Solutions to Problems from Chapter 2 - Rutgers...

Solutions to Problems from Chapter 2

Problem 2.1The signalsu(t)� 0:5sgn(t), u1(t)� 0:5sgn(t), anduh(t)� 0:5sgn(t) are plotted respectively in Figures 2.1a,b,c.Note thatuh(t) � 0:5sgn(t) = 0:5; 8t including t = 0, but u1(0) � 0:5sgn(0) is undefined.1

u(t)-0.5sgn(t)

1

0

t0.5

u (t)-0.5sgn(t)

0

t0.5

u (t)-0.5sgn(t)

0

t0.5

not defined at t=0

1 h

(a) (b) c)(

Figure 2.1

Problem 2.2This signal is presented in Figure 2.2. Dashed lines denote particular signals.

t

u(t+ )

u(-t+ )

0

p (t)

t/2

tt/2

t/2

-t/2

1

Figure 2.2

Problem 2.3This signal is presented in Figure 2.3. The dashed lines denote particular signals.

t

u(t)

sgn(t)

0

p (t)4

1

sgn(t)

-1-2 1 2

-1

-2

-3

-2 (t)D2

Figure 2.3

1 Note that the importance of the usage of the Heaviside unit step function in digital filter design and sampling is discussed in a recentpaper: L. Jackson, “A Correction to Impulse Invariance,”IEEE Signal Processing Letters, Vol. 7, 273–275, 2000.

11

CHAPTER 2

Problem 2.4The required signals are presented respectively in Figures 2.4a,b,c,d.

u(-t+2)

t

0 1 2

u(-t-2)

t

0-1-2

r(-t+3)

t

0 1 2 3

r(-t-1)

t

0-1

(a)

(d)(b)

c)

1

1

(

Figure 2.4

Problem 2.5The graphs of the required signals are plotted respectively in Figures 2.5a,b,c,d.

f (t)

t

-r(-t+1)

1

u(t-3)

0 0

f (t)2

t

-u(t-4)

r(t-2)

1 2 3 43

0 1 2

f (t)3

r(t)

u(t-2)

t

0 1 2 3 4

-u(t-5)-r(-t-2)

f (t)4

(a) (b)

c) (d)

t

(

1

1 1

2

-1

Figure 2.5

12

SOLUTION MANUAL for LINEAR DYNAMIC SYSTEMS AND SIGNALS by ZORAN GAJIC

Problem 2.6The signalf (t) = u(t+ 1) + u(t� 1) � r(t � 1) + r(t� 3) is presented in Figure 2.6 with dashed lines denotingparticular signals.

0 1 2 3-1

t

f(t)=u(t+1)+u(t-1)-r(t-1)+r(t-3)

u(t+1)

u(t-1)

-r(t-1)

r(t-3)

Figure 2.6

Problem 2.7Straight forward solutions are given by

(a) f1(t) = r(t)� 2r(t� 1) + r(t� 2); (b )f2(t) = r(t� 2)� r(t � 3)� u(t� 1)

(c) f3(t) = r(t� 1)� r(t � 2)� u(t� 2)

Plotting these functions we obtain the desired graphs. Note that other solutions are possible since the presentationis not unique.

Problem 2.8See comments made in the solutions to Problem 2.7. In this problem, one of possible several solutions is given by

f(t) = u(t)� r(t� 1) + r(t � 3) + u(t� 5)

The corresponding signals are plotted in Figure 2.7 using dashed lines.

0 1 2 3

t

f(t)=u(t)-r(t-1)+r(t-3)+u(t-5)

u(t)

u(t-5)

-r(t-1)

r(t-3)

4 5

1

Figure 2.7

13

CHAPTER 2

Problem 2.9This signal is plotted using MATLAB and presented in Figure 2.8.Comment: Note that the MATLAB functionstepfun(-k,k0); k=-4:1:6; does not produces the correctresult.

−4 −3 −2 −1 0 1 2 3 4 5 6−2

−1

0

1

2

3

4

Discrete−time k

Sig

nal f

[k]

f[k]p4[k−1]−r[k]u[−k+2]

Figure 2.8

Problem 2.10This signal is plotted in Figure 2.9.

