Discussion on LTI systems, Time domain representation of LTI systems.

1. Signal and Systems-III
Prof: Sarun Soman
Manipal Institute of Technology
Manipal

2. Interconnection of LTI Systems
Parallel Connection of LTI systems
Output
‫ݕ‬ ‫ݐ‬ = ‫ݕ‬ଵ ‫ݐ‬ + ‫ݕ‬ଶ(‫)ݐ‬
= ‫ݔ‬ ‫ݐ‬ ∗ ℎଵ ‫ݐ‬ + ‫ݔ‬ ‫ݐ‬ ∗ ℎଶ(‫)ݐ‬
= ‫ݔ‬ ‫ݐ‬ ∗ {ℎଵ ‫ݐ‬ + ℎଶ(‫})ݐ‬
Proof
Prof: Sarun Soman, MIT, Manipal 2

3. Interconnection of LTI Systems
Distributive Property for CT case:
Distributive property for DT case:
Cascade Connection of LTI systems
Prof: Sarun Soman, MIT, Manipal 3

4. Interconnection of LTI Systems
Interconnection Properties of LTI Systems
Prof: Sarun Soman, MIT, Manipal 4

5. Interconnection of LTI Systems
Relation between LTI system properties and the impulse
response.
Memory less LTI system
The o/p of a DT LTI system
To be memoryless y[n] must depend only on present value of
input
The o/p cannot depend on x[n-k] for ݇ ≠ 0
‫ݕ‬ ݊ = ෍ ℎ ݇ ‫݊[ݔ‬ − ݇]
ஶ
௞ୀିஶ
Prof: Sarun Soman, MIT, Manipal 5

6. Time Domain Representations of LTI Systems
A DT LTI system is memoryless if and only if
ℎ ݇ = ܿߜ[݇] ‘c’ is an arbitrary constant
CT system
Output
A CT LTI system is memoryless if and only if
ℎ ߬ = ܿߜ(߬) ‘c’ is an arbitrary constant
‫ݕ‬ ‫ݐ‬ = න ℎ ߬ ‫ݔ‬ ‫ݐ‬ − ߬ ݀߬
ஶ
ିஶ
Prof: Sarun Soman, MIT, Manipal 6

7. Time Domain Representations of LTI Systems
Causal LTI systems
The o/p of a causal LTI system depends only on past or present
values of the input
DT system
‫ݕ‬ ݊ = ෍ ℎ ݇ ‫݊[ݔ‬ − ݇]
ஶ
௞ୀିஶ
future value of inputs
Prof: Sarun Soman, MIT, Manipal 7

8. Time Domain Representations of LTI Systems
For a DT causal LTI system
ℎ ݇ = 0 ݂‫݇ ݎ݋‬ < 0
Convolution Sum in new form
For CT system
ℎ ߬ = 0 ݂‫߬ ݎ݋‬ < 0
Convolution integral in new form
‫ݕ‬ ݊ = ෍ ℎ ݇ ‫݊[ݔ‬ − ݇]
ஶ
௞ୀ଴
‫ݕ‬ ‫ݐ‬ = න ℎ ߬ ‫ݔ‬ ‫ݐ‬ − ߬ ݀߬
ஶ
଴
Prof: Sarun Soman, MIT, Manipal 8

9. Time Domain Representations of LTI Systems
Stable LTI Systems
A system is BIBO stable if the o/p is guaranteed to be bounded
for every i/p.
Discrete time case:
Bounding convolution sum
Prof: Sarun Soman, MIT, Manipal 9

10. Time Domain Representations of LTI Systems
Assume that i/p is bounded
Condition for impulse response of a stable DT LTI system
Prof: Sarun Soman, MIT, Manipal 10

11. Time Domain Representations of LTI Systems
Continuous time case
Impulse response must be absolutely integrable.
න ℎ ߬ ݀߬ < ∞
ஶ
ିஶ
impulse response must be absolutely summable.
Prof: Sarun Soman, MIT, Manipal 11

12. Tutorial
Determine a DT LTI system
characterized by impulse response
ℎ ݊ = (
ଵ
ଶ
)௡
‫]݊[ݑ‬ is a)causal b)
stable.
Ans:
ℎ ݊ = 0 ݂‫݊ ݎ݋‬ < 0
System is causal
Stable
ℎ ݊ = (0.99)௡‫݊(ݑ‬ + 3)
Ans:
ℎ ݊ ≠ 0, ݊ < 0
Non-causal
Stable
෍ ℎ[݇]
ஶ
௞ୀ଴
= ෍ (
1
2
)௞
ஶ
௞ୀ଴
=
1
1 −
1
2
= 2 < ∞
෍ (0.99)௞
ஶ
௞ୀିଷ
(0.99)ିଷ
1
1 − 0.99
< ∞
Prof: Sarun Soman, MIT, Manipal 12

