Signals and Systems-Unit 1 & 2.pptx

Signals and Systems
School of BME
Jimma institute of Technology

Unit-1: Introduction to Signals & Systems
 Signal, and System?
 Classifications of signals
 Basic signals
 Basic operations on Signals

Signal?
 A detectable physical quantity…by which messages or
information can be transmitted( Merriam-Webster)
 Is a function that maps an independent variable to a
dependent variable.
Examples:
 Speech signals transmit language via acoustic waves
 Radar signals transmit the position and velocity of targets
via electromagnetic waves
 Electrophysiology signals transmit information about
process inside the body

Signal?
 Mathematically, signals are modeled as functions of one
or more independent variables.
Independent variables could be; time, frequency, or
spatial coordinates.
Each pixels is represented by a triplet of numbers {R,G,B} that encode
the color. Thus the signal is represented by 𝑐[𝑘, 𝑚], where 𝑚 and 𝑘 are
the independent variables that specify pixel location and 𝑐 is a color
vector

Systems
 Manipulates/process the information carried by signals
Signal processing involves the theory and application of
 Filtering, coding , transmitting, estimating, detecting,
analyzing, recognizing, synthesizing, recording, and
reproducing signals by digital or analog devices or techniques
 Where signals includes audio, video, speech, image,
communication, geophysical, sonar, radar, medical, musical,
and other signals
( IEEE Signal processing Society Constitutional Amendment, 1994)

Systems
Fig 1.1 Examples of signals and systems. ( a) An Electrical circuit; (c) an audio recording
system; (e) a digital camera; and (g) a digital thermometer. Plot (b), (d), (f) and (h) are output
signals generated respectively by the systems shown in (a), (c), (e) and (g)
(e)
(f)
(g) (h)

Signal processing
 Initially analog signals and systems implemented using
resistors, capacitors, inductors, and transistors.
 Since the 1940s increasingly digital signals and systems
implemented using computers and computer code( C/C++,
Python, Matlab,…)
• Advantages of digital include stability and programmability
• As computers have shrunk, DSP has become ubiquitous

Digital Signal Processing Applications

Computerized Tomography
• Uses x-rays to produce image slices
• Better soft-tissue contrast and no
superimposed images.
• X-ray detectors send the information
to the computer for processing.
• Sources of
malfunction,
error: Electronic
cooling systems,
computer related failures.
• Linear Attenuation Coefficient (μ).

• CT number represent the linear
attenuation coefficient of tissue in
each volume element (Voxel),
relative to 𝜇 of water.
• CT value are relatively stable for a
single organ.
• CT images typically posses 12 bits
of gray scale, for a total of 4,096
shades of gray.
• X-ray detectors send the information
to the computer for processing; e.g.
Back-projection Algorithm.
𝜇𝑣o𝑥e𝑙 − 𝜇w𝑎𝑡e𝑟
𝐶𝑇 𝑁𝑢𝑚𝑏e𝑟 = 1000 ×
𝜇w𝑎𝑡e𝑟

• Back-projection Algorithm.
5 8
3 2
13
5
8 10

Reconstruction
At 00 projection:
13 13
5 5
At 2nd Projection: At 3rd Projection:
20 21
8 12
8 10
28 31
16 22
At 4th Projection:
33 42
27 24
Subtract 1st
projection sum
13+5
15 24
9 6
5 8
3 2
Divide by the # of
remaining projection

1.2 Classification of Signals
A signal is classified into several categories depending upon
the criteria used for its classification.
 Continuous-time and discrete-time signals
 Analog and digital signals
 Periodic and aperiodic( non periodic) signals
 Energy and power signals
 Deterministic and random signals
 Even and odd signals

1.2.1 Continuous-time & Discrete time
signals
Continuous-time signal
 Takes a value at every instant of time t, x(𝑡)
Discrete-time signal
 Is defined only at a particular instants of time, x[n]

1.2.2 Analog and Digital Signals
Analog signal
 A signal whose amplitude can take on any value in a
continuous range, i.e., an analog signal amplitude can take
on an infinite numbers of values
Figure 1.3 (a) analog, continuous time

A Digital signal
 Is signal whose amplitude can take on only a finite numbers
of values.
Figure 1.3 (b) digital, continuous time