−4 −3 −2 −1 0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Discrete−time k

Sig

nal f

[k]

f[k]tri4*u[−k+1]p2[k−1]*u[k−1]

Figure 2.9

14


Problem 2.11

f [3] = sinc[3]� r[�3 + 5] + �6[3]p4[3] = sinc[3]� r[2] + 0� 0 =sin�3

�3� 2 = 0� 2 = �2

Problem 2.12Using the notation from Section 2.1, the analytical representation of the discrete-time triangular pulse is

�m[k] =

8>><>>:0; k � �m21 + 2

mk; �m2 � k � 01� 2

mk; 0 � k � m2

0; k � m2

Note thatm, as defined in formula (2.14) is an even integer. This signal can be represented in terms of discrete-time ramp signals as follows

�m[k] =2

mrhk � m

2

i� 2r[k] + r

hk +

m

2

i

Problem 2.13It is known from (2.29) that

1Z�1

f(t)�(m)(t)dt = (�1)mf (m)(0)

On the other hand, we have

1Z�1

f(t)�(m)(�t)dt = (�1)m1Z

�1f(��)�(m)(�)d� = (�1)2mf (m)(0)

due to the fact thatdm�(�t)=dtm = (�1)mdm�(�)=d�m for t = ��. It follows that form even the right-handsides of the above two expressions are identical, which implies that�(m)(t) = �(m)(�t); m = 2n. Form odd wehave(�1)mf (m)(0) = �(�1)2mf (m)(0) = �f (m)(0), which implies that�(m)(t) = ��(m)(�t);m = 2n+ 1.

Problem 2.14Starting with the integral

1Z�1

f(t)�(1)(at� t0)dt =

1Z�1

f(t)d

dtf�(at � t0)gdt

assuming thata > 0, and using the change of variables as� = at � t0, we obtain

1Z�1

f

�� + t0a

�ad

d��(�)

d�

a=

1Z�1

f

�� + t0a

�d

d��(�)d� =

1Z�1

f

�� + t0a

��(1)(�)d� = (�1)f (1)

�t0a

�; a > 0

For a < 0, the same derivations imply

1Z�1

f(t)�(1)(at � t0)dt =

�1Z1

f

�� + t0a

��(1)(�)d� = �

1Z�1

f

�� + t0a

��(1)(�)d� = f (1)

�t0a

�; a < 0

15

CHAPTER 2

The second integral1Z

�1f (t)�(1)(a(t � t0))dt

for a > 0 and with the change of variables as� = a(t� t0) produces1Z

�1f�t0 +

�

a

��(1)(�)d� = (�1)f (1)(t0); a > 0

Similarly for a < 0, we have1Z

�1f(t)�(1)(a(t � t0))dt = �

1Z�1

f�t0 +

�

a

��(1)(�)d� = f (1)(t0); a < 0

Problem 2.15Starting with the integral

1Z�1

f(t)�(n)(at� t0)dt =

1Z�1

f(t)dn

dtnf�(at� t0)gdt

assuming thata > 0, and using the change of variables as� = at � t0, we obtain1Z

�1f

�� + t0

a

�an

dn

d�n�(�)

d�

a= an�1

1Z�1

f

�� + t0

a

�dn

d�n�(�)d�

= an�11Z

�1f

�� + t0

a

��(n)(�)d� = an�1(�1)nf (n)

�t0a

�; a > 0

For a < 0, the same derivations imply1Z

�1f(t)�(n)(at� t0)dt = an�1

�1Z1

f

�� + t0

a

��(n)(�)d�

= �an�11Z

�1f

�� + t0

a

��(n)(�)d� = an�1(�1)n�1f (n)

�t0a

�; a < 0

The integral1Z

�1f(t)�(n)(a(t� t0))dt =

1Z�1

f(t)dn

dtnf�(a(t� t0))gdt

is evaluated similarly, assuming thata > 0, and using the change of variables as� = a(t� t0), which leads to1Z

�1f��a+ t0

�an

dn

d�n�(�)

d�

a= an�1

1Z�1

f��a+ t0

� dn

d�n�(�)d�

= an�11Z

�1f��a+ t0

��(n)(�)d� = an�1(�1)nf (n)(t0); a > 0

16


For a < 0, the same derivations imply

1Z�1

f (t)�(n)(a(t� t0))dt = �an�11Z

�1f��a+ t0

��(n)(�)d� = an�1(�1)n�1f (n)(t0); a < 0

Problem 2.16Using the formula derived in Problem 2.15 we have

+1Z�1

e�3t�(2)(3t+ 5)dt = 3(�1)2 d2

dt2�e�3t

jt= 5

3= 3� 9� e�3

53 = 27e�5

The second integral according to the formula derived in Problem 2.14 is zero. This can be also justified by directintegration as