13. Tutorial
A CT LTI system is represented by the
impulse response
ℎ ‫ݐ‬ = ݁ିଷ௧‫ݐ(ݑ‬ − 1).Check for
stability and causality.
Ans:
ℎ ‫ݐ‬ = 0, ‫ݐ‬ < 0
Causal
ℎ ‫ݐ‬ = ݁௧‫ݑ‬ −1 − ‫ݐ‬
Ans:
ℎ ‫ݐ‬ ≠ 0, ‫ݐ‬ < 0
Non-causal
stable
න |ℎ ߬ |݀߬
ஶ
ିஶ
= න |ℎ ߬ |݀߬
ஶ
ଵ
= න |݁ିଷఛ
|݀߬
ஶ
ଵ
=
݁ିଷ
3
< ∞ stable
න |݁ఛ
|݀߬
ିଵ
ିஶ
= ݁ିଵ
< ∞
Prof: Sarun Soman, MIT, Manipal 13

14. Tutorial
ℎ(‫)ݐ‬ = ݁ିସ|௧|
Ans:
ℎ ‫ݐ‬ ≠ 0, ݂‫ݐ ݎ݋‬ < 0
Non-causal
න |ℎ(߬)|݀߬
ஶ
ିஶ
= න |݁ସఛ
|݀߬
଴
ିஶ
+ න |݁ିସఛ
|݀߬
ஶ
଴
=
ଵ
ଶ
< ∞
stable
Check for memory ,stability and
causality.
ℎ ݊ = ݁ଶ௡‫݊[ݑ‬ − 1]
ℎ ݊ ≠ 0, ݂‫݊ ݎ݋‬ ≠ 0
Memory system
ℎ ݊ = 0, ݂‫݊ ݎ݋‬ < 0
Causal
݁ଶ
> 1
h[n] is not absolutely summable
unstable
෍ |݁ଶ௡|
ஶ
௡ୀଵ
Prof: Sarun Soman, MIT, Manipal 14

15. Time Domain Representations of LTI Systems
Step Response
Characterizes the response to sudden changes in input.
‫ݏ‬ ݊ = ℎ ݊ ∗ ‫]݊[ݑ‬
u[n-k] exist from −∞ to n
h[k] is the running sum of impulse response.
= ෍ ℎ ݇ ‫݊[ݑ‬ − ݇]
ஶ
௞ୀିஶ
‫]݊[ݏ‬ = ෍ ℎ ݇
௡
௞ୀିஶ
Prof: Sarun Soman, MIT, Manipal 15

16. Time Domain Representations of LTI Systems
CT system
Relation b/w step response and impulse response.
ℎ ‫ݐ‬ =
݀
݀‫ݐ‬
‫)ݐ(ݏ‬
ℎ ݊ = ‫ݏ‬ ݊ − ‫݊[ݏ‬ − 1]
Eg. Step response of an RC ckt
‫ݏ‬ ‫ݐ‬ = න ℎ ߬ ݀߬
௧
ିஶ
Prof: Sarun Soman, MIT, Manipal 16

17. Time Domain Representations of LTI Systems
Prof: Sarun Soman, MIT, Manipal 17

18. Example
Find the step response for LTI system represented by impulse
response ℎ ݊ = (
ଵ
ଶ
)௡‫ݑ‬ ݊ .
Ans:
=
ଵି(
భ
మ
)೙శభ
ଵି
భ
మ
, ݊ ≥ 0
‫]݊[ݏ‬ = ෍ ℎ ݇
௡
௞ୀିஶ
‫]݊[ݏ‬ = ෍(
1
2
)௞
௡
௞ୀ଴
Prof: Sarun Soman, MIT, Manipal 18

19. Example
ℎ ‫ݐ‬ = ‫ݑݐ‬ ‫ݐ‬
Ans:
=
‫ݐ‬ଶ
2
, ‫ݐ‬ ≥ 0
‫ݏ‬ ‫ݐ‬ = න ℎ ߬ ݀߬
௧
ିஶ
‫ݏ‬ ‫ݐ‬ = න ߬݀߬
௧
଴
Prof: Sarun Soman, MIT, Manipal 19

20. Differential and Difference equation
Representations of LTI Systems
Linear constant coefficient differential equation:
Linear constant coefficient difference equation:
Order of the equation is (N,M), representing the number of
energy storage devices in the system.
Often N>M and the order is described using only ‘N’.
෍ ܽ௞
݀௞
݀‫ݐ‬௞
‫ݕ‬ ‫ݐ‬ = ෍ ܾ௞
݀௞
݀‫ݐ‬௞
‫)ݐ(ݔ‬
ெ
௞ୀ଴
ே
௞ୀ଴
෍ ܽ௞‫݊[ݕ‬ − ݇] = ෍ ܾ௞‫݊[ݔ‬ − ݇]
ெ
௞ୀ଴
ே
௞ୀ଴
Prof: Sarun Soman, MIT, Manipal 20

21. Differential and Difference equation
Representations of LTI Systems
Prof: Sarun Soman, MIT, Manipal 21

22. Differential and Difference equation
Representations of LTI Systems
Second order difference equation:
Difference equation are easily arranged to obtain recursive
formulas for computing the current o/p of the system.
(1) Shows how to obtain y[n] from present and past values of
the input.
෍ ܽ௞‫݊[ݕ‬ − ݇] = ෍ ܾ௞‫݊[ݔ‬ − ݇]
ெ
௞ୀ଴
ே
௞ୀ଴
‫ݕ‬ ݊ =
1
ܽ଴
෍ ܾ௞‫݊[ݔ‬ − ݇]
ெ
௞ୀ଴
−
1
ܽ଴
෍ ܽ௞‫ݕ‬ ݊ − ݇
ே
௞ୀଵ
(1)
Prof: Sarun Soman, MIT, Manipal 22

