ch02 time-domain representations of lti systems compatibility mode

TimeTime--Domain Representations of LTI SystemsDomain Representations of LTI SystemsCHAPTER

22..1 1 IntroductionIntroductionObjectives:1. Impulse responses of LTI systems2. Linear constant-coefficients differential or difference equations of LTI

systems3. Block diagram representations of LTI systems4. State-variable descriptions for LTI systems

22..2 2 Convolution SumConvolution Sum11 A bit i l i d i ht d iti f hift dA bit i l i d i ht d iti f hift d11. An arbitrary signal is expressed as a weighted superposition of shifted. An arbitrary signal is expressed as a weighted superposition of shifted

impulses.impulses.Discrete-time signal x[n]: Fig. Fig. 22..11

[ ] [ ] [ ] [ ]0x n n x n =[ ] [ ] [ ] [ ]x n n k x k n k =

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

at time k.

[ ] [ ] [ ] [ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]

2 2 1 1 0

1 1 2 2

x n x n x n x n = + + + + +

+ + +L

Signals and Systems_Simon Haykin & Barry Van Veen

1

[ ] [ ] [ ] [ ]1 1 2 2x n x n + + +L


Figure Figure 22..1 1 (p. (p. 9999))Graphical example illustrating the representation of a signal x[n] as a weighted sum of time-shifted impulses.

[ ] [ ] [ ]k

x n x k n k=

= (2.1)2 Impulse response of LTI system H:2. Impulse response of LTI system H:

LTI systemLTI systemHH

Input x[n] Output y[n]

[ ] [ ]{ } [ ] [ ]y n H x n H x k n k = = HH

Output:

[ ] [ ]{ } [ ] [ ]k

y=

[ ] [ ] [ ]{ }y n H x k n k= Linearity{ }

k=

y[n] x[k]H{ [n k]}= (2.2) Linearity


2

k=


The system output is a weighted sum of the response of the system to time-shifted impulses.For time-invariant system: h[n] = H{ [n]} impulse response = H{ [n k]} h[n k] [ ] { [ ]} p pof the LTI system H(2.3)

y[n] x[k]h[n k]

= (2 4) Convolution process:k

y[n] x[k]h[n k]= (2.4)

3. Convolution sum:

Convolution process: Fig. Fig. 22..22.

[ ] [ ] [ ] [ ]k

x n h n x k h n k

= =

Figure Figure 22..22a (p. a (p. 100100)) Illustration of the convolution sum. (a) LTI system with


3impulse response h[n] and input x[n].


Figure 2.2b (p. 101)(b) The d (b) The decomposition of the input x[n] into a weighted sum of

d

a weighted sum of time-shifted impulses results in an output y[n] p y[ ]given by a weighted sum of time-shifted impulse responses.


4


The output associated with the kth input is expressed as:{ } = H x[k] [n k] x[k]h[n k]

[ ] [ ] [ ]k

y n x k h n k

==

ExampleExample 22 11 Multipath Communication Channel: Direct Evaluation of theExample Example 22..11 Multipath Communication Channel: Direct Evaluation of the Convolution Sum

Consider the discrete-time LTI system model representing a two-path propagation channel described in Section 1 10 If the strength of the indirectpropagation channel described in Section 1.10. If the strength of the indirect path is a = , then [ ] [ ] [ ]1 1

2y n x n x n= +

Letting x[n] = [n], we find that the impulse response is1, 0n =[ ] 1 , 120, otherwise

h n n= =


5

0, otherwise


Determine the output of this system in response to the input2, 04 1

nn= =[ ] 4, 1

2, 20, otherwise

nx n

n= = Input = 0 for n < 0 and n > 0

1. Input: [ ] [ ] [ ] [ ]2 4 1 2 2x n n n n = + 2 Since time shifted imp lse inp t ti hift d i l t t2. Since

[n k]time-shifted impulse input

h [n k]time-shifted impulse response output

3 Output: 0 0n


22..3 3 Convolution Sum Evaluation ProcedureConvolution Sum Evaluation Procedure1. Convolution sum:

k = independent variable[ ] [ ] [ ]k

y n x k h n k=

= 2. Define the intermediate signal: n[k] x[k]h[n k] = (2.5)

k = independent variable

g n[ ] [ ] [ ] ( )n is treated as a constant by writing n as a subscript on w.

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

3. Since

ny[n] [k]

= (2.6) The time shift n determines the time at which we evaluate the nk

y[ ] [ ]= ( )

output of the system.Example Example 22..22 Convolution Sum Evaluation by using Intermediate Signal

3 nConsider a system with impulse response [ ] [ ]34

n

h n u n= Use Eq. (2.6) to determine the output of the system at time n = 5, n = 5, and n =


710 when the input is x [n] = u [n].


Fig. Fig. 22..33 depicts x[k] superimposed on the reflected and time-shifted impulse response h[n k].

1. h [n k]: 65 55

41k [ ]

3 ,40 h i

n k

k nh n k

=

[ ] 5 550

13 4 3 35 3.288

44 3 4 13

k

ky

=

= = =

0, otherwise

[ ]5 0w k =2. Intermediate signal wn[k]:

For n = 5:

For n = 10:

[ ]103 0 10

k

k [ ]5 0w kFor n = 5:

Eq. (2.6) y[ 5] = 0For n = 5:

[ ]10 , 0 1040, otherwise

kw k = E (2 6)

[ ]5

5

3 , 0 54

h i

k

kw k

=

Eq. (2.6)

[ ]11

10 10 1010 1041

3 3 4 3 310k k

0, otherwiseEq. (2.6) [ ] 55 35

4

k

y=

[ ]0 0

31044 4 3 4 13

3 831

k k

y= =

= = =


8

q ( ) [ ]0 4k

y= 3.831=


Figure 2.3 (p. 103) Evaluation of Eq. (2.6) in Example 2.2. (a) The input signal x[k] above the reflected and time-shifted impulse response h[n k], depicted as a function of k. (b) The product signal w5[k] used to evaluate y [5]. (c) The product signal w5[k]


9used to evaluate y[5]. (d) The product signal w10[k] used to evaluate y[10].


k[ ]

3 , 040 otherwise

n k

nk nw k

= 0, otherwiseProcedure Procedure 22..11:: Reflect and Shift Convolution Sum EvaluationReflect and Shift Convolution Sum Evaluation1. Graph both x[k] and h[n k] as a function of the independent variable k. Top [ ] [ ] p

determine h[n k] , first reflect h[k] about k = 0 to obtain h[ k]. Then shift by n.

2. Begin with n large and negative. That is, shift h[ k] to the far left on the timeaxis.

3. Write the mathematical representation for the intermediate signal wn[k].4. Increase the shift n (i.e., move h[n k] toward the right) until the mathematical

f frepresentation for wn[k] changes. The value of n at which the change occursdefines the end of the current interval and the beginning of a new interval.

5. Let n be in the new interval. Repeat step 3 and 4 until all intervals of timeshift d th di th ti l t ti f [k]shifts and the corresponding mathematical representations for wn[k] are

identified. This usually implies increasing n to a very large positive number.6. For each interval of time shifts, sum all the values of the corresponding wn[k]

to obtain y[n] on that intervalSignals and Systems_Simon

Haykin & Barry Van Veen10

to obtain y[n] on that interval.


Example Example 22..33 Moving-Average System: Reflect-and-shift Convolution SumEvaluation

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

[ ] [ ]30

14 k

y n x n k=

= 0k

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

1[ ] [ ] [ ]( )1 44

h n u n u n= Fig. Fig. 22..4 4 (a).(a).Determine the output of the system when the input is the rectangular pulse defined asdefined as

[ ] [ ] [ ]10x n u n u n= Fig. Fig. 22..4 4 (b).(b). 1. Refer to Fig. Fig. 22..44. Five intervals !

1st interval: n < 02nd interval: 0 n 3 3rd interval: 3 < n 9gg

2. 1st interval: wn[k] = 0

[ ]0 1/ 4, 0kw k == 4th interval: 9 < n 125th interval: n > 12

F 03. 2nd interval:

FiFi 22 44 ( )( )Signals and Systems_Simon


[ ]0 0, otherwiseFor n = 0: Fig. Fig. 22..4 4 (c).(c).


Figure 2.4 (p. 106)Evaluation of the convolution sum for Example 2 3sum for Example 2.3. (a) The system impulse response h[n]. (b) The input signal x[n].(b) The input signal x[n]. (c) The input above the reflected and time-shifted impulse response h[n k], p p [ ],depicted as a function of k. (d) The product signal wn[k] for the interval of shifts 0 n 3. (e) The product signal wn[k] for the interval of shifts 3 < n 9. (f) The product signal wn[k] for the interval of shifts 9 < n 12. (g) The output y[n].


12


6 5th interval: n > 12 w [k] = 0For n = 1:

[ ]1 1/ 4, 0,10, otherwisek

w k==

6. 5th interval: n > 12 wn[k] = 07. Output:

The output of the system on each interval n is obtained by summing the values of the,

For general case: n 0:[ ] 1/ 4, 0 k nw k =

obtained by summing the values of the corresponding wn[k] according to Eq. (2.6).

