ch02

Time-Domain Representations ofLTI Systems

1

2. Time2. Time--Domain RepresentationsDomain Representationsof LTI Systemsof LTI Systems

2.1 Introduction2.1 Introduction

2.22.2 Convolution SumConvolution Sum

2.3 Convolution Sum2.3 Convolution Sum EvaluationEvaluation

ProcedureProcedure

2.42.4 Convolution IntegralConvolution Integral

2.5 Convolution Integral2.5 Convolution IntegralEvaluation ProcedureEvaluation Procedure

2.6 Interconnections of LTI2.6 Interconnections of LTISystemsSystems

2.7 Relations between LTI2.7 Relations between LTISystem Properties and theSystem Properties and theImpulse ResponseImpulse Response

2.8 Step Response2.8 Step Response

2.92.9 Differential and DifferenceDifferential and DifferenceEquation RepresentationEquation Representationof LTI Systemsof LTI Systems

2.10 Solving Differential and2.10 Solving Differential andDifference EquationDifference Equation

2.11 Characteristics of Systems2.11 Characteristics of SystemsDescribed by Differential andDescribed by Differential andDifference EquationsDifference Equations

2.12 Block Diagram Representations2.12 Block Diagram Representations


2

2.1 Introduction2.1 Introduction

Objectives:

1. Impulse responses of LTI systems

2. Linear constant-coefficients differential or differenceequations of LTI systems

3. Block diagram representations of LTI systems

4. State-variable descriptions for LTI systems


3

1.1. ExpressedExpressed a signal as a weighteda signal as a weightedsuperposition of shifted impulses.superposition of shifted impulses.Discrete-time signal x[n] :

0x n n x n

x n n k x k n k

2 2 1 1 0

1 1 2 2

x n x n x n x n

x n x n

Fig. 2.1Fig. 2.1

x[n] = entire signal; x[k] =specific value of the signal x[n]

at time k.

2.2 Convolution Sum2.2 Convolution Sum

k

x n x k n k

(2.1)

]2[]2[ nx

]1[]1[ nx

][]0[ nx

]1[]1[ nx

]2[]2[ nx

Shifting Property


4

k

y n H x n H x k n k

LTI systemLTI systemHH

Input x[n] Output y[n]

Output:

k

y n H x k n k

Linearity

2. Impulse Response of LTI System H2. Impulse Response of LTI System H


The system output is a weighted sum of theresponse of the system to time-shifted impulses.

For time-invariant system:

h[n] = H{[n]} impulseresponse of the LTI system H

(2.3)][]}[{ knhknH

(2.2)

k

knHkxny ]}[{][][ Homogeneity

(2.4)

Convolution process:Fig. 2.2Fig. 2.2.

k

knhkxny ][][][


5

k

x n h n x k h n k

2.2 Convolution Sum2.2 Convolution Sum3. Convolution3. Convolution

SumSum]1[]1[ nx

][]0[ nx

]1[]1[ nx

]2[]2[ nx

k

knkxnx ][][][


6

The output associated with the kth input is expressed as:

k

y n x k h n k

Example 2.1Example 2.1：Multipath Communication Channel: DirectEvaluation of the Convolution Sum

Consider the discrete-time LTI system model representing a two-pathpropagation channel described in Section 1.10. If the strength of theindirect path is a = ½, then 1

12

y n x n x n

Letting x[n] = [n], we find thatthe impulse response is

1, 01

, 120, otherwise

n

h n n


Determine the output of thissystem in response to the input

2, 04, 12, 2

0, otherwise

nn

x nn

]1[21

][ nn

][][]}[][{ knhkxknkxH

][][ nhnx

ceiling

][nx

]1[ nx


7

<Sol.><Sol.>1. Input : 2 4 1 2 2x n n n n

Input = 0 for n < 0 and n > 2

2. Since

[n k]

time-shifted impulse input

h [n k]

time-shifted impulse response output

3. Output:



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1. Convolution Sum1. Convolution Sum

2. Convolution Evaluation Procedure:(1) Reflect h[k] to obtain h[k] .

h[nk] = h[(kn)] is a reflected(because of k) and time-shifted(by n) version of h[k].

2.3 Convolution Sum Evaluation Procedure2.3 Convolution Sum Evaluation Procedure

][][][ nhnxny

k

knhkx ][][

k

knxkh ][][ ][][ nxnh Commutative Law

(2) Shift right h[k] n unit to obtain h[nk] .

(3) Multiply x[k] and h[nk] and sum

wn[k]=x[k]h[nk] for all k.

Note: you maybe need to discuss the value of n .

][kh

k0

][ kh

k0

][ knh

kn0

0n

][ knh

kn 0

0n


9

Example 2.2Example 2.2：：Convolution Sum Evaluation by usingIntermediate Signal


Consider a system with impulse response 34

n

h n u n

Find the output of the system when the input is x [n] = u [n].

<Sol.><Sol.>

Find h[nk] :

Fig. 2.3Fig. 2.3 depicts x[k] superimposed on the reflected and time-shifted impulse response h[nk].


