Signals and systems-2

Signals and System-II
Sarun Soman
Assistant Professor
Manipal Institute of Technology
Manipal

Properties of System
• Stability
• Memory
• Causality
• Invertibility
• Time invariance
• Linearity
Prof: Sarun Soman, M.I.T, Manipal

Properties of Systems
Stabililty
A system is said to be bounded input, bounded output(BIBO)
stable iff every bounded input results in a bounded output.
Operator H is BIBO stable if the o/p signal satisfies the condition
‫)ݐ(ݕ‬ ≤ ‫ܯ‬௬ ≤ ∞ for all ‘t’
Whenever input signal x(t) satisfies the condition
‫)ݐ(ݔ‬ ≤ ‫ܯ‬௫ ≤ ∞ for all ‘t’
‫ܯ‬௬ and ‫ܯ‬௫ finite positive numbers

Example of unstable system
• Tacoma Narrows suspension bridge
• Collapsed on November 7, 1940.
• Collapsed due to wind induced vibrations

Example1
Consider the moving average system with input output relation
‫ݕ‬ ݊ =
ଵ
ଷ
‫ݔ‬ ݊ + ‫ݔ‬ ݊ − 1 + ‫ݔ‬ ݊ − 2
Show that the system is BIBO stable.
Ans:
Assume that ‫]݊[ݔ‬ < ‫ܯ‬௫ < ∞ for all ‘n’
‫]݊[ݕ‬ =
ଵ
ଷ
‫ݔ‬ ݊ + ‫ݔ‬ ݊ − 1 + ‫ݔ‬ ݊ − 2
=
ଵ
ଷ
(‫ܯ‬௫ + ‫ܯ‬௫ + ‫ܯ‬௫)
= ‫ܯ‬௫
Bounded input results in bounded o/p

Example2
Consider a DT system whose i/p-o/p relation is defined by
‫ݕ‬ ݊ = ‫ݎ‬௡
‫]݊[ݔ‬
Where ‫ݎ‬ > 1. Show that the system is unstable.
Ans:
Assume that the input signal x[n] satisfies the condition
‫]݊[ݔ‬ ≤ ‫ܯ‬௫ ≤ ∞
‫]݊[ݕ‬ = ‫ݎ‬௡‫]݊[ݔ‬
= ‫ݎ‬௡ ‫]݊[ݔ‬
‫ݎ‬௡ diverges as ‘n’ increases.

Memory
A system is said to posses memory if its o/p signal depends on past or
future values of the input signal.
A system is said to be memoryless if its output signal depends only on
present value of input signal.
Eg.
Resistor
݅ ‫ݐ‬ =
ଵ
ோ
‫)ݐ(ݒ‬ memoryless
Inductor
݅ ‫ݐ‬ =
ଵ
௅
‫׬‬ ‫ݒ‬ ߬ ݀߬
௧
ିஶ
memory

‫ݕ‬ ݊ = ‫ݔ‬ଶ
݊ memoryless
‫ݕ‬ ݊ =
1
3
‫ݔ‬ ݊ + ‫ݔ‬ ݊ − 1 + ‫ݔ‬ ݊ − 2
Memory
‫ݕ‬ 0 =
1
3
‫ݔ‬ 0 + ‫ݔ‬ −1 + ‫ݔ‬ −2

Causality
A system is said to be causal if its present value of the o/p signal depends only
on the present or past values of the input signal.
A system is said to be non causal if its output signal depends on one or more
future values of the input signal.
Eg. Moving average system
‫ݕ‬ ݊ =
ଵ
ଷ
(‫ݔ‬ ݊ + ‫ݔ‬ ݊ − 1 + ‫݊[ݔ‬ − 2])
Moving average system
‫ݕ‬ ݊ =
ଵ
ଷ
(‫ݔ‬ ݊ + ‫ݔ‬ ݊ + 1 + ‫݊[ݔ‬ − 2])
‫ݕ‬ 0 =
ଵ
ଷ
(‫ݔ‬ 0 + ‫ݔ‬ 1 + ‫)]2−[ݔ‬
causal
Non causal

Invertibility
A system is said to be invertible if the input of the system can be
recovered from the o/p.
Note: necessary condition for invertibility distinct i/p must
produce distinct o/p

Show that a square law system described by the input-output
relation ‫ݕ‬ ‫ݐ‬ = ‫ݔ‬ଶ
(‫)ݐ‬ is not invertible.
Ans:
Distinct inputs x(t) and x(-t) produce the same output
Time invariance
A system is said to be time invariant if a delay or time advance
of the input signal lead to an identical time shift in the output
signal.

