time-domain representations of lti systems

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1 Time-Domain Representations of LTI Time-Domain Representations of LTI Systems Systems CHAPTER 2.1 Introduction 2.1 Introduction Objectives: 1. Impulse responses of LTI systems 2. Linear constant-coefficients differential or difference equations of LTI systems 3. Block diagram representations of LTI systems 4. State-variable descriptions for LTI systems 2.2 Convolution Sum 2.2 Convolution Sum 1. An arbitrary signal is expressed as a weighted 1. An arbitrary signal is expressed as a weighted superposition of shifted superposition of shifted impulses. impulses. Discrete-time signal x[n]: 0 xn n x n xn n k xk n k 2 2 1 1 0 1 1 2 2 xn x n x n x n x n x n Fig. 2.1 Fig. 2.1 x[n] = entire signal; x [k] = specific value of the signal x[n] at time k.

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2. CHAPTER. Time-Domain Representations of LTI Systems. 2.1 Introduction. Objectives: Impulse responses of LTI systems Linear constant-coefficients differential or difference equations of LTI systems Block diagram representations of LTI systems State-variable descriptions for LTI systems. - PowerPoint PPT Presentation

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Page 1: Time-Domain Representations of LTI Systems

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

2.1 Introduction2.1 IntroductionObjectives: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 Convolution Sum2.2 Convolution Sum1. An arbitrary signal is expressed as a weighted superposition of shifted1. An arbitrary signal is expressed as a weighted superposition of shifted impulses.impulses.

Discrete-time signal x[n]:

0x n n x n

x n n k x k n k

2 2 1 1 0

1 1 2 2

x n x n x n x n

x n x n

Fig. 2.1Fig. 2.1

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

time k.

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

k

y n H x n H x k n k

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 k

(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 k

k

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

(2.2)

Linearity

Linearity

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

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

For time-invariant system:

H{ [n k]} h[n k]h[n] = H{ [n]} impulse response

of the LTI system H(2.3)

k

y[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].

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

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

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

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

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

11

2y n x n x n

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

1, 0

1, 120, otherwise

n

h n n

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

Determine the output of this system in response to the input

2, 0

4, 1

2, 2

0, otherwise

n

nx n

n

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

Input = 0 for n < 0 and n > 0

2. Since

[n k]

time-shifted impulse input

h [n k]

time-shifted impulse response output

3. Output:

2 4 1 2 2y n h n h n h n

0, 0

2, 0

5, 1

0, 2

1, 3

0, 4

n

n

ny n

n

n

n

(convolution of x[n] and h[n])

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

2.3 Convolution Sum Evaluation Procedure2.3 Convolution 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] (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]

(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 3

4

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].

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

<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)n-k

u[n-k]

3

,4

0, otherwise

n k

k nh n k

5 0w k

5

5

3, 0 5

4

0, otherwise

k

kw k

2. Intermediate signal wn[k]:

For n = 5:

Eq. (2.6) y[ 5] = 0

For n = 5:

Eq. (2.6) 55

0

35

4

k

k

y

For n = 10:

10

10

3, 0 10

4

0, otherwise

k

kw k

Eq. (2.6)

11

10 10 1010 10

0 0

41

3 3 4 3 31044 4 3 4 13

3.831

k k

k k

y

6

5 55

0

41

3 4 3 35 3.28844 3 4 13

k

k

y

(x[k]=u[k]=0, k<=-5=n)

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

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].

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3

, 04

0, otherwise

n k

n

k nw k

Procedure 2.1:Procedure 2.1: Reflect and Shift Convolution Sum Evaluation Reflect 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 time axis.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 occurs defines 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 times 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 interval.

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

Example 2.3Example 2.3 Moving-Average System: Reflect-and-shift Convolution Sum EvaluationThe output y[n] of the four-point moving-average system is related to the input x[n] according to the formula

3

0

1

4 ky n x n k

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

14

4h 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, 0

0, 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).

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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].