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2.3 Convolution Sum Evaluation Procedure2.3 Convolution Sum Evaluation Procedure1. When n < 0 :

2. When n 0 :


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Example 2.3Example 2.3：： Moving-Average System: Reflect-and-shiftConvolution Sum Evaluation

The output y[n] of the four-point moving-average system is related tothe input x[n] according to the formula

3

0

14 k

y n x n k

The impulse response h[n] of this system isobtained by letting x[n] = [n], which yields

14

4h n u n u n


Fig. 2.4 (a).Fig. 2.4 (a).

Fig. 2.4 (b).Fig. 2.4 (b). 10x n u n u n

Determine the output of the system when theinput is the rectangular pulse defined as


12

<Sol.><Sol.>

2. 1’st interval : wn[k] = 01. Refer to Fig. 2.4Fig. 2.4. Five intervals !

1’st interval: n < 02’nd interval: 0 ≤n ≤33’rd interval: 3 < n ≤94th interval: 9 < n ≤125th interval: n > 12



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5. 4th interval: 9 < n ≤12

4. 3’rd interval: 3 < n ≤9

2.3 Convolution Sum Evaluation Procedure2.3 Convolution Sum Evaluation Procedure3. 2’nd interval: 0 ≤n ≤3


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6. 5th interval: n > 12



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Example 2.4Example 2.4：First-order Recursive System:The input-output relationship for the first-order recursive system is given by

1y n y n x n

Let the input be given by 4nx n b u n We use convolution to find the output of this system, assuming that b andthat the system is causal.

<Sol.><Sol.>

Since the system is causal, we have h[n] = 0for n < 0. For n = 0, 1, 2, …, we find that h[0] = 1,h[1]=, h[2]=2, …, or

1. Impulse response:

2. Graph of x[k] and h[nk] : Fig. 2.5 (a).Fig. 2.5 (a).



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3. Intervals of time shifts :


4. For n < 4 :

5. For n 4 :

6. Output y[n]:

1’st interval : n < 4; 2’nd interval : n 4

kb

kn


17

Assuming that = 0.9 and b = 0.8.

Fig. 2.5 (c).Fig. 2.5 (c).



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Example 2.5Example 2.5：： Investment ComputationThe first-order recursive system is used to describe the value of aninvestment earning compound interest at a fixed rate of r % perperiod if we set = 1 + (r/100). Let y[n] be the value of the investmentat the start of period n. If there are no deposits or withdrawals, thenthe value at time n is expressed in terms of the value at the previoustime as y[n] = y[n 1]. Now, suppose x[n] is the amount deposited(x[n] > 0) or withdrawn (x[n] < 0) at the start of period n. In this case,the value of the amount is expressed by the first-order recursiveequation

1y n y n x n

We use convolution to find the value of an investment earning 8 %per year if $1000 is deposited at the start of each year for 10 years andthen $1500 is withdrawn at the start each year for 7 years.



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<Sol.><Sol.>1. Prediction: Account balance to grow for the first 10 year, and to

decrease during next 7 years, and afterwards to continue growing.

2. By using the reflect-and-shift convolution sum evaluationprocedure, we can evaluate y[n] = x[n] h[n], where x[n] isdepicted in Fig. 2.6Fig. 2.6 and h[n] = n u[n] is as shown in Example 2.4with = 1.08.



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3. Graphs of x[k] and h[nk] : Fig. 2.7(a).Fig. 2.7(a).

4. Intervals of time shifts:

1’st interval: n < 02’nd interval: 0 ≤n ≤93’rd interval: 10 ≤n ≤164th interval: 17 ≤n

5. Mathematical representationsfor wn[k] and y[n] :

1) For n < 0 : wn[k] = 0 and y[n] =


2) For 0 ≤n ≤9 :

…


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3) For 9 < n ≤16 :


n

k

knhkxny0

][][][

Fig. 2.7 (c).Fig. 2.7 (c).

1)08.1(18750)08.1(89.7246 9 nn


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2.3 Convolution Sum Evaluation Procedure2.3 Convolution Sum Evaluation Procedure4) For 16 < n :

n

k

knhkxny0

][][][

n)08.1(17.3340Fig. 2.7 (d).Fig. 2.7 (d).


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6. Fig. 2,7(e)Fig. 2,7(e) depicts y[n], the valueof the investment at the startof each period, by combiningthe results for each of the fourintervals.


.16)08.1(17.33401691)08.1(18750)08.1(89.7246

,901)08.1(12500,00

][ 9

1

nnn

n

ny

n

nn

n


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2. Impulse response of LTI system H :

LTI systemLTI systemHH

Input x(t) Output y(t)Output:

y t H x t H x t d

3. h(t) = H{(t)} impulse response of the LTI system HIf the system is also time invariant, then

Fig. 2.9.Fig. 2.9.