Remarks:
 The terms continuous time and discrete time qualify the
nature of a signal along the time axis
 The term analog & digital, on the other hand qualify the
nature of the signal amplitude
Figure 1.3 (c) analog, discrete time Figure 1.3 (d) digital, discrete time

1.2.3 Periodic and aperiodic signals
A periodic Continuous-time signal 𝑥 𝑡 satisfies
(1.1) 𝑥 𝑡 = 𝑥(𝑡 + 𝑘𝑇)
where 𝑘 −is an integer
𝑇 − is the period of the signal
E.g., sinusoid signal
𝑥 𝑡 = 𝐴 sin(𝜔𝑡 + 𝜃) , −∞ < 𝑡 < ∞

1.2.3 Periodic and aperiodic signals
A Discrete-time signal
 Is defined only at a particular instants of time, x[n] is
periodic with period 𝑁 if it satisfies
(1.2) 𝑥 𝑘 = 𝑥[𝑘 + 𝑁]
Figure 1.2 A discrete-time periodic signal with fundamental period 𝑁0 = 3

1.2.4 Energy and Power Signals
Continuous-time signal
 The normalized energy E of 𝑥(𝑡) ( assuming 𝑥 𝑡 is real) is
∞
(1.3) E = ∫ 𝑥(𝑡)2𝑑𝑡
−∞
If 𝑥 𝑡 is complex valued, then (1.3) is generalized
∞
(1.4) 𝐸 = ∫ 𝑥(𝑡) 2𝑑𝑡
−∞
 The normalized power P for complex value 𝑥(𝑡) is
𝑇→∞ 𝑇
𝑇
(1.5) 𝑃 = lim 1
∫
⁄2
𝑥(𝑡) 2𝑑𝑡
−𝑇/2

Discrete-time signal
 The normalized energy E of x[n] is given by
(1.6) E = ∑∞ 𝑥[𝑘] 2
𝑛=−∞
While the normalized power P is
(1.7) P = lim
𝑁→∞ 2𝑁+1
1
∑𝑁
𝑥[𝑘] 2
𝑛=−
𝑁
Note:
 A signal 𝑥(𝑡) or 𝑥[𝑘] is an energy signal if and only if 0 < 𝐸 < ∞
and consequently 𝑃 = 0
 A signal 𝑥(𝑡) or 𝑥[𝑘] is a power signal if and only if 0 < 𝑃 < ∞ and
consequently 𝐸 = ∞

Example 1.1
Determine whether the following signals are energy signals,
power signals, or neither
a) 𝑥 𝑡 = {e−𝑎𝑡,
0,
0 < 𝑡 < ∞
o𝑡ℎe𝑟wi𝑠e
= 𝐴 cos(𝜔𝑡 + 𝜃)
= 10ej2𝑛
b) 𝑥 𝑡
c) 𝑥 𝑘

1.2.5 Even and Odd signals
 A signal is even if
(1.8) x t = x(−t)
A signal is odd if
(1.9) x t = −x(−t)
Remarks:
Any signal 𝑥(𝑡) can be represented as the sum of
even(𝑥e(𝑡)) and odd signal ( 𝑥o(𝑡) ) as
= 𝑥e(𝑡) + 𝑥o(𝑡)
1) 𝑥 𝑡
Proof:
Replacing 𝑡 with −𝑡 in eqn.(1) & using eqns.(1.8) & (1.9) we get

Remarks:
2) 𝑥 −𝑡 = 𝑥e −𝑡 + 𝑥o −𝑡 = 𝑥e 𝑡 − 𝑥o 𝑡
Adding (1) and (2) and diving by 2,
e
3) 𝑥 𝑡 = 1
2
𝑥 𝑡 + 𝑥(−𝑡)
subtracting (2) from (1) and diving by 2,
o
3) 𝑥 𝑡 = 1
2
𝑥 𝑡 − 𝑥(−𝑡)
Thus
(1.10) 𝑥 𝑡 = S 𝑡 +S(−𝑡)
+ S 𝑡 −S(−𝑡)
2 2