23. Differential and Difference equation
Representations of LTI Systems
Recursive evaluation of a difference equation.
Find the first two o/p values y[0] and y[1] for the system
described by ‫ݕ‬ ݊ = ‫ݔ‬ ݊ + 2‫ݔ‬ ݊ − 1 − ‫ݕ‬ ݊ − 1 −
ଵ
ସ
‫݊[ݕ‬ − 2].
Assuming that the input is ‫ݔ‬ ݊ = (
ଵ
ଶ
)௡‫ݑ‬ ݊ and the initial
conditions are y[-1]=1 and y[-2]=-2.
Prof: Sarun Soman, MIT, Manipal 23

24. Solving Differential and Difference
Equations
Output of LTI system described by differential or difference
equation has two components
‫ݕ‬௛
- homogeneous solution
‫ݕ‬௣- particular solution
Complete solution
‫ݕ‬ = ‫ݕ‬௛ + ‫ݕ‬௣
Eg.
Prof: Sarun Soman, MIT, Manipal 24

25. Solving Differential and Difference
Equations
Output due to non zero initial conditions with zero input
Prof: Sarun Soman, MIT, Manipal 25

26. Solving Differential and Difference
Equations
Step response, system initially at rest.
Prof: Sarun Soman, MIT, Manipal 26

27. Solving Differential and Difference
Equations
The Homogenous Solution
CT system
Set all terms involving input to zero
Solution
‫ݎ‬௜are the N roots of the characteristic equation
෍ ܽ௞
݀௞
݀‫ݐ‬௞
‫ݕ‬ ‫ݐ‬ = ෍ ܾ௞
݀௞
݀‫ݐ‬௞
‫)ݐ(ݔ‬
ெ
௞ୀ଴
ே
௞ୀ଴
෍ ܽ௞
݀௞
݀‫ݐ‬௞
‫ݕ‬ ‫ݐ‬ = 0
ே
௞ୀ଴
‫ݕ‬௛(‫)ݐ‬ = ෍ ܿ௜݁௥೔௧
ே
௜ୀଵ
Prof: Sarun Soman, MIT, Manipal 27

28. Solving Differential and Difference
Equations
DT system
Set all terms involving input to zero
Solution
‫ݎ‬௜ are the N roots of the characteristic equation.
෍ ܽ௞‫݊[ݕ‬ − ݇] = ෍ ܾ௞‫݊[ݔ‬ − ݇]
ெ
௞ୀ଴
ே
௞ୀ଴
෍ ܽ௞‫݊[ݕ‬ − ݇] = 0
ே
௞ୀ଴
‫ݕ‬௛[݊] = ෍ ܿ௜‫ݎ‬௜
௡
ே
௜ୀଵ
Prof: Sarun Soman, MIT, Manipal 28

29. Solving Differential and Difference
Equations
If the roots are repeating ‘p’ times then there are ‘p’ distinct
terms
݁௥೔௧, ‫݁ݐ‬௥೔௧, … . ‫ݐ‬௣ିଵ݁௥೔௧
and ‫ݎ‬௜
௡, ݊‫ݎ‬௜
௡ … . . ݊௣ିଵ‫ݎ‬௜
௡
Eg.
RC ckt depicted in figure is described by the differential equation
‫ݕ‬ ‫ݐ‬ + ܴ‫ܥ‬
ௗ
ௗ௧
‫ݕ‬ ‫ݐ‬ = ‫)ݐ(ݔ‬ . Determine the homogenous solution.
Prof: Sarun Soman, MIT, Manipal 29

30. Solving Differential and Difference
Equations
The homogenous equation is
‫ݕ‬ ‫ݐ‬ + ܴ‫ܥ‬
ௗ
ௗ௧
‫ݕ‬ ‫ݐ‬ = 0 (1)
Solution
‫ݕ‬௛
‫ݐ‬ = ܿଵ݁௥భ௧
To determine ‫ݎ‬ଵsubstitute in (1)
‫ݕ‬௛ ‫ݐ‬ + ܴ‫ܥ‬
݀
݀‫ݐ‬
‫ݕ‬௛ ‫ݐ‬ = 0
ܿଵ݁௥భ௧ 1 + ܴ‫ݎܥ‬ଵ = 0
ܿଵ݁௥భ௧ ≠ 0
Characteristic equation
1 + ܴ‫ݎܥ‬ଵ = 0
‫ݎ‬ଵ = −
1
ܴ‫ܥ‬
Homogenous solution of the system
is
‫ݕ‬௛ ‫ݐ‬ = ܿଵ݁ି
ଵ
ோ஼௧
Note: ܿଵ is determined later, in order
that the complete solution satisfy
the initial conditions.
‫ݕ‬௛
(‫)ݐ‬ = ෍ ܿ௜݁௥೔௧
ே
௜ୀଵ
Prof: Sarun Soman, MIT, Manipal 30

