2.2 convolution sum - nctu - mapl

Signals and Systems_Simon Haykin & Barry Van Veen

1

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

2.1 2.1 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

2.2 2.2 Convolution SumConvolution Sum1. 1. An arbitrary signal is expressed as a weighted superposition of An arbitrary signal is expressed as a weighted superposition of shiftedshifted

impulses.impulses.Discrete-time signal x[n]:

[ ] [ ] [ ] [ ]0x n n x nδ δ=

[ ] [ ] [ ] [ ]x n n k x k n kδ δ− = −

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

2 2 1 1 0

1 1 2 2

x n x n x n x n

x n x n

δ δ δ

δ δ

= + − + + − + +

+ − + − +

L

L

Fig. 2.1Fig. 2.1

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

at time k.


2


[ ] [ ]{ } [ ] [ ]k

y n H x n H x k n kδ∞

=−∞

⎧ ⎫= = −⎨ ⎬

⎩ ⎭∑

Figure 2.1 (p. 99)Figure 2.1 (p. 99)Graphical example illustrating the representation of a signal x[n] as a weighted sum of time-shifted impulses.

[ ] [ ] [ ]k

x n x k n kδ∞

=−∞

= −∑ (2.1)

2. Impulse response of LTI system H:

LTI systemLTI systemHH

Input x[n] Output y[n]

Output:

[ ] [ ] [ ]{ }k

y n H x k n kδ∞

=−∞

= −∑

ky[n] x[k]H{ [n k]}δ

∞

=−∞

= −∑ (2.2)

Linearity

Linearity


3


♣ The system output is a weighted sum of the response of the system to time-shifted impulses.For time-invariant system:

δ − = −H{ [n k]} h[n k]h[n] = H{δ [n]} ≡ impulse response

of the LTI system H(2.3)

ky[n] x[k]h[n k]

∞

=−∞

= −∑ (2.4)

3. Convolution sum:

[ ] [ ] [ ] [ ]k

x n h n x k h n k∞

=−∞

∗ = −∑

Convolution process: Fig. 2.2Fig. 2.2.

Figure 2.2a (p. 100)Figure 2.2a (p. 100) Illustration of the convolution sum. (a) LTI system with impulse response h[n] and input x[n].


4


Figure 2.2b (p. 101)(b) The decomposition of the input x[n] into a weighted sum of time-shifted impulses results in an output y[n] given by a weighted sum of time-shifted impulse responses.

d ≡ δ


5


♣ The output associated with the kth input is expressed as:

{ }δ − = −H x[k] [n k] x[k]h[n k]

[ ] [ ] [ ]k

y n x k h n k∞

=−∞

= −∑Example 2.1Example 2.1 Multipath Communication Channel: Direct Evaluation of the

Convolution SumConsider the discrete-time LTI system model representing a two-path propagation channel described in Section 1.10. If the strength of the indirect path is a = ½, then

[ ] [ ] [ ]1 12

y n x n x n= + −

Letting x[n] = δ [n], we find that the impulse response is

[ ]

1, 01 , 120, otherwise

n

h n n

=⎧⎪⎪= =⎨⎪⎪⎩


6


Determine the output of this system in response to the input

[ ]

2, 04, 12, 2

0, otherwise

nn

x nn

=⎧⎪ =⎪= ⎨− =⎪⎪⎩

<<Sol.>Sol.>1. Input: [ ] [ ] [ ] [ ]2 4 1 2 2x n n n nδ δ δ= + − − −

Input = 0 for n < 0 and n > 0

2. Since

γδ [n − k]time-shifted impulse input

γh [n − k]time-shifted impulse response output

3. Output:

[ ] [ ] [ ] [ ]2 4 1 2 2y n h n h n h n= + − − −

[ ]

0, 02, 05, 10, 21, 3

0, 4

nnn

y nnnn

<⎧⎪ =⎪⎪ =

= ⎨ =⎪⎪− =⎪

≥⎩


7


2.3 2.3 Convolution Sum Evaluation ProcedureConvolution Sum Evaluation Procedure1. Convolution sum:

[ ] [ ] [ ]k

y n x k h n k∞

=−∞

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

k = independent variable

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].

3. Since

nk

y[n] [k]ω∞

=−∞

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

output of the system.Example 2.2Example 2.2 Convolution Sum Evaluation by using Intermediate Signal

Consider 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 = 10 when the input is x [n] = u [n].


8


<<Sol.>Sol.> Fig. 2.3Fig. 2.3 depicts x[k] superimposed on the reflected and time-shifted impulse response h[n − k].

1. h [n − k]:

[ ]3 ,4

0, otherwise

n k

k nh n k

−⎧ ⎞⎛ ≤⎪⎜ ⎟− = ⎨⎝ ⎠⎪⎩

[ ]5 0w k− =

[ ]5

5

3 , 0 54

0, otherwise

k

kw k

−⎧ ⎞⎛ ≤ ≤⎪⎜ ⎟= ⎨⎝ ⎠⎪⎩

2. Intermediate signal wn[k]:For n = − 5:

Eq. (2.6)For n = 5:

y[− 5] = 0

Eq. (2.6) [ ]55

0

354

k

k

y−

=

⎞⎛= ⎜ ⎟⎝ ⎠

∑

For n = 10:

[ ]10

10

3 , 0 104

0, otherwise

k

kw k

−⎧ ⎞⎛ ≤ ≤⎪⎜ ⎟= ⎨⎝ ⎠⎪⎩

Eq. (2.6)

[ ]

11

10 10 1010 10

0 0

413 3 4 3 310

44 4 3 4 13

3.831

k k

k k

y−

= =

⎞⎛− ⎜ ⎟⎞ ⎞ ⎞ ⎞⎛ ⎛ ⎛ ⎛ ⎝ ⎠= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎞⎛⎝ ⎝ ⎝ ⎝⎠ ⎠ ⎠ ⎠ − ⎜ ⎟⎝ ⎠

=

∑ ∑

[ ]

6

5 55

0

413 4 3 35 3.288

44 3 4 13

k

k

y=

⎛ ⎞− ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎝ ⎠= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞⎝ ⎠ ⎝ ⎠ ⎝ ⎠ − ⎜ ⎟⎝ ⎠

∑


9


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


10


[ ]3 , 04

0, otherwise

n k

nk nw k

−⎧ ⎞⎛ ≤ ≤⎪⎜ ⎟= ⎨⎝ ⎠⎪⎩

Procedure 2.1:Procedure 2.1: 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. To

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

representation 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 timesshifts and the corresponding mathematical representations for wn[k] areidentified. 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 interval.


11


Example 2.3Example 2.3 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

[ ] [ ]3

0

14 k

y n x n k=

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

[ ] [ ] [ ]( )1 44

h n u n u n= − − Fig. 2.4 (a).Fig. 2.4 (a).

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

[ ] [ ] [ ]10x n u n u n= − − Fig. 2.4 (b).Fig. 2.4 (b).<<Sol.>Sol.> 1. Refer to Fig. 2.4Fig. 2.4.2. 1’st interval: wn[k] = 0

[ ]0

1/ 4, 00, otherwise

kw k

=⎧= ⎨⎩

Five intervals !

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

For n = 0:3. 2’nd interval:

Fig. 2.4 (c).Fig. 2.4 (c).


12


Figure 2.4 (p. 106)Evaluation of the convolution sum for Example 2.3. (a) The system impulse response h[n]. (b) The input signal x[n]. (c) The input above the reflected and time-shifted impulse response h[n – k], 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].