Example 1.2
Determine the even and odd components of
= 10 sin(𝑡) + 5 cos(𝑡) − 2 cos(𝑡) sin(𝑡)
shown below
a) 𝑥 𝑡
b) 𝑦 𝑡

1.3 Basic continuous-time signals
 Unit step function
1, 𝑡 > 0
(1.11) u(t) = { 0, 𝑡 < 0
 Unit impulse function ( Delta function)
𝑑
𝑡
(1.12) δ 𝑡 = 𝑑
𝑢 𝑡
=
0, 𝑡 ≠ 0
{𝑢𝑘𝑑efi𝑘e𝑑, 𝑡 = 0

 The unit impulse may be visualized as a very short duration
pulse of unit area, which is expressed mathematically as
∫0−
0+
ð 𝑡 𝑑𝑡 = 1
 Consider the functions show below
Figure 1.3 Functions that approach the unit step and unit impulse as ∆→ 0

To illustrate how the impulse function affects other function, let
us evaluate the integral
∫𝑎 0 0 0
𝑏
f 𝑡 ð 𝑡 − 𝑡 =
𝑏
f(𝑡 )ð 𝑡 − 𝑡 𝑑𝑡
∫𝑎
∫𝑎
0 0 0
= f(𝑡 )
𝑏
ð 𝑡 − 𝑡 𝑑𝑡 = f(𝑡 ) Sampling
Area = 1
∆→0
ð 𝑡 = ð∆(𝑡) as ∆→ 0 , i.e., ð 𝑡 = lim ð∆(𝑡)
• Zero width
• Infinite height
• Area = 1

 Unit Ramp function
𝑡
(1.13) r(t) = ∫ 𝑢 𝜆 𝑑𝜆 = 𝑡𝑢(𝑡)
−∞
= {𝑡,
0, 𝑡 < 0
𝑡 ≥ 0
Example 1.3
Express the pulse signal below
in terms of the unit step. Calculate
its derivative and sketch it.

 Exponential signal
(1.14) 𝑥 𝑡 = 𝐴e𝑎𝑡
 Sinusoidal* signal
(1.15) 𝑥 𝑡 = sin(𝜔𝑡) , 𝑥 𝑡 = cos(𝜔𝑡)

1.4 Basic Operations on Signals
 Time reversal signal
𝑥 𝑡 𝑥(−𝑡)
Example 1.4:
−1,
𝑥 𝑡 = { 1 − 𝑡,
0,
−1 < 𝑡 < 0
0 < 𝑡 < 2
𝑥 −𝑡 =?
Solution:
𝑥 −𝑡
−1,
= {1 − 𝑡,
0,
−1 < −𝑡 < 0 −1,
0 < −𝑡 < 2 = {1 + 𝑡,
o𝑡ℎe𝑟wi𝑠e 0,
0 < 𝑡 < 1
−2 < 𝑡 < 0
Time
Reversal

1.4 Basic operations on signals
𝑥 𝑡 𝑥(𝑎𝑡)
• the scaled 𝑥(𝑎𝑡) will be compressed if 𝑎 > 1
• 𝑥(𝑎𝑡) will be expanded if 𝑎 < 1
 Time scaling
• involves the compress or expansion of a signal in time.
Time
Scaling

 Time Shifting
𝑥 𝑡 𝑥(𝑡 − 𝑡0) ; 𝑡0 - constant
•If 𝑡0 > 0, then the signal 𝑥(𝑡 − 𝑡0) is delayed and the signal
is shifted to the right relative to 𝑡 = 0
Time
Shifting
Figure 1.4 Time shifted signals

 For any signal 𝑥(𝑡) , the transformation 𝑎𝑡 + 𝑏
independent variable can be performed as follows:
on the
𝑥 𝑎𝑡 + 𝑏 = 𝑥 𝑎 𝑡 + 𝑏
𝑎
Note: The previous three transformations deal with the
independent variable, 𝑡.
 Amplitude transformations
Given a signal 𝑥(𝑡), amplitude transformation take the general
form
𝑦 𝑡 = 𝐴𝑥 𝑡 + 𝐵 ; 𝐴 and 𝐵 - constants