31. Example
Determine the homogenous
solution.
ௗమ
ௗ௧మ ‫ݕ‬ ‫ݐ‬ + 5
ௗ
ௗ௧
‫ݕ‬ ‫ݐ‬ + 6‫ݕ‬ ‫ݐ‬ =
2‫ݔ‬ ‫ݐ‬ +
ௗ
ௗ௧
‫)ݐ(ݔ‬
Ans:
Homogenous equation
ௗమ
ௗ௧మ ‫ݕ‬ ‫ݐ‬ + 5
ௗ
ௗ௧
‫ݕ‬ ‫ݐ‬ + 6‫ݕ‬ ‫ݐ‬ =0 (1)
Homogenous solution
‫ݕ‬௛
‫ݐ‬ = ܿଵ݁௥భ௧
+ ܿଶ݁௥మ௧
To find ‫ݎ‬ଵand ‫ݎ‬ଶ solve CE.
Put
ௗ೙
ௗ௧೙ ‫ݕ‬ ‫ݐ‬ = ‫ݎ‬௡
(1) becomes
‫ݎ‬ଶ + 5‫ݎ‬ + 6 = 0
‫ݎ‬ = −3, −2
‫ݕ‬௛ ‫ݐ‬ = ܿଵ݁ିଷ௧ + ܿଶ݁ିଶ௧
‫ݕ‬௛(‫)ݐ‬ = ෍ ܿ௜݁௥೔௧
ே
௜ୀଵ
Prof: Sarun Soman, MIT, Manipal 31

32. Example
Find the homogenous solution of the
first order recursive system.
‫ݕ‬ ݊ − ߩ‫ݕ‬ ݊ − 1 = ‫]݊[ݔ‬
Ans:
Set i/p to zero
‫ݕ‬ ݊ − ߩ‫ݕ‬ ݊ − 1 = 0 (1)
Substitute in (1)
ܿଵ‫ݎ‬ଵ
௡ − ߩܿଵ‫ݎ‬ଵ
௡ିଵ = 0
‫ݕ‬௛[݊] = ෍ ܿ௜‫ݎ‬௜
௡
ே
௜ୀଵ
‫ݕ‬௛[݊] = ܿଵ‫ݎ‬ଵ
௡
ܿଵ‫ݎ‬ଵ
௡ 1 − ߩ‫ݎ‬ଵ
ିଵ = 0
1 − ߩ‫ݎ‬ଵ
ିଵ = 0
‫ݎ‬ଵ = ߩ
‫ݕ‬௛[݊] = ܿଵߩ௡
Prof: Sarun Soman, MIT, Manipal 32

33. Example
Ans:
Set input to zero
(1) Becomes
1 −
1
4
‫ݎ‬ିଵ −
1
8
‫ݎ‬ିଶ = 0
‫ݎ‬ଶ −
1
4
‫ݎ‬ −
1
8
= 0
‫ݎ‬ =
1
2
, −
1
4
‫ݕ‬௛
݊ = ܿଵ(
1
2
)௡
+ ܿଶ(−
1
4
)௡
‫ݕ‬ ݊ −
1
4
‫ݕ‬ ݊ − 1 −
1
8
‫ݕ‬ ݊ − 2 = ‫݊[ݔ‬ − 1]
N=2
‫ݕ‬ ݊ −
1
4
‫ݕ‬ ݊ − 1 −
1
8
‫ݕ‬ ݊ − 2 = 0 (1)
‫ݕ‬௛
[݊] = ܿଵ‫ݎ‬ଵ
௡+ܿଶ‫ݎ‬ଶ
௡
Prof: Sarun Soman, MIT, Manipal 33

34. Example
‫ݕ‬ ݊ +
9
16
‫ݕ‬ ݊ − 2 = ‫݊[ݔ‬ − 1]
Ans:
‫ݎ‬ଶ +
9
16
= 0
‫ݎ‬ = ±݆
3
4
‫ݕ‬௛ ݊ = ܿଵ(݆
3
4
)௡ + ܿଶ(−݆
3
4
)௡
Prof: Sarun Soman, MIT, Manipal 34

35. Solving Differential and Difference
Equations
The Particular Solution ‫ݕ‬௣
Solution of difference or differential equation for a given input.
Assumption : output is of same general form as the input.
CT System
Input Particular solution
1 c
t ܿଵ‫ݐ‬ + ܿଶ
݁ି௔௧
ܿ݁ି௔௧
cos(⍵‫ݐ‬ + ⏀) ܿଵ cos ߱‫ݐ‬ + ܿଶ sin(߱‫)ݐ‬
DT System
Input Particular solution
1 c
n ܿଵ݊ + ܿଶ
ߙ௡ ܿߙ௡
cos(Ω݊ + ⏀) ܿଵ cos Ω݊ + ܿଶ sin(Ω݊)
Prof: Sarun Soman, MIT, Manipal 35