( )1N c c N M= +[ ] 0, otherwisenw k = 4 3rd interval: 3 < n 9

Fig. Fig. 22..4 4 (d).(d).

( )1k M

c c N M=

+1) For n < 0 and n > 12: y[n] = 0.2) For 0 n 3:4. 3 rd interval: 3 < n 9

[ ] 1/ 4, 30, otherwisen

n k nw k

=

2) For 0 n 3:

[ ]0

11/ 44

n

k

ny n=

+= =FigFig 22 44 (g)(g)

5. 4th interval: 9 < n 12Fig. Fig. 22..4 4 (e).(e).

,3) For 3 < n 9:

[ ] ( )( )3

11/ 4 3 1 14

n

k ny n n n

== = + =

Fig. Fig. 22..4 4 (g)(g)

[ ] 1/ 4, 3 90, otherwisen

n kw k

= FiFi 22 44 (f)(f)

3k n= 4) For 9 < n 12:

[ ] ( )( )9 1 131/ 4 9 3 14 4

ny n n = = + =Signals and Systems_Simon


Fig. Fig. 22..4 4 (f).(f). [ ] ( )( )3 4 4k ny =


Example Example 22..44 First-order Recursive System: Reflect-and-shift Convolution SumEvaluation

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 andWe use convolution to find the output of this system, assuming that b and that the system is causal.

1 Impulse response: [ ] [ ] [ ]1h n h n n = + (2 7)1. Impulse response: [ ] [ ] [ ]1h n h n n +

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

(2.7)

[ ] [ ]nh n u n=2. Graph of x[k] and h[n k]: Fig. Fig. 22..5 5 (a).(a).

[ ] , 40, otherwise

kb kx k

= [ ],

0, otherwise

n k k nh n k

= and

3 I t l f ti hift 1 t i t l < 4 2 d i t l 4Signals and Systems_Simon


3. Intervals of time shifts: 1st interval: n < 4; 2nd interval: n 4


Figure 2.5a&b (p. 109) Evaluation of the convolution sum for Example 2.4. (a) The input signal x[k] depicted above the reflected and time-shifted impulse response h[ k] (b) Th d t i l [k] f 4


15

h[n k]. (b) The product signal wn[k] for 4 n.


4. For n < 4: wn[k] = 0.5. For n 4:

[ ] , 4k n kb k nk

Next, we apply the formula for summing a geometric series of n + 5 terms to obtain

5

1n

b+[ ] ,

0, otherwisenw k

= Fig. Fig. 22..5 5 (b).(b).

[ ] 4 5 541

1

n nn

bby n bbb b

+ +

= = 6. Output:1) For n < 4: y[n] = 0.2) For n 4:

Combining the solutions for each interval of time shifts gives the system output:)

[ ]4

nk n k

ky n b

==

k

[ ] 5 540, 4

, 4n n

ny n bb n

b + +

< = [ ]4

knn

k

by n ==

Let m = k + 4 then

,b

Fig. Fig. 22..5 5 (c).(c).

Assuming that = 0.9 and b = 0.8.

Let m = k + 4, then

[ ]4 44 4

0 0

m mn nn n

m m

b by nb

+ +

= =

= = Signals and Systems_Simon



Figure 2.5c (p. 110)(c) The output y[n] assuming that p = 0 9 and b = 0 8


17

(c) The output y[n] assuming that p 0.9 and b 0.8.


Example Example 22..55 Investment ComputationThe first-order recursive system is used to describe the value of an investment earning compound interest at a fixed rate of r % per period if we set = 1 + ( /100) L t [ ] b th l f th i t t t th t t f i d If th(r/100). Let y[n] be the value of the investment at the start of period n. If there are no deposits or withdrawals, then the value at time n is expressed in terms of the value at the previous time as y[n] = y[n 1]. Now, suppose x[n] is the amo nt deposited ( [n] > 0) or ithdra n ( [n] < 0) at the start of period n Inamount 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 recursive equation [ ] [ ] [ ]1y n y n x n= +[ ] [ ] [ ]y yWe 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 and then $1500 is withdrawn at the start each year for 7 years.withdrawn at the start each year for 7 years.

1. Prediction: Account balance to grow for the first 10 year, and to decrease

during next 7 years and afterwards to continue growingduring next 7 years, and afterwards to continue growing.2. By using the reflect-and-shift convolution sum evaluation procedure, we can

evaluate y[n] = x[n] h[n], where x[n] is depicted in Fig. Fig. 22..66 and h[n] = n u[n]is as shown in Example 2 4 with = 1 08


18

is as shown in Example 2.4 with = 1.08.


Figure 2.6 (p. 111)Cash flow into an investment. Deposits of $1000 are made at the start of

h f th fi t 10 hil ithd l f $1500 d t th t teach of the first 10 years, while withdrawals of $1500 are made at the start of each of the second 10 years.

3. Graphs of x[k] and h[n k]: Fig. Fig. 22..77(a).(a).4. Intervals of time shifts: 1st interval: n < 0

2nd interval: 0 n 9 3rd interval: 10 n 163 rd interval: 10 n 164th interval: 17 n

5. Mathematical representations for wn[k] and y[n]:1) For n < 0: w [k] = 0 and y[n] = 0


19

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

TimeTime--Domain Representations of LTI SystemsDomain Representations of LTI SystemsCHAPTERFigure 2.7a-d (p. 111)Evaluation of gu e a d (p ) a uat o othe convolution sum for Example 2.5. (a) The input signal x[k] depicted above the reflected and time-shifted impulse response p ph(n k). (b The product signal wn[k] for 0 n 9. (c) The product signal wn[k] for 10 n 16. (d) The product signal wn[k] for 17 n.


20


2) For 0 n 9:

[ ] ( )1000 1.08 , 00 otherwise

n k

nk nw k

= 0, otherwiseFig. Fig. 22..7 7 (b).(b).

[ ] ( ) ( ) 1 kn nn k n Apply the formula for summing a geometric[ ] ( ) ( )

0 0

11000 1.08 1000 1.081.08

n k n

k k

y n= =

= = 111

n+

summing a geometric series

[ ] ( ) ( )( )11 1.081000 1.08 12,500 1.08 1111.08

n ny n + = =

3) For 10 n 16:

( )1000 1.08 , 0 9n kk

k [ ] ( )1500 1.08 , 100, otherwise

n knw k k n

= Fig. Fig. 22..7 7 (c).(c).


21


[ ] ( ) ( )90 0

1000 1.08 1500 1.08n

n k n k

k ky n

= ==

9 10 mk

m = k 10Apply the

f l f( ) ( ) =

=

=9

0

10

0

10

08.1108.11500

08.1108.11000

k

n

m

mn

kn formula for

summing a geometric

series 11 910 n series[ ] ( ) ( )

=

11

08.111

08.1150011

08.111

08.11000 10nnny

08.1

108.1

1

( ) ( )( )97246.89 1.08 18,750 1.08 1 , 10 16n n n= 4) For 17 n :

( )1000 1.08 , 0 9n k k [ ] ( )1500 1.08 , 10 160, otherwise

n knw k k

= Fig. Fig. 22..7 7 (d).(d).


22


9 16[ ] ( ) ( )9 160 10

1000 1.08 1500 1.08n k n kk k

y n = =

= ( ) ( )10 71 08 1 1 08 1[ ] ( ) ( ) ( ) ( )

( )9 161.08 1 1.08 11000 1.08 1500 1.08

1.08 1 1.08 13,340.17 1.08 , 17

n n

n

y n

n

= =

6. Fig. Fig. 22,,77(e)(e) depicts y[n], the value ofthe investment at the start of each

i d b bi i th lt fperiod, by combining the results foreach of the four intervals.

Figure 2.7e (p. 113)(e) The output y[n] representing the ( ) p y[ ] p gvalue of the investment immediately after the deposit or withdrawal at the start of year n.


23


22..4 4 The Convolution IntegralThe Convolution Integral11. A continuous. A continuous--time signal can be expressed as a weighted superposition of time signal can be expressed as a weighted superposition of

timetime--shifted impulses.shifted impulses. Th ifti t f th i l !timetime shifted impulses.shifted impulses.

-x(t) x( ) (t - )d = (2.10)

The sifting property of the impulse !

2 Impulse response of LTI system H: Input x(t) Output y(t)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 = = ( ) ( ){ } ( ) ( ){ }y -

y(t) x( )H{ (t - )}d = (2.10) Linearity property

3. h(t) = H{ (t)} impulse response of the LTI system HIf the system is also time invariant, thenH{ (t )} h(t ) (2 11) A time-shifted impulse H{ (t - )} h(t - ) = (2.11) generates a time-shifted

impulse response output

FigFig 22 99-y(t) x( )h(t )d = (2.12)


24

Fig. Fig. 22..99..


Convolution integral:( ) ( ) ( ) ( )x t h t x h t d =

22..5 5 Convolution Integral Evaluation ProcedureConvolution Integral Evaluation Procedure1. Convolution integral:

-y(t) x( )h(t )d = (2.13) 2. Define the intermediate signal:

( ) ( ) ( )h = independent ( ) ( ) ( )tw x h t = variable, t = constanth (t ) = h ( ( t)) is a reflected and shifted (by t) version of h().