Example 1.5 A continuous-time signal is shown below. Sketch
each of the following signals
(a) 𝑥(2𝑡 − 6) (b) 𝑥(𝑡⁄2 + 1)
(c) 𝑦 𝑡 = −1 + 2𝑥(𝑡)
Solution:
(a) 𝑥 2 𝑡 − 3 (b) 𝑥 2
𝑡⁄ + 1 = 𝑥
1
2
𝑡 + 2
Delayed by 3
compressed by x2
Advanced by 2
Expanded by x2

Unit-2: Continuous-time systems
 Classifications of systems
• Causal and non causal systems
• Linear and nonlinear systems
• Time-varying & Time-invariant systems
• Systems with and without memory
 Response of LTI continuous time system in time
domain
 Discrete-time LTI systems and convolution sum
 Properties of LTI systems
 Stability of LTI systems

2.1 Classifications of systems
 Causal and non-causal systems
• A causal ( or non-anticipatory ) system is one whose output
𝑦 𝑡 at the present time depend only on the present and past
values of the input 𝑥 𝑡 .
E.g., 𝑦 𝑡
𝑦 𝑡
= 𝑥 𝑡 + 2
= 𝑥(𝑡 − 1)
Every physical system is causal
X
√

 Linear and nonlinear systems
Linear system  Homogeneity(Scaling)
+
Additive property
(superposition)
System
𝑥1(𝑡) 𝑦1(𝑡)
System
𝑥2(𝑡) 2
𝑦 (𝑡)
System
𝑥1 𝑡 + 𝑥2 𝑡 𝑦1 𝑡 + 𝑦2 𝑡
Figure 2.1(b) Scaling property
Figure 2.1(a)Additive property
System
𝑥(𝑡) 𝑦(𝑡)
System
𝑎𝑥(𝑡) 𝑎𝑦(𝑡)

Example 2.1 Check whether or not the systems listed below
are linear or non-linear
(a) 𝑦 𝑡 = 𝑡. 𝑥(𝑡) (b) 𝑦 𝑡 = 𝑥2(𝑡)
 Time varying & Time-invariant systems
• Conceptually , a system is time invariant if the behavior and
characteristics of the system are fixed over time.
• A time varying system is one whose parameters vary with
Time-invariant
system
𝑥(𝑡 − 𝑟) 𝑦(𝑡 − 𝑟)
System
time
𝑥(𝑡) 𝑦(𝑡)

 Systems with and without memory
• When the output of a system depends on the past and/or
future input, the system is said to have a memory.
E.g., 𝑦 𝑡 = 𝑥 𝑡 − 𝑥 𝑡 − 1 + 𝑥(𝑡 + 2)
Example 2.2 Show that the system represented by the ODE
𝑑𝑦
𝑑𝑡
+ 4𝑦 𝑡 = 2𝑥(𝑡)
Is linear.

2.2 Response of LTI continuous time
systems in time domain
Recall delta function
(2.1) ð∆ 𝑡 = {∆
1
, 0 < 𝑡 < ∆
0, o𝑡ℎe𝑟wi𝑠e
 Any signal can be represented in terms of scaled & shifted
impulses, i.e.,
𝑥 𝑡 ≈ 𝑥 0 ð∆ 𝑡 ∆ + 𝑥 ∆ ð∆ 𝑡 − ∆ + 𝑥 −∆ ð∆ 𝑡 + ∆ + ⋯
∆→0
≈ lim ∑∞ 𝑥(𝑘∆)ð∆(𝑡 − 𝑘∆)∆
k=−∞
(2.2) 𝑥 𝑡 = 𝑥 𝑟 ð 𝑡 − 𝑟 𝑑𝑟
∞
∫−∞
Shifted integral

2.2 Response of LTI continuous time
systems in time domain
−∆ 0 ∆ 2∆
. . . .
𝑥(𝑡)
≈ 𝑥(−2∆)
. . . . . . . .
k∆
𝑥(−2∆)ð∆(𝑡 + 2∆)∆
−∆ 0
−2∆ −∆
≈ 𝑥(−∆)
𝑥(−∆)ð∆(𝑡 + ∆)∆
𝑡
Figure 2.13 Staircase approximation to
a continuous-time signal
𝑡
𝑡
𝑡
≈ 𝑥(0) 𝑥(0)ð∆(𝑡)∆
0 ∆

