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Unformatted text preview: Chapter 6 # 6.1.1 6.12 6.13 The bandwidths of gi(t) and 92(2) are 100 kHz and 150 kHz, respectively. Therefore the Nyquist sampling rates
for 91(1) is 200 kHz and for 92(1) is 300 kHz. A150 912(1) => {737100) # may), and from the width property of convolution the bandwidth of g12(t) is twice
the bandwidth of gi(t) and that. of g23(t) is three times the bandwidth of 92(1) (se also Prob. 4.310). Similarly
the bandwidth of g,(f)g;(t) is the sum of the bandwidth of g1(t) and g2(t). Therefore the Nyquist rate {or 9120)
is 400 kHz. for 92%) is 900 kHz. for mama) is 500 kHz. (8)
sinc1001rt) e=e 0.01rect The bandwidth of this signal is 100 1r rad/s or 50 Hz. The Nyquist rate is 100 Hz (samples/sec).
(b)
sinc2(1001rt) a» 0.01A($) The bandwidth of this signal is 200 1r rad/s or 100 Hz. The Nyquist rate is 200 Hz (samples/sec).
(C)
sinc(1001rt) + sinc(501rt) ==> 0.011'ect0.01 (75%,) + 0.02rect (17%;) The bandwidth of the first term on the righthand side is 50 Hz and the second term is 25 Hz. Clearly the
bandwidth of the composite signal is the higher of the two, that is. 100 Hz. The Nyquist rate is 200 Hz (samples/sec).
(d) sinc(1007rt) + 35inc2(601rt) ¢=s 0.01 recuﬁﬂ + 5% (Mm’iﬂ The bandwidth of recuﬁ‘gﬂ is 50 Hz and that of Aigﬁﬁ) is 60 Hz. The bandwidth of the sum is the higher of
the two. that is. 60 Hz. The Nyquist sampling rate is 120 Hz. (6) sinc(501rr) 4:» 0.02 rectﬁgﬁ)
sinc(1001rt) 4==> 0.01 rect(§ﬁ;) The two signals have bandwidths 25 Hz and 50 Hz respectively. The spectrum of the product of two signals is
1/2r 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(50n!)sinc(1001rt) is
2:3 + 50 = 75 Hz. The Nyquist rate is 150 Hz. The pulse train is a periodic signal with fundamental frequency 28 Hz. Hence, w. = 217(23) = 41rB. The period
is To = 1/28. It is an even function of t. Hence, the Fourier series for the pulse train can be expressed as 1’12“): Co + Z Cn cos my“, as! Using Eqs. (2.72). we obtain 1/168 9 1/168
no=Co=—] di=}, an=Cn=: cosnw.!dt=—2—sin(ﬂ). bn=0
T0 —i/ies 4 T0 —)/168 n" 4 Hence. W) = gitlpmf) Ideal ﬁlm (0 : .
20 9' (Hz) > Practical Filter
\   4 .aiw)‘ 0" Fig. 56.14 6.14 For g(f) = sinc2(51rf) (Fig. 56.14a). the spectrum is C(u) = 0.2 A(§3';) (Fig. 36.14b). The bandwidth of this
signal is 5 Hz ( 101’ rad/s). Consequently. the Nyquist rate is 10 Hz, that is. we must sample the signal at a rate no less than 10 samples/s. The Nyquist interval is T = 1/23 = 0.1 second. Recall that the sampled signal spectrum consists of (1/T)G(u;) = A(5g;) repeating periodically with a period
equal to the sampling frequency f. Hz. We present this information in the following Table [or three sampling rates: I. = 5 Hz (undersampling). 10 Hz (Nyquist rate). and 20 Hz (oversampling). sampling frequency f. sampling interval T
n
—u In the ﬁrst case (undersampling). the sampling rate is 5 Hz (5 samples/sec), and the spectrum +G(..;) repeats every 5 Hz (101: rad/sec). The successive spectra overlap. as shown in Fig. 56.14d, and the spectrum C(w) is not recoverable from 50;). that is. g(!) cannot be reconstructed from its samples 3(1) in Fig. $6.1~4c. If the sampled signal is passed through an ideal lowpass ﬁlter of bandwidth 5 Hz. the output spectrum is rect (TaL
51 6.15 0.16 6.17 and the output signal is 10 sinc(201rt). which is not the desired s_ignal sinc2(51rt). In the second case. we use
the Nyquist sampling rate of 10 Hz (Fig. 56.14e). The spectrum C(w) consists of hicktoback. nonoverlapping
repetitions of +G(LJ) repeating every 10 Hz. Hence, C(w) can be recovered from C(w) using an ideal lowpass
ﬁlter of bandwidth 5 Hz (Fig. 56.14f). The output is losinc2(51.t). Finally, in the last case of oversampling
(sampling rate 20 Hz). the spectrum 5(a)) consists of nonoverlapping repetitions of +0(..:) (repeating every
20 Hz) with empty band between successive cycles (Fig. 36.14h). Hence. 0(9)) can be recover“ from 50”)
using an ideal lowpass ﬁlter or even a practical lowpass ﬁlter (shown dotted in FiS 56.1411) The Ouiput is
20 sinc’wm). This scheme is analyzed fully in Problem 3.41. where we found the bandwidths of 1110). t;2(t}),andy(lt'2 to be 10
kHz, 5 kHz. and 15 kHz. respectively. Hence, the Nyquist rates for the three signals are 20 k z. 10 it 2. and 30
kHz. respectively. (a) When the input to this ﬁlter is «6(1). the output of the summer is h(t) — 6(t — T). This acts as the input to
the integrator. And, Mr). the output of the integrator is: t ' _ T
W) = / wr)  be  mm = no — u(t _ T) = m (' T7)
