信号与系统第二版课后习题解答(6,7,9)奥本海姆

Chap 6

6.1 Consider a continuous-time LTI system with frequency response

H(j ) |H(j )|e H(j )and real impulse response h(t).

Suppose that we apply an input x(t) cos( 0t 0) to this system .The resulting output can be shown to be of the form

y(t) Ax(t t0)

Where A is a nonnegative real number representing an amplitude-scaling factor and t0 is a time delay.

(a)Express A in terms of |H(j )|. (b)Express t0 in terms of H(j 0) Solution:

(a) For y(t) Ax(t t0)

So Y(j ) AX(j )e H(j )

jt0

Y(j )

Ae j t0

X(j )

So A |H(j )|

(b) for H(j ) t0 So t0

H(j )

6.3 Consider the following frequency response for a causal and stable LTI system:

H(j )

1 j

1 j

(a) Show that |H(j )| A,and determine the values of A. (b)Determine which of the following statements is true about

( ),the group delay of the system.(Note

( ) d( H(j ))/d ,where H(j )is expressed in a

form that does not contain any discontinuities.) 1. ( ) 0 for 0 2. ( ) 0 for 0 3 ( ) 0 for 0 Solution:

(a) for |H(j )| 1 So A=1

H(j ) (1 j ) (1 j )

(b) for

arctg( ) arctg( ) 2arctg( )

( )

d H(j )2

2

d 1

So ( ) 0 for 0

6.5 Consider a continuous-time ideal bandpass filter whose frequency

response is

1, | | 3 c

H(j ) c

0,elsewhere

(a) If h(t) is the impulse response of this filter, determine a function

g(t) such that

h(t)

(b) As

sin ct

g(t) t

c is increased, dose the impulse response of the filter get

more concentrated or less concentrated about the origin? Solution

(a) Method 1. Let

1

h(t) x(t)g(t) H(j ) X(j ) G(j )

2

They are shown in the figures,where

x(t)

sin ct

t

1, c

X(j ) {

0, c

So we can get

g(t) 2cos (2t )G j( ) 2 [ (c 2 ) c( c

Method 2. Using the inverse FT definition,it is obtained

3 1

h(t) { ej td ej td }

2 3 11

{sin3 ct sin ct} {sin ct}{2cos2 ct} t t(b) more concentrated.

c

c

c

c

Chap 7

7.1 A real-valued signal x(t) is know to be uniquely determined by its samples when the sampling frequency is values of Solution:

According to the sampling theorem ws 2wM That is wM

s 10,000 .For what

is X(j ) guaranteed to be zero?

110000 ws 5000 22

So if w wM 5000 ,X(jw) 0

7.2 A continuous-time signal x(t) is obtained at the output of an ideal lowpass filter with cutoff frequency

c 1,000 .If impulse-train

sampling is performed on x(t), which of the following sampling

periods would guarantee that x(t) can be recovered from its sampled version using an appropriate lowpass filter? (a) T 0.5 10 (b) T 2 10 (c) T 10 Solution:

4

3 3

wM wc 1000

From the sampling theorem,

,that is Ts ws 2wM 2000

2 2

10 3 2wM2000

the conditions (a) and (c) are satisfied with the sampling

theorem,(b) is not satisfied.

7.3 The frequency which, under the sampling theorem, must be exceeded by the sampling frequency is called the Nyquist rate. Determine the Nyquist rate corresponding to each of the following signals:

(a)x(t) 1 cos(2,000 t) sin(4,000 t) (b)x(t)

sin(4,000 t)

t

Solution:

t

t) sin(4000 t) (a) x(t) 1 cos(2000

wM max(0,2000 ,4000 ) 4000 the Nyquist rate is ws 2wM 8000

(b) x(t)

sin(4000 t)

t

wM 4000

the Nyquist rate is ws 2wM 8000 sin(4000 t)

(c) x(t)

t

1 sin(4000 t)

x(t) 2 2t2(1 cos(8000 t))

t

22

wM 8000

the Nyquist rate is ws 2wM 16000

7.4 Let x(t) be a signal with Nyquist rate 0. Determine the Nyquist rate for each of the following signals: (a)x(t) x(t 1) (b)

dx(t) dt

2

(c)x(t) (d)x(t)cos 0t Solution:

(a) we let y1(t) x(t) x(t 1) So Y1(j ) X(j ) e

j

X(j ) (1 e j )X(j )

So the Nyquist rate of signal (a) is

0.

2dt

So Y2(j ) j X(j ) So the Nyquist rate of signal (b) is (c) we let y3(t) x2(t) So Y3(j )

0.

1

X(j )*X(j ) 2

So the Nyquist rate of signal (c) is 2 0. (d) we let y4(t) x(t)cos 0t

For cos 0t [ ( 0) ( 0)] So Y4(j )

FT

1

(X(j( 0) X(j( 0)) 2

So the Nyquist rate of signal (d) is 3 0

7.9 Consider the signal x(t) (

sin50 t2

) t

Which we wish to sample with a sampling frequency of

s 150

to obtain a signal g(t) with Fourier transform G(j ).Determine the maximum value of

0 for which it is guaranteed that

G(j ) 75X(j ) for | | 0

Where X(j ) is the Fourier transform of x(t). Solution:

1 sin(50 t) x(t) 2 2t2(1 cos(100 t))

t

2

wM 100

But ws 150

the figure about before-sampling and after-sampling of H(jw)is

We can see that only when w0 50 , the before-sampling and after-sampling of H(jw) have the same figure.

