Solution Manual
for Signals and Systems: Continuous and Discrete, 4/E 4th Edition
Rodger E. Ziemer, William H Tranter, Rolla
D. R. Fannin
full chapter at: https://testbankbell.com/product/solution-manualfor-signals-and-systems-continuous-and-discrete-4-e-4th-editionrodger-e-ziemer-william-h-tranter-rolla-d-r-fannin/
CHAPTER I
Problem 1-1
(a) Write the acceleration as
Thus the velocity and position are, respectively, given by
(b) See the figure below for the integrator.
Assume that << R The input impedance to the op-amp integrator is therefore much larger than the output impedance of the previous stage, and
From Example 1-2,
Therefore, Integrating and setting t = t%, we obtain
The second factor on the right is 10 V because of the maximum output limitation on the first integrator. Thus, we require that
For example, from Example 1-2 we have RC = 0.36 s. With t, = 72 s and R, = 10 k ohms, we get R, = 152 ohms
Problem 1-2
(a) Let n = 0, 1, 2, 3, ..., N. Then
v(T) = v(0) + Ta(T) (a)
v(2T) = v(T) + Ta(2T) (b)
NT) = v[(N -)T] + Ta(NT) (c)
Substitute (a) into (b) and so on until (c is reached This gives
ND - v(0) +T) a0D
(b) Let n =0, 1, 2, 3, ..., N. Then
v(T) = v(0) + (T/2)[a(0) + a(T)] (a)
v(2T) = v(T) + (T/2)[a(T) + a(2T)] (b)
(NT) = v[(N - D)T] + (T/2){a[(N - D)T] + a(NT)} (c)
Substitute (a) into (b)and so onuntil (c) is reached. The result is asgiven in theproblem statement.
Problem 1-3
(a) A maximum departure of the weight from equilibrium of 1 cm requires a spring constant of k M max (0 002)(20) = 4 kg/s 2 0.01
(b) For a minimum increment of 0 5 mm= 00005 m, we have Lamin = KA, M = 4(0 0005) = 0.002 1 mis 2
(c) The velocity is given by v,()
Problem 1-4
K is the same asin Example 1-1 because M, x,~ and a,~ arethesame. Also, la,~is the same The velocity profile is
[20d2 - 201,0 < t < 10
0
200, 10 < 1 < 20
200+ [2002. 200 + 20- 20), 20 < t < 30 20
400, t > 30
Problem 1-5
From (1-15) and using the x(t) given in the problem, we have s(0) = cos(6,0) + ap cos[6,(1 - 20)]
= [I +a~cos(26,5)]cos(6,t) + a~sin(26,)sin(6,0)
- A()cos[G,t - 0()]
- A(v)cos6(v)cos(6,0) + A(v)sin0(v)sin(6,D)
Set coefficients of like sin/cos terms equal on each side of the identity to obtain A(v)cos0() 1 + a~cos(26,5)
A()sin0(v) = a~sin(26,)
Square and add to obtain
Divide the second equation bythe first to obtain sin0(v) cos0() = tan0(T) a~sin(26,) 1 + a~cos(26,°)
Problem 1-6
Sketches of the analog and sampled signals for both cases are shown below[(a) top and (b)bottom]:
Problem 1-7
(a) The impulse-sampled signal is
%so,(0 - cos(210)) 6-0 1m)
-) cos(2r06-0 1m)
-) cos(0.2rm)6G -0.1m)
where property (1-59) for the unit impulse has been used to get the last result.
(b) The unit-pulse train sampled signal is
'on rots as0 = cos(210)) 6[ - 01n]
- ) cos(2n)6 0.1n)
-) cos(0 2n)f -01n]
where the fact that the unit pulse is 1 for its argument 0 and 0 otherwise has been used
Problem 1-8
(a)The signal can be developed in terms ofequations as follows:
II(0 10) = {1, [O ltl ~ 1/2 0, otherwise
-{, 1ls 1on2 5 0, otherwise
This is arectangular pulse of amplitude 1 between -5 and 5 and0 otherwise A sketch willbe given atthe end of the problem solution
(b)Following aprocedure similar tothat of (a) one finds that this is arectangular pulse of amplitude 1 between -0.05 and0.05 and0 otherwise. Asketch willbe given attheend ofthe problem solution.
(c ) This is a rectangular pulse of amplitude 1 between 0 and 1 and 0 otherwise. A sketch will be given at the end of the problem solution.
(d) This is a rectangular pulse of amplitude 1 between 0.5 and 4.5 and 0 otherwise A sketch will be given atthe end of the problem solution
(e) The first term of this signal is a rectangular pulse of amplitude 1 between 0 and 2 and 0 otherwise. The second term is a rectangular pulse of amplitude 1 between 0.5 and 1.5 and 0 otherwise. Where both pulses are nonzero, the total amplitude is 2; where only one pulse is nonzero the amplitude is I. A sketch is provided below.
The MATLAB program below uses the special function given in Section I-6 (page 32) of the text to provide the plots.