36. Example
Consider the RC ckt .Find a particular
solution for this system with an input
‫ݔ‬ ‫ݐ‬ = cos(߱଴‫.)ݐ‬
Ans:
From previous example
‫ݕ‬ ‫ݐ‬ + ܴ‫ܥ‬
ௗ
ௗ௧
‫ݕ‬ ‫ݐ‬ = ‫)ݐ(ݔ‬ (1)
‫ݕ‬௣ ‫ݐ‬ = ܿଵcos(⍵଴‫)ݐ‬ + ܿଶ sin(⍵଴‫)ݐ‬
Substitute in (1)
‫ݕ‬௣ ‫ݐ‬ + ܴ‫ܥ‬
݀
݀‫ݐ‬
‫ݕ‬௣
‫ݐ‬ = cos(߱‫)ݐ‬
ܿଵ cos ⍵଴‫ݐ‬ + ܿଶ sin(⍵଴‫)ݐ‬
− RC⍵଴ܿଵ sin ⍵଴‫ݐ‬
+ RC⍵଴ܿଶ cos ⍵଴‫ݐ‬
= cos(⍵଴‫)ݐ‬
Equating the coefficients of ܿଵand ܿଶ
ܿଵ + RC⍵଴ܿଶ = 1
−RC⍵଴ܿଵ + ܿଶ = 0
Solving for ܿଵand ܿଶ
ܿଵ =
1
1 + (RC⍵଴)ଶ
ܿଶ =
RC⍵଴
1 + (RC⍵଴)ଶ
Prof: Sarun Soman, MIT, Manipal 36

37. Example
‫ݕ‬௣ ‫ݐ‬ =
1
1 + (RC⍵଴)ଶ
cos ⍵଴‫ݐ‬
+
RC⍵଴
1 + (RC⍵଴)ଶ
sin ⍵଴‫ݐ‬
Determine the particular solution for
the system described by the
following differential equations.
‫ݕ‬௣ ‫ݐ‬ = ܿ
0 + 10ܿ = 4
ܿ =
2
5
‫ݕ‬௣ ‫ݐ‬ =
2
5
b)‫ݔ‬ ‫ݐ‬ = cos(3‫)ݐ‬
Ans:
‫ݕ‬௣ ‫ݐ‬ = ܿଵcos(3‫)ݐ‬ + ܿଶ sin(3‫)ݐ‬
݀
݀‫ݐ‬
‫ݕ‬௣
‫ݐ‬ = −3ܿଵ sin(3‫)ݐ‬
+ 3 ܿଶcos(3‫)ݐ‬
−15ܿଵ sin 3‫ݐ‬ + 15ܿଶ cos 3‫ݐ‬
+ 10ܿଵ cos 3‫ݐ‬
+ 10ܿଶ sin 3‫ݐ‬ = 2 cos(3‫)ݐ‬
Equating coefficients
−15ܿଵ + 10ܿଶ = 0
10ܿଵ + 15ܿଶ = 2
Prof: Sarun Soman, MIT, Manipal 37

38. Example
ܿଵ =
4
65
, ܿଶ =
6
65
‫ݕ‬௣
‫ݐ‬ =
4
65
cos 3‫ݐ‬ +
6
65
sin(3‫)ݐ‬
Prof: Sarun Soman, MIT, Manipal 38

39. Solving Differential and Difference
Equations
The Complete Solution
Procedure for calculating complete solution
1. Find the form of the homogeneous solution ‫ݕ‬௛from the roots of the CE
equation.
2. Find a particular solution ‫ݕ‬௣by assuming that it is of the same form as
the input, yet is independent of all terms in the homogeneous solution.
3. Determine the coefficients in the homogeneous solution so that the
complete solution ‫ݕ‬ = ‫ݕ‬௛
+ ‫ݕ‬௣
satisfies the initial conditions.
Prof: Sarun Soman, MIT, Manipal 39

40. Example
Find the solution for the first order
recursive system described by the
difference equation.
‫ݕ‬ ݊ −
1
4
‫ݕ‬ ݊ − 1 = ‫]݊[ݔ‬
If the input ‫ݔ‬ ݊ = (
ଵ
ଶ
)௡‫]݊[ݑ‬ and the
initial condition is ‫ݕ‬ −1 = 8.
Ans:
N=1
Homogenous solution
‫ݕ‬ ݊ −
1
4
‫ݕ‬ ݊ − 1 = 0
‫ݕ‬௛
݊ = ܿଵ‫ݎ‬௡
‫ݎ‬௡
−
1
4
‫ݎ‬௡ିଵ
= 0
‫ݎ‬ −
1
4
= 0
‫ݕ‬௛ ݊ = ܿଵ(
1
4
)௡
Particular solution
‫ݕ‬௣ ݊ = ܿଶ(
1
2
)௡
ܿଶ(
1
2
)௡
−
1
4
ܿଶ(
1
2
)௡ିଵ
= (
1
2
)௡
ܿଶ −
1
2
ܿଶ = 1
ܿଶ = 2
Prof: Sarun Soman, MIT, Manipal 40

41. Example
‫ݕ‬௣ ݊ = 2 (
1
2
)௡
Complete solution
‫]݊[ݕ‬ = ‫ݕ‬௛ ݊ + ‫ݕ‬௣[݊]
‫]݊[ݕ‬ = ܿଵ
ଵ
ସ
௡
+ 2 (
ଵ
ଶ
)௡ (1)
ܶ‫ܿ ݂݀݊݅ ݋‬ଵ-use initial conditions
Particular solution exist only for
n≥ 0
Translate the initial condition to
n≥ 0
Rewrite the system equation in
recursive form
‫ݕ‬ ݊ = ‫ݔ‬ ݊ +
1
4
‫ݕ‬ ݊ − 1
‫ݕ‬ 0 = ‫ݔ‬ 0 +
1
4
‫]1−[ݕ‬
= 1 +
1
4
∗ 8 = 3
Substitute y[0]in (1)
‫ݕ‬ 0 = ܿଵ(
1
4
)଴
+2 (
1
2
)଴
3 = ܿଵ + 2
ܿଵ = 1
‫]݊[ݕ‬ =
1
4
௡
+ 2 (
1
2
)௡, ݊ ≥ 0
Prof: Sarun Soman, MIT, Manipal 41