So if

G(jw) 75X(jw)..for w0

The maximum value of w0 is 50 .

Chap 9

9.2 Consider the signal x(t) e 5tu(t 1) and denote its Laplace transform by X(s).

(a)Using eq.(9.3),evaluate X(s) and specify its region of convergence. (b)Determine the values of the finite numbers A and t0 such that the Laplace transform G(s) of g(t) Ae 5tu( t t0) has the same algebraic form as X(s).what is the region of convergence corresponding to G(s)? Solution:

(a). According to eq.(9.3), we will get

X(s)

1

x(t)edt

st

e 5tu(t 1)e stdt

e (s 5)tdt

e (s 5)t e (s 5)e (s 5)

(s 5)1 (s 5)(s 5)

Re{s}>-5

ROC:

(b). g(t) Ae 5tu( t t0)

LT G(s)

A(s 5)t0

e, Re{s}<-5 s 5

If G(s) X(s)

then it’s obviously that A=-1, t0 1, Re{s}<-5.

9.5 For each of the following algebraic expressions for the Laplace transform of a signal, determine the number of zeros located in the finite s-plane and the number of zeros located at infinity:

11 s 1s 3s 1(b) 2

s 1

(a)

s3 1(c) 2

s s 1

Solution: (a).1, 1

112s 4

s 1s 3(s 1)(s 3)

it has a zero in the finite s-plane, that is s 2

And because the order of the denominator exceeds the order

of the numerator by 1

X(s) has 1 zero at infinity. (b). 0, 1

s 1s 11

2

(s 1)(s 1)s 1s 1

it has no zero in the finite s-plane.

And because the order of the denominator exceeds the order

of the numerator by 1 X(s) has 1 zero at infinity. (c). 1, 0

s3 1(s 1)(s2 s 1)

s 1 2

s s 1s2 s 1

it has a zero in the finite s-plane, that is s 1

And because the order of the denominator equals to the order

of the numerator X(s) has no zero at infinity.

9.7 How many signals have a Laplace transform that may be expressed as

(s 1)

in its region of 2

(s 2)(s 3)(s s 1)

convergence? Solution: There are 4 poles in the expression, but only 3 of them have different real part. The s-plane will be divided into 4 strips which parallel to the jw-axis and have no cut-across. There are 4 signals having the same Laplace transform expression.

9.8 Let x(t) be a signal that has a rational Laplace transform with exactly two poles located at s=-1 and s=-3. If

g(t) e2tx(t) and G(j )[ the Fourier transform of g(t)]

converges, determine whether x(t) is left sided, right sided, or two sided. Solution:

g(t) e2tx(t)

G(s) X(s 2) ROC: R(x)+Re{2}

And x(t) have three possible ROC strips:

( , 3),( 3, 1),( 1, )

g(t) have three possible ROC strips: ( , 1),( 1,1),(1, )

IF G(jw) G(s)|s jw Then the ROC of G(s) is (-1,1)

x(t) is two sides.

9.9 Given that

e atu(t)

1

,Re{s} Re{ a} s a

s

Determine the inverse Laplace transform of

X(s)

2(s 2)

,Re{s} 3

s2 7s 12

Solution:

It is obtained from the partial-fractional expansion:

X(s)

2(s 2)2(s 2)4 2

,

s2 7s 12(s 4)(s 3)s 4s 3

Re{s} 3

We can get the inverse Laplace transform from given formula and linear property.

x(t) 4e 4tu(t) 2e 3tu(t)

9.10 Using geometric evaluation of the magnitude of the Fourier transform from the corresponding pole-zero plot ,determine, for each of the following Laplace transforms, whether the magnitude of the corresponding Fourier transform is approximately lowpass, highpass, or bandpass. (a): H1(s) (b): H2(s)

1

,.........Re{s} 1

(s 1)(s 3)

s1

, e{s}

s2 s 12

s2

, e{s} 1 (c): H3(s) 2

s 2s 1

Solution: (a). H1(s)

1

,.........Re{s} 1

(s 1)(s 3)

It’

s lowpass.

(b).H2(s)

s1

,.........Re{s}

2s2 s 1

It’

s bandpass.

s2

,..........Re{s} 1 (c). H3(s) 2

s 2s 1

It’

s highpass.

t

9.13 Let g(t) x(t) x( t) ,Where x(t) eu(t). And

the Laplace transform of g(t) is G(s) Determine the values of the constants Solution:

s

, 1 e{s} 1. 2

s 1

and

t

g(t) x(t) x( t),and x(t) eu(t)

The Laplace transform : G(s) X(s) X( s) and

X(s)

s 1

,Re{s} 1

From the scale property of Laplace transform,

X( s)

So

s 1

,Re{s} 1

G(s) X(s) X( s)

s 1 s 1

(1 )s (1 )

s2 1

,

1 Re{s} 1

s

, 1 Re{s} 1 2

s 1

1

We can determine : 1,

2

From given G(s)

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