Ex Signal and System

8/3/2019 Ex Signal and System

1/25

Chapter 1 Elementary Signals

1-20 Signals and Systems with MATLAB Applications, Second EditionOrchard Publications

1.9 Exercises

1. Evaluate the following functions:

a.

b.

c.

d.

e.

f.

2.

a. Express the voltage waveform shown in Figure 1.24, as a sum of unit step functions for

the time interval .

b. Using the result of part (a), compute the derivative of , and sketch its waveform.

Figure 1.24. Waveform for Exercise 2

tsin t 6---

2tcos t 4---

t2 t

2---

cos

2ttan t 8---

t2e

t t 2( ) td

t2 1 t

2---

sin

v t( )

0 t 7s<

1 mH

S

t 0=

iL t( ) +L

32V

10

20

R1

R2

S t 0=

vc t( ) t 0>

S

t 0=

+

72V

6 K

C

+60 K

30 K 20 K

10 K40

9------F

vC t( )

R1

R2

R3 R4

R5

i1 t( ) i2 t( )

iL(0

) 0= vc(0

) 0=

+

C

+

1

3

1F

i1 t( )+

1H

v1 t( ) u0 t( )=

v2 t( ) 2u0 t( )=

2H

i2 t( )

L1

R1

R2

L2


8/25

Signals and Systems with MATLAB Applications, Second Edition 4-19Orchard Publications

Exercises

4. For the circuit of Figure 4.25,

a. compute the admittance

b. compute the value of when , and all initial conditions are zero.

Figure 4.25. Circuit for Exercise 4

5. Derive the transfer functions for the networks (a) and (b) of Figure 4.26.

Figure 4.26. Networks for Exercise 5




s domain

Y s( ) I1 s( ) V1 s( )=

t domain i1 t( ) v1 t( ) u0 t( )=

R1

R2+

1

+

R3

1

3 1 s

V1 s( )

VC s( )I1 s( )

+

V2 s( ) 2VC s( )=2 R4

R

C

+

+

Vin s( ) Vou t s( )R

L

+

Vin s( )

+

Vou t s( )

(a) (b)

R

C

+

+

Vin s( ) Vou t s( )

R

L

+

Vin s( )

+

Vou t s( )

(a) (b)


9/25

Chapter 4 Circuit Analysis with Laplace Transforms

4-20 Signals and Systems with MATLAB Applications, Second Edition

Orchard Publications


8. Derive the transfer function for the networks (a) and (b) of Figure 4.29.


9. Derive the transfer function for the network of Figure 4.30. Using MATLAB, plot versus

frequency in Hertz, on a semilog scale.

Figure 4.30. Network for Exercise 9

R

C

+

+

Vin s( ) Vou t s( )

RL

+

Vin s( )

+

Vou t s( )

(a) (b)

L

R2R1

C

R1

Vin s( )Vou t s( )

R2

C

Vin s( ) Vou t s( )

(a) (b)

G s( )

R1 R2

R3

C1

C2

Vou t s( )Vin s( )

R4 R1 = 11.3 k

R2 = 22.6 k

R3=R4 = 68.1 k

C1=C2 = 0.01 F


10/25

Chapter 6 The Impulse Response and Convolution



6.7 Exercises

1. Compute the impulse response in terms of and for the circuit of Figure 6.36.

Then, compute the voltage across the inductor.


2. Repeat Example 6.4 by forming instead of , that is, use the convolution integral

3. Repeat Example 6.5 by forming instead of .

4. Compute given that

5. For the series circuit shown in Figure 6.37, the response is the current . Use the convolu-

tion integral to find the response when the input is the unit step .


6. Compute for the network of Figure 6.38 using the convolution integral, given that

.

h t( ) iL t( )= R L

vL t( )

+

R

L

iL t( )

t( )

h t ( ) u t ( )

u ( )h t ( ) d

h t ( ) u t ( )

v1 t( )*v2 t( )

v1 t( )

4t t 0

0 t 0


11/25


Exercises


7. Compute for the circuit of Figure 6.39 given that .


Hint: Use the result of Exercise 6.