1Z�1

4�(1)(�t+ 2)dt = �41Z

�14�(1)(t� 2)dt = �4

1Z�1

d

d�f�(�)gd� = �4

1Z�1

d�(�) = �4�(�)jt=1t=�1 = 0

Problem 2.17By the sifting property of the delta impulse function, for a given functiong(f) (note we use heref to denotefrequency), we have

1Z�1

g(f)�(f)df = g(0)

If instead of�(f) we use�(!), with ! = 2�f , we obtain from (2.27), in this case the frequency scaling property that

1Z�1

g(f)�(2�f )df =1

2�g(0)

This leads to the conclusion that�(!) = �(2�f ) = 12� �(f).

Problem 2.18The current in this electrical circuit for the constant voltageE and with no initial voltage on the capacitor isgiven by

i(t) =E

Re�

1RC t; t � 0

In the case whenR ! 0, the magnitude of the current tends to infinity att = 0 and the duration of the current

tends to zero sincelimR!0

ne�

1RC t

o! 0; t > 0. Hence, the circuit current tends to the delta impulse function.

Problem 2.19The current in this electrical circuit is given by

i(t) = CdvC(t)

dt=

�C 1"; 0 < t < "

0; t > "

When " ! 0, the magnitude of the current tends to infinity and its with tends to zero. Hence, the current tendsto the delta impulse function.

17

CHAPTER 2

Problem 2.201Z

�1

�t2 + sin(t) + e�2t

�h�(t) + 3�(2t� 1) + 4�(1)(t � 2)

idt =

1Z�1

f(t)h�(t) + 3�(2t� 1) + 4�(1)(t� 2)

idt

= f(0) + 31

2f

�1

2

�+ 4(�1)f (1)(2) = 1 +

3

2

�1

4+ sin

�1

2

�+ e�1

�� 4

�4 + cos (2)� 2e�4

�

Problem 2.211Z

�1

�tsin(t) + e�t

�h�(3)(t)� �(2)(t � 2)

idt =

1Z�1

f(t)h�(3)(t)� �(2)(t� 2)

idt

= (�1)3f (3)(0)� (�1)2f (2)(2) = ��3 sin (0)� 0 cos (0)� e0��

2 cos (2)� 2 sin (2) + e�2�

= 1� 2 cos (2) + 2 sin (2)� e�2

Problem 2.22

(a)

5Z�3

e�4tu(t)�(t� 4)dt = e�4�4u(4) = e�16

(b)

5Z�3

e�t�(t� 6)dt = 0 since �(t� 6) is outside of the integration limits

(c)

5Z�3

e�2t sin (t� 3)�(t� 5)dt =

5Z�3

f(t)�(t� 5)dt =1

2f(5) =

1

2e�10 sin (2)

(d)

5Z�3

e�t�(�t+ 3)dt =

5Z�3

f(t)�(�t + 3)dt = f(3) = e�3

Problem 2.23

2 cos (�t)�(t � 1) +

1Z�1

e�3th�(2)(t� 1) + �(2t� 3)

idt+

+2Z�2

tan (2t)�(t� 3)dt+

4Z�3

5�(t+ 2)dt

= 2 cos (�)�(t� 1) + (�1)29e�3 + 1

2e�3(

32 ) + 0 + 5 = �2�(t� 1) + 9e�3 +

1

2e�(

92 ) + 5

Problem 2.24*

e�2t�(2t� 1) +

1Z�1

sin (�(t� 1))h�(2)(t� 1) + �(t� 2)

idt+

+2Z�2

tan (2t)�(2t� 4)dt+

4Z�3

5�(t+ 2)dt

= e�212 �(2t� 1) +

��2�(�1)2 sin (�(1� 1))+1

2sin (�(1� 1)) +

1

4tan (4)+ 5 = e�1�(2t� 1)+

1

4tan (4)+ 5

18


Note that the first term was evaluated using the result established in (2.30), that is

f (t)�(at� t0) = f

�t0a

��(at� t0)

and that we used also the following result (see (2.22))

t0=aZ�1

f(bt)�(at� t0)dt =

�=0Z�=�1

f

�b� + t0

a

��(�)

d�

a=

1

a

1

2f

�bt0a

�; a > 0

Problem 2.25

(a) e�5t sin (2t+ 4)�(t� 1) = f(t)�(t� 1) = f(1)�(t� 1) = e�5 sin (6)�(t� 1)

