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Unformatted text preview: E5 31‘] HW 7 Soluh'ons 6.12 (3)
1 smc(21007rt) 4:) mil (f/2100) The bandwidth of this signal is 2100 1r rad/s or 1050 Hz. The Nyquist rate is 2100 Hz (samples/s). (b)
55inc2 (200m 4=> 0.02m (f/400) The bandwidth of this signal is 400 1r rad/s or 200 Hz. The Nyquist rate is 400 Hz (samples/s). (c) ‘ _
sinc (210mm) + sinc2(2001rt) «l» gang/2100) + 0.005A(f/400)
qr." " .. if: in" L _
The bandwidth of the ﬁrst term on the right—hand side is 1050 Hz, and the bandwidth of the second term is 200
Hz. Clearly the bandwidth of the composite signal is the higher of the two. that is, 1050 Hz. The Nyquist rate is 2100 Hz (samples/s). (d) sinc(2007rt) 4:) 0.005 U (f/200) . l
smc (21001rt) <=> m H(f/2100) The two signals have bandwidths 100 Hz, and 1050 Hz, respectively. The spectrum of the product of two signals is
1/27r times the convolution of their spectra. Rom width property of the convolution, the width of the convoluted
signal is the sum of the widths of the signals convolved. Therefore, the bandwidth of sinc (2007rt)sinc (21007rt) is 100 + 1050 = 1150 Hz. The Nyquist rate is 2300 Hz. 6.18 Assume that a signal g(t) is simultaneously timelimited and bandlimited. Let C(f) = 0 for If! > B denote
its bandlimited characteristics. Therefore G'(f)H (T27) =G(f) for 3' >8. Therefore from the timeconvolution property
g(t) = g(t) * 2B’sinc (0nB’t)
= 2B’g(t) at sinc (27rB’t)
Because g(t) is timelimited, there exists a T such that g(t) = 0 for [t] > T. But 9(6) is equal to convolution of g(t) with sinc (27rB’t) which is not timelimited. The convolution lasts forever. It is impossible to obtain a time~limited
signal from the convolution of a timelimited signal with a nontime—limited signal. 6.14 I,
l
(a) 1 x
10(1) = H (Ti — 5) 4:» P(f) = T, sine (ﬁnkinfra
(b) t 1 T ”T ,
p“) = H (Ts/2 ' 5) 4:» P(f) = —23sinc ( 2 3) e—wm
(C) 1 i
pm = sin (27,153) M) — u<t — T./2>1«=> Pm = a 717T: f2 (1+ e—mn)
Based on Section 6.1.1, the equalizer ﬁlter E(F) should be
T,/P(f), lfl S B
E(f)= ﬂexible, B<lfl<(1/T,—B)
0» IfIZU/TsB) Substituting P( f ) of parts (a) and (b) into the foregoing equation, we can get their corresponding equalizer ﬁlter
EU). 6.2—2
(a) The Nyquist rate is 2 x 4.5 = 9 MHz. The actual sampling rate is 1.2 x 9 = 10.8 MHz. (b) Since 1024 = 21°, 10 bits or binary pulses are needed to encode each sample. (c) 10.8 x 10 = 108 Mbit/s. The minimal bandwidth is 108/2 = 54 MHz. ...
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 Spring '09
 Yee

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