13


For n = 1:

[ ]1

1/ 4, 0,10, otherwise

kw k

=⎧= ⎨⎩

For general case: n ≥ 0:

[ ] 1/ 4, 00, otherwisen

k nw k

≤ ≤⎧= ⎨⎩

4. 3’rd interval: 3 < n ≤ 9

[ ] 1/ 4, 3 90, otherwisen

n kw k

− ≤ ≤⎧= ⎨⎩

Fig. 2.4 (d).Fig. 2.4 (d).

5. 4th interval: 9 < n ≤ 12Fig. 2.4 (e).Fig. 2.4 (e).

[ ] 1/ 4, 30, otherwisen

n k nw k

− ≤ ≤⎧= ⎨⎩

Fig. 2.4 (f).Fig. 2.4 (f).

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 corresponding wn[k] according to Eq. (2.6).

( )1N

k M

c c N M=

= − +∑1) For n < 0 and n > 12: y[n] = 0.2) For 0 ≤ n ≤ 3:

[ ]0

11/ 44

n

k

ny n=

+= =∑

3) For 3 < n ≤ 9:

[ ] ( )( )3

11/ 4 3 1 14

n

k n

y n n n= −

= = − − + =∑4) For 9 < n ≤ 12:

[ ] ( )( )9

3

1 131/ 4 9 3 14 4k n

ny n n= −

−= = − − + =∑

Fig. 2.4 (g)Fig. 2.4 (g)


14


Example 2.4Example 2.4 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 nρ− − =

Let the input be given by [ ] [ ]4nx n b u n= +We use convolution to find the output of this system, assuming that b ≠ ρ and that the system is causal.<<Sol.>Sol.>1. Impulse response: [ ] [ ] [ ]1h n h n nρ δ= − + (2.7)

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] = ρ, h[2] = ρ 2, …, or

[ ] [ ]nh n u nρ=

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

[ ] , 40, otherwise

kb kx k

⎧ − ≤= ⎨⎩

[ ] ,0, otherwise

n k k nh n k

ρ −⎧ ≤− = ⎨

⎩and

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


15


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


16


Assuming that ρ = 0.9 and b = 0.8.

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

[ ] , 40, otherwise

k n k

nb k n

w kρ −⎧ − ≤ ≤

= ⎨⎩

Fig. 2.5 (b).Fig. 2.5 (b).6. Output:1) For n < − 4: y[n] = 0.2) For n ≥ − 4:

[ ]4

nk n k

k

y n b ρ −

=−

= ∑

[ ]4

knn

k

by n ρρ=−

⎞⎛= ⎟⎜

⎝ ⎠∑

Let m = k + 4, then

[ ]4 44 4

0 0

m mn nn n

m m

b by nbρρ ρ

ρ ρ

−+ +

= =

⎞ ⎞⎛ ⎛⎞⎛= = ⎜ ⎟⎟ ⎟⎜ ⎜⎝ ⎠⎝ ⎝⎠ ⎠∑ ∑

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

[ ]

5

4 5 54

1

1

n

n nn

bby n bbb b

ρρ ρρρ

ρ

+

+ +−

⎞⎛− ⎟⎜ ⎞⎛ −⎞⎛ ⎝ ⎠= = ⎟⎜ ⎟ ⎜ −⎝ ⎠ ⎝ ⎠−

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

[ ] 5 54

0, 4

, 4n n

ny n bb n

bρρ

+ +−

< −⎧⎪= ⎞⎛ −⎨ − ≤⎟⎜⎪ −⎝ ⎠⎩

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


17


Figure 2.5c (p. 110)(c) The output y[n] assuming that p = 0.9 and b = 0.8.


18


Example 2.5Example 2.5 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 ρ = 1 + (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] = ρy[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 recursive equation [ ] [ ] [ ]1y n y n x nρ= − +

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 and then $1500 is withdrawn at the start each year for 7 years.<<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 evaluation procedure, we can

evaluate y[n] = x[n] ∗ h[n], where x[n] is depicted in Fig. 2.6Fig. 2.6 and h[n] = ρ n u[n]is as shown in Example 2.4 with ρ = 1.08.


19


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

each 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. 2.7(a).Fig. 2.7(a).4. Intervals of time shifts: 1’st interval: n < 0

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

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


20


Figure 2.7a-d (p. 111)Evaluation of the convolution sum for Example 2.5. (a) The input signal x[k] depicted above the reflected and time-shifted impulse response h(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.


21


2) For 0 ≤ n ≤ 9:

[ ] ( )1000 1.08 , 00, otherwise

n k

nk nw k

−⎧ ≤ ≤⎪= ⎨⎪⎩

Fig. 2.7 (b).Fig. 2.7 (b).

[ ] ( ) ( )0 0

11000 1.08 1000 1.081.08

kn nn k n

k k

y n −

= =

⎞⎛= = ⎜ ⎟⎝ ⎠

∑ ∑

[ ] ( ) ( )( )1

1

111.081000 1.08 12,500 1.08 1111.08

n

n ny n

+

+

⎞⎛− ⎜ ⎟⎝ ⎠= = −−

Apply the formula for summing a geometric

series

3) For 10 ≤ n ≤ 16:

[ ]( )( )

1000 1.08 , 0 9

1500 1.08 , 100, otherwise

n k

n kn

k

w k k n

−

−

⎧ ≤ ≤⎪⎪= − ≤ ≤⎨⎪⎪⎩

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


22


[ ] ( ) ( )9

0 0

1000 1.08 1500 1.08n

n k n k

k k

y n − −

= =

= −∑ ∑

( ) ( )∑ ∑=

−

=

− ⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

9

0

10

0

10

08.1108.11500

08.1108.11000

k

n

m

mn

kn

m = k − 10

Apply the formula for summing a geometric

series

[ ] ( ) ( )⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

⎟⎠⎞

⎜⎝⎛−

−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

⎟⎠⎞

⎜⎝⎛−

=

−

−

08.111

08.111

08.11500

08.111

08.111

08.11000

9

10

10 n

nnny

( ) ( )( )97246.89 1.08 18,750 1.08 1 , 10 16n n n−= − − ≤ ≤

4) For 17 ≤ n :

[ ]( )( )

1000 1.08 , 0 9

1500 1.08 , 10 160, otherwise

n k

n kn

k

w k k

−

−

⎧ ≤ ≤⎪⎪= − ≤ ≤⎨⎪⎪⎩

Fig. 2.7 (d).Fig. 2.7 (d).


23


[ ] ( ) ( )9 16

0 10

1000 1.08 1500 1.08n k n k

k k

y n − −

= =

= −∑ ∑

[ ] ( ) ( ) ( ) ( )

( )

10 79 161.08 1 1.08 1

1000 1.08 1500 1.081.08 1 1.08 1

3,340.17 1.08 , 17

n n

n

y n

n

− −− −= −

− −

= ≤

6. Fig. 2,7(e)Fig. 2,7(e) depicts y[n], the value ofthe investment at the start of eachperiod, by combining the results foreach of the four intervals.

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


24


2.4 2.4 The Convolution IntegralThe Convolution Integral1. 1. A continuousA 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.

-x(t) x( ) (t - )dτ δ τ τ

∞

∞= ∫ (2.10)

The sifting property of the impulse !

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 dτ δ τ τ∞

−∞= = −∫

-y(t) x( )H{ (t - )}dτ δ τ τ

∞

∞= ∫ (2.10)

Linearity property

3. h(t) = H{δ (t)} ≡ impulse response of the LTI system HIf the system is also time invariant, thenH{ (t - )} h(t - )δ τ τ= (2.11)

A time-shifted impulse generates a time-shifted impulse response output

Fig. 2.9.Fig. 2.9.-y(t) x( )h(t )dτ τ τ

∞

∞= −∫ (2.12)


25


♣ Convolution integral:

( ) ( ) ( ) ( )x t h t x h t dτ τ τ∞

−∞∗ = −∫

2.5 2.5 Convolution Integral Evaluation ProcedureConvolution Integral Evaluation Procedure1. Convolution integral:

-y(t) x( )h(t )dτ τ τ

∞

∞= −∫ (2.13)

2. Define the intermediate signal:( ) ( ) ( )tw x h tτ τ τ= −

τ = independent variable, t = constant

h (t − τ) = h (− (τ − t)) is a reflected and shifted (by − t) version of h(τ).