2.4 Convolution Integral2.4 Convolution Integral1. Weighted Superposition of Time1. Weighted Superposition of Time--Shifted FunctionsShifted Functions

4. Convolution integral: x t h t x h t d

(2.11) A time-shifted impulse generates atime-shifted impulse response output

)()}({ thtH

(2.10) The shifting property of the impulse !

dtxtx )()()(

(2.10)Linearity property

dtHxty )}({)()(

(2.12)

dthxty )()()(


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2. Convolution Evaluation Procedure :

2.5 Convolution Integral Evaluation Procedure2.5 Convolution Integral Evaluation Procedure

(2.13)

1. Convolution Integral :1. Convolution Integral :

dthxthtxty )()()()()(

)()()()( txthdtxh

Commutative Law

1) Graph both x() and h() .2) Reflect h() about = 0 to obtain h() and then shift h() by t

unit to obtain h(t ) .3) Multiply x() and h(t ) .4) Increase the shift t (i.e., move h(t ) toward the right) until the

mathematical form for x()h(t ) changes.5) Discuss the values of t to obtain y(t) .


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2.5 Convolution Integral Evaluation Procedure2.5 Convolution Integral Evaluation ProcedureExample 2.6Example 2.6：： Reflect-and-shift Convolution EvaluationGiven 1 3x t u t u t 2h t u t u t and as depicted in Fig. 2Fig. 2--1010,Evaluate the convolution integral y(t) = x(t) h(t).

<Sol.><Sol.>1. Graph of x() and h(t ) :

Fig. 2.11 (a).Fig. 2.11 (a).

2. )()()( thtxty

dthx )()(

1

)( th

t2t

1

1 3

)(x

t2t t2t 2t


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0, 11, 1 3

5 , 3 50, 5

tt t

y tt t

t



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2.5 Convolution Integral Evaluation Procedure2.5 Convolution Integral Evaluation ProcedureExample 2.7Example 2.7：： RC Circuit OutputFor the RC circuit in Fig. 2.12Fig. 2.12, assume that the circuit’s time constantis RC = 1 sec. Ex. 1.21 shows that the impulse response of thiscircuit is h(t) = e t u(t).Use convolution to determine thecapacitor voltage, y(t), resulting froman input voltage x(t) = u(t) u(t 2).

<Sol.><Sol.>RC circuit is LTI system, so y(t) = x(t) h(t).

1. Graph of x() and h(t ) : Fig. 2.13 (a).Fig. 2.13 (a).

)(x

)( th)( te


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4. Second interval of time shifts : 0 ≤t < 2

For t > 0, Fig. 2.13 (b).Fig. 2.13 (b).

3. First interval of time shifts : t < 0 y(t ) =

5. Third interval : 2 ≤t

2. Intervals of time shifts : Three intervalsThree intervals1’st interval: t < 02’nd interval: 0 ≤t < 23’rd interval: 2 ≤t


Fig. 2.13 (c).Fig. 2.13 (c).

)(tw

)()()( thxwt


30

2

0, 01 , 0 2

1 , 2

t

t

ty t e t

e e t


Fig. 2.13 (d).Fig. 2.13 (d).


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2.5 Convolution Integral Evaluation Procedure2.5 Convolution Integral Evaluation ProcedureExample 2.8Example 2.8：： Another Reflect-and-Shift Convolution EvaluationSuppose that the input x(t) and impulse response h(t) of an LTIsystem are, respectively, given by

1 1 3x t t u t u t and 1 2 2h t u t u t

Find the output of the system.<Sol.><Sol.>

2. Intervals of time shifts : Five intervalsFive intervals1’st interval: t < 02’nd interval: 0 ≤t < 23’rd interval: 2 ≤t < 34th interval: 3 ≤t < 55th interval: t 5

3. First interval of time shifts : t < 0 y(t ) =

1. Graph of x() and h(t ) : Fig. 2.14 (a).Fig. 2.14 (a).

)(x

2

1

)(h

1

1 )( th


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5. Third interval of time shifts : 2 ≤t < 3

1, 1 3

0, otherwisetw

4. Second interval of time shifts : 0 ≤t < 2

1, 1 1

0, otherwiset

tw


6. Fourth interval of time shifts : 3 ≤t < 5

1 , 1 2

2 31,0, otherwise

t

tw t

Fig. 2.14 (b).Fig. 2.14 (b).

Fig. 2.14 (c).Fig. 2.14 (c).

Fig. 2.14 (d).Fig. 2.14 (d).

)(tw

)(tw

)(tw


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2

2

0, 0

, 0 222, 2 36 7, 3 52, 5

t

tt

y tt

t t tt

Fig. 2.14 (f).Fig. 2.14 (f).

7. Fifth interval of time shifts : t 5

1 , 1 30, otherwisetw

Fig. 2.14 (e).Fig. 2.14 (e).

)(tw


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1. Parallel Connection of LTI Systems1. Parallel Connection of LTI Systems

(2) Output :

1 2

1 2

( ) ( ) ( )

( ) ( ) ( ) ( )

y t y t y t

x t h t x t h t

1 2( ) ( ) ( ) ( ) ( )y t x h t d x h t d

1 2( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

y t x h t h t d

x h t d x t h t

where h(t) = h1(t) + h2(t)

2.6 Interconnection of LTI Systems2.6 Interconnection of LTI Systems

(1) Two LTI systems : Fig. 2.18(a).Fig. 2.18(a).