(b) (�2)k cos [2k � 1]�[k � 4] = f [k]�[k � 4] = f [4]�[k � 4] = (�2)4 cos [7]�[k � 4]

Problem 2.26(a) The scaled and shifted rectangular pulse signals are defined by

p2(3t) =

�1; �1 � 3t � 1,�1=3 � t � 1=30; elsewhere

; p2(3t� 2) =

�1; �1 � 3t� 2 � 1, 1=3 � t � 10; elsewhere

p4(4(t� 5)) =

�1; �2 � 4(t� 5) � 2,�0:5 + 5 � t � 0:5 + 5, 4:5 � t � 5:50; elsewhere

The plots of these signals are presented in Figure 2.10.

1

p (3t)2

t

1

p (3t-2)2

t

1/3 1

p (4(t-5))4

t

4 5 61/3-1/3 0

Figure 2.10

(b) The analytical expressions for the scaled and shifted step signals are given by

u(2t� 3) =

�1; 2t� 3 � 0, t � 3=20; elsewhere

; u(�3t+ 2) =

�1; �3t+ 2 � 0, t � 2=30; elsewhere

Graphical presentations of these signals are given in Figure 2.11.

u(2t-3)

t

1

1 20

u(-3t+2)

t

1

12/30

Figure 2.11

19

CHAPTER 2

Problem 2.27The ramp signal is defined byr(t) = tu(t); t > 0 andr(t) = 0; t < 0. Using this definition, we have

r(�4(t+ 2)) =

��4(t + 2)u(�4(t+ 2)); �4(t+ 2) > 0, t < �20; elsewhere

=

��4t� 8; t < �20; elsewhere

Note also that

u(�4(t+ 2)) =

�1; �4(t+ 2) > 0, t < �20; elsewhere

= u(�t � 2)

The scaled and shifted step and rectangular pulse signals are analytically given by

u(�3t� 1) =

�1; �3t� 1 � 0, t � �1=30; elsewhere

; p2(�2t� 4) =

�1; �1 � �2t� 4 � 1,�5=2 � t � �3=20; elsewhere

The above signals are plotted in Figure 2.12.

0-2-4

4

t

r(-4(t+2))

2 1t

u(-3t-1)

-1/3-2-4 0-2-4 -3 -1

1t

p (-2t-4)2

(a) (b) c)

-3

-1 0

(

Figure 2.12

Problem 2.28Generalized derivatives for the signals defined in Problem 2.5 are given by

Df1(t)

Dt=

8>>><>>>:1; t < 1unde�ned; t = 10; 1 < t < 3�(t � 3); t = 30; t > 3

;Df2(t)

Dt=

8>>><>>>:0; t < 2unde�ned; t = 21; 2 < t < 4��(t� 4); t = 41; t > 4

Df3(t)

Dt=

8>>><>>>:0; t < 0unde�ned; t = 01; 0 < t < 2�(t� 2); t = 21; t > 2

;Df4(t)

Dt=

8>>><>>>:1; t < �2unde�ned; t = �20; �2 < t < 5��(t� 5); t = 50; t > 5

The generalized derivative for the signal defined in Problem 2.6 is given by

Df(t)

Dt=

8>>>>>>><>>>>>>>:

0; t < �1�(t+ 1); t = �10; �1 < t < 1�(t� 1); t = 1�1; 1 < t < 3unde�ned; t = 30; t > 3

20


Problem 2.29Using the derivative product rule and the property of the delta impulse function, which states thatf (t)�(t� t0) =f(t0)�(t� t0), we obtain

Df(t)

Dt= u(t� 2) + (t+ 2)�(t � 2)� sin (t)u(t� 3) + cos (t)�(t� 3)� 4e�4t sin (t) + e�4t cos (t)

= u(t� 2) + 4�(t� 2)� sin (t)u(t� 3) + cos (3)�(t� 3) + e�4t(cos (t)� 4 sin (t))

Problem 2.30The graphs are presented in Figures 2.13 and 2.14

0 1 2-1

t

f (t)=r(t-2)-r(t)u(-t+2)+p (t)

u(-t+2)

-r(t)u(-t+2)