3. Output:

τ τ∞

∞= ∫ t-

y(t) w ( )d (2.14) The time shift t determines the time at which we evaluate the

output of the system.


26


Procedure 2.2:Procedure 2.2: Reflect and Shift Convolution Integral EvaluationReflect and Shift Convolution Integral Evaluation1. Graph both x(τ) and h(t − τ) as a function of the independent variable τ . To

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

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

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

representation for wt (τ) 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 (τ) are identified. Thisusually implies increasing t to a very large positive number.

6. For each sets of shifts t, integrate wt (τ) from τ = − ∞ to τ = ∞ to obtain y(t). Example 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).


27


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

<<Sol.>Sol.>1. Graph of x(τ) and h(t − τ): Fig. 2.11 (a).Fig. 2.11 (a).2. Intervals of time shifts: Four intervalsFour intervals

1’st interval: t < 12’nd interval: 1 ≤ t < 3 3’rd interval: 3 ≤ t < 54th interval: 5 ≤ t

3. First interval of time shifts: t < 1

( )1, 10, otherwiset

tw

ττ

< <⎧= ⎨⎩

Fig. 2.11 (b).Fig. 2.11 (b).

4. Second interval of time shifts: 1 ≤ t < 3wt(τ) = 0


28


Figure 2.11 (p. 118)Evaluation of the convolution integral for Example 2.6. (a) The input x(τ) depicted above the reflected and time-shifted impulse response. (b) The product signal wt(τ) for 1 ≤ t < 3. (c) The product signal wt(τ) for 3 ≤ t < 5. (d) The system output y(t).

t t ≡≡ ττ


29


5. Third interval: 3 ≤ t < 5

( )1, 2 30, otherwiset

tw

ττ

− < <⎧= ⎨⎩

Fig. 2.11 (c).Fig. 2.11 (c).

6. Fourth interval: 5 ≤ t wt(τ) = 07. Convolution integral:

1) For t < 1 and t ≥ 5: y(t) = 02) For second interval 1 ≤ t < 3, y(t) = t − 13) For third interval 3 ≤ t < 5, y(t) = 3 − (t − 2)

( )

0, 11, 1 3

5 , 3 50, 5

tt t

y tt t

t

<⎧⎪ − ≤ <⎪= ⎨ − ≤ <⎪⎪ ≥⎩

Example 2.7Example 2.7 RC Circuit OutputFor the RC circuit in Fig. 2.12Fig. 2.12, assume that the circuit’s time constant is RC = 1 sec. Ex. 1.21 shows that the impulse response of this circuit is h(t) = e − t u(t).Use convolution to determine the capacitor voltage, y(t), resulting from an input voltage x(t) = u(t) − u(t − 2).

Figure 2.12 (p. 119)RC circuit system with the

voltage source x(t) as input and the voltage measured across the

capacitor y(t), as output.


30


<<Sol.>Sol.>1. Graph of x(τ) and h(t − τ): Fig. 2.13 (a).Fig. 2.13 (a).

2. Intervals of time shifts: Three intervalsThree intervals

1’st interval: t < 02’nd interval: 0 ≤ t < 2 3’rd interval: 2 ≤ t

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

( )1, 0 20, otherwise

xτ

τ< <⎧

= ⎨⎩

and

3. First interval of time shifts: t < 04. Second interval of time shifts: 0 ≤ t < 2

wt(τ) = 0

For t > 0, ( )( ) , 00, otherwise

t

te tw

τ ττ− −⎧ < <⎪= ⎨

⎪⎩Fig. 2.13 (b).Fig. 2.13 (b).

5. Third interval: 2 ≤ t

( )( ) , 0 20, otherwise

t

tew

τ ττ− −⎧ < <⎪= ⎨

⎪⎩Fig. 2.13 (c).Fig. 2.13 (c).

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

tt e th t e u t

ττ ττ τ

− −− − ⎧ <⎪− = − = ⎨

⎪⎩


31


Figure 2.13 (p. 120)Evaluation of the convolution integral for Example 2.7. (a) The input x(τ) superimposed over the reflected and time-

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

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

t t ≡≡ ττ


32


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

3) For third interval 2 ≤ t:

( ) ( ) ( )001

t tt t ty t e d e e eτ ττ− − − −= = = −∫

( ) ( ) ( ) ( )2 2 2

001t t ty t e d e e e eτ ττ− − − −= = = −∫

( )( )2

0, 01 , 0 2

1 , 2

t

t

ty t e t

e e t

−

−

⎧ <⎪⎪= − ≤ <⎨⎪ − ≥⎪⎩

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

Example 2.8Example 2.8 Another Reflect-and-Shift Convolution EvaluationSuppose that the input x(t) and impulse response h(t) of an LTI system 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.


33


<<Sol.>Sol.>1. Graph of x(τ) and h(t − τ): Fig. 2.14 (a).Fig. 2.14 (a).2. Intervals of time shifts: Five intervalsFive intervals

1’st interval: t < 02’nd interval: 0 ≤ t < 2 3’rd interval: 2 ≤ t < 34th interval: 3 ≤ t < 5

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

( )1, 1 1

0, otherwiset

tw

τ ττ

− < < +⎧= ⎨⎩

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

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

( )1, 1 3

0, otherwisetwτ τ

τ− < <⎧

= ⎨⎩

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

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


34


Figure 2.14 (p. 121) Evaluation of the convolution integral for Example 2.8. (a) The input x(τ) superimposed on the reflected and

time-shifted impulse response h(t – τ), depicted as a function of τ. (b) The product signal wt(τ) for 0 ≤ t < 2. (c) The product signal wt(τ) for 2 ≤ t < 3. (d) The product signal wt(τ) for 3 ≤ t < 5. (e) The product signal

wt(τ) for t ≥ 5. The system output y(t).t t ≡≡ ττ


35


( )( )1 , 1 2

2 31,0, otherwise

t

tw t

τ ττ ττ

− −⎧ < < −⎪= − < <−⎨⎪⎩

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

7. Fifth interval of time shifts: t ≥ 5

( ) ( )1 , 1 30, otherwisetwτ τ

τ− − < <⎧

= ⎨⎩

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

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

( ) ( )2 21 1

111

2 2t t ty t d ττ τ τ+ + ⎞⎛

= − = − =⎟⎜⎝ ⎠

∫3) For third interval 2 ≤ t < 3: y(t) = 24) For third interval 3 ≤ t < 5:

5) For third interval t ≥ 5: y(t) = − 2

( ) ( ) ( )2 3 2

1 21 1 6 7

t

ty t d d t tτ τ τ τ

−

−= − − + − = − + −∫ ∫


36


( )

2

2

0, 0

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

tt t

y tt

t t tt

<⎧⎪⎪ ≤ <⎪

∴ = ⎨ ≤ <⎪⎪− + − ≤ <⎪ − ≥⎩

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

Example 2.9Example 2.9 Radar range Measurement: Propagation ModelWe identify an LTI system describing the propagation of the pulse. Let the transmitted RF pulse be given by

( ) ( ) 0sin , 00, otherwise

ct t Tx t

ω ≤ ≤⎧= ⎨⎩

as shown in Fig. 2.16 (a).Fig. 2.16 (a).Suppose we transmit an impulse from the radar to determine the impulse response of the round-trip propagation to the target. The impulse is delay in time and attenuated in amplitude, which results in the impulse response h(t) = aδ (t − β ), where a represents the attenuation factor and β the round-trip time delay. Use the convolution of x(t) with h(t) to verify this result.