Fig. 2.18(b)Fig. 2.18(b)

(3) Distributive property :

(2.15)

(2.16)]}[][{][][][][][ nhnhnxnhnxnhnx 2121

)}()({)()()()()( ththtxthtxthtx 2121


35


(1) Two LTI systems : Fig. 2.19(a).Fig. 2.19(a).

Substituting Eq. (2.19) for z(t) into Eq. (2.18) gives

1 2( ) ( ) ( ) ( )y t x v h v h t dvd

(2.19)

Define h(t) = h1(t) h2(t), then

1 2( ) ( ) ( )h t v h h t v d

Fig. 2.19(b).Fig. 2.19(b).

2. Cascade Connection of LTI Systems2. Cascade Connection of LTI Systems

(2.17)(2) The output is expressed in terms of z(t) as

)()()( thtzty 2

(2.18) dthzty

2 )()()(

Change of variable=

Since z(t) is the output of the first system, so it can be expressed as

11 dhxhxz )()()()()(

(2.20)

21 ddthhxty )()()()(

(2.21))()()()()( thtxdthxty


36


(4) Commutative property:

Write h(t) = h1(t) h2(t) as the integral

1 2( ) ( ) ( )h t h h t d

Change of variable= t

Fig. 2.19(c).Fig. 2.19(c).

Interchanging the order of the LTI systems in the cascade withoutaffecting the result:

1 2 2 1( ) ( ) ( ) ( ) ( ) ( ) ,x t h t h t x t h t h t

(3) Associative property :

(2.22)

(2.25)

)}()({)()()}()({ 2121 ththtxththtx

]}[][{][][]}[][{ 2121 nhnhnxnhnhnx

(2.23))()()()()( 1221 ththdhthth

Commutative property forcontinuous-time case :

(2.24))()()()( 1221 thththth

Commutative property fordiscrete-time case :

(2.26)][][][][ 1221 nhnhnhnh


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Example 2.11Example 2.11：： Equivalent System to Four Interconnected SystemsConsider the interconnection of four LTI systems, as depicted in Fig.Fig.2.202.20. The impulse responses of the systems are

1[ ] [ ],h n u n 2[ ] [ 2] [ ],h n u n u n 3[ ] [ 2],h n n 4[ ] [ ].nh n u nand

Find the impulse response h[n] of the overall system.

<Sol.><Sol.>1. Parallel combination of h1[n] and

h2[n] : h12[n] = h1[n] + h2[n]

Fig. 2.21 (a).Fig. 2.21 (a).



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2. h12[n] is in series with h3[n] : h123[n] = h12[n] h3[n]

h123[n] = (h1[n] + h2[n]) h3[n] Fig. 2.21 (b).

3. h123[n] is in parallel with h4[n] : h[n] = h123[n] h4[n]

1 2 3 4[ ] ( [ ] [ ]) [ ] [ ],h n h n h n h n h n Fig. 2.21 (c).

Thus, substitute the specified forms of h1[n] and h2[n] to obtain

12[ ] [ ] [ 2] [ ]

[ 2]

h n u n u n u n

u n

[ ] 1 [ ].nh n u n


123[ ] [ 2] [ 2]

[ ]

h n u n n

u n

Convolving h12[n] with h3[n] gives


39(2.27)

Table 2.1Table 2.1：：summarizes the interconnection properties presented insummarizes the interconnection properties presented inthis section.this section.



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(1) The output of a discrete-time LTI system :

[ ] [ ] [ ] [ ] [ ]k

y n h n x n h k x n k

2.7 Relations Between LTI2.7 Relations Between LTISystem Properties and the Impulse ResponseSystem Properties and the Impulse Response1.1. MemorylessMemoryless LTI SystemsLTI Systems

(2) To be memoryless, y[n] must depend only on x[n] andtherefore cannot depend on x[nk] for k 0.A discrete-time LTI system is memoryless if and only if

[ ] [ ]h k c k c is an arbitrary constant

Continuous-time system :(1) Output:

( ) ( ) ( ) ,y t h x t d

(2) A continuous-time LTI system ismemoryless if and only if

( ) ( )h c c is an arbitrary constant

(2.27)

]2[]2[]1[]1[][]0[]1[]1[]2[]2[][ nxhnxhnxhnxhnxhny


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Discrete-time system :

The output of a causal LTI system depends only on past or presentvalues of the input.

1. Convolution sum : [ ] [ 2] [ 2] [ 1] [ 1] [0] [ ][1] [ 1] [2] [ 2] .

y n h x n h x n h x nh x n h x n

2.7 Relations Between LTI2.7 Relations Between LTISystem Properties and the Impulse ResponseSystem Properties and the Impulse Response2. Causal LTI Systems2. Causal LTI Systems

2. For a discrete-time causalcausal LTIsystem, [ ] 0 for 0h k k

3. Convolution sum in new form:

0

[ ] [ ] [ ].k

y n h k x n k

Continuous-time system:1. Convolution integral:

( ) ( ) ( ) .y t h x t d

2. For a continuous-time causalcausal LTI system,( ) 0 for 0h

3. Convolution integral innew form:

0( ) ( ) ( ) .y t h x t d


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A system is BIBO stable if the output is guaranteed to be bounded forevery bounded input.Discrete-time case : Input [ ] xx n M [ ] yy n M Output :