0 The impulse response 11(1) is shown in Fig. 56.163.
(b) The transfer function of this circuit is: H(w == Tsinc {WT/2 s‘c m 2 The amplitude response of the ﬁlter is shown in Fig. 86.1—6b. Observe that the ﬁlter is a lowpass ﬁlter of
bandwidth 27r/T rad/s or l/T Hz. '
The impulse response of the circuit is a rectangular pulse. When a sampled signal is applied at the in ut. each
sample generates a recmngular pulse at the output. proportional to the corresponding sample value. ence the
output is a staircase approximation of the input as s own in Fig. 56.16c. and
WW)! = T (a) Figure 56.1Ta shows the si 'nal reconstruction from its samples using the ﬁrstorder hold circuit. Each
sum le generates a triangle of wi th 27' and centered at the sampling instant . The height of the triangle is equal
to t 9 sample value. The resulting signal consists of straight line segments Joming the sample tops. (h) The transfer function of this circuit is: T
up) = sum} = f {A = Tuna Because H (.4) is positive for all u. it also represents the amplitude response. Fig. 86.17b shows the impulse response I: (t) = A(;‘T ). The corresponding amplitude responSe H(;u) and the ideal amplitude response (lowpsss) required for signal reconstruction is shown in Fig. 86.17c.
(c) A minimum of T secs delay is r uired to make Mt) causal (realizable). Such a delay would cause the reconstructed signal in Fi . $6178. to e delayed by T secs.
(d) When the input to t e ﬁrst ﬁlter is Mt). then as shown in Prob. 6.14. its output is a rectangular pulse p(r) = 11(I) — u(t — T) shown in Fig. 56.14a. This pulse p(t) is applied to the input of the second identical ﬁlter.
The output of the summer of the second filter is p(t)  p(t — T) = u(t)  2u(t. — T) + "(t  2T). which is applied
to the integrator. The output h(t) of the integrator is the area under p(t)  p(t  '1‘). which. as Mr) = / [11(1) — 2n(r — T) + "(1 — 2T)Idr = tu.(t)  2(t  T)u(t  T) + (t — 2T)u(t  21”) = A
0 shown in Fig. 56.17b. 6.18 Assume a signal git) that is simultaneously timelimited and bandlimited. Let 9(a) = O for wl > 21rB. Therefore g(s')rect(ﬁr) = _q(..;) for B' > B. Therefore from the timeconvolution property (3.43) 9(1) = 9(1) ~ (2B'sinc(21tB't)]
= QB'g(t) n sinc(21rB't) 5‘2 .,« “a. 6.21 6.22 6.23 6.24 6.26 ._T c
(bx Figure 86.17 Because 9(1) is timelimited. 9(1) = 0 for It! > T. But 9(a) is equal to convolution of 9(2) with sinc(21r B’t) which
is not timelimited. It is impossible to obtain a timelimited signal from the convolution of a timelimited signal with a nontimelimited signal. (a) Since 128 = 2". we need 7 bits/character.
(b)For 100,000 characters/second . we need 700 kbits/second.
(a) 8 bits/character and 800 kbits/second. (a) The bandwidth is 15 kHz. The Nyquist rate is 30 kHz. (b) 65536 = 2”. so that 16 binary digits are needed to encode each sample.
(c) 30000 x 16 = 480000 bits/s. (d) 44100 x 16 = 705600 bits/s, (a) The Nyquist rate is 2 x 4.5 x 10‘ = 9 MHz. The actual sampling rate = 1.2 x 9 = 10.8 MHz.
(b) 1024 2 2‘0. so that 10 bits or binary pulses are needed to encode each sample.