3. Output:

= ty(t) w ( )d (2.14) The time shift t determines the time at which we evaluate the - output of the system.


25


P dP d 22 22 R fl t d Shift C l ti I t l E l tiR fl t d Shift C l ti I t l E l tiProcedure Procedure 22..22:: Reflect and Shift Convolution Integral EvaluationReflect and Shift Convolution Integral Evaluation1. Graph both x() and h(t ) as a function of the independent variable . To

obtain h(t ), reflect h() about = 0 to obtain h( ) and then h( ) shift by t.

2. Begin with the shift t large and negative. That is, shift h( ) to the far left onthe time axis.the time axis.

3. Write the mathematical representation for the intermediate signal wt ().4. Increase the shift t (i.e., move h(t ) toward the right) until the mathematical

representation for w ( ) changes The value of t at which the change occursrepresentation for wt () changes. The value of t at which the change occursdefines the end of the current set and the beginning of a new set.

5. Let t be in the new set. Repeat step 3 and 4 until all sets of shifts t and thecorresponding mathematical representations for wt () are identified. Thisusually implies increasing t to a very large positive number.

6. For each sets of shifts t, integrate wt () from = to = to obtain y(t). , g t ( ) y( )Example Example 22..66 Reflect-and-shift Convolution EvaluationGiven ( ) ( ) ( )1 3x t u t u t= ( ) ( ) ( )2h t u t u t= and as depicted in Fig. Fig. 22--1010,


26Evaluate the convolution integral y(t) = x(t) h(t).


Figure 2 10 (p 117)Figure 2.10 (p. 117)Input signal and LTI system impulse response for Example 2.6.

1. Graph of x() and h(t ): Fig. Fig. 22..11 11 (a).(a).2. Intervals of time shifts: Four intervalsFour intervals

1st interval: t < 12nd interval: 1 t < 3 3rd interval: 3 t < 54th interval: 5 t

3 First interval of time shifts: t < 1 wt() = 03. First interval of time shifts: t < 1

( ) 1, 10 h it

tw

<


Figure 2.11 (p. 118)Evaluation of the

t t Evaluation of the convolution integral for Example 2.6. (a) The input x()The input x() depicted above the reflected and time-shifted impulse presponse. (b) The product signal wt() for 1 t < 3. (c) The product signal wt() for 3 t < 5. (d) The system output y(t).


28


5 Thi d i t l 3 t < 55. Third interval: 3 t < 5

( ) 1, 2 30, otherwiset

tw

<


1. Graph of x() and h(t ): Fig. Fig. 22..13 13 (a).(a).

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

( ) 1, 0 2<

TimeTime--Domain Representations of LTI SystemsDomain Representations of LTI SystemsCHAPTERFigure 2.13 (p. 120)Figure 2.13 (p. 120)

Evaluation of the convolution integral for Example 2.7. (a) The input x() superimposed over the reflected and time-

shifted impulse response h(t ), depicted as a function of . p p ( ), p(b) The product signal wt() for 0 t < 2. (c) The product

signal wt() for t 2. (d) The system output y(t).t t


31


6. Convolution integral: 1) For t < 0: y(t) = 02) For second interval 0 t < 2:

3) For third interval 2 t:

( ) ( ) ( )00 1t tt t ty t e d e e e = = = 3) For third interval 2 t:

( ) ( ) ( ) ( )2 2 200 1t t ty t e d e e e e = = =

( )( )2

0, 01 , 0 2

1 2

t

t

ty t e t

e e t


SS1. Graph of x() and h(t ): Fig. Fig. 22..14 14 (a).(a).2. Intervals of time shifts: Five intervalsFive intervals

1st interval: t < 02nd interval: 0 t < 2 3rd interval: 2 t < 34th interval: 3 t < 5

5th interval: t 53 First interval of time shifts: t < 0 wt() = 03. First interval of time shifts: t < 0 wt() 04. Second interval of time shifts: 0 t < 2

( ) 1, 1 1tw < < += FiFi 22 1414 (b)(b)( ) 0, otherwisetw = Fig. Fig. 22..14 14 (b).(b).5. Third interval of time shifts: 2 t < 3

( ) 1, 1 30, otherwiset

w <

TimeTime--Domain Representations of LTI SystemsDomain Representations of LTI SystemsCHAPTERFigure 2.14 (p. 121) Evaluation of the convolution integral for

Example 2.8. (a) The input x() superimposed on the reflected and time-shifted impulse response h(t ), depicted as a function of . (b) The product signal wt() for 0 t < 2. (c) The product signal wt() for 2 t < 3 (d) The product signal w ( ) for 3 t < 5 (e) The product signal t < 3. (d) The product signal wt() for 3 t < 5. (e) The product signal

wt() for t 5. The system output y(t).t t


34


( )( )1 , 1 2

2 31,0 otherwise

t

tw t

< < = <


( )2

0, 0

, 0 22

tt t


SS1. Find h(t ):

Reflecting h(t) = a (t ) about = 0 gives h() = a ( + ), since the impulse has even symmetry.

2. Shift the independent variable by t to obtain h(t ) = a ( (t )). 3. Substitute this equation for h(t ) into the convolution integral of Eq. (2.12),

and use the shifting property of the impulse to obtain the received signal as

( ) ( ) ( )( ) ( )r t x a t d ax t = =

Figure 2.16 (p. 124)Radar range measurement. (a) Transmitted RF pulse. (b) The received echo

i tt t d d d l d i f th t itt d lSignals and Systems_Simon


is an attenuated and delayed version of the transmitted pulse.


E lE l 22 1010 R d M t ( ti d) Th M t h d FiltExample Example 22..1010 Radar range Measurement (continued): The Matched FilterIn Ex. 2.9, the received signal is contaminated with noise (e.g., the thermal noise, discussed in section 1.9) and may weak. For these reasons, the time delay is determined by passing the received signal through an LTI system commonly referred to as a matched filter. An important property of this system is that it optimally discriminates against certain types of noise in the received waveform. Th i l f th t h d filt i fl t d ti dThe impulse response of the matched filter is a reflected, or time-reversed, version of the transmitted signal x(t). That is, hm(t) = x( t), so

( ) ( ) 0sin , 0ct T th ( ) ( ) 0,0, otherwisecmh t = As shown in Fig. Fig. 22..17 17 (a).(a). The terminology matched filter refers to the fact th t th i l f th d i i t h d t th t itt dthat the impulse response of the radar receiver is matched to the transmitted signal.To estimate the time delay from the matched filter output, we evaluate the

l ticonvolution ( ) ( ) ( )my t r t h t=

1. Intermediate signal: ( ) ( ) ( )t mw r h t =


38

g ( ) ( ) ( )


Figure 2.17a (p. 125)Figure 2.17a (p. 125)(a) Impulse response of

the matched filter for processing the received p g

signal.

Figure 2 17b (p 126)

t t

Figure 2.17b (p. 126)(b) The received signal r()

superimposed on the reflected and time shifted matched filterand time-shifted matched filter

impulse response hm(t ), depicted as functions of . (c)

Matched filter output x(t)


39

Matched filter output x(t).


2 Th i d i l ( ) d th fl t d ti hift d i l2. The received signal r() and the reflected, time-shifted impulse response hm(t ) are shown in Fig. Fig. 22..1717(b).(b).

hm() = reflected version of x(t) hm(t ) = x(t )m( ) ( ) m( ) ( )3. Intervals of time shifts: Three intervalsThree intervals

1st interval: t < T02nd interval: T0 < t 3rd interval: < t + T0

4th interval: t + T0 04. First interval of time shifts: t < T0 wt() = 0 and y(t) = 05. Second interval of time shifts: T0 < t

( ) ( )( ) ( )( ) 0sin sin ,0, otherwise

c ct

a w w t t Tw

< < += ( ) ( ) ( )( ) ( ) ( )( )0 / 2 cos / 2 cos 2t T c cy t a w t a w t d + = +

( ) ( )( )[ ] ( ) ( )( ) 02sin4/cos2/ 0 TttwaTtwa +++= Signals and Systems_Simon


( ) ( )( )[ ] ( ) ( )( )2sin4/cos2/ 0 ccc twaTtwa ++


( ) ( )( ) ( ) ( ) ( )( ) ( )( )0 0( ) / 2 cos / 4 sin 2 sinc c c cy t a w t t T a w t T w t = + + 6. 3rd interval of time shifts: < t + T0

( ) ( ) 0sin ( ) sin ( ) ,( )0, otherwise

c ct

a t tw

< < += + ( ) ( )0( ) [( / 2)cos ( ) ( / 2)cos (2 ) ]c cty t a t a t d += + ( )[ ] ( ) 00( / 2)cos ( ) ( / 4 )sin (2 )c c c ta t t a t += + +

( )[ ] ( ) ( ) ( )[ ] ( ) ( )0( / 2) cos ( ) ( / 4 ) sin ( 2 ) sin ( )c c c ca t t a t t = + + + 7. 4th interval of time shifts: t + T0 wt() = 0 and y(t) = 08 The output of matched filter:8. The output of matched filter:

[ ] ( )[ ] ( )0 00 0

( / 2) ( ) cos ( ) ,( ) ( / 2) ) cos ( ) ,

c

c

a t t ty t a t t t

< = + < < +[ ] ( )0 0

0, otherwisec


41


22..6 6 Interconnection of LTI SystemsInterconnection of LTI Systems22..66..1 1 Parallel Connection of LTI SystemsParallel Connection of LTI Systems1. Two LTI systems: Fig.Fig. 22..1818(a).(a).