2.7 Relations Between LTI2.7 Relations Between LTISystem Properties and the Impulse ResponseSystem Properties and the Impulse Response3. Stable LTI Systems3. Stable LTI Systems

(1) The magnitude of output :

[ ] [ ] [ ] [ ] [ ]k

y n h n x n h k x n k

[ ] [ ] [ ]k

y n h k x n k

a b a b [ ] [ ] [ ]k

y n h k x n k

ab a b

(2) Assume that the input is bounded, i.e.,

[ ] xx n M [ ] xx n k M

Hence, the output is bounded,or y[n] ≤for all n, providedthat the impulse response ofthe system is absolutelysummable.

and it follows that(2.28)

k

x khMny ][][


43

Continuous-time case :Condition for impulse response of a stable continuous-time LTIsystem :

0( ) .h d

Example 2.12Example 2.12：： Properties of the First-Order Recursive SystemThe first-order system is described by the difference equation

[ ] [ 1] [ ]y n y n x n

and has the impulse response[ ] [ ]nh n u n

Is this system causal, memoryless, and BIBO stable?

[ ] .k

h k

(3) Condition for impulse response of a stable discrete-time LTI system :

2.7 Relations Between LTI2.7 Relations Between LTISystem Properties and the Impulse ResponseSystem Properties and the Impulse Response


44

<Sol.><Sol.>1. The system is causal, since h[n] = 0 for n < 0.2. The system is not memoryless, since h[n] 0 for n > 0.


3. Stability : Checking whether the impulse response is absolutelysummable?

0 0

[ ] kk

k k k

h k

if and only if < 1

◆ Special case : A system can be unstable even though the impulseresponse has a finite value.

Input : x() = (), then theoutput is y(t) = h(t) = u(t).

h(t) is not absolutely integrableIdeal integrator is not stable!

2. Ideal accumulator :

[ ] [ ]n

k

y n x k

Impulse response : h[n] = u[n]

h[n] is not absolutely summableIdeal accumulator is not stable!

1. Ideal integrator :

(2.29)

tdxty )()(


45

A system is invertible if the input to the system can be recovered fromthe output except for a constant scale factor.(1) h(t) = impulse response of LTI system,(2) hinv(t) = impulse response of LTI inverse system

Fig. 2.24.Fig. 2.24.

(3) The process of recovering x(t) from h(t) x(t) is termeddeconvolution.

(4) An inverse system performs deconvolution.in( ) ( ( ) ( )) ( ).vx t h t h t x t

( ) ( ) ( )invh t h t t (2.30)

Continuous-timecase

2.7 Relations Between LTI2.7 Relations Between LTISystem Properties and the Impulse ResponseSystem Properties and the Impulse Response4. Invertible Systems and4. Invertible Systems and DeconvolutionDeconvolution

(5) Discrete-time case: (2.31)][][][ nnhnh inv


46

Example 2.13Example 2.13：： Multipath Communication Channels: Compensationby means of an Inverse System

Consider designing a discrete-time inverse system to eliminate thedistortion associated with multipath propagation in a data transmissionproblem. Assume that a discrete-time model for a two-pathcommunication channel is [ ] [ ] [ 1].y n x n ax n Find a causal inverse system that recovers x[n] from y[n]. Checkwhether this inverse system is stable.<Sol.><Sol.>1. Impulse response:

2. The inverse system hinv[n] must satisfy h[n] hinv[n] = [n].

1) For n < 0, we must have hinv[n] = 0 in order to obtain a causalinverse system



47

3) For n > 0, [n] = 0, and eq. (2.32) implies that

3. Since hinv[0] = 1, Eq. (2.33) implies that hinv[1] = a, hinv[2] = a2,hinv[3] = a3, and so on. The inverse system has the impulseresponse

4. To check for stability, we determine whether hinv[n] is absolutelysummable, which will be the case if

[ ] kinv

k k

h k a

is finite.

ForFor aa< 1< 1, the system is stable., the system is stable.


2) For n = 0, [n] = 1, and eq. (2.32) implies that


48

★ Table 2.2 summarizes the relation between LTI system propertiesand impulse response characteristics.



49

1.The step response is defined as the output dueto a unit step input signal.

2. Discrete-time LTI system:Let h[n] = impulse response and s[n] = step response.

[ ] [ ]* [ ] [ ] [ ].k

s n h n u n h k u n k

2.8 Step Response2.8 Step Response

3. Since u[nk] = 0 for k > n and u[nk] = 1for k ≤n, we have

[ ] [ ].n

k

s n h k

The step response is the running sum of the impulseresponse.


50

The step response s(t) is the running integral of the impulseresponse h(t).

◆ Express the impulse response in terms of the stepresponse as

[ ] [ ] [ 1]h n s n s n ( ) ( )d

h t s tdt

and

Example 2.14Example 2.14：： RC Circuit: Step Response

The impulse response of the RC circuit depicted in Fig. 2.12Fig. 2.12 is

1( ) ( )

tRCh t e u t

RC

Find the step response of the circuit.