R

L

+

1 H +

1

vin t( )vou t t( )

vou t t( ) vin t( ) u0 t( ) u0 t 1( )=

R

L

1 H

+ +

vin t( ) vou t t( )1


12/25


Exercises

7.14 Exercises

1. Compute the first 5 components of the trigonometric Fourier series for the waveform of Figure

7.47. Assume .


2. Compute the first 5 components of the trigonometric Fourier series for the waveform of Figure

7.48. Assume .


3. Compute the first 5 components of the exponential Fourier series for the waveform of Figure

7.49. Assume .



7.50. Assume .


1=

0t

A

f t( )

1=

0t

A

f t( )

1=

0t

A

f t( )

1=

0 t

f t( )

A 2

A 2


13/25

Chapter 7 Fourier Series




7.51. Assume .



7.52. Assume .


1=

0t

A

f t( )

1=

0t

A

A

f t( )


14/25


Exercises

8.10 Exercises

1. Show that

2. Compute

3. Sketch the time and frequency waveforms of

4. Derive the Fourier transform of



7. For the circuit of Figure 8.21, use the Fourier transform method to compute .


8. The input-output relationship in a certain network is

Use the Fourier transform method to compute given that .

u0 t( ) t( ) td

1 2=

F teat

u0 t( ){ } a 0>

t( ) 0tcos u0 t T+( ) u0 t T( )[ ]=

t( ) A u0 t 3T+( ) u0 t T+( ) u0 t T( ) u0 t 3T( )+[ ]=

t( ) AT---t u0 t T+( ) u0 t T( )[ ]=

t( ) AT--- t A+

u0 t T+( ) u0 t( )[ ]A

T--- t A+

u0 t( ) u0 tT

2---

+=

vC t( )

+

+C

1 Fvin t( )

vC t( )

1

0.5

R1

R2

vin t( ) 50 4tucos 0 t( )=

d2

dt2

-------vou t t( ) 5d

dt-----vou t t( ) 6vou t t( )+ + 10v in t( )=

vou t t( ) vin t( ) 2etu0 t( )=


15/25

Chapter 8 The Fourier Transform



9. In a bandpass filter, the lower and upper cutoff frequencies are , and respec-

tively. Compute the energy of the input, and the percentage that appears at the output, if the

input signal is volts.

10. In Example 8.4, we derived the Fourier transform pair

Figure 8.22. Figure for Exercise 10

Compute the percentage of the energy of contained in the interval of

.

1 2Hz= 2 6Hz=

1

vin t( ) 3e2t

u0 t( )=

A u0 t T+( ) u0 t T( )[ ] 2ATTsin

T---------------

0 T2

2

A

T T

t0

f t( )

F ( )

1 t( ) T T

F ( )


16/25

Chapter 9 Discrete Time Systems and the Z Transform

9-52 Signals and Systems with MATLAB Applications, Second EditionOrchard Publications

9.10 Exercises

1. Find the Z transform of the discrete time pulse defined as

2. Find the Z transform of where is defined as in Exercise 1.

3. Prove the followingZ transform pairs:

a.

b.

c.

d.

e.

4. Use the partial fraction expansion to find given that

5. Use the partial fraction expansion method to compute the InverseZ transform of

6. Use the Inversion Integral to compute the Inverse Z transform of

7. Use the long division method to compute the first 5 terms of the discrete time sequence whose Z

transform is

p n[ ]

p n[ ]1 n 0 1 2 m 1, , , ,=

0 otherwise

=

anp n[ ] p n[ ]

n[ ] 1

n 1[ ] z m

na

n

u0 n[ ]

az

z a( )2------------------

n2a

nu0 n[ ]

az z a+( )

z a( )3----------------------

n 1+[ ]u0 n[ ]z

2

z 1( )2------------------

n[ ] Z1

F z( )[ ]=

F z( ) A

1 z1

( ) 1 0.5z 1( )--------------------------------------------------=

F z( ) z2

z 1+( ) z 0.75( )2-------------------------------------------=

F z( ) 1 2z1

z3

+ +

1 z1

( ) 1 0.5z 1( )--------------------------------------------------=

F z( ) z1

z2

z3

+

1 z1

z2

4z3

+ + +

----------------------------------------------=


17/25


Exercises

8. a. Compute the transfer function of the difference equation

b. Compute the response when the input is

9. Given the difference equation

a. Compute the discrete transfer function

b. Compute the response to the input

10. A discrete time system is described by the difference equation

where

a. Compute the transfer function

b. Compute the impulse response

c. Compute the response when the input is

11. Given the discrete transfer function

write the difference equation that relates the output to the input .