Figure 2.18 (p. 128)Interconnection of two LTI systems1. Two LTI systems: Fig. Fig. 22..1818(a).(a). Interconnection of two LTI systems.

(a) Parallel connection of two systems. (b) Equivalent system.

2. Output:1 2( ) ( ) ( )y t y t y t= +1 2

1 2( ) ( ) ( ) ( )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)

FiFi 22 1818(b)(b)Signals and Systems_Simon


( ) ( ) ( ) ( )x h t d x t h t = = Fig. Fig. 22..1818(b)(b)


Distributive property for Continuous-time case:1 2 1 2x(t) h (t) x(t) h (t) x(t) {h (t) h (t)} + = + (2.15)

Distributive property for Discrete-time case: Distributive property for Discrete-time case:1 2 1 2x[n] h [n] x[n] h [n] x[n] {h [n] h [n]} + = + (2.16)

22..66..22 Cascade Connection of LTI SystemsCascade Connection of LTI Systems22..66..2 2 Cascade Connection of LTI SystemsCascade Connection of LTI Systems1. Two LTI systems: Fig. Fig. 22..1919(a).(a).

Figure 2.19 (p. 128)Interconnection of two LTI systems. (a) Cascade connection of two systems.

(b) Equivalent system. (c) Equivalent system: Interchange system order.


43


2. The output is expressed in terms of z(t) as2y(t) z(t) h (t)= (2.17)

(t) ( )h (t )d (2 18)2-y(t) z( )h (t )d = (2.18)

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

( ) ( ) h ( ) ( )h ( )d (2 19)1 1-z( ) x( ) h ( ) x( )h ( )d = = (2.19)

Substituting Eq. (2.19) for z(t) into Eq. (2.18) gives Change of variable1 2( ) ( ) ( ) ( )y t x v h v h t dvd =

1 2-y(t) x( ) h ( )h (t )d d = (2.20)

Change of variable =

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

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

y(t) x( )h(t )d x(t) h(t) = = (2.21) 3 Associative property for continuous time case:

Fig. Fig. 22..1919(b).(b).


44

3. Associative property for continuous-time case:


1 2 1 2{x(t) h (t)} h (t) x(t) {h (t) h (t)} = (2.22) 4. Commutative property:

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

1 2( ) ( ) ( )h t h h t d = Change of variable = t 1 2 2 1-

h(t) h (t )h ( )d h (t) h (t) = = (2.23) Fig. Fig. 22..1919(c).(c).Interchanging the order of the LTI systems in the cascade without affecting th ltthe result:

{ } { }1 2 2 1( ) ( ) ( ) ( ) ( ) ( ) ,x t h t h t x t h t h t = Commutative property for continuous time case:Commutative property for continuous-time case:

1 2 2 1h (t) h (t) h (t) h (t) = (2.24) 5. Associative property for discrete-time case:p p y

1 2 1 2{x[n] h [n]} h [n] x[n] {h [n] h [n]} = (2.25) Commutative property for discrete-time case:

h [ ] h [ ] h [ ] h [ ]Signals and Systems_Simon


1 2 2 1h [n] h [n] h [n] h [n] = (2.26)


E lE l 22 1111 E i l t S t t F I t t d S tExample Example 22..1111 Equivalent System to Four Interconnected SystemsConsider the interconnection of four LTI systems, as depicted in Fig. Fig. 22..2020. 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 n=andFind the impulse response h[n] of the overall system.

Figure Figure 22..20 20 (p(p 131131))(p. (p. 131131))

Interconnection of systems for

Example 2 11

1. Parallel combination of h1[n] and h2[n]:

h12[n] = h1[n] + h2[n] FigFig 22 2121 (a)(a)

Example 2.11.


46

h12[n] = h1[n] + h2[n] Fig. Fig. 22..21 21 (a).(a).


Figure 2.21 (p. 131)(a) Reduction of(a) Reduction of parallel combination of LTI systems in upper branch of Figupper branch of Fig. 2.20. (b) Reduction of cascade of systems in uppersystems in upper branch of Fig. 2.21(a). (c) Reduction of ( )parallel combination of systems in Fig. 2.21(b) to obtain an ( )

equivalent system for Fig. 2.20.


47


2 h [ ] i i i ith h [ ]2. h12[n] is in series with h3[n]:

h123[n] = h12[n] h3[n]h [n] = (h [n] + h [n]) h [n] Fig 2 21 (b)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 nu n

= + + = +

Convolving h12[n] with h3[n] gives

123[ ] [ 2] [ 2][ ]

h n u n nu n

= + [ ]u n=

{ }[ ] 1 [ ].nh n u n= Signals and Systems_Simon


Table Table 22..1 1 summarizes the interconnection properties presented in this section.summarizes the interconnection properties presented in this section.


22..7 7 Relation Between LTI System Properties and theRelation Between LTI System Properties and theImpulse ResponseImpulse Response

22..77..1 1 Memoryless LTI SystemsMemoryless LTI Systems1. The output of a discrete-time LTI system:

[ ] [ ] [ ] [ ] [ ]k

y n h n x n h k x n k=

= = y[n] h[ 2]x[n 2] h[ 1]x[n 1] h[0]x[n] h[1]x[n 1] h[2]x[n 2]= + + + + + + + +L L


49(2.27)


2. To be memoryless, y[n] must depend only on x[n] and therefore cannotdepend on x[n k] 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 is memoryless if and only if

( ) ( )h c = c is an arbitrary constant22..77..2 2 Causal LTI SystemsCausal LTI SystemsThe output of a causal LTI system depends only on past or present values of the input. Discrete-time system:

p

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

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

= + + + + ++ + +

L


50

[1] [ 1] [2] [ 2] .h x n h x n+ + +L


2. For a discrete-time causalcausal LTI system,

[ ] 0 for 0h k k= n and u[n k] = 1 for k n we haveSignals and Systems_Simon


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


[ ] [ ].n

ks n h k

==

The step response is the running sum of the impulse response. Continuous-time LTI system:

t( ) h( )ds(t) h( )d = (2.34)

The step response s(t) is the running integral of the impulse response h(t). Express the impulse response in terms of the step response as Express the impulse response in terms of the step response as

[ ] [ ] [ 1]h n s n s n= ( ) ( )dh t s tdt

=and Figure 2.12 (p. 119)

Example Example 22..1414 RC Circuit: Step ResponseThe impulse response of the RC circuit depicted in Fig. Fig. 22..1212 is

1 t

RC circuit system with the voltage source x(t) as input and1( ) ( )RCh t e u t

RC=

Find the step response of the circuit.

x(t) as input and the voltage measured across the capacitor y(t),


59

as output.


1. Step response:1( ) ( ) .

tRCs t e u d

RC

= RC0, 0

( ) 1 tt

s t


22..9 9 Differential and Difference Equation Representations of Differential and Difference Equation Representations of LTILTI SystemsSystems

1 Linear constant coefficient differential equation:1. Linear constant-coefficient differential equation:k kN M

k kk kk 0 k 0

d da y(t) b x(t)dt dt= =

= (2.35) Input = x(t), output = y(t)k 0 k 0= =

2. Linear constant-coefficient difference equation:N M

k ka y[n k] b x[n k] = (2 36) Input = x[n], output = y[n]k kk 0 k 0

a y[n k] b x[n k]= = (2.36)

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

Ex. RLC circuit depicted in Fig. Fig. 22..2626.1. Input = voltage source x(t), output = loop current

Often, N M, and the order is described

using only N2. KVL Eq.:

( ) ( ) ( ) ( )1 tdRy t L y t y d x tdt C

+ + =using only N.


61

dt C


Figure 2.26 (p. 141)

Example of an RLC circuit described by a differential equation.

( ) ( ) ( ) ( )221 d d dy t R y t L y t x tC dt dt dt+ + = N = 2C dt dt dtEx. Accelerator modeled in Section 1.10:

( ) ( ) ( ) ( )22 n d dt t t t+ + N 2( ) ( ) ( ) ( )2 2nn y t y t y t x tQ dt dt + + =where y(t) = the position of the proof mass, x(t) = external acceleration.

N = 2


62


Ex. Second-order difference equation:1y[n] y[n 1] y[n 2] x[n] 2x[n 1]4

+ + = + (2.37) N = 2 Difference equations are easily rearranged to obtain recursive formulas for

computing the current output of the system from the input signal and thepast outputs.past outputs.

Ex. Eq. (2.36) can be rewritten as

[ ] [ ] [ ]1 1M Nb k k [ ] [ ] [ ]0 10 0

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 the second-order difference

equation (2.37).