2.8 Step Response2.8 Step Response◆ Continuous-time LTI system :

(2.34)

tdhdtuhtuthts )()()()()()(


51

1. Step response:1

( ) ( ) .t

RCs t e u dRC

Fig. 2.25Fig. 2.25

<Sol.><Sol.>2.8 Step Response2.8 Step Response


52

The order of the differential or difference equation is (N, M),representing the number of energy storage devices in the system.

Ex. RLC circuit depicted in Fig. 2.26Fig. 2.26.

2. KVL Eq. :

1. Input = voltage source x(t),output = loop current

1 tdRy t L y t y d x t

dt C

Often, N M, and theorder is described

using only N.

2.9 Differential and Difference2.9 Differential and DifferenceEquation Representations ofEquation Representations of LTILTI SystemsSystems

1. Linear constant-coefficient differential equation :

(2.35)Input = x(t), output = y(t)

M

kk

k

k

N

kk

k

k txdtd

btydtd

a00

)()(

2. Linear constant-coefficient difference equation :

(2.36)Input = x[n], output = y[n]

M

kk

N

kk knxbknya

00

][][


53

2

2

1 d d dy t R y t L y t x t

C dt dt dt

N = 2

Ex. Accelerator modeled in Section 1.10 :

2

22

nn

d dy t y t y t x t

Q dt dt

where y(t) = the position of the proof mass, x(t) = externalacceleration.

N = 2


Difference equations are easily rearranged to obtain recursiveformulas for computing the current output of the system from theinput signal and the past outputs.

Ex. Second-order difference equation :(2.37) N = 2]1[2][]2[

41

]1[][ nxnxnynyny


54

Ex. Eq. (2.36) can be rewritten as

0 10 0

1 1M N

k kk k

y n b x n k a y n ka a

Ex. Consider computing y[n] for n 0 from x[n] for thesecond-order difference equation (2.37).

<Sol.><Sol.>

2. Computing y[n] for n 0:


12 2 2 1 1 0

4y x x y y

13 3 2 2 2 1

4y x x y y

1. Eq. (2.37) can be rewritten as

(2.38)]2[41

]1[]1[2][][ nynynxnxny

(2.39)]2[41

]1[]1[2]0[]0[ yyxxy

(2.40)]1[41

]0[]0[2]1[]1[ yyxxy


55

◆ Initial conditions : y[1] and y[2].

◆ The initial conditions for Nth-order difference equation are the Nvalues

, 1 ,..., 1 ,y N y N y

◆ The initial conditions for Nth-order differential equation are the Nvalues

2 1

0 , 2 10 00

, , ...,N

t Nt tt

d d dy t y t y t y t

dt dt dt



56

Example 2.15Example 2.15：： Recursive Evaluation of a Difference EquationFind the first two output values y[0] and y[1] for the system describedby Eq. (2.38), assuming that the input is x[n] = (1/2)nu[n] and the initialconditions are y[1] = 1 and y[2] = 2.<Sol.><Sol.>1. Substitute the appropriate values into Eq. (2.39) to obtain

2. Substitute for y[0] in Eq. (2.40) to find



57

Example 2.16Example 2.16：：Evaluation of a Difference Equation by means of aComputer

A system is described y the difference equation

1.143 1 0.4128 2 0.0675 0.1349 1 0.675 2y n y n y n x n x n x n

Write a recursive formula that computes the present output from thepast outputs and the current inputs.

Use a computer to determine the step response of the system, thesystem output when the input is zero and the initial conditions are y[1]= 1 and y[2] = 2, and the output in response to the sinusoidal inputsx1[n] = cos(n/10), x2[n] = cos(n/5), and x3[n] = cos(7n/10),assuming zero initial conditions.

Last, find the output of the system if the input is the weekly closingprice of Intel stock depicted in Fig. 2.27Fig. 2.27, assuming zero initialconditions.



58

1. Recursive formula for y[n] :

1.143 1 0.4128 2 0.0675 0.1349 1 0.675 2y n y n y n x n x n x n

<Sol.><Sol.>



59


2. Stepresponse:Fig. 2.28 (a).Fig. 2.28 (a).

3. Zero inputresponse:Fig. 2.28 (b).Fig. 2.28 (b).


60


4. The outputs due to the sinusoidal inputs x1[n], x2[n], and x3[n] :Fig.Fig. 2.28 (c), (d), and (e).2.28 (c), (d), and (e).


61


5. Fig. 2.28(f)Fig. 2.28(f) shows the systemoutput for the Intel stock priceunit. A comparison of peaks inFigs. 2.27Figs. 2.27 and 2.28 (f)2.28 (f)


62

Complete solution: y = y (h) + y (p), y (h) = homogeneous solution,y (p) = particular solution

1. The Homogeneous Solution1. The Homogeneous SolutionContinuous-time case:

1. Homogeneous differential equation : 0

0kN

hk k

k

da y t

dt

Discrete-time case:1. Homogeneous differential equation :

0

0N

hk

k

a y n k

2.10 Solving2.10 SolvingDifferential and Difference EquationsDifferential and Difference Equations

2. Homogeneous solution :

(2.41)Coefficients ci is

determined by I.C.