2.2 Response of LTI CT Systems in
time domain
 For Continuous time LTI system
∴ 𝑦 𝑡
∞
= lim ∑ 𝑥 𝑘∆ ℎ(𝑡 − 𝑘∆)∆
∆→0
k=−∞
(2.3)
∞
𝑦 𝑡 = ∫
−∞ 𝑥 𝑟 ℎ 𝑡 − 𝑟 𝑑𝑟
LTI Sys.
ð∆(𝑡) h(𝑡)
LTI Sys.
ð∆(𝑡 − 𝑘∆) h(𝑡 − 𝑘∆)
LTI Sys.
𝑥(𝑘∆)ð∆(𝑡 − 𝑘∆)∆ 𝑥 𝑘∆ ℎ(𝑡 − 𝑘∆)∆
Convolution
integral

time domain
Example 2.3 For an LTI continuous time system with the unit
impulse response ℎ 𝑡 = e−2𝑡𝑢(𝑡), determine the response 𝑦(𝑡)
for the input
𝑥 𝑡 = e−𝑡𝑢(𝑡)
Solution:
Note both ℎ(𝑡) and 𝑥(𝑡) are causal.
𝑡 𝑡
𝑦 𝑡 = ∫ 𝑥 𝑟 ℎ 𝑡 − 𝑟 𝑑𝑟 = ∫ e−𝑐 e−2(𝑡−𝑐)𝑑𝑟
0 0
= e−2𝑡
∫0
𝑡
e𝑐𝑑𝑟 = e−2𝑡(e𝑡 − 1)
= e−𝑡 − e−2𝑡
𝑡 𝑟
e−2(𝑡−𝑐) e−𝑐

time domain
Remarks The convolution integral ( Eqn. 1.18) can be
evaluated in three different ways:
1. Analytical method, which involves performing the integration
by hand when 𝑥(𝑡) and ℎ(𝑡) are specified analytically.
2. Graphical method, which is appropriate when 𝑥(𝑡) and ℎ(𝑡)
are provided in graphical form
3. Numerical method, where we approximate 𝑥(𝑡) and ℎ(𝑡) by
numerical sequence and obtain 𝑦(𝑡) by discrete convolution
using a digital computer.

time domain
Example 2.4 Let 𝑥(𝑡) be the input to an LTI system with unit
impulse response ℎ(𝑡), where
𝑥 𝑡 = e−𝑎𝑡𝑢(𝑡), 𝑎 > 0 and ℎ 𝑡 = 𝑢 𝑡
Example 2.5 Consider the convolution of the following two
signals:
𝑥 𝑡 ℎ 𝑡
= {
1, 0 < 𝑡 < 𝑇
= {
𝑡, 0 < 𝑡 < 2𝑇
0, o𝑡ℎe𝑟wi𝑠e 0, o𝑡ℎe𝑟wi𝑠e
Example 2.6 Obtain the convolution of the two signals shown
below.

2.3 Discrete-time LTI Systems and
convolution sum
Any discrete-time signal 𝑥[𝑘] can be represented in terms of
scaled & shifted impulses.
(2.4) x 𝑘 = ∑∞ 𝑥 𝑘 ð[𝑘 − 𝑘]
k=−∞
Where
ð 𝑘 − 𝑘 = {
0, 𝑘 ≠ 𝑘
1, 𝑘 = 𝑘
To understand eqn.(2.4) , consider the signal 𝑥 𝑘
below
depicted

convolution sum
Figure 2.2 Decomposition of a discrete time signal into weighted sum of
shifted impulses

convolution sum
 The response of a LTI system to 𝑥[𝑘] will be the
superposition of the scaled response of the system to each
of these shifted impulses.
 For LTI discrete-time system
And from superposition property of LTI
Discrete
LTI Sys.
ð[n] ℎ[𝑘]
Discrete
LTI Sys.
ð[𝑘 − 𝑘] h[𝑘 − 𝑘]
Discrete
LTI Sys.
𝑥[𝑘] 𝑦[𝑘] =
∞
∑ 𝑥 𝑘 ℎ[𝑘 − 𝑘]
k=−∞

convolution sum
Thus, response of discrete LTI system for an arbitrary input
signal 𝑥[𝑘] is given by
(2.5) 𝑦 𝑘 = 𝑥 𝑘 ∗ ℎ 𝑘 = ∑∞ 𝑥 𝑘 ℎ[𝑘 − 𝑘]
k=−∞
Implication of eqn.(2.5)
 If we know the response of a LTI system to the set of shifted
impulses, we can find the response to an arbitrary input.
i.e., a LTI system is completely characterized by its response
to a unit impulse.