42. Example
‫ݕ‬ ݊ −
1
4
‫ݕ‬ ݊ − 1 −
1
8
‫ݕ‬ ݊ − 2
= ‫ݔ‬ ݊ + ‫ݔ‬ ݊ − 1
Input ‫ݔ‬ ݊ = (
ଵ
ଶ
)௡
,
Ans:
Homogenous solution
N=2
‫ݕ‬ ݊ −
1
4
‫ݕ‬ ݊ − 1 −
1
8
‫ݕ‬ ݊ − 2 = 0
‫ݕ‬௛
݊ = ܿଵ‫ݎ‬௡
+ ܿଶ‫ݎ‬௡
To find ‘r’
‫ݎ‬௡
−
1
4
‫ݎ‬௡ିଵ
−
1
8
‫ݎ‬௡ିଶ
= 0
‫ݎ‬ =
1
2
, −
1
4
‫ݕ‬௛
݊ = ܿଵ(
1
2
)௡
+ܿଶ(−
1
4
)௡
Particular Solution
‫ݕ‬௣
݊ = ܿଷ(
1
2
)௡
Same form as homogeneous solution
‫ݕ‬௣
݊ = ܿଷ݊(
1
2
)௡
‫]݊[ݑ‬
ܿଷ݊(
1
2
)௡
−
1
4
ܿଷ(݊ − 1)(
1
2
)௡ିଵ
−
1
8
ܿଷ(݊ − 2)(
1
2
)௡ିଶ
= (
1
2
)௡
+ (
1
2
)௡ିଵ
Prof: Sarun Soman, MIT, Manipal 42

43. Example
ܿଷ݊ −
1
4
ܿଷ(݊ − 1)(
1
2
)ିଵ
−
1
8
ܿଷ(݊
− 2)(
1
2
)ିଶ
= 1 + (
1
2
)ିଵ
ܿଷ݊ −
1
2
ܿଷ(݊ − 1) −
1
2
ܿଷ(݊ − 2) = 3
1
2
ܿଷ + ܿଷ = 3
ܿଷ = 2
‫ݕ‬௣
݊ = 2݊(
1
2
)௡
‫]݊[ݑ‬
Complete solution
‫ݕ‬ = ܿଵ(
1
2
)௡
+ܿଶ(−
1
4
)௡
+2݊(
1
2
)௡
‫]݊[ݑ‬
Determine the o/p of the system
‫ݕ‬ ݊ −
1
9
‫ݕ‬ ݊ − 2 = ‫݊[ݔ‬ − 1]
‫ݔ‬ ݊ = ‫ݑ‬ ݊ , ‫ݕ‬ −1 = 1, ‫ݕ‬ −2 = 0
Ans:
Homogenous solution
N=2
‫ݕ‬ ݊ −
1
9
‫ݕ‬ ݊ − 2 = 0
‫ݕ‬௛
݊ = ܿଵ‫ݎ‬ଵ
௡
+ ܿଶ‫ݎ‬ଶ
௡
Find ‘r’
‫ݎ‬௡
−
1
9
‫ݎ‬௡ିଶ
= 0
‫ݎ‬ଶ
−
1
9
= 0
Prof: Sarun Soman, MIT, Manipal 43

44. Example
‫ݎ‬ = ±
1
3
‫ݕ‬௛ ݊ = ܿଵ
1
3
௡
+ ܿଶ −
1
3
௡
Particular solution
‫ݕ‬௣ ݊ = ܿଷ‫]݊[ݑ‬
Substitute in system eqn.
ܿଷ −
1
9
ܿଷ = 1
ܿଷ =
9
8
‫ݕ‬௣ ݊ =
ଽ
଼
‫]݊[ݑ‬
Complete solution
‫ݕ‬ = ܿଵ
1
3
௡
+ ܿଶ −
1
3
௡
+
9
8
‫]݊[ݑ‬
To find ܿଵand ܿଶ translate initial
conditions to ݊ ≥ 0
Rewrite system eqn. in recursive
form.
‫ݕ‬ ݊ = ‫ݔ‬ ݊ − 1 +
1
9
‫ݕ‬ ݊ − 2
‫ݕ‬ 0 = ‫ݔ‬ −1 +
1
9
‫]2−[ݕ‬
‫ݕ‬ 0 = 0
‫ݕ‬ 1 = ‫ݔ‬ 0 +
1
9
‫]1−[ݕ‬
Prof: Sarun Soman, MIT, Manipal 44

45. Example
‫ݕ‬ 1 =
10
9
Substitute y[0] and y[1] in complete
solution y[n].
ܿଵ = −
7
12
, ܿଶ = −
13
24
‫ݕ‬ ݊ =
9
8
‫ݑ‬ ݊ −
7
12
1
3
௡
−
13
24
−
1
3
௡
Prof: Sarun Soman, MIT, Manipal 45