Eg. Thermistor
Input output relation of thermistor
‫ݕ‬ଵ ‫ݐ‬ =
௫భ(௧)
ோ(௧)
Response ‫ݕ‬ଶ ‫ݐ‬ of the thermistor to input ‫ݐ(ݔ‬ − ‫ݐ‬଴)
‫ݕ‬ଶ ‫ݐ‬ =
௫భ(௧ି௧బ)
ோ(௧)

Shifting the output ‫ݕ‬ଵ ‫ݐ‬
‫ݕ‬ଵ ‫ݐ‬ − ‫ݐ‬଴ =
௫భ(௧ି௧బ)
ோ(௧ି௧బ)
For a thermistor ܴ(‫ݐ‬ − ‫ݐ‬଴) ≠ ܴ(‫)ݐ‬ for ‫ݐ‬ ≠ 0
‫ݕ‬ଵ(‫ݐ‬ − ‫ݐ‬଴) ≠ ‫ݕ‬ଶ(‫)ݐ‬ for ‫ݐ‬ ≠ 0
Thermistor is time variant
Linearity
A system is linear in terms of the system input x(t) and the
system output y(t) if it satisfies two properties.

ܽଵ‫ݔ‬ଵ ‫ݐ‬ + ܽଶ‫ݔ‬ଶ ‫ݐ‬ = ܽଵ‫ݕ‬ଵ ‫ݐ‬ + ܽଶ‫ݕ‬ଶ(‫)ݐ‬

Consider a discrete time system described by the input-output
relation ‫ݕ‬ ݊ = ݊‫.]݊[ݔ‬ Show that the system is linear.
Ans:
Let the input signal x[n] be expressed as the weighted sum
‫ݔ‬ ݊ = ∑ ܽ௜‫ݔ‬௜[݊]ே
௜ୀଵ
Output of the system
= ݊ ∑ ܽ௜
ே
௜ୀଵ ‫ݔ‬௜[݊]
= ∑ ܽ௜݊ே
௜ୀଵ ‫ݔ‬௜[݊]
= ∑ ܽ௜
ே
௜ୀଵ ‫ݕ‬௜[݊]
The system is linear

Tutorial
‫ݕ‬ ݊ = 2‫ݔ‬ ݊ + 3 check for linearity
Ans:
‫ݕ‬ଵ ݊ = 2‫ݔ‬ଵ ݊ + 3
‫ݕ‬ଶ ݊ = 2‫ݔ‬ଶ ݊ + 3
Let ‫ݔ‬ଷ ݊ = ‫ݔ‬ଵ[݊] + ‫ݔ‬ଶ[݊]
‫ݕ‬ଷ ݊ = 2 ‫ݔ‬ଵ ݊ + ‫ݔ‬ଶ ݊ + 3
= 2‫ݔ‬ଵ ݊ + 2‫ݔ‬ଶ ݊ + 3
≠ ܽ‫ݕ‬ଵ ݊ + ܾ‫ݕ‬ଶ[݊]

Tutorial
‫ݕ‬ ‫ݐ‬ = ‫ݔ‬
‫ݐ‬
2
Memory or memoryless system?
Ans: ‫ݕ‬ 1 = ‫)5.0(ݔ‬
Output depends on past value of i/p
Memory system
‫ݕ‬ ‫ݐ‬ = cos ‫ݔ‬ ‫ݐ‬ invertible or non-invertible?
Ans: Distinct i/p must give distinct o/p.
‫ݔ‬ ‫ݐ‬ ܽ݊݀ ‫)ݐ(ݔ‬ + 2ߨ) gives same o/p.
Non-invertible

Tutorial
‫ݕ‬ ݊ = ‫]݊−[ݔ‬ time invariant or time variant?
Ans:
Delay in input ‫݊[ݔ‬ − ݉]
Corresponding output ‫ݕ‬ଵ ݊ = ‫݊−[ݔ‬ − ݉]
Delay the output
‫ݕ‬ଶ ݊ = ‫ݕ‬ ݊ − ݉ = ‫݊(−[ݔ‬ − ݉)]
= ‫݊−[ݔ‬ + ݉]
‫ݕ‬ଶ[݊] ≠ ‫ݕ‬ଵ ݊
Time variant

Tutorial
‫ݕ‬ ‫ݐ‬ = ‫ݔ‬
‫ݐ‬
2
Linearity, time invariance, memory, causality, stability?
Ans:
Memory system
Linearity
‫ݔ‬ଵ ‫ݐ‬ → ‫ݕ‬ଵ ‫ݐ‬ = ‫ݔ‬ଵ
௧
ଶ
‫ݔ‬ଶ ‫ݐ‬ → ‫ݕ‬ଶ ‫ݐ‬ = ‫ݔ‬ଶ
௧
ଶ
Let ‫ݔ‬ଷ ‫ݐ‬ = ܽ‫ݔ‬ଵ ‫ݐ‬ + ܾ‫ݔ‬ଶ ‫ݐ‬
‫ݕ‬ଷ ‫ݐ‬ = ܽ‫ݔ‬ଵ
௧
ଶ
+ ܾ‫ݔ‬ଶ
௧
ଶ
= ܽ‫ݕ‬ଵ ‫ݐ‬ + ܾ‫ݕ‬ଶ ‫ݐ‬
Linear as it satisfies principle superposition and homogeneity

Tutorial
Causality
‫ݕ‬ −1 = ‫ݔ‬ −0.5
o/p depends on future value of input
Non-causal
Stability
Let ‫)ݐ(ݔ‬ ≤ ‫ܤ‬௫ ≤ ∞
Then ‫)ݐ(ݕ‬ = ‫ݔ‬
௧
ଶ
≤ ‫ܤ‬௬ ≤ ∞
The system is stable.