1. Eq. (2.37) can be rewritten as

= + 1y[n] x[n] 2x[n 1] y[n 1] y[n 2]4

(2.38)

2 Computing y[n] for n 0:Signals and Systems_Simon


2. Computing y[n] for n 0:


1y[0] x[0] 2x[ 1] y[ 1] y[ 2]4

= + (2.39) 1[1] [1] 2 [0] [0] [ 1]1y[1] x[1] 2x[0] y[0] y[ 1]4

= + (2.40)

[ ] [ ] [ ] [ ] [ ]12 2 2 1 1 0y x x y y= + [ ] [ ] [ ] [ ] [ ]4

y y y

[ ] [ ] [ ] [ ] [ ]13 3 2 2 2 14

y x x y y= + 4

Initial conditions: y[ 1] and y[ 2]. The initial conditions for Nth-order difference equation are the N values

[ ] [ ] [ ], 1 ,..., 1 ,y N y N y + The initial conditions for Nth-order differential equation are the N values

( ) ( ) ( ) ( )2 10 , 2 10 00

, , ...,N

t Nt tt

d d dy t y t y t y tdt dt dt

= = = =


64


Example Example 22..1515 Recursive Evaluation of a Difference EquationFind the first two output values y[0] and y[1] for the system described by Eq. (2.38), assuming that the input is x[n] = (1/2)nu[n] and the initial conditions are ( ) g ( )y[ 1] = 1 and y[ 2] = 2.

1. Substitute the appropriate values into Eq. (2.39) to obtain1. Substitute the appropriate values into Eq. (2.39) to obtain

[ ] ( )1 10 1 2 0 1 24 2

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

[ ] ( )1 1 1 31 2 1 1 12 2 4 4

y = + =[ ] ( )2 2 4 4

Example Example 22..1616 Evaluation of a Difference Equation by means of a ComputerA system is described y the difference equationy y q

[ ] [ ] [ ] [ ] [ ] [ ]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 the past


65outputs and the current inputs. Use a computer to determine the step response


of the system, the system 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 inputs x1[n] = cos(n/10), x2[n] = cos(n/5), and x3[n] = cos(7n/10), p 1[ ] ( ), 2[ ] ( ), 3[ ] ( ),assuming zero initial conditions. Last, find the output of the system if the input is the weekly closing price of Intel stock depicted in Fig. Fig. 22..2727, assuming zero initial conditionsinitial conditions.

1. Recursive formula for y[n]:

[ ] [ ] [ ] [ ] [ ] [ ]1 143 1 0 4128 2 0 0675 0 1349 1 0 675 2[ ] [ ] [ ] [ ] [ ] [ ]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= + + + 2. Step response: Fig. Fig. 22..28 28 (a).(a).3. Zero input response: Fig.Fig. 22..2828 (b).(b).3. Zero input response: Fig. Fig. 22..28 28 (b).(b).4. The outputs due to the sinusoidal inputs x1[n], x2[n], and x3[n]: Fig. Fig. 22..28 28 (c),(c),

(d), and (e).(d), and (e).5 FigFig 22 2828(f)(f) shows the system output for the Intel stock price unit5. Fig. Fig. 22..2828(f)(f) shows the system output for the Intel stock price unit.

A comparison of peaks in Figs. Figs. 22..2727 and 22..28 28 (f)(f) Slightly delay!Slightly delay!


66


Figure 2.27 (p. 144)W klWeekly

closing price of Intel stock.


67


FiFi 22 2828 ( )( )Fig. Fig. 22..28 28 (a).(a).


68


Fig. Fig. 22..28 28 (b).(b).


69


Fig. Fig. 22..28 28 (c).(c).gg ( )( )

FigFig 22 2828 (d)(d)Signals and Systems_Simon


Fig. Fig. 22..28 28 (d).(d).


FigFig 22 2828 (e)(e)

Figure 2.28a (p. 145)Illustration of the solution to Example 2.16.

Fig. Fig. 22..28 28 (e).(e).

Illustration of the solution to Example 2.16. (a) Step response of system.

(b) Output due to nonzero initial conditions with zero input. (c) Output due to x1[n] = cos (1/10n). ( ) p 1[ ] ( )(d) Output due to x2[n] = cos (1/5n). (e) Output due to x3[n] = cos(7/10n).


71


Figure 2.28f (p. 146)Output associated with the weekly l i iclosing price

of Intel stock.


72


22 1010 S l i Diff ti l d Diff E tiS l i Diff ti l d Diff E ti22..10 10 Solving Differential and Difference EquationsSolving Differential and Difference EquationsComplete solution: y = y (h) + y (p)

y (h) = homogeneous solution, y (p) = particular solutiony g , y p22..1010..1 1 The Homogeneous SolutionThe Homogeneous Solution Continuous-time case:1 Homogeneous differential equation: ( ) ( ) 0kN hda y t =1. Homogeneous differential equation: ( )

00k k

ka y t

dt==

2. Homogeneous solution:

i

Nr t(h)

iy (t) c e= (2 41) Coefficients ci is d t i d b I Cii 0

y (t) c e= (2.41)

3. Characteristic eq.: N

kk

k 0a r 0

== (2.42)

Di t ti

determined by I.C.

Discrete-time case:1. Homogeneous differential equation:2. Homogeneous solution:

( ) [ ]0

0N

hk

ka y n k

= =

N(h) n

i ii 1

y [n] c r=

= (2.43) 3 Characteristic eq :

NN ka r 0 = (2 44)

Coefficients ci is determined by I.C.


73

3. Characteristic eq.: kk 0

a r 0=

= (2.44)


If t i t d ti i h t i ti th di If a root rj is repeated p times in characteristic eqs., the correspondingsolutions are

1, , ...,j j jr r rt t p te te t eContinuous-time case:1, , ...,n n p nj j jr nr n rDiscrete-time case:

Example Example 22..1717 RC Circuit: Homogeneous SolutionThe RC circuit depicted in Fig. Fig. 22..3030 is described by the differential equation

( ) ( ) ( )dy t RC y t x tdt

+ =( ) ( ) ( )dt

Determine the homogeneous solution of this equation.

( ) ( )d1. Homogeneous Eq.: ( ) ( ) 0dy t RC y tdt

+ =2. Homo. Sol.: ( ) ( ) 11 Vh r ty t c e=

Figure 2.30 (p. 148)RC circuit.( ) 1y

3. Characteristic eq.: 11 0RCr+ = r1 = 1/RC4. Homogeneous solution: ( ) ( ) Vth RCt


74

( ) ( ) 1 Vh RCy t c e=


E lE l 22 1818 Fi t O d R i S t H S l tiExample Example 22..1818 First-Order Recursive System: Homogeneous SolutionFind the homogeneous solution for the first-order recursive system described by the difference equation

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

1 Homogeneous Eq : [ ] [ ]1 0y n y n =1. Homogeneous Eq.: [ ] [ ]1 0y n y n =2. Homo. Sol.:

( ) [ ] 1 1h ny n c r=[ ] 1 1y3. Characteristic eq.: 1 0r = r1 = 4. Homogeneous solution:

( ) [ ] 1h ny n c =22..1010..2 2 The Particular SolutionThe Particular SolutionA particular solution is usually obtained by assuming an output of the same general form as the input.

TableTable 22 33Signals and Systems_Simon


Table Table 22..33


Example Example 22..1919 First-Order Recursive System (Continued): Particular SolutionFind a particular solution for the first-order recursive system described by the difference equation

[ ] [ ] [ ]1y n y n x n =if the input is x[n] = (1/2)n.

1. Particular solution form: ( ) ( )12[ ] np py n c=2 Substituting y(p)[n] and x[n] into the given difference Eq :


76

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

TimeTime--Domain Representations of LTI SystemsDomain Representations of LTI SystemsCHAPTERBoth sides of above eq are( ) ( ) ( )11 1 12 2 2n n np pc c =

c (1 2 ) 1 = (2 45)

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

pc (1 2 ) 1 (2.45) 3. Particular solution:

( ) [ ] 1 1 np Figure 2.30

(p 148)( ) [ ] 1 11 2 2

py n =

ExampleExample 22..2020 RC Circuit (continued): Particular Solution

(p. 148)RC circuit.

Example Example 22..2020 RC Circuit (continued): Particular SolutionConsider the RC circuit of Example Example 22..1717 and depicted in Fig. Fig. 22..3030. Find a particular solution for this system with an input x(t) = cos(0t).Sol.Sol.1. Differential equation: ( ) ( ) ( )dy t RC y t x t

dt+ =

2. Particular solution form:( )

1 2( ) cos( ) sin( )py t c t c t = +

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


g y ( ) ( ) ( 0 ) g q


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

0 1 2 0RC c c + =4. Coefficients c1 and c2:

( )1 201

1c

RC= + ( )0

2 201

RCcRC= +and

5. Particular solution:

( ) ( ) ( ) ( ) ( ) ( )0

0 02 21 cos sin V

1 1p RCy t t t

RC RC = ++ +( ) ( )0 01 1RC RC + +

22..1010..3 3 The Complete SolutionThe Complete SolutionComplete solution: y = y (h) + y (p)p y y yy (h) = homogeneous solution, y (p) = particular solution

The procedure for finding complete solution of differential or difference equations is summarized as follows:


78

equations is summarized as follows:


P dP d 22 33 S l i Diff ti l Diff tiProcedure Procedure 22..33:: 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 the homogeneous solution.3. Determine the coefficients in the homogeneous solution so that the complete3. Determine the coefficients in the homogeneous solution so that the complete

solution y = y(h) + y(p) satisfies the initial conditions. Note that the initial translation is needed in some cases.Note that the initial translation is needed in some cases.Example Example 22..2121 First-Order Recursive System (Continued): Complete SolutionFind the complete solution for the first-order recursive system described by the difference equation

1y[n] y[n 1] x[n]4

= (2.46) if the input is x[n] = (1/2)n u[n] and the initial condition is y[ 1] = 8if the input is x[n] = (1/2) u[n] and the initial condition is y[ 1] = 8.