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For n = 1:

1

1/ 4, 0,1

0, otherwise

kw k

For general case: n 0:

1/ 4, 0

0, otherwisen

k nw k

4. 3’rd interval: 3 < n ≤ 9

1/ 4, 3 9

0, otherwisen

n kw k

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

5. 4th interval: 9 < n ≤ 12

Fig. 2.4 (e).Fig. 2.4 (e).

1/ 4, 3

0, 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/ 4

4

n

k

ny n

3) For 3 < n ≤ 9:

3

11/ 4 3 1 1

4

n

k n

y n n n

4) For 9 < n ≤ 12:

9

3

1 131/ 4 9 3 1

4 4k n

ny n n

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

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Example 2.4Example 2.4 First-order Recursive System: Reflect-and-shift Convolution Sum EvaluationThe 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 and that the system is causal.<Sol.><Sol.>1. Impulse response: 1h n h n n

Since the system is causal, we have h[n] = 0 for n < 0 (why?). 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. 2.5 (a).Fig. 2.5 (a).

, 4

0, otherwise

kb kx k

,

0, otherwise

n k k nh n k

and

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

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

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.

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

Assuming that = 0.9 and b = 0.8.

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

, 4

0, otherwise

k n k

n

b k nw k

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

4

knn

k

by n

Let m = k + 4, then

4 44 4

0 0

m mn nn n

m m

b by n

b

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

b

by n b

bb b

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

5 54

0, 4

, 4n n

n

y n bb n

b

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

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

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

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 = 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] = y[n 1]. Now, suppose x[n] is the amount deposited (x[n] > 0) or withdrawn (x[n] < 0) at the start of period n. In this case, the value of the amount is expressed by the first-order recursive equation

1y n y n x n

We use convolution to find the value of an investment earning 8 % per year if $1000 is deposited at the start of each year for 10 years 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] =

n u[n]

is as shown in Example 2.4 with = 1.08.

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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 < 02’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

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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.

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2) For 0 ≤ n ≤ 9:

1000 1.08 , 0

0, otherwise

n k

n

k nw k

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

0 0

11000 1.08 1000 1.08

1.08

kn nn k n

k k

y n

1

1

11

1.081000 1.08 12,500 1.08 11

11.08

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 , 10

0, otherwise

n k

n k

n

k

w k k n

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

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9

0 0

1000 1.08 1500 1.08n

n k n k

k k

y n

9

0

10

0

10

08.1

108.11500

08.1

108.11000

k

n

m

mn

kn

m = k 10

Apply the formula for summing a geometric

series

08.1

11

08.1

11

08.11500

08.1

11

08.1

11

08.11000

9

10

10 n

nnny

97246.89 1.08 18,750 1.08 1 , 10 16

n nn

4) For 17 ≤ n :

1000 1.08 , 0 9

1500 1.08 , 10 16

0, otherwise

n k

n k

n

k

w k k

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

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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 of the investment at the start of each period, by combining the results for each 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.

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2.4 The Convolution Integral2.4 The Convolution Integral1. A continuous-time signal can be expressed as a weighted superposition of 1. A continuous-time signal can be expressed as a weighted superposition of time-shifted impulses.time-shifted impulses.

-x(t) x( ) (t - )d

(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 d

-y(t) x( )H{ (t - )}d

(2.10)

Linearity property

3. h(t) = H{ (t)} impulse response of the LTI system H

If the system is also time invariant, then

H{ (t - )} h(t - ) (2.11) A time-shifted impulse

generates a time-shifted impulse response output

Fig. 2.9.Fig. 2.9.-

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

(2.12)

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Convolution integral:

x t h t x h t d

2.5 Convolution Integral Evaluation Procedure2.5 Convolution Integral Evaluation Procedure1. Convolution integral:

-y(t) x( )h(t )d

(2.13)

2. Define the intermediate signal: tw x h t

= independent variable, t = constant

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

3. Output:

t-

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

output of the system.

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Procedure 2.2:Procedure 2.2: Reflect and Shift Convolution Integral Evaluation Reflect 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 on 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 wt () changes. The value of t at which the change occurs

defines 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 the

corresponding mathematical representations for wt () are identified. This

usually implies increasing t to a very large positive number.