Tutorial
‫ݕ‬ ‫ݐ‬ = ‫)ݐ(ݔݐ‬ stable or unstable?
Let ‫)ݐ(ݔ‬ be bounded.
‫ݕ‬ ‫ݐ‬ will be unbounded , ‘t’ is unbounded
Unstable system
‫ݕ‬ ‫ݐ‬ = ݁௫(௧)
Check for linearity, Stability, Time Invariance, Memory.
‫ݔ‬ଵ ‫ݐ‬ → ‫ݕ‬ଵ ‫ݐ‬ = ݁௫భ(௧)
‫ݔ‬ଶ ‫ݐ‬ → ‫ݕ‬ଶ ‫ݐ‬ = ݁௫మ(௧)
Let ‫ݔ‬ଷ ‫ݐ‬ = ܽ‫ݔ‬ଵ ‫ݐ‬ + ܾ‫ݔ‬ଶ ‫ݐ‬

Tutorial
‫ݕ‬ଷ ‫ݐ‬ = ݁௔௫భ ௧ ା௕௫మ ௧
≠ ܽ‫ݕ‬ଵ ‫ݐ‬ + ܾ‫ݕ‬ଶ ‫ݐ‬
The system is Non-linear
Time Invariance
delay i/p ‫ݐ(ݔ‬ − ‫ݐ‬଴)
‫ݕ‬ଵ ‫ݐ‬ = ݁௫(௧ି௧బ) (1)
Delay o/p
‫ݕ‬ଶ ‫ݐ‬ − ‫ݐ‬଴ = ݁௫(௧ି௧బ) (2)
(1)=(2) system is time invariant
The system is memoryless.

Tutorial
The i/p and o/p of the system
is shown. Determine if the
system could be memoryless &
causal.
Memory, Causal
Non-Causal
Memory
1 20 t
y(t)
1
10 t
x(t)
1
10 t
y(t)
1
-1
10 t
x(t)
1

Tutorial
‫ݕ‬ ‫ݐ‬ = [sin 6‫)ݐ(ݔ]ݐ‬ Check for Memory, Time invariance,
Linearity, Stability and Causality.
Ans:
Present value of i/p depends on present value of o/p
Memoryless System
Delay in i/p ‫ݐ(ݔ‬ − ‫ݐ‬଴)
‫ݕ‬ଵ ‫ݐ‬ = [sin 6‫ݐ(ݔ]ݐ‬ − ‫ݐ‬଴)
Delay o/p
‫ݕ‬ଶ ‫ݐ‬ − ‫ݐ‬଴ = [sin 6(‫ݐ‬ − ‫ݐ‬଴]‫ݐ(ݔ‬ − ‫ݐ‬଴)
‫ݕ‬ଵ ‫ݐ‬ ≠ ‫ݕ‬ଶ ‫ݐ‬
Time variant system

Tutorial
‫ݔ‬ଵ ‫ݐ‬ → ‫ݕ‬ଵ ‫ݐ‬ = sin 6‫ݐ‬ ‫ݔ‬ଵ ‫ݐ‬
‫ݔ‬ଶ ‫ݐ‬ → ‫ݕ‬ଶ ‫ݐ‬ = sin 6‫ݐ‬ ‫ݔ‬ଶ ‫ݐ‬
Let ‫ݔ‬ଷ ‫ݐ‬ = ܽ‫ݔ‬ଵ ‫ݐ‬ + ܾ‫ݔ‬ଶ ‫ݐ‬
‫ݕ‬ଷ ‫ݐ‬ = [sin 6‫ݔ]ݐ‬ଷ(‫)ݐ‬
= ܽ sin 6‫ݐ‬ ‫ݔ‬ଵ ‫ݐ‬ + ܾ sin 6‫ݐ‬ ‫ݔ‬ଶ ‫ݐ‬
= ܽ‫ݕ‬ଵ ‫ݐ‬ + ܾ‫ݕ‬ଶ(‫)ݐ‬
Linear System
Causal – o/p doesn't depend on future value of i/p.

Tutorial
Let ‫)ݐ(ݔ‬ ≤ ‫ܤ‬௫ ≤ ∞
‫)ݐ(ݕ‬ = sin 6‫ݐ‬ ‫)ݐ(ݔ‬
sin 6‫ݐ‬ ≤ 1
‫)ݐ(ݕ‬ is bounded , system is stable.
Check for linearity, causality, time-invariance, stability
1.‫ݕ‬ ݊ = ݊ଶ‫]݊[ݔ‬
Linear , timevariant, causal, unstable
2. ‫ݕ‬ ݊ = ‫]݊2[ݔ‬
Linear, time-variant, non-causal

Sketch the waveform
‫ݕ‬ ‫ݐ‬ = ‫ݎ‬ ‫ݐ‬ + 1 − ‫ݎ‬ ‫ݐ‬ + ‫ݐ(ݎ‬ + 2)
0-1 1 2 t
r(t+1)
0 1 2 t
-r(t)
0 1 2
t
r(t-2)
-1 0 1 2
y(t)
t

