y CE)2g(t) +2yCt) = xC)
a Ce ,9Ce) Aoy(t) = bo x(t)
bo1
a 2 a :3 Ao2
da,C6

d
- a , - 3
l e
daple)
o t
C 2
o, -3Ce -2 (e)-+ xlE
oll
dCe)
CE)
dyle) -3
-E
Ce)
ylt)= To
cy, (e) +ToT xCe)
A 6 - c =
[o ] DO
- 3 -2
A
LI

FreauanKeapon
j o qy(co) =Ayw)texw
Ylw) cl)+x/w)
Ju
y(cu)-AC)+RAo)
Swq(o) -
ACw) = e x(w)
Juwl- A) () = 6 x(w) Idenl:l
-
lcw)=3wT -A7 x/u)
V (w) = Cqlw) -+ Dxl
Ylo) C Guwg-A)s x + x)
(Sw7-A) B+
Yo) c[hw2 -AJ D
x(o)
y
H(c0) freqpenoy Raspons

ocbio
TYonslex
nstanel
SyCs) AyCs) x(s)
y () Cyl) DXC)
Cs)=ST -A Bxls)
Ys)
c s-ATBxls) 4Dxs)
Yca e o-]'8 +
b} xs)
H(s) c(s3-A +D
XCs)
Funct'on, difexenbal
Lctieh Tepresenlad
1 Find tho tranader
Copouhon of tho Sqstam
stda Cauabbn.
A: =
[o 3b-
A
H c s-A +D
2-)
ST A )
s-2
O -)
S-2
s 2

-)
S2
(s1-A (s-- S-2
P
R
C-2
s-AT
s-4S +y-) 1 -z -
-R P
PS-R
S-2
S-4S43 S-2
HCs) Ce
csT-A s+D
o s-s+3
-1 S-2
O+s-2
sus+3
S - 2
-O(S- 2)1
sus+3 s St3
S-2
Hs) = YG)
x(s) S + 3
SyCs)-4 S Ys) + 3 y/s) S Xs) -
2x s)
deCE)2x(
dytt -Adylt) + 3y(
olt 2
t

Waveform Synthesis Using Laplace transforms
AIM:
Finding the Laplace transform Inverse Laplace transform of some signals.
Software Required:
Matlab software
Theory:
Bilateral Laplace transforms:
The Laplace transform of a signal f(t) can be defined as follows:
Inverse Laplace transform
The inverse Laplace transform is given by the following formula :

Program:
clc;
clear all;
close all;
%representation of symbolic variables
syms f t w s;
%laplace transform of t
f=t;
z=laplace(f);
disp('the laplace transform of f = ');
disp(z);
% laplace transform of a signal
%f1=sin(w*t);
f1=-1.25+3.5*t*exp(-2*t)+1.25*exp(-2*t);
v=laplace(f1);
disp('the laplace transform of f1 = ');
disp(v);
lv=simplify(v);
pretty(lv)
%inverse laplace transform
y1=ilaplace(z);
disp('the inverse laplace transform of z = ');
disp(y1);
y2=ilaplace(v);
disp('the inverse laplace transform of v = ');
disp(y2);
ezplot(y1);
figure;
ezplot(y2)

UNIT- III – Post – Test – MCQ
1. Which among the following are the interconnected units of state diagram
representation?
a.Scalars
b. Adders
c. Integrators
d. All of the above
2. State space analysis is applicable even if the initial conditions are _____
a. Zero
b. Non-zero
c. Equal
d. Not equal
3. What is the overall impulse response h(t) when two systems with impulse response h1(t)
and h2(t) are in parallel?
a. h(t) = h1(t) /h2(t)
b. h(t) = h1(t) *h2(t)
c. h(t) = h1(t) h2(t)
d. h(t) = h1(t) +h2(t)
4. What is the transfer function of a system whose poles are at -0.3± j 0.4 and a zero at –
0.2?
a.
𝑠+2
(𝑆2+0.6𝑆+0.25)
b.
𝑠
(𝑆2+0.6𝑆+0.25)
c.
𝑠+0.2
(𝑆2+0.6𝑆+0.25)
d.
𝑠+0.2
(𝑆2+0.3𝑠+0.4)
5. The particular solution for the input 𝑥(𝑡) = cos(𝜔𝑡 + 𝛷) is
a) K
b) 𝑘𝑒−𝑎𝑡
c) 𝑘1cos 𝜔𝑡 + 𝑘2 sin 𝜔𝑡
d) 𝑒−𝑎𝑡
[k1 cos 𝜔𝑡 + k2 sin 𝜔𝑡 ]

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