1. Homogeneous sol.: ( ) [ ] ( )11 4 nhy n c=


79

( )



( ) [ ] 122

npy n =

3. Complete solution:n n

11 1y[n] 2( ) c ( )2 4

= + (2.47) [ ]2

4. Coefficient c1 determined by I.C.:

I C : [ ] [ ] [ ]0 0 1 4 1y x y= + [ ] [ ]0 0 (1 4) 8 3y x= + =I.C.: [ ] [ ] [ ]0 0 1 4 1y x y= + [ ] [ ]0 0 (1 4) 8 3y x= + =We substitute y[0] = 3 into Eq. (2.47), yielding

0 01 1 1

1 13 22 4

c = + c1 = 15. Final solution:

1 1n n [ ] 1 122 4

y n = + for n 0Example Example 22..2222 RC Circuit (continued): Complete Responsepp ( ) p pFind the complete response of the RC circuit depicted in Fig. Fig. 22..3030 to an input x(t) = cos(t)u(t) V, assuming normalized values R = 1 and C = 1 F and assuming that the initial voltage across the capacitor is y(0) = 2 V


80

assuming that the initial voltage across the capacitor is y(0 ) = 2 V.


1. Homogeneous sol.: ( ) ( ) Vth RCy t ce=2. Particular solution:

0 = 1( ) ( ) ( ) ( ) ( ) ( )2 2

1 cos sin V1 1

p RCy t t tRC RC

= ++ +3. Complete solution:

( ) 1 1cos sin V2 2

ty t ce t t= + +R = 1 , C = 1 F

4. Coefficient c1 determined by I.C.:2 2

y(y(00) = y() = y(00++))

0 1 1 12 0 i 0+ + + 3/202 cos0 sin 0

2 2 2ce c + += + + = + c = 3/2

5. Final solution:

( ) 3 1 1cos sin V2 2 2

ty t e t t= + +


81


E lE l 22 2323 Fi i l C t ti L R tExample Example 22..2323 Financial Computations: Loan RepaymentThe following difference equation describes the balance of a loan if x[n] < 0 represents the principal and interest payment made at the beginning of each period and y[n] is the balance after the principal and interest payment is credited. As before, if r % is the interest rate per period, then = 1 + r/100.

[ ] [ ] [ ]1y n y n x n= +[ ] [ ] [ ]y yUse the complete response of the first-order difference equation to find the payment required to pay off a $20,000 loan in 10 periods. Assume equal payments and a 10% interest rate.payments and a 10% interest rate.

1. We have = 1.1 and y[ 1] = 20,000, and we assume that x[n] = b is the

payment each period.payment each period.2. The first payment is made when n = 0. The loan balance is to be zero after 10

payments, thus we seek the payment b for which y[9] = 0.3 Homogeneous sol :3. Homogeneous sol.:

( ) [ ] ( )1.1 nh hy n c=4 Particular solution:


82



( ) [ ]p py n c=Substituting y(p)[n] = cp and x[n] = b into the difference equation y[n] 1.1y[n 1]

[ ] bt i= x[n], we obtain10pc b=

3 Complete solution:3. Complete solution:= nhy[n] c (1.1) 10b, n 0 (2.48)

4. Coefficient ch determined by I.C.:I.C.: [ ] [ ] [ ]0 1.1 1 0 22,000y y x b= + = +

( )022,000 1.1 10hb c b+ = ( ), h22,000 11hc b= +

[ ] ( )( )22 000 11 1 1 10ny n b b= + Fig. Fig. 22..3131..[ ] ( )( )22,000 11 1.1 10y n b b= + 5. Payment b: By setting y[9] = 0, we have

( )( )90 22 000 11 1 1 10b b= + ( )( )9

9

22,000 1.13,254.91

11 1 1 10b

= = Signals and Systems_Simon


( )( )0 22,000 11 1.1 10b b= + ( )11 1.1 10


Figure 2.31 (p. 155)Balance on a $20,000 loan for Example 2.23. Assuming 10% interest per


84period, the loan is paid off with 10 payments of $3,254.91.


22 1111 Ch t i ti f S t D ib d b Diff ti lCh t i ti f S t D ib d b Diff ti l22..11 11 Characteristics of Systems Described by DifferentialCharacteristics of Systems Described by Differentialand Difference Equationsand Difference EquationsComplete solution: y = y (n) + y (f)Complete solution: y y ( ) + y ( )

y (n) = natural response, y (f) = forced response22..1111..1 1 The Natural ResponseThe Natural ResponseE lE l 22 2424 RC Ci it ( ti d) N t l RExample Example 22..2424 RC Circuit (continued): Natural ResponseThe system In Example 2.17 is described by the differential equation

( ) ( ) ( )dFind the natural response of the this system, assuming that y(0) = 2 V, R = 1

d C 1 F

( ) ( ) ( )dy t RC y t x tdt

+ =

and C = 1 F.

1. Homogeneous sol.: ( ) ( ) 1 Vh ty t c e=2 I C (0) 2 V2. I.C.: y(0) = 2 V

y (n) (0) = 2 V c1 = 23. Natural Response: ( ) ( ) 2 Vn ty t e


85

p ( ) ( ) 2 Vy t e=


Example Example 22..2525 First-Order Recursive System (Continued): Natural ResponseThe system in Example Example 22..2121 is described by the difference equation

[ ] [ ] [ ]1 1[ ] [ ] [ ]1 14

y n y n x n =Find the natural response of this system.

1. Homogeneous sol.: ( ) [ ] 1 14

nhy n c

= 2. I.C.: y[ 1] = 81

1184

c= c1 = 2

3. Natural Response:

( ) [ ] 12 , 14

nny n n=

The forced response is valid only for t 0 or n 04

22..1111..2 2 The Forced ResponseThe Forced ResponseThe forced response is the system output due to the input signal assuming


86zero 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, , N 1 before solving for theundetermined coefficients, such as when one is determining the complete

l tisolution.Example Example 22..2626 First-Order Recursive System (Continued): Forced ResponseThe system in Example Example 22..2121 is described by the difference equation

[ ] [ ] [ ]1 14

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

1. Complete solution: [ ] 11 12 , 02 4

n n

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

[ ] [ ] [ ]10 0 14

y x y= + y[0] = 1 + (1/4) 0 =1[ ] [ ] [ ]4

y[ ] ( )

3. Finding c1: 0 01

1 11 22 4

c = + c1 = 1Signals and Systems_Simon


2 4


4. Forced response:

( ) [ ] 1 12 , 02 4

n nfy n n = 2 4

Example Example 22..2727 RC Circuit (continued): Forced ResponseThe system In Example 2.17 is described by the differential equationThe system In Example 2.17 is described by the differential equation

( ) ( ) ( )dy t RC y t x tdt

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

1 C l t l ti ( ) 1 1 i Vt

From Example 2.221. Complete solution: ( ) cos sin V

2 2ty t ce t t= + +

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

3. Forced response:( ) ( ) 1 1 1cos sin V

2 2 2f ty t e t t= + +


88

2 2 2


22..1111..3 3 The Impulse ResponseThe Impulse Response Relation between step response and impulse response1. Continuous-time case: 2. Discrete-time case:

( ) ( )dh t s tdt

= [ ] [ ] [ 1]h n s n s n= 22 1111 44 Li it d Ti I iLi it d Ti I i22..1111..4 4 Linearity and Time InvarianceLinearity and Time Invariance

Input Forced response

Forced response LinearityInitial Cond. Natural response

Natural response LinearityInput Forced response

x1 y1(f)

x2 y2(f)

Initial Cond. Natural responseI1 y1(n)

I2 y2(n)2 y2 x1 + x2 y1(f) + y2 (f)

2 y2 I1 + I2 y1(n) + y2 (n)

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

22..1111..5 5 Roots of the Characteristic EquationRoots of the Characteristic Equation


89

qq


Roots of characteristic equationForced response, natural response, stability, and response time. BIBO Stable:BIBO Stable: BIBO Stable:BIBO Stable:1. Discrete-time case: boundednir 1, for allir i = Block diagram:

Fig 2 37(N k )

N M(N k)

k kk 0 k 0

a y (t) b x (t)

= == (2.55)

Ex Second-order system: (1) (2) (1) (2)y(t) a y (t) a y (t) b x(t) b x (t) b x (t)= + + +(2.56)

Fig. 2.37


94

Ex. Second-order system: 1 0 2 1 0y(t) a y (t) a y (t) b x(t) b x (t) b x (t)= + + +


Figure 2.37 (p. 166)Block diagram representations

of a continuous-time LTI system described by a second-

order integral equation. (a) Direct form I. (b) Direct form II.