37


<<Sol.>Sol.>1. Find h(t − τ):

Reflecting h(t) = a δ (t − β ) about τ = 0 gives h(−τ) = aδ (τ + β ), since the impulse has even symmetry.

2. Shift the independent variable τ by − t to obtain h(t −τ ) = a δ (τ − (t − β )). 3. Substitute this equation for h(t − τ ) 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 tτ δ τ β τ β∞

−∞= − − = −∫

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

is an attenuated and delayed version of the transmitted pulse.


38


Example 2.10Example 2.10 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. The 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 , 00, otherwise

cm

t T th t

ω− − ≤ ≤⎧= ⎨⎩

As shown in Fig. 2.17 (a).Fig. 2.17 (a). The terminology “matched filter” refers to the fact that 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 convolution ( ) ( ) ( )my t r t h t= ∗<<Sol.>Sol.>1. Intermediate signal: ( ) ( ) ( )t mw r h tτ τ τ= −


39


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

the matched filter for processing the received

signal.

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

superimposed on the reflected and time-shifted matched filter

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

Matched filter output x(t).

t t ≡≡ ττ


40


2. The received signal r(τ) and the reflected, time-shifted impulse response hm(t − τ) are shown in Fig. 2.17(b).Fig. 2.17(b).

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

0, otherwisec c

t

a w w t t Tw

τ β τ β ττ

⎧ − − < < +⎪= ⎨⎪⎩

♣ hm(τ) = reflected version of x(t) ⇒ hm(t − τ) = x(t − τ)3. Intervals of time shifts: Three intervalsThree intervals

1’st interval: t < β − T02’nd interval: β − T0 < t ≤ β3’rd interval: β < t ≤ β + T0

4th interval: t ≥ β + T0

4. First interval of time shifts: t < β − T0 wt(τ) = 0 and y(t) = 05. Second interval of time shifts: β − T0 < t ≤ β

( ) ( ) ( )( ) ( ) ( )( )0 / 2 cos / 2 cos 2t T

c cy t a w t a w t dβ

β τ β τ+⎡ ⎤= − + − −⎣ ⎦∫

( ) ( )( )[ ] ( ) ( )( ) 02sin4/cos2/ 0Tt

ccc twaTtwa +−−+−+−=β

βτωββ


41


( ) ( )( ) ( ) ( ) ( )( ) ( )( )0 0( ) / 2 cos / 4 sin 2 sinc c c cy t a w t t T a w t T w tβ β ω β β⎡ ⎤∴ = − − − + + − − −⎡ ⎤⎣ ⎦ ⎣ ⎦6. 3’rd interval of time shifts: β < t ≤ β + T0

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

0, otherwisec c

ta t t

wω τ β ω τ τ β

τ⎧ − − < < +Τ

= ⎨⎩

( ) ( )0( ) [( / 2)cos ( ) ( / 2)cos (2 ) ]c cty t a t a t d

βω β ω τ β τ

+Τ= − + − −∫

( )[ ] ( ) 00( / 2)cos ( ) ( / 4 )sin (2 )c c c ta t t a t βω β β ω ω τ β +Τ= − +Τ − + − −

( )[ ] ( ) ( )0( / 2)cos ( ) ( / 4 ) sin ( 2 ) sin ( )c c c ca t t a t tω β β ω ω β τ ω β⎡ ⎤= − − +Τ + + − − −⎣ ⎦7. 4th interval of time shifts: t ≥ β + T0 wt(τ) = 0 and y(t) = 08. The output of matched filter:

[ ] ( )[ ] ( )

0 0

0 0

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

0, otherwise

c

c

a t t ty t a t t t

β ω β β ββ ω β β β

⎧ − −Τ − −Τ < ≤⎪= − +Τ − < < +Τ⎨⎪⎩


42


2.6 2.6 Interconnection of LTI SystemsInterconnection of LTI Systems2.6.1 2.6.1 Parallel Connection of LTI SystemsParallel Connection of LTI Systems1. Two LTI systems: Fig. 2.18(a).Fig. 2.18(a).

Figure 2.18 (p. 128)Interconnection of two LTI systems.

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

2. Output:1 2

1 2

( ) ( ) ( )( ) ( ) ( ) ( )

y t y t y tx t h t x t h t

= += ∗ + ∗

1 2( ) ( ) ( ) ( ) ( )y t x h t d x h t dτ τ τ τ τ τ∞ ∞

−∞ −∞= − + −∫ ∫

{ }1 2( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

y t x h t h t d

x h t d x t h t

τ τ τ τ

τ τ τ

∞

−∞

∞

−∞

= − + −

= − = ∗

∫∫

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

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


43


♣ 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:

1 2 1 2x[n] h [n] x[n] h [n] x[n] {h [n] h [n]}∗ + ∗ = ∗ + (2.16)

2.6.2 2.6.2 Cascade Connection of LTI SystemsCascade Connection of LTI Systems1. Two LTI systems: Fig. 2.19(a).Fig. 2.19(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.


44


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

2-y(t) z( )h (t )dτ τ τ

∞

∞= −∫ (2.18)

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

1 1-z( ) x( ) h ( ) x( )h ( )dτ τ τ ν τ ν ν

∞

∞= ∗ = −∫ (2.19)

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

1 2( ) ( ) ( ) ( )y t x v h v h t dvdτ τ τ∞ ∞

−∞ −∞= − −∫ ∫

1 2-y(t) x( ) h ( )h (t )d dν η ν η η ν

∞ ∞

∞ −∞

⎡ ⎤= − −⎢ ⎥⎣ ⎦∫ ∫ (2.20)

Change of variableη = τ − ν

Define h(t) = h1(t) ∗ h2(t), then

1 2( ) ( ) ( )h t v h h t v dη η η∞

−∞− = − −∫

-y(t) x( )h(t )d x(t) h(t)ν ν ν

∞

∞= − = ∗∫ (2.21)

3. Associative property for continuous-time case:

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


45


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 integral

1 2( ) ( ) ( )h t h h t dτ τ τ∞

−∞= −∫

1 2 2 1-h(t) h (t )h ( )d h (t) h (t)ν ν ν

∞

∞= − = ∗∫ (2.23)

Change of variableν = t − τ

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

Interchanging the order of the LTI systems in the cascade without affecting the result:

{ } { }1 2 2 1( ) ( ) ( ) ( ) ( ) ( ) ,x t h t h t x t h t h t∗ ∗ = ∗ ∗

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:

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

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


46


Example 2.11Example 2.11 Equivalent System to Four Interconnected SystemsConsider the interconnection of four LTI systems, as depicted in Fig. 2.20Fig. 2.20. The impulse responses of the systems are

1[ ] [ ],h n u n= 2[ ] [ 2] [ ],h n u n u n= + − 3[ ] [ 2],h n nδ= − 4[ ] [ ].nh n u nα=andFind 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).

Figure 2.20 Figure 2.20 (p. 131)(p. 131)

Interconnection of systems for

Example 2.11.


47


Figure 2.21 (p. 131)(a) Reduction of parallel combination of LTI systems in upper branch of Fig. 2.20. (b) Reduction of cascade of systems 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.


48


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 nu n

= + + −= +

Convolving h12[n] with h3[n] gives

123[ ] [ 2] [ 2][ ]

h n u n nu n

δ= + ∗ −=

{ }[ ] 1 [ ].nh n u nα= −

♣♣ Table 2.1 summarizes the interconnection properties presented inTable 2.1 summarizes the interconnection properties presented in this section.this section.