-r(t)

r(t-2)

-2

p (t)4

41

Figure 2.13

0 1 2-1

t

f (t)

u(t+1)

-r(t-4)

r(-t-2)

-2

2

3 4-3

1

-1-u(-t+2)

Figure 2.14

The corresponding generalized derivatives are given by

Df1(t)

Dt=

8>>>>>>><>>>>>>>:

0; t� 2�(t+ 2); t = �20; �2 < t < 0unde�ned; t = 0�1; 0 < t < 2unde�ned; t = 21; t > 2

;Df2(t)

Dt=

8>>>>>>>>>>><>>>>>>>>>>>:

�1; t < �2unde�ned; t = 20; �2 < t < �1�(t+ 1); t = �10 0 < t < 2�(t� 2); t = 20; 2 < t < 4unde�ned; t = 4�1 t > 4

21

CHAPTER 2

Problem 2.31Signals(a) and (g) are anticausal since they are different from zero for negative time. Signals (b), (c), (d), (e), (f),and (h) are causal since all of them are equal to zero fort < 0.

Problem 2.32The required generalized derivatives of the signals in FIGURE 2.212 are given by

(a)Df1(t)

Dt=

8>>>>>>>>>>><>>>>>>>>>>>:

0; t < 0unde�ned; t = 01; 0 < t < 1f1(1+) � f1(1�) = (0� 1)�(t� 1) = ��(t� 1); t = 11; 1 < t < 2f1(2+) � f1(2�) = (�1� 1)�(t � 2) = �2�(t� 2); t = 20; 2 < t < 3f1(3+) � f1(3�) = (0� (�1))�(t� 3) = �(t� 3); t = 30; 3 < t

;

(b)Df2(t)

Dt=

8>>>>>>>>>>><>>>>>>>>>>>:

0; t < 0unde�ned; t = 01; 0 < t < 1f2(1

+) � f2(1�) = (�1� 1)�(t� 1) = �2�(t� 1); t = 1

1; 1 < t < 3unde�ned; t = 30; 3 < t < 4f2(4+) � f2(4�) = (0� 1)�(t� 4) = ��(t� 4); t = 40; 4 < t

These generalized derivatives are graphically represented in Figure 2.15.

0 1 2 3

t

Df (t)

Dt1

(t-3)

-2 (t-1)- (t-1)d d

1 d

Df (t)

Dt2

0 1 2 3

-2 (t-1)d - (t-4)d

(a) (b)

t

Figure 2.15

Problem 2.33The regular derivativedf(t)=dt is presented in Figure 2.16. The generalized derivativeDf(t)=Dt contains, inaddition, the impulse delta functions��(t� 0:3); �(t� 0:9); ��(t� 1) .

2 Note that “FIGURE x.y” denotes a figure from the text book and “Figure x.y” denotes a figure from the solution manual.

22


0.1 0.40.2 0.3 0.5 0.6 0.7 0.8 0.9 10

-1

1

2

df(t)

dt

t

Figure 2.16

Problem 2.34Since an integral is equal to the area between the function (signal) and the horizontal axis, we have

1Z0

f (t)dt =0:1� 0:2

2+ 0:1� 0:1� 0:1� 0:1 + 0:1� 0:1 = 0:02

Hence, the signal integration has a smoothing effect.

Problem 2.35

D

Dtfsin (t)u(t)g = cos (t)u(t) + sin (t)�(t) = cos (t)u(t) + 0�(t) = cos (t)u(t)

D2

Dt2fsin (t)u(t)g = D

Dt

�D

Dtfsin (t)u(t)g

�=

D

Dtfcos (t)u(t)g = � sin (t)u(t) + �(t)

D3

Dt3fsin (t)u(t)g = D

Dt

�D2

Dt2fsin (t)u(t)g

�=

D

Dtf� sin (t)u(t) + �(t)g = � cos (t)u(t) + �(1)(t)

D4

Dt4fsin (t)u(t)g =

D

Dt

�D3

Dt3fsin (t)u(t)g

�=

D

Dt

n� cos (t)u(t) + �(1)(t)

o= sin (t)u(t)� �(t) + �(2)(t)

Problem 2.36

D

Dtftu(t)g = t�(t) + u(t) = 0�(t) + u(t) = u(t)

D2

Dt2ftu(t)g = D

Dt

�D

Dtftu(t)g

�=

D

Dtfu(t)g = �(t)