LTI Systems
• Linear Time Invariant Systems
• Tools for analyzing and predicting the behavior of LTI systems.
• Characterizing an LTI system
– Impulse Response
– Linear constant-coefficient differential equation or difference equation
• Gives insight into system’s behavior.
• Useful in analyzing and implementing the system
i/p o/p
System

Impulse Response
Impulse response is defined as the o/p of an LTI
system due to a unit impulse signal input applied at
time t=0 or n=0.
Symbol ℎ ‫ݐ‬ , ℎ[݊]
Impulse response completely characterizes the
behavior of any LTI system.
Given the impulse response, output can be
determined for any arbitrary input signal.
Convolution Sum – DT LTI systems
Convolution Integral – CT LTI systems

Convolution Sum
Consider an LTI system
ߜ ݊ → ℎ ݊
Apply time invariance property
ߜ[݊ − ݇] → ℎ ݊ − ݇
Apply principle of Homogeneity
‫݊[ߜ]݇[ݔ‬ − ݇] → ‫]݇[ݔ‬ℎ ݊ − ݇
Apply principle of superposition
∑ ‫݊[ߜ]݇[ݔ‬ − ݇]ஶ
௞ୀିஶ →
Input is weighted superposition of time shifted impulse
Output is weighted super position of time shifted impulse
responses.
෍ ‫ݔ‬ ݇ ℎ[݊ − ݇]
ஶ
௞ୀିஶ

Convolution Sum
‫ݕ‬ ݊ = ∑ ‫]݇[ݔ‬ℎ ݊ − ݇ஶ
௞ୀିஶ - Convolution Sum
‫ݕ‬ ݊ = ‫ݔ‬ ݊ ∗ ℎ[݊]

Example
‫ݔ‬ ݊ = −1,0, 1, 2, 3
ℎ ݊ = 1, −0.5
Change of variable ݊ → ݇
Starting index of o/p (-1+0=-1)
‫ݕ‬ −1 = ෍ ‫]݇[ݔ‬ℎ −1 − ݇
ஶ
௞ୀିஶ

= −1
‫ݕ‬ 0 = ෍ ‫]݇[ݔ‬ℎ −݇
ஶ
௞ୀିஶ

= 0.5
-3 -2 -1 0 1 2 3 k
0 0 -1 0 1 2 3 x[k]
0 0 0 1 -0.5 0 0 h[k]
0 0 -0.5 1 0 0 0 h[-k]
0 -0.5 1 0 0 0 0 h[-1-k]
0 0 0 -0.5 1 0 0 h[1-k]
0 0 0 0 -0.5 1 0 h[2-k]
0 0 0 0 0 -0.5 1 h[3-k]
0 0 0 0 0 0 -0.5 h[4-k]
-3
‫ݕ‬ 1 = 1, ‫ݕ‬ 2 = 1.5, ‫ݕ‬ 3 = 2,
‫ݕ‬ 4 = −1.5
‫ݕ‬ ݊ = {−1, 0.5, 1, 1.5, 2, −1.5}

Example
Assuming that the impulse response of an LTI system is given by
ℎ ݊ = 0.5௡‫.]݊[ݑ‬ Determine the o/p response y[n] to the i/p
sequence.‫ݔ‬ ݊ = 0.8௡‫ݑ‬ ݊ .
Ans:
Change of variable x[k] and h[k]
‫ݕ‬ ݊ = ∑ ‫]݇[ݔ‬ℎ ݊ − ݇ஶ
௞ୀିஶ
= ∑ 0.8௞‫ݑ‬ ݇ 0.5௡ି௞‫ݑ‬ ݊ − ݇ஶ
௞ୀିஶ
= 0.5௡ ∑ (
଴.଼
଴.ହ
)௞‫ݑ‬ ݇ ‫݊[ݑ‬ − ݇]ஶ
௞ୀିஶ

Example
For ݊ < 0
‫ݑ‬ ݇ ‫ݑ‬ ݊ − ݇ = 0
‫ݕ‬ ݊ = 0
For ݊ ≥ 0
‫ݕ‬ ݊ = 0.5௡
෍(
0.8
0.5
)௞
௡
௞ୀ଴
GP series
∑ ‫ݎ‬௡ =
ଵି௥ಿశభ
ଵି௥
ே
௡ୀ଴
‫ݕ‬ ݊ = 0.5௡
1 − (
0.8
0.5
)௡ାଵ
1 −
0.8
0.5
1
●●●
0 2 4-1
u[k]
k
_ _ _ _
1
● ● ●
-2 0 n-4
u[-k+n]
k
_ _ _ _

Example
= 0.5௡
0.5
0.5௡ାଵ
1
0.3
0.8௡ାଵ − 0.5௡ାଵ
=
10
3
[0.8௡ାଵ − 0.5௡ାଵ]
‫ݕ‬ ݊ = ൝
0, ݊ < 0
10
3
[0.8௡ାଵ
− 0.5௡ାଵ
], n ≥ 0