y n[ ] y n 1[ ] Tx n 1[ ]=

y n[ ] x n[ ] ena T

=

y n[ ] y n 1[ ]T

2--- x n[ ] x n 1[ ]+{ }=

H z( )

x n[ ] ena T

=

y n[ ] y n 1

[ ]+

x n[ ]=

y n[ ] 0for n 0


18/25


Exercises

10.8 Exercises

1. Compute the DFT of the sequence ,

2. A square waveform is represented by the discrete time sequence

and

Use MATLAB to compute and plot the magnitude of this sequence.

3. Prove that

a.

b.

4. The signal flow graph of Figure 10.6 is a decimation in time, natural-input, shuffled-output type

FFT algorithm. Using this graph and relation (10.69), compute the frequency component .

Verify that this is the same as that found in Example 10.5.

Figure 10.6. Signal flow graph for Exercise 4

5. The signal flow graph of Figure 10.7 is a decimation in frequency, natural input, shuffled output

type FFTalgorithm. There are two equations that relate successive columns. The first is

and it is used with the nodes where two dashed lines terminate on them.

x 0[ ] x 1[ ] 1= = x 2[ ] x 3[ ] 1= =

x 0[ ] x 1[ ] x 2[ ] x 3[ ] 1= = = = x 4[ ] x 5[ ] x 6[ ] x 7[ ] 1= = = =

X m[ ]

x n[ ]2kn

N-------------cos

1

2--- X m k [ ] X m k +[ ]+{ }

x n[ ]2kn

N-------------sin

1

j2----- X m k [ ] X m k +[ ]+{ }

X 3[ ]

W0

W0

W0

W0

W4

W4

W4

W4

W0

W4

W2

W1

W0

W4

W4

W2

W2

W6

W6

W0

W6

W5

W3

W7

x 0[ ]

x 1[ ]

x 2[ ]

x 3[ ]

x 4[ ]

x 5[ ]

x 6[ ]

x 7[ ] X 7[ ]

X 3[ ]

X 5[ ]

X 1[ ]

X 0[ ]

X 4[ ]

X 2[ ]

X 6[ ]

Ydash R C,( ) Ydash Ri C 1,( ) Ydash Rj C 1,( )+=


19/25

Chapter 10 The DFT and the FFT Algorithm



The second equation is

and it is used with the nodes where two solid lines terminate on them. The number inside the cir-

cles denote the power of , and the minus () sign below serves as a reminder that the brack-

eted term of the second equation involves a subtraction. Using this graph and the above equa-

tions, compute the frequency component . Verify that this is the same as in Example 10.5.

Figure 10.7. Signal flow graph for Exercise 5

Ysol R C,( ) Wm

Yso l Ri C 1,( ) Yso l Rj C 1,( )[ ]=

WN

X 3[ ]

W0

W1

W2

W3

W0

W0

W2

W0

W2

W0

W0

W0

x 0[ ]

x 1[ ]

x 2[ ]

x 3[ ]

x 4[ ]

x 5[ ]

x 6[ ]

x 7[ ] X 7[ ]

X 3[ ]

X 5[ ]

X 1[ ]

X 0[ ]

X 4[ ]

X 2[ ]

X 6[ ]


20/25


Exercises

11.10 Exercises

1. The circuit of Figure 11.39 is a VCVS second-order high-pass filter whose transfer function is

and for given values of , , and desired cutoff frequency , we can calculate the values of

to achieve the desired cutoff frequency .


For this circuit,

and the gain is

Using these relations, compute the appropriate values of the resistors to achieve the cutoff fre-

quency . Choose the capacitors as and . Plot versus

frequency.