D3

Dt3ftu(t)g =

D

Dt

�D2

Dt2ftu(t)g

�=

D

Dtf�(t)g = �(1)(t)

23

CHAPTER 2

D4

Dt4ftu(t)g =

D

Dt

�D3

Dt3ftu(t)g

�=

D

Dt

n�(1)(t)

o= �(2)(t)

Problem 2.37

D

Dtfcos (t)u(t)g = � sin (t)u(t) + cos (t)�(t) = � sin (t)u(t) + �(t)

D2

Dt2fcos (t)u(t)g =

D

Dt

�D

Dtfcos (t)u(t)g

�=

D

Dtf� sin (t)u(t) + �(t)g = � cos (t)u(t) + �(1)(t)

D3

Dt3fcos (t)u(t)g =

D

Dt

�D2

Dt2fcos (t)u(t)g

�=

D

Dt

n� cos (t)u(t) + �(1)(t)

o= sin (t)u(t)� �(t) + �(2)(t)

D4

Dt4fcos (t)u(t)g =

D

Dt

�D3

Dt3fcos (t)u(t)g

�=

D

Dt

nsin (t)u(t) � �(t) + �(2)(t)

o= cos (t)u(t)��(1)(t)+�(3)(t)

Problem 2.38

D

Dt

�e�tu(t)

= �e�tu(t) + e�t�(t) = �e�tu(t) + �(t)

D2

Dt2�e�tu(t)

=

D

Dt

�D

Dt

�e�tu(t)

�=

D

Dt

��e�tu(t) + �(t)

= �2e�tu(t) + ��(t) + �(1)(t)

D3

Dt3�e�tu(t)

=

D

Dt

�D2

Dt2�e�tu(t)

�=

D

Dt

n�2e�tu(t) + ��(t) + �(1)(t)

o= �3e�tu(t) + �2�(t) + ��(1)(t) + �(2)(t)

D4

Dt4�e�tu(t)

=

D

Dt

�D3

Dt3�e�tu(t)

�=

D

Dt

n�3e�tu(t) + �2�(t) + ��(1)(t) + �(2)(t)

o= �4e�tu(t) + �3�(t) + �2�(1)(t) + ��(2)(t) + �(3)(t)

Following the same pattern, we have

Dn

Dtn�e�tu(t)

= �ne�tu(t) + �n�1�(t) + �n�2�(1)(t) + � � �+ ��(n�2)(t) + �(n�1)(t); � > 0

Problem 2.39

sat(t) = �r(t)� �r(t� t0) = r�(t) � r�(t� t0); t0 =�

�

24


MATLAB PROBLEMS

Problem 2.40The saturation signal is presented in Figure 2.17. This figure is obtained via the following MATLAB code.

>> alp=1; bet=2; t0=bet/alp;>> t1=-10:0; t2=0:t0; t3=t0:10; t=[t1 t2 t3];>> sat=[zeros(1,11),t2,bet*ones(1,9)]>> plot(t,sat)>> xlabel(’Time’); ylabel(’Saturation signal’)>> grid; axis([-5 5 —0.5 2.5])

−5 −4 −3 −2 −1 0 1 2 3 4 5−0.5

0

0.5

1

1.5

2

2.5

Time

Sat

urat

ion

func

tion

Figure 2.17

Problem 2.41The shifted discrete-time impulse delta and unit step signals are presented in Figure 2.18.

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

1.5

Discrete−time

Shi

fted

delta

impu

lse

sign

al

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

1.5

Discrete−time

Shi

fted

step

sig

nal

Figure 2.18

This figure is obtained via the following MATLAB code.

>> k1=0; k2=10; k0=3;>> k=k1:k2

25

CHAPTER 2

>> del3=[(k-k0)==0]>> subplot(211); plot(k,del3,’*’)>> xlabel(’Discrete-time’)>> ylabel(’Shifted delta impulse signal’)>> grid; axis([0 10 -0.5 1.5])>> k00=1; step1=[(k-k00)>=0]>> subplot(212)>> plot(k,step1,’*’)>> xlabel(’Discrete-time’); ylabel(’Shifted step signal’)>> grid; axis([0 10 -0.5 1.5])