Example
‫ݕ‬ ݊ = ‫ݑ‬ ݊ ∗ ‫݊[ݑ‬ − 3]
Ans:
Change of variable
‫ݔ‬ ݇ = ‫ݑ‬ ݇ , ℎ ݇ = ‫݇[ݑ‬ − 3]
‫ݕ‬ ݊ = ෍ ‫]݇[ݔ‬ℎ ݊ − ݇
ஶ
௞ୀିஶ
‫ݕ‬ ݊ = ෍ ‫ݑ‬ ݇ ‫ݑ‬ ݊ − ݇ − 3
ஶ
௞ୀିஶ
For ݊ − 3 < 0
݊ < 3
‫ݕ‬ ݊ = 0
1
●●●
0 2 4-1
u[k]
k
_ _ _ _
1
● ● ●
-2 0 n-3-4
u[-k+n-3]
k
_ _ _ _

Example
For ݊ − 3 ≥ 0
݊ ≥ 3
‫ݕ‬ ݊ = ෍ 1
௡ିଷ
௞ୀ଴
‫ݕ‬ ݊ = ݊ − 3 + 1
= ݊ − 2
‫ݕ‬ ݊ = ൜
0, ݊ < 3
݊ − 2, ݊ ≥ 3

Example
‫ݕ‬ ݊ = (
1
2
)௡‫ݑ‬ ݊ − 2 ∗ ‫]݊[ݑ‬
Ans:
Change of variable
‫ݔ‬ ݇ = (
ଵ
ଶ
)௞‫ݑ‬ ݇ − 2
ℎ ݇ = ‫]݇[ݑ‬
‫ݕ‬ ݊ = ෍ ‫]݇[ݔ‬ℎ ݊ − ݇
ஶ
௞ୀିஶ
‫ݕ‬ ݊ = ෍ (
1
2
)௞‫ݑ‬ ݇ − 2 ‫ݑ‬ ݊ − ݇
ஶ
௞ୀିஶ
For ݊ < 2
‫ݕ‬ ݊ = 0
1
4
●●●
2 4 60
x[k]
k
_ _ _ _
1
8 1
16
1
● ● ●
-2 0 n-4
h[-k+n]
k
_ _ _ _

Example
For ݊ ≥ 2
‫ݕ‬ ݊ = ෍(
1
2
)௞
௡
௞ୀଶ
GP series
∑ ‫ݎ‬௡
=
ଵି௥ಿశభ
ଵି௥
ே
௡ୀ଴
Let k-2=m
‫ݕ‬ ݊ = ෍ (
1
2
)௠ାଶ
௡ିଶ
௠ୀ଴
‫ݕ‬ ݊ = (
1
2
)ଶ ෍ (
1
2
)௠
௡ିଶ
௠ୀ଴
‫ݕ‬ ݊ = (
ଵ
ଶ
)ଶ
ଵି(
భ
మ
)೙షభ
ଵି
భ
మ
‫ݕ‬ ݊ =
ଵ
ଶ
− (
ଵ
ଶ
)௡
‫ݕ‬ ݊ = ቊ
0, ݊ < 2
ଵ
ଶ
− (
ଵ
ଶ
)௡, ݊ ≥ 2

Example
‫ݔ‬ ݊ = 2௡‫ݑ‬ −݊ , ℎ ݊ = ‫.]݊[ݑ‬
Find the convolution sum.
Ans:
Change of variable
‫ݔ‬ ݇ = 2௞
‫ݑ‬ −݇ , ℎ ݇ = ‫]݇[ݑ‬
‫ݕ‬ ݊ = ෍ 2௞‫ݑ‬ −݇
ஶ
௞ୀିஶ
‫݊[ݑ‬ − ݇]
For ݊ < 0
‫ݕ‬ ݊ = ෍ 2௞
௡
௞ୀିஶ
1
2
●●●
-2 20
x[k]
k
_ _ _ _
1
41
8
-4
1
1
● ● ●
-2 0 n-4
u[-k+n]
k
_ _ _ _
GP series
∑ ‫ݎ‬௡
=
ଵି௥ಿశభ
ଵି௥
ே
௡ୀ଴

Example
Let ݈ = −݇
‫ݕ‬ ݊ = ෍ 2ି௟
ି௡
௟ୀஶ
Let ݈ + ݊ = ݉
‫ݕ‬ ݊ = ෍ 2ି(௠ି௡)
଴
௠ୀஶ
‫ݕ‬ ݊ = ෍ 2௡ି௠
ஶ
௠ୀ଴
‫ݕ‬ ݊ = 2௡ ෍ (
1
2
)௠
ஶ
௠ୀ଴

Infinite sum
෍ ܽ௞
ஶ
௞ୀ଴
=
1
1 − ܽ
, ܽ < 1
‫ݕ‬ ݊ = 2௡
1
1 −
1
2
= 2௡ାଵ
For ݊ ≥ 0
‫ݕ‬ ݊ = ෍ 2௞
଴
௞ୀିஶ

Example
Let ݈ = −݇
‫ݕ‬ ݊ = ෍ 2ି௟
ஶ
௟ୀ଴
‫ݕ‬ ݊ = ෍(
1
2
)௟
ஶ
௟ୀ଴
‫ݕ‬ ݊ =
1
1 −
1
2
= 2
‫ݕ‬ ݊ = ൜
2௡ାଵ, ݊ < 0
2, ݊ ≥ 0