convolution sum
Example 2.7 consider an LTI system with impulse response
ℎ[𝑘] and input 𝑥[𝑘], as shown in the figure below. Find 𝑦 𝑘 =
𝑥 𝑘 ∗ ℎ[𝑘]
Answer:
𝑦 0 = 0.5, 𝑦 1 = 2.5,
= 0 for 𝑘 < 0 and 𝑘 > 3
𝑦 2 = 2.5, 𝑦 3 = 2
𝑦 𝑘

convolution sum
Alternative method of discrete-time convolution
Let 𝑥 𝑘 = 0
ℎ 𝑘 = 0
for all 𝑘 < 𝑁
for all 𝑘 < 𝑀
- Sample value 𝑁 is the first non-zero value of 𝑥[𝑁]
- Sample value 𝑀 is the first non-zero value of ℎ[𝑘]
∴ 𝑦 𝑘 = 𝑥 𝑘 ∗ ℎ[𝑘]
= {
0, 𝑘 < 𝑀 + 𝑁
𝑀
∑i=𝑁 𝑥 i ℎ 𝑘 − i , 𝑘 ≥ 𝑀 + 𝑁

convolution sum
...Values of 𝑥[𝑘]
…values of ℎ[𝑘]
To compute the convolution, use the following array:
𝑥[𝑁] 𝑥[𝑁 + 1] 𝑥[𝑁 + 2] 𝑥[𝑁 + 3]
ℎ[𝑀] ℎ[𝑀 + 1] ℎ[𝑀 + 2] ℎ[𝑀 + 3]
𝑥[𝑁] ℎ[𝑀] 𝑥[𝑁 + 1] ℎ[𝑀] 𝑥[𝑁 + 2] ℎ[𝑀] 𝑥[𝑁 + 3] ℎ[𝑀] …
…
…
. 𝑥[𝑁] ℎ[𝑀 + 1] 𝑥[𝑁 + 1] ℎ[𝑀 + 1] 𝑥[𝑁 + 2] ℎ[𝑀 + 1]
𝑥[𝑁] ℎ[𝑀 + 2] 𝑥[𝑁 + 1] ℎ[𝑀 + 2]
.
.
.
𝑦[𝑁 + 𝑀] 𝑦 𝑁 + 𝑀 + 1 𝑦 𝑁 + 𝑀 + 2 𝑦 𝑁 + 𝑀 + 3 …

convolution sum
Exercise: Re-do example 2.7 using the short-cut method.
Solution:
- 𝑁 = 0: index of the 1st non-zero value of 𝑥[𝑘]
- 𝑀 = 0: index of the 1st non-zero value of ℎ[𝑘]
Next , write the array
0.5 2
1 1 1
0.5 2
+ 0.5 2
0.5 2
2
𝑦 𝑘 = 0.5ð 𝑘
0.5 2.5 2.5
+ 2.5ð 𝑘 − 1 + 2.5ð 𝑘 − 2 + 2ð 𝑘 − 3

convolution sum
Useful relationship for computing convolution sum:
2.6) (a) 𝑁−1
∑ ∝𝑛 =
𝑛=0
𝑁
{1−∝𝑁
1 −∝
, ∝= 1
, for any complex
number ∝≠ 1
( Finite Sum Formula )
∞
∑ ∝𝑛 =
𝑛=0
(b) 1
1 −∝
, ∝ < 1
( Infinite Sum Formula )
(c) ∝
(1 −∝)2
, ∝ < 1
(d)
=
∞
∑ 𝑘 ∝𝑛 =
𝑛=0
∞
∑ ∝𝑛
𝑛=k
∞
∝k ∑ ∝𝑛
𝑛=0
∝k
=
1 −∝
; ∝ < 1