% Sketches for Problem 1-8
% t=-6:0.0015:6;
xa = pls fn(0.10);
xb = pls fn( 1 0*t);
xc = pls fn(t- 0.5);
xd =pls fn((t- 2)/5);
xe =pls fn((t- 1/2) +pls fn(t - l);
subplot(3,2, 1),plot(t, xa,'-w'), axis([-6 60 1.5]),xlabel('t'),ylabel('xa(t)')
subplot(3,2,2),plot(t, xb,'-w'), axis([-.I .1 0 1.5)),xlabel('t'),ylabel('xb(t')
subplot(3,2,3),plot(t, xc,'-w'), axis([-1 2 0 1 5]),xlabel('t'),ylabel('xc(t)')
subplot(3,2,4),plot(t, xd,'-w'), axis([-I 5 0 1 5]),xlabel('t'),ylabel('xd(t)')
subplot(3,2,5),plot(t, xe,'-w'), axis([-1 3 0 25]),xlabel('t'),ylabel('xe(t)')
Problem 1-9
(a) 2nf= 50n, so T,,= 1/f,= 1/25 =004 s (b) 2nf, = 60, so T,= 1/f, = 1/30 = 00333 s
(c)2nf,= 707, so T,= 1/f,= 1/35 =00286 s.(d)We have 50n = 2mf, and 60n = 2nf,, where m and n are integers andf isthe largest constant that satisfies these equations The largest f is 5Hz with m = 5and n = 6. (e) We have 50 = 27mf and 70 = 27nf, where m and n are integers and f% is the largest constant that satisfies these equations The largest f is 5Hz with m = 5 and n =7
Problem 1-10
(a) IAI = 4.2426; angle(A) = 0.7854 radians; B = 5.0 +j 8.6603, so Re(B) = 5 and lm(B) = 8.6603.
(b) = 8.0 + 1.6603. (c) -
= -2.0- (d) = -10.9808 +j40.9808. (e) = 04098 - j0 I098
Problem 1-11
(a) 2nf,= 10n, so T,,= 1/f, = 1/5 =02 s (b) 2nf, = 17, so T,= 1/f = 1/85 = 01176 s.
(c)2nf = 19n, so T,= 1/f, = 1/9 5 = 01053 s (d) We have 10 = 2mmf, and 177 = 2nf, where m and n are integers andf is the largest constant that satisfies these equations The largest f, is05 Hz with m = 10 and n = 17 (e)We have 10 = 2mf and 19 = 2rnf, where m and n are integers andf, is the largest constant that satisfies these equations. The largest f is 0.5 Hz with m = 10 and n = 19.(f) We have 17 = 2mf and 19 = 2nf,, where m and n are integers andf is the largest constant that satisfies these equations. The largest f is 0.5 Hz with m = 17 and n = 19.
Problem 1-12
(a) Written as the real part of rotating phasors:
x,() = Re[2e/om W@]; x,(D) = Re[5ei7-Wj
x,(t) = Re[3eiuO-3 -mW)] = Re[3eon 5an/6
x,(t) = Re[2el0¢·ml6) , 5%iu7m -WP]; x,(t) = Re[2e/Ou·I6) , 3%i(om5/6)
x,(0) = Re[5el7mu -» , 3%iuo-Sm/6
(b) Interms of counter rotating phasors, the signals are:
x,(0 = [eiuow w6) , , Jou @]; x,(t) = [25el7#-W , 25e7-!9)J ,( [1.5ei0-51/6) 1.5a ion -5/6)
x(0)= [ai(or• mW6) , a ioru /6) + 25ei7-/9) + 2 5e i7mi - 1/)
x,(t) = [ei(om·mW6) , , i(or w6) , 1.5%ion-5r6) , 1.5e i0-51/6);
x,(0) [2.5ei7-W9 + 2.5e 7 W , 1 5au0-56) , 1 5% i045/61
(c ) Single-sided spectra are plotted below Double-sided amplitude spectra are obtained by halving the lines and taking mirror image; phase spectra are obtained by taking antisymmetric mirror image.
Problem 1-13
(a) Written as the real part of rotating phasors:
x,(0) = Re[e5-a)]; x,(D = Re[e]; x,(0 = Re[ei"9
x,() Re[ei~50a 2) , ]; x,(t) = Re[el5 M2) , %mo
(b) In terms of counter rotating phasors, the signals are:
x,(0) = Re[05i50 w2) , 0 5% Ks0 mW]; x,(0) = Re[05e' + 0 5% 60
x,() Re[0.5el0 +0.5e m0
x,(t) = Re[0.5ei50-w2) , 0 5% 050 mW2) , 0 5%0 +0 5 60u
x,(t) Re[0.5ei5002) ,0 5e 50 ) ,0 5%i0, +0 5e m70
(c ) The single-sided amplitude and phase spectra areshown below See Prob 1-12c for comments on obtaining double-sided spectra from single-sided spectra.
(a) A sketch is given below:
From the figure, it is evident that x(t) =
(b)The derivative of x() is