N

i

tri

h iecty0

)( )(

3. Characteristic eq. : (2.42)

N

k

kkra

0

0


(2.43)

N

i

nii

h rcny1

)( ][ 3. Characteristic eq. : (2.44)

N

k

kNkra

0

0


63

If a root rj is repeated p timesin characteristic eqs., thecorresponding solutions are

1, , ...,j j jr r rt t p te te t e

1, , ...,n n p nj j jr nr n r

Continuous-time case :

Discrete-time case:

Example 2.17Example 2.17：： RC Circuit: Homogeneous SolutionThe RC circuit depicted in Fig. 2.30Fig. 2.30 is described by the differentialequation

dy t RC y t x t

dt

Determine the homogeneous solution of this equation.<Sol.><Sol.>1. Homogeneous Eq. :

2. Homo. Sol. :

3. Characteristic eq. :

4. Homogeneous solution:



64

Example 2.18Example 2.18：： First-Order Recursive System: Homogeneous Solution

Find the homogeneous solution for the first-order recursive systemdescribed by the difference equation 1y n y n x n

<Sol.><Sol.>1. Homogeneous Eq. :

2. Homo. Sol. :

3. Characteristic eq. :




65

2. The Particular Solution2. The Particular SolutionA particular solution is usually obtained by assuming an output ofthe same general form as the input. Table 2.3Table 2.3



66

Example 2.19Example 2.19：： First-Order Recursive System (Continued):Particular Solution

Find a particular solution for the first-order recursive system describedby the difference equation 1y n y n x n

if the input is x[n] = (1/2)n.<Sol.><Sol.>

1. Particular solution form :


2. Substituting y(p)[n] and x[n] into the given difference Eq. :

3. Particular solution :

(2.45)

Both sides of above eq. aremultiplied by (1/2)n

1)21( pc


67


1. Differential equation :

2. Particular solution form :

Example 2.20Example 2.20：： RC Circuit (continued): Particular SolutionConsider the RC circuit of Example 2.17Example 2.17 and depicted in Fig. 2.30Fig. 2.30. Finda particular solution for this system with an input x(t) = cos(0t).<Sol.><Sol.>

3. Substituting y(p)(t) and x(t) = cos(0t) into the given differential Eq. :

1 2 0 1 0 0 2 0 0cos( ) sin( ) sin( ) cos( ) cos( )c t c t RC c t RC c t t

4. Coefficients c1 and c2 :

5. Particular solution :


68

3. The Complete Solution3. The Complete SolutionComplete solution: y = y(h) + y(p)

y(h) = homogeneous solution, y(p) = particular solutionThe procedure for finding complete solution of differential ordifference equations is summarized as follows :


Procedure 2.3:Procedure 2.3: Solving a Differential or Difference equation1. Find the form of the homogeneous solution y(h) from the roots of

the characteristic equation.2. Find a particular solution y(p) by assuming that it is of the same

form as the input, yet is independent of all terms in thehomogeneous solution.

3. Determine the coefficients in the homogeneous solution so thatthe complete solution y = y(h) + y(p) satisfies the initial conditions.★ Note that the initial translation is needed in some cases.


69


Example 2.21Example 2.21：： First-Order Recursive System : Complete SolutionFind the complete solution for the first-order recursive systemdescribed by the difference equation

(2.46)][]1[41

][ nxnyny

if the input is x[n] = (1/2)n u[n] and the initial condition is y[1] = 8.<Sol.><Sol.>1. Homogeneous sol. : 2. Particular solution :

3. Complete solution : (2.47)

4. Coefficient c1 determined by I.C. :

I.C. : 0 0 1 4 1y x y We substitute y[0] = 3into Eq. (2.47), yielding

5. Final solution :for n 0


70

Example 2.22Example 2.22：： RC Circuit (continued): Complete Response

Find the complete response of the RC circuit depicted in Fig. 2.30Fig. 2.30 toan input x(t) = cos(t)u(t) V, assuming normalized values R = 1 and C = 1 F and assuming that the initial voltage across thecapacitor is y(0) = 2 V.<Sol.><Sol.>


1. Homogeneous sol. :

3. Complete solution :

2. Particular solution:

2 2

1cos sin V

1 1p RC

y t t tRC RC

0 = 1

R = 1 , C = 1 F

4. Coefficient c1 determined by I.C. : yy(0(0) =) = yy(0(0++))

0 1 1 12 cos0 sin 0

2 2 2ce c

c = 3/2

5. Final solution:


71

Complete solution: y = y(n) + y(f)

y(n) = natural response, y(f) = forced response

1. The Natural Response (1. The Natural Response (Zero InputZero Input))Example 2.24Example 2.24：： RC Circuit (continued): Natural Response

Find the natural response of the this system, assuming thaty(0) = 2 V, R = 1 and C = 1 F.

The system In Example 2.17 is described by the differential equation

dy t RC y t x t

dt

<Sol.><Sol.> 1. Homogeneous sol. :2. I.C.: y(0) = 2 V

3. Natural Response :

2.11 Characteristics of Systems Described2.11 Characteristics of Systems Describedby Differential and Difference Equationsby Differential and Difference Equations


72

Example 2.25Example 2.25：： First-Order Recursive System : Natural ResponseThe system in Example 2.21Example 2.21 is described by the difference equation

11

4y n y n x n

Find the natural response of this system.