46. Characteristics of Systems Described by
Differential and Difference Equations
It is informative to express the o/p of a system as sum of two
components.
1) Only with initial conditions
2) Only with input
‫ݕ‬ = ‫ݕ‬௡ + ‫ݕ‬௙
Natural response ࢟࢔ (zero i/p response)
System o/p for zero i/p
Describes the manner in which the system dissipates any stored
energy or memory.
Prof: Sarun Soman, MIT, Manipal 46

47. Characteristics of Systems Described by
Differential and Difference Equations
Procedure to calculate ‫ݕ‬௡
1. Find the homogeneous solution.
2. From the homogeneous solution find the coefficients ܿ௜ using initial
conditions.
Note: homogeneous solutions apply for all time, the natural
response is determined without translating initial conditions.
Eg.
A system is described by the difference equation
‫ݕ‬ ݊ −
ଵ
ସ
‫ݕ‬ ݊ − 1 = ‫ݔ‬ ݊ , ‫ݕ‬ −1 = 8. Find the natural response
of the system.
Prof: Sarun Soman, MIT, Manipal 47

48. Characteristics of Systems Described by
Differential and Difference Equations
Homogenous solution
‫ݕ‬௛
݊ = ܿଵ
1
4
௡
Use initial conditions to find ܿଵ
Translation not required
‫ݕ‬ −1 = ܿଵ
1
4
ିଵ
ܿଵ = 2
‫ݕ‬௡
= 2
1
4
௡
, ݊ ≥ −1
Prof: Sarun Soman, MIT, Manipal 48

49. Characteristics of Systems Described by
Differential and Difference Equations
The Forced Response ‫ݕ‬௙
System o/p due to the i/p signal assuming zero initial conditions.
The forced response is of the same form as the complete
solution.
Eg.
‫ݕ‬ ݊ −
ଵ
ସ
‫ݕ‬ ݊ − 1 = ‫ݔ‬ ݊ .Find the forced response of this
system if the i/p is ‫ݔ‬ ݊ =
ଵ
ଶ
௡
‫]݊[ݑ‬
Ans:
Homogenous solution
Prof: Sarun Soman, MIT, Manipal 49

50. Characteristics of Systems Described by
Differential and Difference Equations
‫ݕ‬ ݊ −
1
4
‫ݕ‬ ݊ − 1 = 0
‫ݕ‬௛
݊ = ܿଵ
1
4
௡
Particular solution
‫ݕ‬௣
݊ = ܿଶ
1
2
௡
‫]݊[ݑ‬
‫ݕ‬௣
݊ = 2
1
2
௡
‫]݊[ݑ‬
Complete solution
‫]݊[ݕ‬ = 2
ଵ
ଶ
௡
‫ݑ‬ ݊ + ܿଵ
ଵ
ସ
௡
(1)
To find ܿଵassume zero initial conditions
‫ݕ‬ −1 = 0
Translate initial conditions
‫ݕ‬ ݊ = ‫ݔ‬ ݊ +
1
4
‫ݕ‬ ݊ − 1
‫ݕ‬ ݊ =
1
2
௡
‫]݊[ݑ‬ +
1
4
‫ݕ‬ ݊ − 1
‫ݕ‬ 0 = 1 + 0
Substitute in (1)
‫ݕ‬ 0 = 2
1
2
଴
‫ݑ‬ 0 + ܿଵ
1
4
଴
ܿଵ = −1
‫ݕ‬௙
݊ = 2
1
2
௡
−
1
4
௡
, ݊ ≥ 0
Prof: Sarun Soman, MIT, Manipal 50

51. Example
Identify the natural and forced
response.
‫ݕ‬ ݊ −
3
4
‫ݕ‬ ݊ − 1 +
1
8
‫ݕ‬ ݊ − 2
= 2‫ݔ‬ ݊
‫ݕ‬ −1 = 1, ‫ݕ‬ −2 = −1 , ‫ݔ‬ ݊
= 2‫]݊[ݑ‬
Ans:
Homogeneous solution
N=2
‫ݕ‬௛
݊ = ܿଵ‫ݎ‬ଵ
௡
+ ܿଶ‫ݎ‬ଶ
௡
‫ݎ‬ =
1
2
,
1
4
‫ݕ‬௛
݊ = ܿଵ
1
2
௡
+ ܿଶ
1
4
௡
Natural response
Find ܿଵand ܿଶusing initial conditions
No translation required.
2ܿଵ + 4ܿଶ = 1
4ܿଵ + 16ܿଶ = −1
࢟࢔ ࢔ =
૞
૝
૚
૛
࢔
−
૜
ૡ
૚
૝
࢔
Forced response.
Particular solution
‫ݕ‬௣ ݊ = ܿଷ‫]݊[ݑ‬
Prof: Sarun Soman, MIT, Manipal 51