(a)


95(b)


22..13 13 StateState--Variable Description of Variable Description of LTILTI SystemsSystems The state of a system may be defined as a minimal set of signals that

represent the systems entire memory of the past.y yGiven initial point ni (or ti) and the input for time n ni (or t ti), we can determine the output for all times n ni (or t ti).

22 1313 11 The StateThe State Variable DescriptionVariable Description22..1313..1 1 The StateThe State--Variable DescriptionVariable Description1. Direct form II of a second-order LTI system: Fig. Fig. 22..3939..2. Choose state variables: q1[n] and q2[n].3. State equation:

1 1 1 2 2q [n 1] a q [n] a q [n] x[n]+ = + (2.57) 2 1q [n 1] q [n]+ = (2 58)2 1q [n 1] q [n]+ (2.58)

4. Output equation:= + +1 1 2 2 1 1 2 2y[n] x[n] a q [n] a q [n] b q [n] b q [n],

1 1 1 2 2 2y[n] (b a )q [n] (b a )q [n] x[n]= + + (2.59) 1 11 2q [n 1] q [n]a a 1 x[n][ 1] [ ]1 0 0+ = + (2.60)

5. Matrix Form of state equation:


962 2

[ ]q [n 1] q [n]1 0 0 + ( )


Figure 2.39 (p. 167)Direct form II representation of a second-order discrete-time LTI system

depicting state variables q [n] and q [n]


97

depicting state variables q1[n] and q2[n].


6. Matrix form of output equation:

11 1 2 2

q [n]y[n] [b a b a ] [1]x[n]

q [n] = + (2.61) 2q [n]

Define state vector as the column vector = 1q [n]

[n]q = 2[n]

q [n]q

We can rewrite Eqs. (2.60) and (2.61) as[n 1] [n] x[n]+ = +q Aq b (2 62)[n 1] [n] x[n]+ = +q Aq b (2.62)

y[n] [n] Dx[n]= +cq (2.63) where matrix A, vectors b and c, and scalar D are given by

1 2

1 0A

a a = 10

b = [ ]1 1 2 2c b a b a= 1D =Example Example 22..2828 State-Variable Description of a Second-Order SystemFind the state-variable description corresponding to the system depicted in Fig. Fig. 22..4040 by choosing the state variable to be the outputs of the unit delays.


98


Figure 2.40 (p. 169)Block diagram of LTI system for Example 2.28.

1. State equation:[ ] [ ] [ ]1 1 11q n q n x n + = +[ ] [ ] [ ] [ ]1q n q n q n x n + + +


99

[ ] [ ] [ ] [ ]2 1 2 21q n q n q n x n + = + +


2. Output equation:[ ] [ ] [ ]1 1 2 2y n q n q n = +

3 Define state vector as3. Define state vector as

[ ] [ ][ ]12qq n

nq n = [ ]2q

In standard form of dynamic equation:+ = +[n 1] [n] x[n]q Aq b (2.62) = +y[n] [n] Dx[n]cq (2.63)

0A

= 1b =

[ ]1 2c = [ ]2D = 2 [ ] [ ] State-variable description for continuous-time systems:

d (t) (t) (t)A b (2 64)(t) (t) x(t)dt

= +q Aq b (2.64) y(t) (t) Dx(t)= +cq (2.65)


100


E lE l 22 2929 St t V i bl D i ti f El t i l Ci itExample Example 22..2929 State-Variable Description of an Electrical CircuitConsider the electrical circuit depicted in Fig. Fig. 22..4242. Derive a state-variable description of this system if the input is the applied voltage x(t) and the output i th t (t) th h th i tis the current y(t) through the resistor.

Figure 2.42 (p. 171)Circuit diagram of LTI

system for Example 2.29.

1. State variables: The voltage acrosseach capacitor.

2. KVL Eq. for the loop involving x(t), R1, and C1:

( ) ( ) ( )1 1x t y t R q t= +1

1 1y(t) q (t) x(t)R R

= + (2.66)

Output equation

11 1

y( ) q ( ) ( )R R (2.66)

3. KVL Eq. for the loop involving C1, R2, and C2:( ) ( ) ( )q t R i t q t= +


101

( )1 2 2 2( ) ( )q t R i t q t= +


1 12 1 2

2 2

1 1i (t) q (t) q (t)R R

= (2.67) 4. The current i2(t) through R2:

Use Eq. (2.67) to eliminate i2(t)2( ) g 2

( )2 2 2 ( )di t C q tdt= ( )2 1 22 2 2 21 1( ) ( )d q t q t q t

dt C R C R= (2.68)

5. KCL Eq. between R1 and R2:( ) ( ) ( )1 2y t i t i t= + Current through C1 = i1(t) where dwhere ( ) ( )1 1 1di t C q tdt=

1 1 1 1d ( )1 1 21 1 2 2 1 2 1 1

1 1 1 1( ) ( ) ( )d q t q t q t x tdt C R C R C R C R

= + + + (2.69) Eqs. (2.66), (2.68), and (2.69) = State-Variable Description. q ( ), ( ), ( ) p

= +d (t) (t) x(t)dt

q Aq b (2.64)

(t) (t) D (t)Signals and Systems_Simon


= +y(t) (t) Dx(t)cq (2.65)


1 1 1 2 1 2

1 1 1

,1 1

AC R C R C R

+ = 1 11

b C R =

1 0 ,cR

= 1DR

=and

2 2 2 2

1 1C R C R

0

1R 1R

ExampleExample 22 3030 State Variable Description from a Block DiagramExample Example 22..3030 State-Variable Description from a Block DiagramDetermine the state-variable description corresponding to the block diagram in Fig. Fig. 22..4444. The choice of the state variables is indicated on the diagram.

Figure 2.44 (p. 172)Block diagram of LTI system for Example 2.30.


103

Block diagram of LTI system for Example 2.30.


1 St t ti 3 St t i bl d i ti1. State equation:

( ) ( ) ( ) ( )1 1 22d q t q t q t x tdt = +3. State-variable description:

2 1,

1 0A

= 1

,0

b = ( ) ( )2 1d q t q tdt =

2 O t t ti

1 0 0 [ ]3 1 ,c = [ ]0D =

2. Output equation:

( ) ( ) ( )1 23y t q t q t= +22..1313..2 2 Transformations of The StateTransformations of The StateThe transformation is accomplished by defining a new set of state variables that are a weighted sum of the original ones.

The input-output characteristic of the system is not changed.1. Original state-variable description:

q Aq bx= +& (2 70)q Aq bx= + (2.70) cqy Dx= + (2.71)

2. Transformation: q = Tq

T = state-transformation matrix

q = T1 q


104


3. New state-variable description:.q TAq Tbx = +&1) State equation: q Tq =& &

1q TAT q Tb +& T 1 1 .q TAT q Tbx = + q = T1 q 2) Output equation:

1 .cT qy Dx= + .cT qy Dx+3) If we set

1 1, , , andA TAT b Tb c cT D D = = = =then q A q b x = +& and c qy D x = +

Ex Consider ExampleExample 22 3030 again Let us define new statesEx. Consider Example Example 22..3030 again. Let us define new states

2 1 1 2( ) ( ) and ( ) ( )q t q t q t q t = =Find the state-variable description.p

1. State equation: 1 1 2 2 2 1( ) 2 ( ) ( ) ( ) ( ) 2 ( ) ( ) ( )

ld d i ti N d i ti

d dq t q t q t x t q t q t q t x tdt dt

= + = +1444442444443 1444442444443


105

old description New description


3 St t i bl d i ti2 1 1 2( ) ( ) ( ) ( )

Old description New description

d dq t q t q t q tdt dt

= =1442443 1442443

3. State-variable description:

0 1,

1 2A =

0,

1b = Old description New description

1 2 2 13 ( ) ( ) 3 ( ) ( )y q t q t y q t q t = + = +1442443 14424432. Output equation:

1 2 1 [ ]1 3 ,c = [ ]0 .D =

Old description New description2 2

Example Example 22..3131 Transforming The StateA discrete-time system has the state-variable descriptionA discrete time system has the state variable description

1 41A ,4 110 =

2,

4b = [ ]

1 1 1 ,2

c = 2.D =and Find the state-variable description A, b, c, D corresponding to the new states

1 11 1 22 2[ ] [ ] [ ]q n q n q n = + 1 12 1 22 2[ ] [ ] [ ]q n q n q n = +and

1. Transformation: q = Tq, where 1 11 .

1 12T

= Signals and Systems_Simon


1 12


1 1 1 1 1 .1 1

T =

2 New state-variable description:2. New state variable description:12

3

0,

0A

= 1

,3

b = [ ]0 1 ,c = 2.D =and 10 This choice for T results in A being a diagonal matrix and thus separates

the state update into the two decoupled first-order difference equations

[ ] [ ] [ ]1 111 2q n q n x n+ = + [ ] [ ] [ ]2 231 310

q n q n x n+ = +and 22 1414 Exploring Concepts with MATLABExploring Concepts with MATLAB22..14 14 Exploring Concepts with MATLABExploring Concepts with MATLAB Two limitations: 1. MATLAB is not easily used in the continuous-time case. 2. Finite memory or storage capacity and nonzero computation times. Both the MATLAB Signal Processing Toolbox and Control System Toolbox

are use in this section.


