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Unformatted text preview: Midterm l
ECElOl
Fall 2008 Problem 1 (12 points) X(f) = 45inc(4f) X“) Problem 2 (12 points)
.5
f_c —B —f_c +B f_c —B f_c f_c +3
I
m(t)cos(21rfct)
Problem 3 (25 points)
0..)
Y6“)
l I I
i ' V2, 2
' f
(f,+ 3) f1 0 «5l (ﬂm) b) ;> To obtain the spectrum Z(f), we need to shift the positive half of Y(f) down by
f, + B and the negative half of Y(f) up by f1+ B. This can be achieved by setting
the frequency f2 of the second mixer to be :L(fl + B).
b) '9 Since ﬁlm. tiplication by two cosines results in a. amplitude attenuation of 0.25, the transfer ‘ C') Theoretically, one could recover the original signal by passing the scrambled signal
through the scrambler again. C x) a”, I 7,) Another acceptable answer would be to multiply z(t) by another cosine of frequency
B. Lowpass ﬁltering and amphﬁcation by 2 would also yield the original spectrum XUl
Problem 4 (25 points)
) We’re given that = NosLd  we can ﬁnd the power spectral density by
a" taking Sy( f) = TlRyU m We also know that, since the phase is 0,
Ho) = lH(f)le‘"g(H"” = IH me“ = H(f) At this point, we’re ready to ﬁnd H(f): Sn(f)H(f)l2;Tn(f)H2(f) = Syo) = NOTTectUT)
=> H20”) = ———° T) = ———N°TT:,:t(fT) = 2Trect(fT) <=> H( f) = y/2Trect(fT) = x/Z—T—TecztUT) The last step comes from the fact that the rect function raised to any power simply
equals the rect function. So, we’ve found H ( f), and we can now ﬁnd h(t): sinc(t/T) H Trect(fT) <=> 1/Tsinc(t/T) H rect(fT)
4=> «ﬁ/Tsincu/T) H ﬁreaur) <=> \/—1—2—:sinc(t/T) H ﬁreagr) 5,) Our ﬁnal answer is thuslh(t) = @sincU/T). I C.) The signal y(t) is uncorrelated when = 0. Since = Nosinc(T/T), we
see that condition for uncorrelation occurs whenever T = nT Where 'n. = i1, i2, :l:3...
(since the sine function is O at the even and negative integers). 4,) When T increases, the autocorrelation function widens, so the signal changes more
slowly. In other words, when the autocorrelation function widens it means that signal
samples stay correlated for a longer time period, and so that implies slower change in the signal.
Problem 5 (16 points) a.) The sampling interval is T5=111.11 usec. There are 96 channels plus I sync pulse, so
the time allotted to each channel is Tc=Tsl97=l.145 usec. The pulse duration 15 1
usec, so the time between pulses is 0.145 usec. b.) / If sampled at the Nyquist rate;
and the time betWeen pulses is O . 7,17,“ sgL . l v“_ V >  I V V
rs Hz ) :5 , Havana , 71 — 13/97 Luz/155‘ Problem 6 (10 points) 4.) Demodulation of standard AM is accomplished simply by using an envelope detector
or squarelaw detector. Thus, the receivers are cheaper and less complex than those
of FM systems where differentiation followed by envelope detection or zerocrossing
detection must be performed. However, AM is more sensitive to amplitude distortion, Modern digital modulation techniques are more spectrally efﬁcient than analog tech
niques. Furthermore, most information today is already in the form of bits, digital
components are often cheaper and faster, and one is able to perform encryption and
error correction upon digital data. ...
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This note was uploaded on 02/27/2012 for the course CHEMISTRY/ CH/ECE/PH/ taught by Professor Faculty during the Spring '08 term at Cooper Union.
 Spring '08
 Faculty

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