Example
‫ݕ‬ ݊ = (‫ݑ‬ ݊ + 10 − 2‫ݑ‬ ݊
+ ‫ݑ‬ ݊ − 4 ) ∗ ‫݊[ݑ‬ − 2]
Ans:
ℎ ݇ = ‫݇[ݑ‬ − 2]
ℎ −݇ = ‫ݑ‬ −݇ − 2
ℎ ݊ − ݇ = ‫݇−[ݑ‬ − 2 + ݊]
= ‫݇−[ݑ‬ + ݊ − 2]
-2
2 40
-2u[k]
k
_ _ _ _
1
●●● 6 84
u[k-4]
k
_ _ _ _
●
20
-4 0
1
-8 -6-10
u[k+10]
k
_ _ _ _
-2
● 42
4
-4
0
1
-8 -6-10
x[k]
k-2
●
2
● ●
-1
-4 0
1
-6
h[-k+n]
k-2 n-2
● ●
_ _ _ _

Example
Let k+10=m
‫ݕ‬ ݊ = ∑ 1௡ା଼
௠ୀ଴ = ݊ + 9
For 0 ≤ ݊ − 2 < 4, 2 ≤ ݊ < 6
‫ݕ‬ ݊ = ෍ 1
ିଵ
௞ୀିଵ଴
+ ෍ −1
௡ିଶ
௞ୀ଴
= ෍ 1
ଽ
௠ୀ଴
+ ෍ −1
௡ିଶ
௞ୀ଴
= 10 − ݊ − 1 = 11 − ݊
For ݊ − 2 ≥ 4, ݊ ≥ 6
‫ݕ‬ ݊ = ෍ 1
ିଵ
௞ୀିଵ଴
+ ෍ −1
ଷ
௞ୀ଴
= 10 + −4 = 6
For ݊ − 2 < −10 , ݊ < −8
‫ݕ‬ ݊ = 0
For −10 ≤ ݊ − 2 < 0, −8 ≤ ݊ < 2
‫ݕ‬ ݊ = ෍ 1
௡ିଶ
௞ୀିଵ଴
4
-4
0
1
-8 -6-10
x[k]
k-2
●
2
● ●
-1
-4 0
k-2 n-2
● ●
_ _ _ _
0
k-2 n-2
● ●
_ _ _ _
-4 0-6
k-2 n-2
● ●
_ _ _

Example
Find the convolution sum of
two sequences.
‫ݔ‬ ݊ = 1, 2, 3 ℎ ݊ = 2, 1, 4
‫ݕ‬ ݊ = {2, 5, 12, 11, 12}
x[n]
h[n]
1 2 3
2 2 4 6
1 1 2 3
4 4 8 12

Additional example
For the two signals shown in
Fig.Q5. obtain the expression for
(a)y(t) in terms of x(t)
(a) ‫ݕ‬ ‫ݐ‬ = ‫ݔ‬ ܽ‫ݐ‬ − ܾ (1)
a and b are the two unknowns
From Fig.
‫ݕ‬
ଷ
ଶ
= ‫ݔ‬ −
ଵ଼
ହ
Using eq.(1)
ଷ
ଶ
= −ܽ
ଵ଼
ହ
− ܾ (2)
Similarly ‫ݕ‬ 2 = ‫−(ݔ‬
ଵ଺
ହ
)
2 = −ܽ
ଵ଺
ହ
− ܾ (3)
Solve (2)& (3)
ܽ =
5
4
, ܾ = −6
‫ݕ‬ ‫ݐ‬ = ‫ݔ‬
5
4
‫ݐ‬ + 6

Additional Example
The o/p y[n] of an LTI system is given
as
Where x[n] is the i/p and ݃ ݊ =
‫ݑ‬ ݊ − ‫݊[ݑ‬ − 3]. Determine y[n]
when ‫ݔ‬ ݊ = ߜ[݊ − 2].
Ans:
Change of variable ‫ݔ‬ ݇ = ߜ[݇ − 2]
ߜ ݇ − 2 exist only at k=2
‫ݕ‬ ݊ = ݃[݊ + 4]
݃ ݊ + 4 = ‫ݑ‬ ݊ + 4 − ‫݊[ݑ‬ + 1]
‫ݕ‬ ݊ = ෍ ‫]݇[ݔ‬
ஶ
௞ୀିஶ
݃[݊ + 2݇]
‫ݕ‬ ݊ = ෍ ߜ[݇ − 2]
ஶ
௞ୀିஶ
݃[݊ + 2݇]

Convolution Integral
Consider an LTI system with impulse response ℎ(‫)ݐ‬
ߜ ‫ݐ‬ → ℎ(‫)ݐ‬
ߜ ‫ݐ‬ − ߬ → ℎ(‫ݐ‬ − ߬) time invariance
‫ߜ)߬(ݔ‬ ‫ݐ‬ − ߬ → ‫)߬(ݔ‬ℎ(‫ݐ‬ − ߬) homogeneity
super position
Convolution integral
න ‫ߜ)߬(ݔ‬ ‫ݐ‬ − ߬ ݀
ஶ
ିஶ
߬ → න ‫)߬(ݔ‬ℎ(‫ݐ‬ − ߬)
ஶ
ିஶ
݀߬
‫ݕ‬ ‫ݐ‬ = න ‫)߬(ݔ‬ℎ(‫ݐ‬ − ߬)
ஶ
ିஶ
݀߬