49


2.7 2.7 Relation Between LTI System Properties and theRelation Between LTI System Properties and theImpulse ResponseImpulse Response

2.7.1 2.7.1 MemorylessMemoryless LTI SystemsLTI 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

(2.27)


50


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 kδ= c is an arbitrary constant

♣ Continuous-time system:1. Output:

( ) ( ) ( ) ,y t h x t dτ τ τ∞

−∞= −∫

2. A continuous-time LTI system is memoryless if and only if

( ) ( )h cτ δ τ= c is an arbitrary constant

2.7.2 2.7.2 Causal LTI SystemsCausal LTI Systems

♣ Discrete-time system:

The output of a causal LTI system depends only on past or present values 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

= + − + + − + ++ − + − +

L

L


51


2. For a discrete-time causalcausal LTI system,

[ ] 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 dτ τ τ∞

−∞= −∫

2. For a continuous-time causalcausal LTI system,

( ) 0 for 0h τ τ= <

3. Convolution integral in new form:

0( ) ( ) ( ) .y t h x t dτ τ τ

∞= −∫

2.7.3 2.7.3 Stable LTI SystemsStable LTI SystemsA system is BIBO stable if the output is guaranteed to be bounded for every bounded input.♣ Discrete-time case: Input [ ] xx n M≤ ≤ ∞ [ ] yy n M≤ ≤ ∞Output:


52


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− ≤

and it follows that

xk

y[n] M h[k]∞

=−∞

≤ ∑ (2.28)

Hence, the output is bounded, or |y[n]| ≤ ∞ for all n, provided that the impulse response of the system is absolutely summable.

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


53


[ ] .k

h k∞

=−∞

< ∞∑♣ Continuous-time case:

Condition for impulse response of a stable continuous-time LTI system:

0( ) .h dτ τ

∞< ∞∫

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 nρ= − +and has the impulse response

[ ] [ ]nh n u nρ=Is this system causal, memoryless, and BIBO stable?<<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 absolutely summable?


54


0 0[ ] kk

k k kh k ρ ρ

∞ ∞ ∞

=−∞ = =

= = < ∞∑ ∑ ∑ if and only if |ρ| < 1

◆ Special case:A system can be unstable even though the impulse response has a finite value.1. Ideal integrator:

ty(t) x( )dτ τ

−∞= ∫ (2.29)

Input: x(τ) = δ (τ), then the output is y(t) = h(t) = u(t).h(t) is not absolutely integrableIdeal integrator is not stable!

2. Ideal accumulator:

[ ] [ ]n

ky n x k

=−∞

= ∑Impulse response: h[n] = u[n]

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


55


2.7.4 2.7.4 Invertible Systems and Invertible Systems and DeconvolutionDeconvolutionA system is invertible if the input to the system can be recovered from the 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.

Figure 2.24 (p. 137)Cascade of LTI system with impulse response h(t) and inverse system with

impulse response h-1(t).3. The process of recovering x(t) from h(t) ∗ x(t) is termed deconvolution.4. An inverse system performs deconvolution.

in( ) ( ( ) ( )) ( ).vx t h t h t x t∗ ∗ =

( ) ( ) ( )invh t h t tδ∗ = (2.30)

Continuous-time case

5. Discrete-time case: invh[n] h [n] [n]δ∗ = (2.31)


56


Example 2.13Example 2.13 Multipath Communication Channels: Compensation by means ofan Inverse System

Consider designing a discrete-time inverse system to eliminate the distortion associated with multipath propagation in a data transmission problem. Assume that a discrete-time model for a two-path communication channel is

[ ] [ ] [ 1].y n x n ax n= + −Find a causal inverse system that recovers x[n] from y[n]. Check whether this inverse system is stable.<<Sol.>Sol.>1. Impulse response:

1, 0[ ] , 1

0, otherwise

nh n a n

=⎧⎪= =⎨⎪⎩

2. The inverse system hinv[n] must satisfy h[n] ∗ hinv[n] = δ[n]. [ ] [ 1] [ ].inv invh n ah n nδ+ − = (2.32)

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


57


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

[ ] [ 1] 0,inv invh n ah n+ − =

(2.33) 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 impulse response

[ ] ( ) [ ]inv nh n a u n= −

inv invh [n] ah [n 1]= − −

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

[ ] kinv

k kh k a

∞ ∞

=−∞ =−∞

=∑ ∑ is finite.

♣♣ For For ||aa|| < 1, the system is stable.< 1, the system is stable.★ Table 2.2 summarizes the relation between LTI system properties and

impulse response characteristics.


58


2.8 2.8 Step ResponseStep Response1. The step response is defined as the output due to 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∞

=−∞

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


59


[ ] [ ].n

ks n h k

=−∞

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

ts(t) h( )dτ τ

−∞= ∫ (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

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

=

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

and

Example 2.14Example 2.14 RC Circuit: Step Response

1( ) ( )t

RCh t e u tRC

−=

Find the step response of the circuit.<<Sol.>Sol.>

Figure 2.12 (p. 119)RC circuit system with the voltage source x(t) as input and the voltage measured across the capacitor y(t), as output.


60


1. Step response:1( ) ( ) .

tRCs t e u d

RC

τ

τ τ−

−∞= ∫

0, 0( ) 1 ( ) 0

tRC

ts t

e u d tRC

τ

τ τ−

−∞

<⎧⎪= ⎨

≥⎪⎩ ∫

0

0, 0( ) 1 0

0, 0

1 , 0

tRC

tRC

ts t

e d tRC

t

e t

τ

τ−

−

<⎧⎪= ⎨

≥⎪⎩<⎧⎪= ⎨

⎪ − ≥⎩

∫

Fig. 2.25Fig. 2.25

Figure 2.25 (p. 140)RC circuit step response for RC = 1 s.


61


2.9 2.9 Differential and Difference Equation Representations of Differential and Difference Equation Representations of LTILTI SystemsSystems

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)

2. Linear constant-coefficient difference equation:N M

k kk 0 k 0

a y[n k] b x[n k]= =

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

♣ The order of the differential or difference equation is (N, M), representing thenumber 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 tdt C

τ τ−∞

+ + =∫

Often, N ≥ M, and the order is described

using only N.


62


Figure 2.26 (p. 141)Example of an RLC circuit described by a differential equation.

( ) ( ) ( ) ( )2

2

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

+ + = N = 2

Ex. Accelerator modeled in Section 1.10:

( ) ( ) ( ) ( )2

22

nn

d dy t y t y t x tQ dt dtωω + + =

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

N = 2


63


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

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 the second-order differenceequation (2.37).

<<Sol.>Sol.>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:


64


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

= + − − − − − (2.39)

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

= + − − − (2.40)

[ ] [ ] [ ] [ ] [ ]12 2 2 1 1 04

y x x y y= + − −

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

y x x y y= + − −

♣ 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 1

0 , 2 10 00

, , ...,N

t Nt tt

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

−

= − −= − = −= −


65


Example 2.15Example 2.15 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 y[ − 1] = 1 and y[ − 2] = − 2.<<Sol.>Sol.>1. 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 = + × − − × =

Example 2.16Example 2.16 Evaluation of a Difference Equation by means of a ComputerA 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 the past outputs and the current inputs. Use a computer to determine the step response


66


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(πn/10), x2[n] = cos(πn/5), and x3[n] = cos(7πn/10), 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. 2.27Fig. 2.27, assuming zero initial conditions.<<Sol.>Sol.>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= − − − + + − + −

2. Step response: Fig. 2.28 (a).Fig. 2.28 (a).3. Zero input response: Fig. 2.28 (b).Fig. 2.28 (b).4. The outputs due to the sinusoidal inputs x1[n], x2[n], and x3[n]: Fig. 2.28 (c),Fig. 2.28 (c),

(d), and (e).(d), and (e).5. Fig. 2.28(f)Fig. 2.28(f) shows the system output for the Intel stock price unit.