convolution sum
Proof:
(a) For ∝= 1:
For ∝≠ 1:
∑𝑁−1 ∝𝑛 = 𝑁 − 1 + 1 = 𝑁
𝑛=0
(1 −∝) ∑𝑁−1 ∝𝑛 = ∑𝑁−1 ∝𝑛 − ∑𝑁−1 ∝𝑛+1
𝑛=0 𝑛=0 𝑛=0
= 1 +∝ +∝2 + ⋯ +∝𝑁−1 − (∝ +∝2 + ⋯ +∝𝑁)
= 1−∝𝑁
∝ =
1−∝𝑁
1−∝
 ∑𝑁−1 𝑛
𝑛=0
b) For ∝ < 1: lim ∝𝑁= 0
𝑁→∞
𝑁→∞
𝑛=0 𝑛=0
∴ lim ∑𝑁−1 ∝𝑛 = ∑∞ ∝𝑛 =
1
1−∝

convolution sum
Proof(cont’d):
c) Differentiating both sides of the result of (b) w.r.t ∝ , gives
𝑑
𝑑∝
∑∞ ∝𝑛
𝑛=0 = =
𝑑 1 1
𝑑∝ 1−∝ (1−∝)2
∑∞ 𝑘 ∝𝑛−1 =
𝑛=0
1
1−∝ 2
⇒ ∑∞ 𝑘 ∝𝑛=
𝑛=0
∝
Example 2.8 Consider
1−∝ 2
an input 𝑥[𝑘] and a unit impulse
response ℎ[𝑘] given by 𝑥 𝑘 =∝𝑛 𝑢 𝑘 ; 0 <∝< 1
ℎ 𝑘 = 𝑢[𝑘]. Find 𝑦 𝑘 = 𝑥 𝑘 ∗ ℎ[𝑘]

convolution sum
Example 2.9 Consider the two sequences
𝑥 𝑘 = {
1, 0 ≤ 𝑘 ≤ 4
and
ℎ 𝑘
= {∝𝑛, 0 ≤ 𝑘 ≤ 6 ; ∝> 1
Example 2.10 Consider an LTI system with input 𝑥[𝑘] and with
unit impulse response ℎ[𝑘] specified as follows:
𝑥 𝑘 = 2𝑛𝑢[−𝑘]
ℎ 𝑘 = 𝑢[𝑘]

2.4 Properties of LTI Systems
k=−∞
= ∑∞ ℎ 𝑘 𝑥[𝑘 − 𝑘]
Commutative property
DT: 𝑥 𝑘 ∗ ℎ 𝑘 = ℎ 𝑘 ∗ 𝑥 𝑘
CT: 𝑥 𝑡 ∗ ℎ 𝑡 = ℎ 𝑡 ∗ 𝑥 𝑡 =
Distributive property
ℎ 𝑟 𝑥 𝑡 − 𝑟 𝑑𝑟
∞
∫−∞
DT: 𝑥 𝑘 ∗ ℎ1 𝑘 + ℎ2 𝑘 = 𝑥 𝑘 ∗ ℎ1 𝑘
CT: 𝑥 𝑡 ∗ ℎ1 𝑡
+ x n ∗ ℎ2 𝑘
+𝑥 𝑡 ∗ ℎ2 𝑡
ℎ1(𝑡)
ℎ2(𝑡)
+
𝑥(𝑡) y(𝑡)
𝑦2(𝑡)
+ ℎ2 𝑡 = 𝑥 𝑡 ∗ ℎ1 𝑡
𝑦1(𝑡)
(𝑎)
𝑥(𝑡) y(𝑡)
ℎ1 𝑡 + ℎ2 𝑡
(𝑏)
Figure 2.3 System interconnection

Causality for LTI system
 A causal (or non anticipatory) system is one whose output
𝑦[𝑘] at present time depend only on the present and past
values of the input 𝑥 𝑘 .
 𝑦[𝑘] must not depend on 𝑥 𝑘 for 𝑘 > 𝑘. From Eqn.(2.5), we
see that for this to be true, all of the coefficients ℎ 𝑘 − 𝑘 that
multiply values of 𝑥[𝑘] for 𝑘 > 𝑘 must be zero.
The impulse response of a causal discrete-time LTI system
satisfy the condition
(2.7) ℎ 𝑘 = 0 for 𝑘 < 0
ℎ 𝑘 − 𝑘
Let 𝑚 = 𝑘 − 𝑘
ℎ 𝑚 = 0, 𝑚 < 0

Causality for LTI system(cont’d)
Note:
 Every physical system is causal
 Causality is a necessary condition for a system to be
realized in the real world we live in.
 However, causality is not often an essential constraint in
application in which the independent variable is not time,
such as IMAGE PROCESSING.