Example
‫ݕ‬ ‫ݐ‬ = ‫ݑ‬ ‫ݐ‬ + 1 ∗ ‫ݐ(ݑ‬ − 2)
Ans:
Change of variable
‫ݔ‬ ߬ = ‫ݑ‬ ߬ + 1 , ℎ ߬ = ‫߬(ݑ‬ − 2)
ℎ −߬ = ‫߬−(ݑ‬ − 2)
ℎ ‫ݐ‬ − ߬ = ‫ݑ‬ −߬ + ‫ݐ‬ − 2
For ‫ݐ‬ − 2 < −1, ‫ݐ‬ < 1
‫ݕ‬ ‫ݐ‬ = 0
For ‫ݐ‬ − 2 ≥ −1, ‫ݐ‬ ≥ 1 = ‫ݐ‬ − 2 + 1
= ‫ݐ‬ − 1
‫ݕ‬ ‫ݐ‬ = ൜
0, ‫ݐ‬ < 1
‫ݐ‬ − 1, ‫ݐ‬ ≥ 1
10 ߬
x(߬)
1
-1
10 ߬
h(t-߬)
1
-1
‫ݐ‬ − 2
‫ݕ‬ ‫ݐ‬ = න ‫)߬(ݔ‬ℎ(‫ݐ‬ − ߬)
ஶ
ିஶ
݀߬
‫ݕ‬ ‫ݐ‬ = න 1
௧ିଶ
ିଵ
݀߬

Example
‫ݕ‬ ‫ݐ‬ = ‫ݑ‬ ‫ݐ‬ + 2 − ‫ݑ‬ ‫ݐ‬ − 1
∗ ‫ݐ−(ݑ‬ + 2)
Ans:
Change of variable
‫ݔ‬ ߬ = ‫ݑ‬ ߬ + 2 − ‫߬(ݑ‬ − 1)
ℎ ߬ = ‫߬−(ݑ‬ + 2)
ℎ −߬ = ‫߬(ݑ‬ + 2)
ℎ ‫ݐ‬ − ߬ = ‫߬(ݑ‬ + ‫ݐ‬ + 2)
For ‫ݐ‬ + 2 < −2 , ‫ݐ‬ < −4
‫ݕ‬ ‫ݐ‬ = 3
For−2 ≤ ‫ݐ‬ + 2 < 1, −4 ≤ ‫ݐ‬ < −1
= 1 − (‫ݐ‬ + 2)
= −‫ݐ‬ − 1
For ‫ݐ‬ + 2 ≥ 1, ‫ݐ‬ ≥ −1
‫ݕ‬ ‫ݐ‬ = 0
10 ߬
x(߬)
1
-1-2
10 ߬
h(t-߬)
1
t+2
‫ݕ‬ ‫ݐ‬ = න 1݀߬
ଵ
௧ାଶ
‫ݕ‬ ‫ݐ‬ = න 1݀
ଵ
ିଶ
߬

Example
Exponential Function
‫ݔ‬ ‫ݐ‬ = ݁ି௧‫)ݐ(ݑ‬
‫ݔ‬ ‫ݐ‬ − 1 = ݁ି ௧ିଵ ‫ݐ(ݑ‬ − 1)
= ݁ି௧ାଵ‫ݐ(ݑ‬ − 1)
ℎ ߬ = ݁ିఛ‫)߬(ݑ‬
Plot ℎ(‫ݐ‬ − ߬)
Ans:
ℎ ‫ݐ‬ − ߬ = ݁ି ௧ିఛ ‫ݐ(ݑ‬ − ߬)
= ݁ఛି௧‫߬−(ݑ‬ + ‫)ݐ‬
10 ‫ݐ‬
x(t)
1
10 ‫ݐ‬
x(t-1)
1
10 ߬
h(-߬)
1
t0 ߬
h(t-߬)
1

Example
‫ݔ‬ ‫ݐ‬ = ݁ିଷ௧[‫ݑ‬ ‫ݐ‬ − ‫ݐ(ݑ‬ − 2)]
ℎ ‫ݐ‬ = ݁ି௧‫)ݐ(ݑ‬
Ans:
Change of variable
For‫ݐ‬ < 0, ‫ݕ‬ ‫ݐ‬ = 0
For 0≤ ‫ݐ‬ < 2
= −݁ି௧
݁ିଶ௧ − 1
−2
=
݁ି௧ − ݁ିଷ௧
2
For ‫ݐ‬ > 2
210 ߬
x(߬)
1
t0 ߬
h(t-߬)
1
‫ݕ‬ ‫ݐ‬ = න ݁ିଷఛ
݁ି(௧ିఛ)
݀߬
௧
଴
= ݁ି௧ න ݁ିଶఛ
݀߬
௧
଴
‫ݕ‬ ‫ݐ‬ = ݁ି௧
න ݁ିଶఛ
݀߬
ଶ
଴
=
௘ష೟
ଶ
1 − ݁ିସ