A comparison of peaks in Figs. 2.27Figs. 2.27 and 2.28 (f)2.28 (f) Slightly delay!Slightly delay!


67


Figure 2.27 (p. 144)Weekly

closing price of Intel stock.


68


Fig. 2.28 (a).Fig. 2.28 (a).


69


Fig. 2.28 (b).Fig. 2.28 (b).


70


Fig. 2.28 (d).Fig. 2.28 (d).

Fig. 2.28 (c).Fig. 2.28 (c).


71


Figure 2.28a (p. 145)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 (1/10πn). (d) Output due to x2[n] = cos (1/5πn). (e) Output due to x3[n] = cos(7/10πn).

Fig. 2.28 (e).Fig. 2.28 (e).


72


Figure 2.28f (p. 146)Output associated with the weekly closing price of Intel stock.


73


2.10 2.10 Solving Differential and Difference EquationsSolving Differential and Difference EquationsComplete solution: y = y (h) + y (p)

y (h) = homogeneous solution, y (p) = particular solution2.10.1 2.10.1 The Homogeneous SolutionThe Homogeneous Solution♣ Continuous-time case:1. Homogeneous differential equation: ( ) ( )

0

0kN

hk k

k

da y tdt=

=∑2. Homogeneous solution:

i

Nr t(h)

ii 0

y (t) c e=

= ∑ (2.41)

3. Characteristic eq.: N

kk

k 0a r 0

=

=∑ (2.42)

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

( ) [ ]0

0N

hk

ka y n k

=

− =∑N

(h) ni i

i 1y [n] c r

=

= ∑ (2.43)

3. Characteristic eq.: N

N kk

k 0a r 0−

=

=∑ (2.44)

Coefficients ci is determined by I.C.

Coefficients ci is determined by I.C.


74


♣ 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 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 differential equation

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

+ =

Determine the homogeneous solution of this equation.<<Sol.>Sol.>1. Homogeneous Eq.: ( ) ( ) 0dy t RC y t

dt+ =

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

3. Characteristic eq.: 11 0RCr+ = r1 = − 1/RC4. Homogeneous solution:

( ) ( ) 1 Vt

h RCy t c e−

=

Figure 2.30 (p. 148)RC circuit.


75


Example 2.18Example 2.18 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 nρ− − =

<<Sol.>Sol.>1. Homogeneous Eq.: [ ] [ ]1 0y n y nρ− − =2. Homo. Sol.:

( ) [ ] 1 1h ny n c r=

3. Characteristic eq.: 1 0r ρ− = r1 = ρ4. Homogeneous solution:

( ) [ ] 1h ny n c ρ=

2.10.2 2.10.2 The Particular SolutionThe Particular SolutionA particular solution is usually obtained by assuming an output of the same general form as the input.

Table 2.3Table 2.3


76


Example 2.19Example 2.19 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 nρ− − =

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

2[ ]

nppy n c=

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


77


( ) ( ) ( )11 1 12 2 2

n n n

p pc cρ−

− =

pc (1 2 ) 1ρ− = (2.45)

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

3. Particular solution:

( ) [ ] 1 11 2 2

npy n

ρ⎛ ⎞= ⎜ ⎟− ⎝ ⎠

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. Find a particular solution for this system with an input x(t) = cos(ω0t).

Figure 2.30 (p. 148)

RC circuit.

<<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 tω ω= +

3. Substituting y(p)(t) and x(t) = cos(ω0t) into the given differential Eq.:


78


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

1 0 2 1c RC cω+ =

0 1 2 0RC c cω− + =

4. Coefficients c1 and c2:

( )1 20

11

cRCω

=+ ( )

02 2

01RCcRCωω

=+

and


( ) ( )( )

( )( )

( )00 02 2

0 0

1 cos sin V1 1

p RCy t t tRC RC

ωω ωω ω

= ++ +

2.10.3 2.10.3 The Complete SolutionThe Complete SolutionComplete solution: y = y (h) + y (p)

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


79


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 the homogeneous solution.3. 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 2.21Example 2.21 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] = 8.<<Sol.>Sol.>1. Homogeneous sol.: ( ) [ ] ( )1

1 4

nhy n c=


80



( ) [ ] 122

npy n ⎞⎛= ⎜ ⎟

⎝ ⎠

3. Complete solution:n n

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

= + (2.47)

4. Coefficient c1 determined by I.C.:

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), yielding0 0

11 13 22 4

c⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

c1 = 1

5. Final solution:[ ] 1 12

2 4

n n

y n ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

for n ≥ 0

Example 2.22Example 2.22 RC Circuit (continued): Complete ResponseFind the complete response of the RC circuit depicted in Fig. 2.30Fig. 2.30 to an input x(t) = cos(t)u(t) V, assuming normalized values R = 1 Ω and C = 1 F and assuming that the initial voltage across the capacitor is y(0−) = 2 V.


81


<<Sol.>Sol.>1. Homogeneous sol.: ( ) ( ) V

th RCy t ce

−=

2. Particular solution:( ) ( )

( )( )

( )( )2 2

1 cos sin V1 1

p RCy t t tRC RC

= ++ +

4. Coefficient c1 determined by I.C.:

3. Complete solution:

( ) 1 1cos sin V2 2

ty t ce t t−= + +

ω0 = 1

R = 1 Ω, C = 1 F

y(0y(0−−) = y(0) = y(0++))

0 1 1 12 cos 0 sin 02 2 2

ce c+− + += + + = + c = 3/2

5. Final solution:

( ) 3 1 1cos sin V2 2 2

ty t e t t−= + +


82


Example 2.23Example 2.23 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 ρ = 1 + r/100.

[ ] [ ] [ ]1y n y n x nρ= − +

Use 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.<<Sol.>Sol.>1. We have ρ = 1.1 and y[− 1] = 20,000, and we assume that x[n] = b is the

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.:

( ) [ ] ( )1.1 nhhy n c=



83


( ) [ ]ppy n c=

Substituting y(p)[n] = cp and x[n] = b into the difference equation y[n] − 1.1y[n − 1] = x[n], we obtain

10pc b= −

3. Complete solution:= − ≥n

hy[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+ = −

22,000 11hc b= +

[ ] ( )( )22,000 11 1.1 10ny 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

−= = −

−

Fig. 2.31.Fig. 2.31.


84


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

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


85


2.11 2.11 Characteristics of Systems Described by DifferentialCharacteristics of Systems Described by Differentialand Difference Equationsand Difference EquationsComplete solution: y = y (n) + y (f)

y (n) = natural response, y (f) = forced response2.11.1 2.11.1 The Natural ResponseThe Natural ResponseExample 2.24Example 2.24 RC Circuit (continued): Natural Response

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

The system In Example 2.17 is described by the differential equation

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

+ =

<<Sol.>Sol.>1. Homogeneous sol.: ( ) ( ) 1 Vh ty t c e−=2. I.C.: y(0) = 2 V

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


86


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

[ ] [ ] [ ]1 14

y n y n x n− − =

Find the natural response of this system.<<Sol.>Sol.>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⎞⎛= ≥ −⎜ ⎟

⎝ ⎠2.11.2 2.11.2 The Forced ResponseThe Forced ResponseThe forced response is the system output due to the input signal assuming zero initial conditions.