Stability for LTI Systems
A system is stable if every bounded input produces a bounded
output(Eqns.(2.8) & (2.9))
(2.8) 𝑥[𝑘] < 𝐵 for all 𝑘, then
(2.9) 𝑦[𝑘] = ∑∞
ℎ 𝑘 𝑥[𝑘 − 𝑘]
k=−∞
Since the magnitude of the sum of a set of numbers is no larger
than the sum of the magnitudes of the numbers, it follows
(2.10) 𝑦[𝑘] ≤ ∑∞ ℎ[𝑘] 𝑥[𝑘 − 𝑘]
k=−∞

From eqn.(2.8), 𝑥[𝑘 − 𝑘] < 𝐵 for all values of 𝑘 and 𝑘.
(2.11) 𝑦[𝑘] ≤ 𝐵 ∑∞ ℎ[𝑘] for all n
k=−∞
From eqn.(2.11), we can conclude that if the impulse response
is absolutely summable, that is, if
(2.12) ∑∞ ℎ[𝑘] < ∞
k=−∞
then 𝑦[𝑘] is bounded in magnitude, and hence the system is
stable.

𝑥(𝑡) < 𝐵 for all t,
Similarly, in continuous-time LTI system if
then
𝑦(𝑡) = ∫ ℎ 𝑟 𝑥 𝑡 − 𝑟 𝑑𝑟
≤
∞
ℎ 𝑟 𝑥 𝑡 − 𝑟 𝑑𝑟 ≤ 𝐵 ℎ(𝑟) 𝑑𝑟
∞
∫−∞ ∫−∞
Therefore, the system is stable if the impulse response is
absolutely integrable, i.e., if
(2.13) ℎ(𝑟) 𝑑𝑟 < ∞
∞
∫−∞

Exercise:
Determine whether each system is causal and/or stable. Justify
your answers
(a) ℎ 𝑘 =
1
5
𝑛
𝑢[𝑘] (b) ℎ 𝑘 = 0.8𝑛𝑢[𝑘 + 2]
(c) ℎ 𝑘 = 5𝑛𝑢[3 − 𝑘] (d) ℎ 𝑘 = − 1
2
𝑛
𝑢 𝑘 + 1.01 𝑛𝑢[𝑘 − 1]
= e−4𝑡𝑢(𝑡 − 2) (f)ℎ 𝑡
= e2𝑡𝑢(−1 − 𝑡)
= e−2𝑡𝑢(𝑡 + 50)
(e) ℎ 𝑡
(g) ℎ 𝑡
Solution:
(a)causal: ℎ 𝑘 = 0 fo𝑟 𝑘 < 0
Stable: ∑∞ 1
5
𝑛
=
1−
1
5
1
= 5
4
𝑛=0

Exercise:
(b) Not causal: ℎ 𝑘
Stable: ∑∞
𝑛=−2
(c) Not Causal: ℎ 𝑘
≠ 0 fo𝑟 𝑘 < 0
0.8𝑛 < ∞
≠ 0 fo𝑟 𝑘 < 0
5𝑛 = 625
Stable: ∑3
4
𝑛=−∞
(e) Causal: ℎ 𝑡 = 0 fo𝑟 𝑡 < 0
Stable: 4
−8
ℎ(𝑡) 𝑑𝑡 = e
< ∞
∞
∫−∞
(g) Not Causal: ℎ 𝑡 ≠ 0 fo𝑟 𝑡 < 0
Stable: ℎ(𝑡) 𝑑𝑡 = e−2⁄2 < ∞
∞
∫−∞

Signals and Systems-Unit 1 & 2.pptx

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