Example
The input signal ‫ݔ‬ ‫ݐ‬ = ݁ି௧
‫ݑ‬ ‫ݐ‬ to
an LTI system whose impulse
response is given by
ℎ ‫ݐ‬ = ൜
1 − ‫,ݐ‬ 0 ≤ ‫ݐ‬ ≤ 1
0, ‫ݐ݋‬ℎ݁‫݁ݏ݅ݓݎ‬
Calculate the o/p.
Ans:
0 ߬
x(߬)
1
0 ߬
h(߬)
1
1
0 ߬
h(−߬)
-1
1
t ߬
h(−߬+t)
t-1
1
0

Example
For ‫ݐ‬ < 0
‫ݕ‬ ‫ݐ‬ = 0
For ‫ݐ‬ ≥ 0 and ‫ݐ‬ − 1 ≤ 0
0 ≤ ‫ݐ‬ ≤ 1
‫ݕ‬ ‫ݐ‬ = න ‫ݔ‬ ߬ ℎ ‫ݐ‬ − ߬ ݀߬
௧
଴
= න ݁ିఛ
‫߬()߬(ݑ‬ + 1 − ‫߬݀)ݐ‬
௧
଴
= න ߬݁ିఛ
݀߬ + (1 − ‫)ݐ‬ න ݁ିఛ
݀߬
௧
଴
௧
଴
Indefinite integral
න ‫݁ݐ‬௔௧݀‫ݐ‬ =
݁௔௧
ܽଶ
(ܽ‫ݐ‬ − 1)
= −߬݁ିఛ
− ݁ିఛ
| + 1 − ‫ݐ‬ −݁ିఛ
|
= 1 − ‫݁ݐ‬ି௧
− ݁ି௧
+ 1 − ‫ݐ‬ −݁ି௧
‫ݕ‬ ‫ݐ‬ = 2 − ‫ݐ‬ − 2݁ି௧
0 ߬
x(߬)
1
t-1 t

Example
For ‫ݐ‬ − 1 > 0
‫ݐ‬ > 1
‫ݕ‬ ‫ݐ‬ = න ‫ݔ‬ ߬ ℎ ‫ݐ‬ − ߬ ݀߬
௧
௧ିଵ
= න ݁ିఛ
(߬ + 1 − ‫߬݀)ݐ‬
௧
௧ିଵ
= න ߬݁ିఛ
݀߬ + (1 − ‫)ݐ‬ න ݁ିఛ
݀߬
௧
௧ିଵ
௧
௧ିଵ
= −߬݁ିఛ − ݁ିఛ | + 1 − ‫ݐ‬ −݁ିఛ |
‫ݕ‬ ‫ݐ‬ = ݁ି(௧ିଵ) − 2݁ି௧
0 ߬
x(߬)
1
t-1 t

Example
Calculate the o/p for the following
i/p signal & impulse
‫ݔ‬ ‫ݐ‬ = ൜
1.5, −2 ≤ ‫ݐ‬ ≤ 3
0, ‫ݐ݋‬ℎ݁‫݁ݏ݅ݓݎ‬
ℎ ‫ݐ‬ = ൜
2, −1 ≤ ‫ݐ‬ ≤ 2
0, ‫ݐ݋‬ℎ݁‫݁ݏ݅ݓݎ‬
Ans:
10 ߬
x(߬)
1.5
-1-2 3
10 ߬
h(߬)
2
-1-2 2
0-1 ߬
h(−߬)
2
-2 1
0 ߬
h(−߬+t)
2
-2+t 1+t

Example
For 1 + ‫ݐ‬ > −2
‫ݐ‬ < −3
‫ݕ‬ ‫ݐ‬ = 0
For ‫ݐ‬ + 1 ≥ −2 & ‫ݐ‬ + 1 < 0
−3 ≤ ‫ݐ‬ < −1
For ‫ݐ‬ + 1 ≥ 0 & − 2 + ‫ݐ‬ < 0
−1 ≤ ‫ݐ‬ < 2
t+1 0 ߬
1.5
t-2 -2 3
‫ݕ‬ ‫ݐ‬ = න ‫ݔ‬ ߬ ℎ ‫ݐ‬ − ߬ ݀߬
௧ାଵ
ିଶ
= න 3݀߬
௧ାଵ
ିଶ
= 3(‫ݐ‬ + 3)
‫)ݐ(ݕ‬ = න 3݀߬
௧ାଵ
௧ିଶ
= 9
t+10 ߬
1.5
t-2-2 3

Example
For‫ݐ‬ + 1 ≥ 3 and ‫ݐ‬ − 2 ≤ 3
2 ≤ ‫ݐ‬ ≤ 5
for‫ݐ‬ ≥ 5
‫ݕ‬ ‫ݐ‬ = 0
‫)ݐ(ݕ‬ = න 3݀߬
ଷ
௧ିଶ
= 3(5 − ‫)ݐ‬ t+10 ߬
1.5
t-2-2 3

Signals and systems-2

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