Problem 2.42For real and negative exponents� < 0, the damped sinusoidal signal decays oscillatory to zero as time increases.Such a signal for� = �2 is plotted in Figure 2.19(a) in the time interval from 0 to 5. For positive values of theexponent, the damped sinusoidal signal increases to infinity ast ! 1. The considered signals is also presentedin Figure 2.19(a) for� = 1.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.1

0

0.1

0.2

0.3

0.4

Time

Rea

l and

neg

ativ

e ex

pone

nt

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−100

−50

0

50

100

Time

Rea

l and

pos

itive

exp

onen

t

Figure 2.19a

For complex conjugate values of�, the signal is a complex function that has its real and imaginary parts. Forsuch signals we plot their magnitude and phase (angle). The corresponding plots for� = �1 + j1, obtained viaMATLAB, are presented in Figure 2.19(b). It can be observed that for this complex signal the magnitude decaysto zero, but the phase is a periodic function of time. In Chapter 3, where the Fourier transform of signals willbe presented, we will study complex signals in detail. The figures in this problem are obtained via the followingMATLAB code.>> t=0:0.01:5>> % Sinusiodal signal with a real and negative exponent>> lam=-2>> f=exp(lam*t).*sin(2*t)>> figure(1)>> subplot(211); plot(t,f)>> xlabel(’Time’); ylabel(’Real and negative exponent’)>> % Sinusiodal signal with a real and positive exponent>> lam=1>> f=exp(lam*t).*sin(2*t)>> subplot(212); plot(t,f)

26


>> xlabel(’Time’); ylabel(’Real and positive exponent’)>> % Sinusoidal signal with a complex conjugate exponent>> t=0:0.1:15>> lam=-1+j*1>> fcc=exp(lam*t).*sin(2*t)>> figure (2)>> subplot(211); plot(t,abs(fcc))>> xlabel(’Time’); ylabel(’Magnitude’)>> subplot(212); plot(t,angle(fcc))>> xlabel(’Time’); ylabel(’Phase’)

0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Time

Mag

nitu

de

0 5 10 15−4

−2

0

2

4

Time

Pha

se

Figure 2.19b

Problem 2.43The train of rectangular pulses is plotted in Figure 2.20. This figure is obtained via the following MATLAB code.>> tau=0.1;T0=0.5;t=-0.3:0.001:4.3;>> ptau=stepfun(t,-tau/2)-stepfun(t,tau/2)>> trainptau=ptau>> for i=1:8>> trainptau=trainptau+stepfun(t,i*T0-tau/2)-stepfun(t,i*T0+tau/2)>> end>> plot(t,trainptau)>> xlabel(’Time’); ylabel(’Train of rectangular pulses’)>> axis([-0.3 4.3 -0.5 1.5]); grid

Problem 2.44The continuous- and discrete-time sinc signals are plotted in Figure 2.21. This figure is obtained via the followingMATLAB code.>> t=-5:0.01:5;>> contsinc=sinc(2*t-3)>> subplot(211); plot(t,contsinc)>> xlabel(’Continuous time’); ylabel(’Continuous sinc signal’); grid>> T=0.1; kT=-5:T:5;>> discsinc=sinc(2*kT+1)>> subplot(212); plot(kT,discsinc,’o’)>> xlabel(’Discrete time’); ylabel(’Discrete sinc signal’); grid

27

CHAPTER 2

0 0.5 1 1.5 2 2.5 3 3.5 4−0.5

0

0.5

1

1.5

Time

Tra

in o

f rec

tang

ular

pul

ses

Figure 2.20

−5 −4 −3 −2 −1 0 1 2 3 4 5−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Continuous time

Con

tinuo

us s

inc

sign

al

−5 −4 −3 −2 −1 0 1 2 3 4 5−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Discrete time

Dis

cret

e si

nc s

igna

l

Figure 2.21

Problem 2.45The considered signal is plotted in Figure 2.22. This figure is obtained via the following MATLAB code.>> t=-1:0.001:7;>> f=stepfun(t,0)+stepfun(t,5)-(t-1).*stepfun(t,1)+(t-3).*stepfun(t,3);>> plot(t,f)>> xlabel(’Continuous-time’)>> ylabel(’Continuous-time signal from FIGURE 2.8’); axis([-1 7 -2 2])

28


−1 0 1 2 3 4 5 6 7−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Continuous−time

Con

tinuo

ustim

e si

gnal

from

FIG

UR

E 2

.8

Figure 2.22

29

Solutions to Problems from Chapter 2 - Rutgers...

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