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


87


♣ 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 completesolution.

Example 2.26Example 2.26 First-Order Recursive System (Continued): Forced ResponseThe system in Example 2.21Example 2.21 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].<<Sol.>Sol.>1. Complete solution:

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

[ ] [ ] [ ]10 0 14

y x y= + −

[ ] 11 12 , 02 4

n n

y n c n⎞ ⎞⎛ ⎛= + ≥⎜ ⎟ ⎜ ⎟⎝ ⎝⎠ ⎠

y[0] = 1 + (1/4) × 0 =1

3. Finding c1: 0 0

11 11 22 4

c⎞ ⎞⎛ ⎛= +⎜ ⎟ ⎜ ⎟⎝ ⎝⎠ ⎠

c1 = − 1


88


4. Forced response:

( ) [ ] 1 12 , 02 4

n nfy n n⎞ ⎞⎛ ⎛= − ≥⎜ ⎟ ⎜ ⎟

⎝ ⎝⎠ ⎠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 Ω and C = 1 F.

The system In Example 2.17 is described by the differential equation

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

+ =

<<Sol.>Sol.>1. Complete solution: ( ) 1 1cos sin V

2 2ty t ce t t−= + +

From Example 2.22

2. I.C.:y(0−) = y(0+) = 0 c = − 1/2

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

2 2 2f ty t e t t−= − + +


89


2.11.3 2.11.3 The Impulse ResponseThe Impulse Response♣ Relation between step response and impulse response1. Continuous-time case:

( ) ( )dh t s tdt

=

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

2.11.4 2.11.4 Linearity and Time InvarianceLinearity and Time Invariance

Input Forced response

x1 y1(f)

x2 y2(f)

α x1 + β x2 αy1(f) + β y2

(f)

♣ Forced response ⇒ Linearity

Initial Cond. Natural response

I1 y1(n)

I2 y2(n)

α I1 + β I2 αy1(n) + β y2

(n)

♣ Natural response ⇒ Linearity

♣ 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.

2.11.5 2.11.5 Roots of the Characteristic EquationRoots of the Characteristic Equation


90


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

★★ BIBO Stable:BIBO Stable:1. Discrete-time case: boundedn

ir 1, for allir i<

2. Continuous-time case: boundedir te { } 0ie rℜ <

and1ir = { } 0ie rℜ = ⇒ The system is said to be on the verge of instability.

2.12 2.12 Block Diagram RepresentationsBlock Diagram Representations♣ A block diagram is an interconnection of the elementary operations 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 a scalar.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 dτ τ

−∞= ∫; and a time shift for discrete-time LTI

system: y[n] = x[n − 1].Fig. 2.32.Fig. 2.32.


91

(a)

(b)

( ) ( )t

y t x dτ τ−∞

= ∫

(c)

Figure 2.32 (p. 162)Symbols for elementary operations in block diagram descriptions of systems. (a) Scalar multiplication. (b)

Addition. (c) Integration for continuous-time systems and time shifting for discrete-time systems.

∫


Ex. A discrete-time LTI system: Fig. 2.33.Fig. 2.33.

1. In dashed box:0 1 2w[n] b x[n] b x[n 1] b x[n 2]= + − + − (2.49)

2. y[n] in terms of w[n]:

♣♣ Direct Form I:Direct Form I:


92


1 2y[n] w[n] a y[n 1] a y[n 2]= − − − − (2.50) 3. System output y[n] in terms of input x[n]:

= − − − − + + − + −1 2 0 1 2y[n] a y[n 1] a y[n 2] b x[n] b x[n 1] b x[n 2]

1 2 0 1 2y[n] a y[n 1] a y[n 2] b x[n] b x[n 1] b x[n 2]+ − + − = + − + − (2.51)

Figure 2.33 (p. 162)Block diagram representation of a discrete-time LTI system described by a second-order difference equation.

Cascade Form(Direct Form I)


93


♣♣ Direct Form II:Direct Form II:1. Interchange the order of Direct Form I.2. Denote the output of the new first system as f[n].

1 2f[n] a f[n 1] a f[n 2] x[n]= − − − − + (2.52) Input: x[n]

3. The signal is also the input to the second system. The output of the second system is

0 1 2y[n] b f[n] b f[n 1] b f[n 2]= + − + − (2.53) Fig. 2.35.Fig. 2.35.

Figure 2.35 (p. 164)Direct form II representation of an LTI system described by a second-order

difference equation.


94


♣ Block diagram representation for continuous-time LTI system:1. Differential Eq.:

k kN M

k kk kk 0 k 0

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

=∑ ∑ (2.54)

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

( ) ( ) ( ) ( )1 , 1, 2, 3, ...tn nt d nν ν τ τ−

−∞= =∫

v(n)(t) is the n-fold integral of v(t) with respect to time3. Integrator with initial condition:

( ) ( ) ( ) ( )1 , 0 and 1, 2, 3, ...n nd t t t ndtν ν −= > =

( ) ( ) ( ) ( ) ( ) ( )1

00 , 1, 2, 3, ...

tn n nt d nν ν τ τ ν−= + =∫

(N k )N M

(N k)k k

k 0 k 0a y (t) b x (t)

−−

= =

=∑ ∑ (2.55)

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

(2.56)

Block diagram:Fig. 2.37


95

∫

∫

∫

∫

∫

∫


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)

(b)


96


2.13 2.13 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 system’s entire memory of the past.Given 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).

2.13.1 2.13.1 The StateThe State--Variable DescriptionVariable Description1. Direct form II of a second-order LTI system: Fig. 2.39.Fig. 2.39.2. 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) 4. Output equation:

1 1 1 2 2 2y[n] (b a )q [n] (b a )q [n] x[n]= − + − + (2.59)

1 11 2

2 2

q [n 1] q [n]a a 1x[n]

q [n 1] q [n]1 0 0+ − −⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥+ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦(2.60)

= − − + +1 1 2 2 1 1 2 2y[n] x[n] a q [n] a q [n] b q [n] b q [n],

5. Matrix Form of state equation:


97


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

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


98


6. Matrix form of output equation:

11 1 2 2

2

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

q [n]⎡ ⎤

= − − +⎢ ⎥⎣ ⎦

(2.61)

Define state vector as the column vector⎡ ⎤

= ⎢ ⎥⎣ ⎦

1

2

q [n][n]

q [n]q

We can rewrite Eqs. (2.60) and (2.61) as[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 2.28Example 2.28 State-Variable Description of a Second-Order SystemFind the state-variable description corresponding to the system depicted in Fig. 2.40Fig. 2.40 by choosing the state variable to be the outputs of the unit delays.<<Sol.>Sol.>


99


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

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

[ ] [ ] [ ] [ ]2 1 2 21q n q n q n x nγ β δ+ = + +


100


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

3. Define state vector as

[ ] [ ][ ]

1

2

qq n

nq n⎡ ⎤

= ⎢ ⎥⎣ ⎦

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

(2.62)

0A

αγ β⎡ ⎤

= ⎢ ⎥⎣ ⎦

1

2

bδδ⎡ ⎤

= ⎢ ⎥⎣ ⎦

[ ]1 2c η η= [ ]2D =

♣ State-variable description for continuous-time systems:d (t) (t) x(t)dt

= +q Aq b (2.64)

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


101


Example 2.29Example 2.29 State-Variable Description of an Electrical CircuitConsider the electrical circuit depicted in Fig. 2.42Fig. 2.42. Derive a state-variable description of this system if the input is the applied voltage x(t) and the output is the current y(t) through the resistor.<<Sol.>Sol.>

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= +

11 1

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

= − + (2.66)

Output equation

3. KVL Eq. for the loop involving C1, R2, and C2:( )1 2 2 2( ) ( )q t R i t q t= +


102


2 1 22 2

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

= − (2.67)

4. The current i2(t) through R2:

( )2 2 2 ( )di t C q tdt

= ( )2 1 22 2 2 2

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

= − (2.68)

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

5. KCL Eq. between R1 and R2:( ) ( ) ( )1 2y t i t i t= + Current through C1 = i1(t)

where( ) ( )1 1 1

di t C q tdt

=

( )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.