52. Example
‫ݕ‬௣ ݊ =
32
3
‫]݊[ݑ‬
Complete solution
‫ݕ‬ ݊ = ‫ݕ‬௛ ݊ + ‫ݕ‬௣[݊]
‫]݊[ݕ‬ = ܿଵ
ଵ
ଶ
௡
+ ܿଶ
ଵ
ସ
௡
+
ଷଶ
ଷ
‫]݊[ݑ‬
(1)
Forced response
Find the coefficients ܿଵand ܿଶ using
zero initial conditions.
‫ݕ‬ −1 = 0, ‫ݕ‬ −2 = 0
Translate the initial conditions
‫ݕ‬ 0 = 4, ‫ݕ‬ 1 = 7
Substitute y[0] and y[1] in (1)
ܿଵ + ܿଶ = −
20
3
1
2
ܿଵ +
1
4
ܿଶ = −
11
3
࢟ࢌ ࢔ =
૜૛
૜
࢛ ࢔ − ૡ
૚
૛
࢔
+
૝
૜
૚
૝
࢔
, ࢔ ≥ ૙
Prof: Sarun Soman, MIT, Manipal 52

53. Example
Find the forced response
‫ݕ‬௛ ݊ = ܿଵ
1
2
௡
‫ݕ‬௣ ݊ = ܿଶ −
1
2
௡
‫]݊[ݑ‬
ܿଶ = 1
Complete solution
‫ݕ‬ ݊ = ܿଵ
1
2
௡
+ −
1
2
௡
‫]݊[ݑ‬
Use zero initial conditions
Translate the initial conditions
‫ݕ‬ 0 = 2
ܿଵ = 1
Prof: Sarun Soman, MIT, Manipal 53

54. Example
‫ݕ‬ ݊ + ‫ݕ‬ ݊ − 1 +
1
4
‫ݕ‬ ݊ − 2
= ‫ݔ‬ ݊ + ‫݊[ݔ‬ − 1]
I/p ‫ݔ‬ ݊ = −
ଵ
ଶ
௡
‫ݑ‬ ݊ , ‫ݕ‬ −1 =
− 1, ‫ݕ‬ −2 = 2. Find the forced
response.
Ans:
Homogeneous solution
‫ݕ‬ ݊ + ‫ݕ‬ ݊ − 1 +
1
4
‫ݕ‬ ݊ − 2 = 0
N=2
‫ݕ‬௛ ݊ = ܿଵ‫ݎ‬ଵ
௡ + ܿଶ‫ݎ‬ଶ
௡ (1)
‫ݎ‬௡ + ‫ݎ‬௡ିଵ +
1
4
‫ݎ‬௡ିଶ = 0
1 + ‫ݎ‬ିଵ
+
1
4
‫ݎ‬ିଶ
= 0
‫ݎ‬ଶ
+ ‫ݎ‬ +
1
4
= 0
‫ݎ‬ = −
1
2
, −
1
2
Roots are repeating
If a root ‫ݎ‬௝ is repeated p times then
there are p distinct terms
‫ݎ‬௝
௡, ݊‫ݎ‬௝
௡, … … . ݊௣ିଵ‫ݎ‬௝
௡
Eqn (1) is modified
‫ݕ‬௛
݊ = ܿଵ‫ݎ‬ଵ
௡ + ܿଶ݊‫ݎ‬ଶ
௡
Prof: Sarun Soman, MIT, Manipal 54

55. Example
‫ݕ‬௛
݊ = ܿଵ −
1
2
௡
+ ܿଶ݊ −
1
2
௡
Particular solution
‫ݕ‬௣
݊ = ܿଷ −
1
2
௡
‫]݊[ݑ‬
‫ݕ‬௣
[݊] must be independent of ‫ݕ‬௛
[݊]
‫ݕ‬௣
݊ = ܿଷ݊ଶ
−
1
2
௡
‫]݊[ݑ‬
Substitute ‫ݕ‬௣
[݊] in system equation
ܿଷ = −
1
2
‫ݕ‬௣
݊ = −
1
2
݊ଶ
−
1
2
௡
‫]݊[ݑ‬
Complete solution
‫ݕ‬ = ‫ݕ‬௛
݊ + ‫ݕ‬௣
[݊]
‫ݕ‬ = ܿଵ −
ଵ
ଶ
௡
+ ܿଶ݊ −
ଵ
ଶ
௡
−
ଵ
଺
݊ଶ
−
ଵ
ଶ
௡
‫]݊[ݑ‬ (2)
Forced response
Use zero initial conditions ‫ݕ‬ −1 =
0, ‫ݕ‬ −2 = 0.
Translate the initial conditions to ݊ ≥ 0
‫ݕ‬ ݊ = −‫ݕ‬ ݊ − 1 −
1
4
‫ݕ‬ ݊ − 2
+ ‫ݔ‬ ݊ + ‫݊[ݔ‬ − 1]
Prof: Sarun Soman, MIT, Manipal 55

56. Example
‫ݕ‬ 0 = −‫ݕ‬ −1 −
1
4
‫ݕ‬ −2 + ‫ݔ‬ 0
+ ‫]1−[ݔ‬
‫ݕ‬ 0 = 1,‫ݕ‬ 1 = −
ଵ
ଶ
Substitute y[0] and y[1] in (2)
ܿଵ = 1, ܿଶ =
1
2
‫ݕ‬௙ ݊ = −
1
2
௡
+
1
2
݊ −
1
2
௡
−
1
2
݊ଶ −
1
2
௡
‫ݑ‬ ݊ , ݊ ≥ 0
Prof: Sarun Soman, MIT, Manipal 56