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Unformatted text preview: Problem 13 (a‘) The signal can be developed in terms of equations as follows: _ 1, mm s 112
mu“) * {0, otherwise _ 1, r g 10:2 = 5
_ 0, otherwise This is a rectangular pulse of amplitude 1 between 5 and 5 and 0 otherwise. A sketch will be given
at the end of the problem solution. (b) Following a procedure similar to that of (a) one ﬁnds that this is a rectangular pulse of amplitude
1 between 0.05 and 0.05 and 0 otherwise. A sketch will be given at the end of the problem solution.
(c )Tliis is a rectangular pulse of amplitude 1 between 0 and l and 0 otherwise. A sketch will be
given at the end of the problem solution. ((1) This is a rectangular pulse of amplitude 1 between 0.5 and 4.5 and 0 otherwise. A sketch will
be given at the 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 1. A sketch is provided below. The MATLAB program below uses the special function given in Section 16 (page 32) of the text
to provide the plots. % Sketches for Problem 18 % t = —6:0.0015:6; xa = pls_fn(0. l *t); xb == pls_fn(10*t); xc = pls_fn(t  0.5); xd = pls_fn((t  2V5); xe = pls_fn((t  W2) + pls_fn(t  l); subplot(3,2, l),plot(t, xa,'w'), axis([6 6 0 l.5]),xlabel(‘t'),ylabel(‘xa(t)')
subplot(3,2,2),plot(t, xb,‘w‘), axis([.] .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([~l S 0 l.5]),xlabel('l'),ylabel('xd(t)')
subplot(3.2,5),plot(t, xe,‘w‘), axis([—l 3 0 2.5]),xlabel('t'),ylabel(‘xe(t)') —5 0 5 —%.1 —0.05 O 0.05 0.1
t t
15 1.5
E 1 E 1
9 O 1 2 0 0 2 4 (c ) Singlesided spectra are plotted below. Doublesided amplitude spectra are obtained by halving
the lines and taking mirror image; phase spectra are obtained by taking antisymmetric mirror image. 4 _ 2
a
$2 ‘ 0
j —2
O
5 1O 0 ph rad 5 10 Problem 19 (3)2193 = 501:, so T0 = lltfu = “25 = 0.04 s. (b) 211‘)“0 = 601:, so To = 1% = 1130 = 0.0333 s (c ) 211:}; =_ 701:, so To = llﬁ, = N35 = 0.0286 s. (d) We have 50a = anfo and 6011: = 2mg" Iwhere m
and n are integers and fo is the largest constant that satisﬁes these equations. The largesto’f is 5 Hz
with m = S and n = 6. (e) We have 5011: = 21tmﬁ, and 70a = 21tnf0, where m and n are inteogers and
f0 is the largest constant that satisﬁes these equations. The largest f0 is 5 Hz with m = 5 and n = 1 Problem 110 (3) IA! 24.2426: angletA) = 0.7854 radians; B = 5.0 +J 3.6601 so Rem) = 5 and MB) = 8 6603
()A+ =8.0+j11.6603. (c)A»B=2.0'5.6603. dA*B=1090 .  _ Problem 114
(a) A sketch is given below:
From the figure, it is evident that x0) = u(r) + u(r  3)  u(r  5)  u(r  6). 3 o
1 o 1 2 a 4 5
t (b) The derivative of 1:0) is dx(r)}'dr = 5(0 + 6 (r  3)  (‘5 (1 5)  6 (t  6) MAW—HQ (a) The integral is zero because the delta function is outside the range of integration.
(b) The integral evaluates as follows: 5
fcos(2m)6(:  2)dt = cos(41't) = l
0 (c ) This integral can be evaluated as 5
feosam) 6(1 — 0.5)d: = (20501:) 0 

—— (d) The value of this integral is 0: II
O In — 2)26(:  2)dr = (2 — 2F (c) This integral evaluates to [125(I2)dt = 22 = 4 W (a) Only (1) is pv.‘.'.riodi£:,‘f1 = 2.5 Hz= 0.5m andf2= 3 Hz = 0.511 where the integers m and n are 5 and
6. respectively. The fundamental frequency is 0.5 Hz and the period is 2 s. (b) Signals (1) and (2) are power signals. Their powers are both 1 W. (c) Only signal (3) is an energy signal; its energy is 1/20 I. Signal (4) is neither energy nor power. ...
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This note was uploaded on 09/15/2008 for the course EE 308 taught by Professor Atkin during the Spring '08 term at Illinois Tech.
 Spring '08
 ATKIN

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