6. For each sets of shifts t, integrate wt () from = to = to obtain y(t). Example 2.6Example 2.6 Reflect-and-shift Convolution Evaluation

Given 1 3x t u t u t 2h t u t u t and as depicted in Fig. 2-10Fig. 2-10,

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

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

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

0, otherwiset

tw

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

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

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

t t

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5. Third interval: 3 ≤ t < 5

1, 2 3

0, otherwiset

tw

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

6. Fourth interval: 5 ≤ t wt() = 0

7. Convolution integral: 1) For t < 1 and t 5: y(t) = 02) For second interval 1 ≤ t < 3, y(t) = t 1 3) For third interval 3 ≤ t < 5, y(t) = 3 (t 2)

0, 1

1, 1 3

5 , 3 5

0, 5

t

t ty t

t 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 capacit

or y(t), as output.

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<Sol.><Sol.>1. Graph of x() and h(t ): 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 2

0, otherwisex

and

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

wt() = 0

For t > 0, , 0

0, otherwise

t

t

e tw

Fig. 2.13 (b).Fig. 2.13 (b).

5. Third interval: 2 ≤ t

, 0 2

0, otherwise

t

t

ew

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

,

0, otherwise

tt e t

h t e u t

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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 . (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

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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 e

2 2 2

001t t ty t e d e e e e

2

0, 0

1 , 0 2

1 , 2

t

t

t

y 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.

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<Sol.><Sol.>1. Graph of x() and h(t ): 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 5

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

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

1, 1 1

0, otherwiset

tw

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

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

1, 1 3

0, otherwisetw

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

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

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Figure 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 wt() for 3 t < 5. (e) The product signal wt() for t 5.

The system output y(t).

t t

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1 , 1 2

2 31,

0, otherwiset

t

w t

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

7. Fifth interval of time shifts: t 5

1 , 1 3

0, otherwisetw

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 2

t t ty t d

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 t

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2

2

0, 0

, 0 222, 2 3

6 7, 3 5

2, 5

t

tt

y tt

t t t

t

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

Example 2.9Example 2.9 Radar range Measurement: Propagation Model

We identify an LTI system describing the propagation of the pulse. Let the transmitted RF pulse be given by

0sin , 0

0, otherwisect t T

x t

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) = a (t ), where a represents the attenuation factor and the round-trip time delay. Use the convolution of x(t) with h(t) to verify this result.

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<Sol.><Sol.>1. 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

is an attenuated and delayed version of the transmitted pulse.

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Example 2.10Example 2.10 Radar range Measurement (continued): The Matched Filter

In 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 , 0

0, otherwisec

m

t T th t

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 t

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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() superimposed on the reflected and time-shifted matched filter impulse response hm(t – ), depicted as functions of . (c) Matched fil

ter output x(t).

t t

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2. The received signal r() and the reflected, time-shifted impulse response

hm(t ) are shown in Fig. 2.17(b).Fig. 2.17(b).

0sin sin ,

0, otherwise

c ct

a w w t t Tw

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

1’st interval: t < T0

2’nd interval: T0 < t ≤

3’rd interval: < t ≤ + T0

4th interval: t + T0

4. First interval of time shifts: t < T0 wt() = 0 and y(t) = 0

5. Second interval of time shifts: T0 < t ≤

0

/ 2 cos / 2 cos 2t T

c cy t a w t a w t d

02sin4/cos2/ 0

Tt

ccc twaTtwa

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0 0( ) / 2 cos / 4 sin 2 sinc c c cy t a w t t T a w t T w t 6. 3’rd interval of time shifts: < t ≤ + T0

0sin ( ) sin ( ) ,( )

0, otherwisec c

t

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) = 0

8. The output of matched filter:

0 0

0 0

( / 2) ( ) cos ( ) ,

( ) ( / 2) ) cos ( ) ,

0, otherwise

c

c

a t t t

y t a t t t