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

q Aq b (2.64)

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


103


1 1 1 2 1 2

2 2 2 2

1 1 1

,1 1

AC R C R C R

C R C R

⎡ ⎤⎞⎛− +⎢ ⎥⎟⎜⎝ ⎠⎢ ⎥=

⎢ ⎥−⎢ ⎥

⎣ ⎦

1 1

1

0b C R

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

1

1 0 ,cR

⎡ ⎤= −⎢ ⎥⎣ ⎦ 1

1DR

=and

Example 2.30Example 2.30 State-Variable Description from a Block DiagramDetermine the state-variable description corresponding to the block diagram in Fig. 2.44Fig. 2.44. 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.

∫ ∫

<<Sol.>Sol.>


104


1. State equation:

( ) ( ) ( ) ( )1 1 22d q t q t q t x tdt

= − +

( ) ( )2 1d q t q tdt

=

2. Output equation:

( ) ( ) ( )1 23y t q t q t= +

3. State-variable description:2 1

,1 0

A−⎡ ⎤

= ⎢ ⎥⎣ ⎦

1,

0b ⎡ ⎤= ⎢ ⎥⎣ ⎦

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

2.13.2 2.13.2 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)

cqy Dx= + (2.71) 2. Transformation: q’ = Tq

T = state-transformation matrix

q = T−1 q’


105


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

q Tq′ =& &

1 .q TAT q Tbx−′ ′= +& q = T−1 q’2) Output equation:

1 .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 Example 2.30Example 2.30 again. Let us define new states

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

<<Sol.>Sol.>1. State equation: 1 1 2 2 2 1( ) 2 ( ) ( ) ( ) ( ) 2 ( ) ( ) ( )

old description New description

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

′ ′ ′= − + ⇒ = − +1444442444443 1444442444443


106


2 1 1 2( ) ( ) ( ) ( )

Old description New description

d dq t q t q t q tdt dt

′ ′= ⇒ =1442443 1442443

1 2 2 13 ( ) ( ) 3 ( ) ( )Old description New description

y q t q t y q t q t′ ′= + ⇒ = +1442443 1442443

2. Output equation:

3. State-variable description:

0 1,

1 2A ⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

0,

1b ⎡ ⎤′ = ⎢ ⎥

⎣ ⎦

[ ]1 3 ,c′ = [ ]0 .D′ =

Example 2.31Example 2.31 Transforming The StateA 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 1

1 1 22 2[ ] [ ] [ ]q n q n q n′ = − + 1 12 1 22 2[ ] [ ] [ ]q n q n q n′ = +

<<Sol.>Sol.>1. Transformation: q′ = Tq, where

and

1 11 .1 12

T−⎡ ⎤

= ⎢ ⎥⎣ ⎦


107


1 1 1.

1 1T− −⎡ ⎤

= ⎢ ⎥⎣ ⎦

2. New state-variable description:12

310

0,

0A

−⎡ ⎤′ = ⎢ ⎥

⎢ ⎥⎣ ⎦

1,

3b ⎡ ⎤′ = ⎢ ⎥

⎣ ⎦[ ]0 1 ,c′ = 2.D′ =and

♣ 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 1112

q n q n x n+ = − + [ ] [ ] [ ]2 231 3

10q n q n x n+ = +and

2.14 2.14 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.


108


2.14.1 2.14.1 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 inx and h, minus one.

<<pf.>pf.>1) 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) 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 2.1Impulse 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: Ly = ly – ky + 1 = 4MATLAB Program: >> h = [1, 0.5];

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

y =

2 5 0 -1


109


Repeat Example 2.3

Given [ ] ( ) [ ] [ ]( )1 4 4h n u n u n= − − [ ] [ ] [ ]10 .x n u n u n= − −and

Impulse response Input

Find the convolution sum x[n] ∗ h[n].<<Sol.>Sol.>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:>> n = 0:12;>> y = conv(x, h);>> stem(n, y); xlabel('n'); ylabel('y[n]')

Fig. 2.45.Fig. 2.45.


110


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


111


2.14.2 2.14.2 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.3. 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.1) 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 by

[ ] ( ) [ ]nh n u nρ=

with ρ = − 0.9, by using MATLAB program.<<Sol.>Sol.>1. MATLAB Commands:


112


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

2. Step response: Fig. 2.47.Fig. 2.47.

Figure 2.47 (p. 178)Step response

computed using MATLAB.

2.14.3 2.14.3 Simulating Difference equationsSimulating Difference equations

1. Difference equation:= =

− = −∑ ∑N M

k kk 0 k 0

a y[n k] b x[n k] (2.36) Command:Command:filter


113


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) 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− − + − = + − + −(2.73)Determine the output in response to zero input and initial condition

y[−1] = 1 and y[−2] = 2.<<Sol.>Sol.>


114


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

2. Output: Fig. 2.28(b).Fig. 2.28(b).

0 10 20 30 40 50-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

3. System response to aninput consisting of theIntel stock price data Intc:

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

♣ We have assume that theIntel stock price data arein the file Intc.mat.

♣ The command [h, t] = impz(b, a, n) evaluates n values of the impulse responseof a system described by a different equation.


115


2.14.4 2.14.4 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 − 1♣ System manipulation:1. sys = sys1 + sys2 Parallel combination of sys1 and sys2.2. 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: impulse

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 transformation1. Command form: sysT = ss2ss(sys, T), where T = Transformation matrix


116


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

1 41 ,4 110

A−⎡ ⎤

= ⎢ ⎥−⎣ ⎦

2,

4b ⎡ ⎤= ⎢ ⎥⎣ ⎦

[ ]1 1 1 ,2

c = 2,D =and

2. State-transformation matrix:1 11 .

1 12T

−⎡ ⎤= ⎢ ⎥

⎣ ⎦3. MATLAB Program:

>> a = [-0.1, 0.4; 0.4, -0.1]; b = [2; 4];>> 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)

4. Result:


117


a = x1 x2

x1 -0.5 0x2 0 0.3

b = u1

x1 1x2 3

c = x1 x2

y1 0 1

d = u1

y1 2

Sampling time: unspecifiedDiscrete-time model.

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

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

>> subplot(2, 1, 1)>> stem([0:9], h)>> title ('Original System Impulse Response');>> xlabel('Time'); ylabel('Amplitude')>> subplot(2, 1, 2)>> stem([0:9], hT)>> title('Transformed System Impulse Response');>> xlabel('Time'); ylabel('Amplitude')


118


2. Simulation results:Fig. 2.48.Fig. 2.48.

0 2 4 6 8 100

1

2

3Original Sys te m Impulse Re sponse

TimeA

mpl

itude

0 2 4 6 8 100

1

2

3Transformed Syste m Impulse Re sponse

Time

Am

plitu

de

Figure 2.48 (p. 181)Impulse responses associated with the original and transformed state-variable descriptions computer using MATLAB.

♣ We may verify thatthe original andtransformed systemshave the (numerically)identical impulse response by computingthe error, err = h – hT.


119


0 2 4 6 8 10-6

-5

-4

-3

-2

-1

0

1x 10

-17

Time

Am

plitu

de e

rr

Plot for err = h Plot for err = h −− hThT


120


2.2 convolution sum - nctu - mapl

Documents