107


22..1414..1 1 ConvolutionConvolution1. MATLAB command: y = conv(x, h)

x and h are signal vectors.

2. The number of elements in y is given by the sum of the number of elements in2. The number of elements in y is given by the sum of the number of elements inx and h, minus one.

1) Elements in vector x: from time n = kx to n = lx1) Elements in vector x: from time n kx to n lx2) Elements in vector h: from time n = kh to n = lh3) Elements in vector y: from time n = ky = kx + kh to n = ly = lx + ly4) Th l th f [ ] d h[ ] L l k + 1 d L l k +14) The length of x[n] and h[n] are Lx = lx kx + 1 and Lh = lh kh +15) The length of y[n] is Ly = Lx + Lh 1 Ex. Repeat Example 2.1p pImpulse and Input : From time n = kh = kx = 0 to n = lh = 1 and n = lx =2Convolution sum: From time n = ky = kx + kh = 0 to n = ly = lx + lh = 3 The length of convolution sum: L = l k + 1 = 4The length of convolution sum: Ly ly ky + 1 4MATLAB Program: >> h = [1, 0.5];

>> x = [2, 4, -2];( h)

y =

2 5 0 1Signals and Systems_Simon


>> y = conv(x,h) 2 5 0 -1


Repeat Example 2.3

Given [ ] ( ) [ ] [ ]( )1 4 4h n u n u n= [ ] [ ] [ ]10 .x n u n u n= andImpulse response Input

[ ] ( ) [ ] [ ]( ) [ ] [ ] [ ]Find the convolution sum x[n] h[n].

1 In this case kh = 0 lh = 3 k = 0 and l = 9

1. In this case, kh 0, lh 3, kx 0 and lx 92. y starts at time n = ky = 0, ends at time n = ly =12, and has length Ly = 13.3. Generation for vector h with MATLAB:

>> h = 0.25*ones(1, 4);>> x = ones(1, 10);

4 Output and its plot:4. Output and its plot:>> n = 0:12;>> y = conv(x, h);

t ( ) l b l(' ') l b l(' [ ]')>> stem(n, y); xlabel('n'); ylabel('y[n]')

Fig. Fig. 22..4545..


109


Figure 2.45 (p. 177)Convolution sum computed using MATLAB


110

Convolution sum computed using MATLAB.


22..1414..2 2 The Step ResponseThe Step Response1. Step response = the output of a system in response to a step input2. In general, step response is infinite in duration.g , p p3. We can evaluate the first p values of the step response using the conv

function if h[n] = 0 for n < kh by convolving the first p values of h[n] with afinite-duration step of length p.p g p1) Vector h = the first p nonzero values of the impulse response.2) Define step: u = ones(1, p).3) convolution: s = conv(u, h).) ( , )

Ex. Repeat Problem 2.12Determine the first 50 values of the step response of the system with impulse response given byresponse given by

[ ] ( ) [ ]nh n u n=with = 0.9, by using MATLAB program.

1. MATLAB Commands:


111


>> h = (-0.9).^[0:49];>> u = ones(1, 50);>> s = conv(u, h); s conv(u, h);>> stem([0:49], s(1:50))

2. Step response: Fig. Fig. 22..4747..

Figure 2.47 (p. 178)g (p )Step response

computed using MATLAB.

22..1414..3 3 Simulating Difference equationsSimulating Difference equations1. Difference equation: = N Mk ka y[n k] b x[n k] (2.36) Command:Command:f


112= = k kk 0 k 0

y[ ] [ ] (2.36) filter


2. Procedure:1) Define vectors a = [a0, a1, , aN] and b =[b0, b1, , bM] representing the

coefficients of Eq. (2.36).2) Input vector: x3) y = filter(b, a, x) results in a vector y representing the output of the system

for zero initial conditions.4) filt (b i) lt i t ti th t t f th4) y = filter(b, a, x, zi) results in a vector y representing the output of the

system for nonzero initial conditions zi. The initial conditions used by filter are not the past values of the output. Command zi = filtic(b, a, yi), where yi is a vector containing the initial

conditions in the order [y[1], y[2], , y[N]], generates the initialconditions obtained from the knowledge of the past outputs.

Ex. Repeat Example 2.16The system of interest is described by 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 + = + + [ ] [ ] [ ] [ ] [ ] [ ]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+ = + +(2.73)Determine the output in response to zero input and initial condition

y[1] = 1 and y[2] = 2.


113


1. MATLAB Program:>> a = [1, -1.143, 0.4128]; b = [0.0675, 0.1349, 0.675];>> x = zeros(1, 50); 0 5( , );>> zi = filtic(b, a, [1, 2]);>> y = filter(b, a, x, zi);>> stem(y)

0.4

0.5

(y)2. Output: Fig. Fig. 22..2828(b).(b).

0.2

0.3

3. System response to aninput consisting of the

0

0.1

input consisting of theIntel stock price data Intc:

>> load Intc;

-0.1

0>> load Intc;>> filtintc = filter(b, a, Intc);

We have assume that theI t l t k i d t

0 10 20 30 40 50-0.2Intel stock price data arein the file Intc.mat.

The command [h, t] = impz(b, a, n) evaluates n values of the impulse responsef ff


114

of a system described by a different equation.


22..1414..4 4 StateState--Variable DescriptionsVariable Descriptions MATLAB command: ss1. Input MATLAB arrays: a, b, c, d

Representing the matices A,b,c, and D.

2. Command: sys = ss(a, b, c, d, -1) produces an LTI object sys that representsthe discrete-time system in state-variable form. Continuous-time case: sys = ss(a, b, c, d) No 1No 1 System manipulation:1. sys = sys1 + sys2 Parallel combination of sys1 and sys2.2 sys = sys1 sys2 Cascade combination of sys1 and sys22. sys = sys1 sys2 Cascade combination of sys1 and sys2. MATLAB command: lsim1. Command form: y = lsim(sys, x)2. Output = y, input = x. MATLAB command: impulse1 Command form: h = impulse(sys N)2. This command places the first N values of the impulse response in h.1. Command form: h = impulse(sys, N)

MATLAB routine: ss2ss Perform the state transformation


1151. Command form: sysT = ss2ss(sys, T), where T = Transformation matrix


Ex. Repeat Example 2.31.1. Original state-variable description:

1 41 2 11 41 ,4 110

A = 2

,4

b = [ ]1 1 1 ,2

c = 2,D =and 2. State-transformation matrix:

1 11 .1 12

T =

3. MATLAB Program:>> a = [-0.1, 0.4; 0.4, -0.1]; b = [2; 4];>> c = [0 5 0 5]; d = 2;>> c [0.5, 0.5]; d 2;>> sys = ss(a, b, c, d, -1); % define the state-space object sys>> T = 0.5*[-1, 1; 1, 1];>> sysT = ss2ss(sys, T) sysT ss2ss(sys, T)

4. Result:


116


a = x1 x2

x1 -0.5 0

b = u1

x1 1c =

x1 x2d =

u1x2 0 0.3 x2 3 y1 0 1 y1 2

Sampling time: unspecifiedDiscrete-time modelDiscrete-time model.

Ex. Verify that the two systems represented by sys and sysT have identicalinput-output characteristic by comparing their impulse responses .

1. MATLAB Program: >> h = impulse(sys, 10); hT = impulse(sysT, 10);

>> subplot(2, 1, 1)>> t ([0 9] h)>> stem([0:9], h)>> title ('Original System Impulse Response');>> xlabel('Time'); ylabel('Amplitude')>> subplot(2 1 2)>> subplot(2, 1, 2)>> stem([0:9], hT)>> title('Transformed System Impulse Response');>> xlabel('Time'); ylabel('Amplitude')


117

>> xlabel( Time ); ylabel( Amplitude )

TimeTime--Domain Representations of LTI SystemsDomain Representations of LTI SystemsCHAPTEROriginal System Impulse Response

2. Simulation results:Fig. Fig. 22..4848..

2

3g y p p

d

e

We may verify that

1

A

m

p

l

i

t

u

d

y ythe original andtransformed systemshave the (numerically)

0 2 4 6 8 100

Time

identical impulse response by computingthe error, err = h hT.

3Transformed System Impulse Response

Figure 2.48 ( 181)

1

2

A

m

p

l

i

t

u

d

e(p. 181)Impulse responses associated with the original and transformed

0 2 4 6 8 100

original and transformed state-variable descriptions computer using MATLAB


118Timeusing MATLAB.

TimeTime--Domain Representations of LTI SystemsDomain Representations of LTI SystemsCHAPTERx 10

-17

0

1x 10

-1r

r

3

-2

p

l

i

t

u

d

e

e

-4

-3

A

m

p

-5

0 2 4 6 8 10-6

TimePlot for err = hPlot for err = h hThT


119

Plot for err = h Plot for err = h hThT



120

ch02 time-domain representations of lti systems compatibility mode

Documents

n n x n

n x n x n x n

y n x n x n

otherwiseh n n

n x n ltimetime

kx n h n x

y n h x n h x

ky n x