Solution using MATLAB is highly recommended.

G s( )Vou t s( )

Vin

s( )-----------------

Ks2

s

2

a b( )Cs 1 b( )C2

+ +

---------------------------------------------------------------= =

a b C

C1 C2 R1 R2 R3 and R4, , , , , C

vin vout

C1 C2

R2

R1

R3

R4

R24b

C1 a a2

8b K 1( )+[ ]+

C

---------------------------------------------------------------------------=

R1b

C12R2C

2--------------------=

R3KR2

K 1------------- K 1,=

R4 KR2=

KK 1 R4 R3+=

C 1KHz= C1 10 fC F= C2 C1= G s( )


21/25

Chapter 11 Analog and Digital Filters



2. The circuit of Figure 11.40 is a VCVS second-order band-pass filter whose transfer function is


Let , , ,

, and

We can calculate the values of to achieve the desired centered frequency

and bandwidth . For this circuit,

Using these relations, compute the appropriate values of the resistors to achieve center frequency

, , and .

Choose the capacitors as . Plot versus frequency.


G s( )Vou t s( )

Vin s( )-----------------

K BW[ ]s

s2

BW[ ]s 02

+ +

----------------------------------------= =

vin voutC1

R5

C2

R3

R2

R1

R4

0 centerfrequency= 2 uppercutofffrequency= 1 lowercutofffrequency=

BandwidthBW 2 1= QualityFactor Q 0 BW=

C1 C2 R1 R2 R3 and R4, , , , ,

0 BW

R12Q

C10K-----------------=

R22Q

C10 1 K 1( )2

8Q2

++

--------------------------------------------------------------------------=

R31

C120

2-------------

1

R1-----

1

R2-----+

=

R4 R5 2R3= =

0 1KHz= GainK 10= Q 10=

C1 C2 0.1F= = G s( )


22/25


Exercises

3. The circuit of Figure 11.41 is a second-order band elimination filter whose transfer func-

tion is


Let , , ,

, , and gain

We can calculate the values of to achieve the desired centered fre-

quency and bandwidth . For this circuit,

The gain must be unity, but can be up to 10. Using these relations, compute the appropriate

values of the resistors to achieve center frequency , and .

Choose the capacitors as and . Plot versus frequency.


VCVS

G s( )Vou t s( )

Vin s( )-----------------

K s2 02

+( )

s2

BW[ ]s 02

+ +

----------------------------------------= =

vinvout

C2C1R3

R1 R2

C3

0 centerfrequency= 2 uppercutofffrequency= 1 lowercutofffrequency=

BandwidthBW 2 1= Quality Factor Q 0 BW= K 1=

C1 C2 R1 R2 R3 and R4, , , , ,

0 BW

R1 120QC1--------------------=

R22Q

0C1------------=

R32Q

C10 4Q2

1+( )------------------------------------- =

K Q

0 1KHz= Gain K 1= Q 10=

C1 C2 0.1F= = C3 2C1= G s( )


23/25

Chapter 11 Analog and Digital Filters



4. The circuit of Figure 11.42 is a MFB second-order all-pass filterwhose transfer function is

where the gain , , and the phase is given by


The coefficients and can be found from

For arbitrary values of , we can compute the resistances from

For , we compute the coefficient from

G s( )Vou t s( )

Vin s( )-----------------

K s2 a0s b0

2+( )

s2

a0s b02

+ +

----------------------------------------------= =

K cons t tan= 0 K 1<


24/25


Exercises

and for , from

Using these relations, compute the appropriate values of the resistors to achieve a phase shiftat with .

Choose the capacitors as and plot phase versus frequency.


5. The Bessel filterof Figure 11.43 has the same configuration as the low-pass filter of Example

11.3, and achieves a relatively constant time delay over a range . The second-order

transfer function of this filter is


where is the gain and the time delay at is given as

We recognize the transfer function above as that of a low-pass filter where

and the substitution of . Therefore, we can use a low-pass filter circuit such as that of

Figure 11.43, to achieve a constant delay by specifying the resistor and capacitor values of the

circuit.

The resistor values are computed from

180 0 0

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