<Sol.><Sol.>1. Homogeneous sol. :2. I.C.: y[1] = 8

3. Natural Response :


and the initial condition is y[1] = 8.


73

2. The Forced Response2. The Forced Response ((only valid for t 0 or n 0))The forced response is the system output due to the input signalassuming zero initial conditions.


The at-rest conditions for a discrete-time system, y[N] = 0, …,y[1] = 0, must be translated forward to times n = 0, 1, …, N1before solving for the undetermined coefficients, such as whenone is determining the complete solution.

Example 2.26Example 2.26：： First-Order Recursive System : Forced ResponseThe system in Example 2.21Example 2.21 is described by the difference equation

11

4y n y n x n

Find the forced response of this system if the input isx[n] = (1/2)n u[n].


74

<Sol.><Sol.>

2. I.C.: Translate the at-rest condition y[1] to time n = 0

1. Complete solution :

3. Finding c1 :


4. Forced response:


75

Example 2.27Example 2.27：： RC Circuit (continued): Forced Response

Find the forced response of the this system, assuming that x(t) =cos(t)u(t) V, R = 1 and C = 1 F.

The system In Example 2.17 is described by the differential equation

dy t RC y t x t

dt

<Sol.><Sol.>1. Complete solution : 1 1

cos sin V2 2

ty t ce t t From Example 2.22

2. I.C. : y(0) = y(0+) = 0 c = 1/2

3. Forced response:



76

1. Continuous-time case :

( ) ( )d

h t s tdt

2. Discrete-time case : [ ] [ 1]h n s n s n

4. Linearity and Time Invariance4. Linearity and Time Invariance

y1(f) + y2

(f)x1 + x2

y2(f)x2

y1(f)x1

Forcedresponse

Input

Forced response Linearity

y1(n) + y2

(n)I1 + I2

y2(n)I2

y1(n)I1

Naturalresponse

Initial Cond.

Natural response Linearity

The complete response of an LTI system is notnot time invariant.Response due to initial condition will not shift with a time shiftof the input.


Relation between step response and impulse response3. The Impulse Response3. The Impulse Response


77

Roots of characteristic equationForced response, natural response, stability, and response time.

★★ BIBO Stable:BIBO Stable:

1. Discrete-time case : boundednir 1, for allir i

2. Continuous-time case : boundedir te

0ie r and1ir 0ie r

The system is said to be on the verge of instability.

5. Roots of the Characteristic Equation5. Roots of the Characteristic Equation



78

A block diagram is an interconnection of the elementaryoperations that act on the input signal.

Three elementary operations for block diagram :

1. Scalar multiplication: y(t) = cx(t) or y[n] = cx[n], where c is ascalar.

2. Addition: y(t) = x(t) + w(t) or y[n] = x[n] + w[n].3. Integration for continuous-time LTI system : ;( ) ( )

ty t x d

and a time shift for discrete-time LTI system: y[n] = x[n1].


Fig. 2.32.Fig. 2.32.

(a)

(b)

( ) ( )t

y t x d

(c)


79

Direct Form I :Direct Form I :

2.12 Block Diagram Representations2.12 Block Diagram RepresentationsEx. A discrete-time LTI system: Fig. 2.33.Fig. 2.33.

Cascade Form(Direct Form I)

1. In dashed box :(2.49)]2[]1[][][ 210 nxbnxbnxbnw

2. y[n] in terms of w[n] : (2.50)]2[]1[][][ 21 nyanyanwny


80


Direct Form II :Direct Form II :1. Interchange the order of Direct Form I.

Fig. 2.35.Fig. 2.35.

3. System output y[n] in terms of input x[n] :]2[]1[][]2[]1[][ 21021 nxbnxbnxbnyanyany

(2.51)]2[]1[][]2[]1[][ 21021 nxbnxbnxbnyanyany

2. Denote the output of the new first system as f[n].(2.52) Input: x[n]][]2[]1[][ 21 nxnfanfanf

3. The signal is also the input tothe second system. The outputof the second system is

(2.53)]2[]1[][][ 210 nfbnfbnfbny


81

Block diagram representation for continuous-time LTIsystem:

2. Let v(0)(t) = v(t) be an arbitrary signal, and set 1 , 1, 2, 3, ...

tn nt d n

v(n)(t) is the n-fold integral of v(t) with respect to time

1 , 0 and 1, 2, 3, ...n ndt t t n

dt

3. Integrator with initial condition : 1

00 , 1, 2, 3, ...

tn n nt d n


1. Differential Eq. :

(2.54))()(00

txdtd

btydtd

aM

kk

k

k

N

kk

k

k

(2.55)

M

k

kNk

N

k

kNk txbtya

0

)(

0

)( )()(


82

(a)


(b)

Ex. Second-order system:

(2.56)Block diagram:Fig. 2.37

)()()()()()( )2(0

)1(12

)2(0

)1(1 txbtxbtxbtyatyaty


83

End ofEnd of Chapter 2Chapter 2

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