(c) 10.8 x 10‘5 x 10 = 108 x 10° or 103 Mbits/n. if m” is the peak sample amplitude. then quantization error 3 (0.233;. ) = 1?. it follows that . . . . 2
Because the maXimum quantization error is 921 = 1? = IL" 3 [it = Because L should be a power of 2. we choose L = 512 = 29. This requires a 9bit. binary code per sample The
Nyquist rate is 2 x 1000 = 2000 Hz. 20% above this rate is 2000 x 1.2 = 2400 Hz. Thus. each signal has 2400
samples/second. and each sample is encoded by 9 bits. Therefore. each signal uses 9 x 2400 = 21.6 kbits/second.
Five such signals are multiplexed. hence. we need a total of 5 x 21.6 = 108 kBits/sccond data bits. Framing
and synchronization requires additional 0.5% bits. that is. 108, 000 x 0.005 = 540 bits. yielding a total of 108540
bits/second. The minimum transmission bandwidth is = 54.27 kHz. Nyquist rate for each signal is 200 Hz. The sampling rate I, = 2 x Nyquist rate = 400 Hz Total number of samples for 10 signals = 400 x 10 = 4000 samples/second.
Quantization error 5 %’£ = 3% Moreorer. quantization error = A} = 212? = 3? = % = L = 400
Because L is a power of 2. we select L = 512 = 2°. that is, 9 bits/sample.
Therefore. the minimum bit rate = 9 x 4000 = 36 kbit.s/second. The minimum cable bandwidth is 36/2=18 kHz. W
For a sinusoid. as” = 0.5. The SNR = 47 dB =50119. From Eq. (6.16) P 2"
5" = 31.29;“; = 3L’(o.5) = 50119 = L =182.8
No "1;,
Because L is a power of 2. we select L = 256 = 2‘. The SNR for this value of L is
2
ﬂ = 31.21'1—(1l = 3(256)’(0.5) = 98304 = 49.43 dB
No m; ' 6.27 6.28 6.29 6.210 6.41 Fig. 56.27 For this periodic mu). each quarter cycle takes on the same set of amplitude values. Hence, each quarter cycle
contributes identical energy. Consequently, we can compute the power for this signal by averaging its energy
over a quarter cycle. The equation of the ﬁrst quarter cycle as shown in Fig. 36.27 is m(t) = 4A/To. where A is
the peak amplitude and To is the period of m(t). The power or the mean squared value (energy averaged over a quarter cycle) is
W Tol‘ 2 2
mz(t) = —1— dt = é—
To/4 0 To 3
2 2 1
Hence. 1;? = = p The rest of the solution is identical to that of Prob. 6.26. From Eq. (6.16). SNR of 47 dB is a ratio of 50119. is So gmsu) 2
70 = 3L ~711 = 3L (1/3) = 50119 ==> L = 223.87
" r Because L is a power of ‘2. we select L = 256 = 25. The SNR for this value of L is 50 “115(1)
— = 3L‘——5— = 3(256)’(1/3) = 65536 = 48.16 dB
No 1",, Here (I = 100 and the SNR = 45 dB: 31,622.77. From Eq. (6.18) g 3L3 N0 = (“1le = I. = 473.83 = 31,622.77
Berause L is a power of 2. we select L = 512 = 2°. The SNR for this value of L is So _ 3(512)2 _ 9 _
N0 — ————(1n 1 01), .. 3692.84 .. 45.67 dB (21) Nyquist rate = 2 x 106 Hz. The actual sampling rate is 1.5 x (2 x 10°) == 3 x 10" Hz. Moreover. L = 256
and p = 255. From Eq. (6.18) so _ 3L2 _ 3(256)’ __
No — [ln(u + 1)]2  (ln 256)2 6394  38.06 dB
(b) If we reduce the sampling rate and increase the value of L so that the same number of bits/second is
maintained. we can improve the SNR (because of increased L) with the same bandwidth. In part (a). the
sampling rate is 3 x 10° Hz and each sample is encoded by 8 bits (L = 256). Hence. the transmission rate is s x 3 x 10‘ = 24 Mbits/second.
If we reduce the sampling rate to 2.4 x 106 (20% above the Nyquist rate). then for the same transmission rate
(24 Mbits/s). we can have (24 x 10°)/(2.4 x 10") = 10 bits/sample. This results in L = 21° = 1024. Hence. the new SNR is
so _ 3L2 _ 3(1024)2 _ 9 _
‘17., ‘ """pmu +1)‘,= ‘ (ln256)'~' “10300 “5411‘”; Clearly. the SNR is increased by more than 10 dB. (6.23) shows that increasing n by one bit increases the SNR by 6 dB. Hence. an increase in the SNR Equation
(from 30 to 42) can be accomplished by increasing 1: from 10 to 12, that is increasing by “20%. by 12 dB
(a) From Eq. (6.33) Am“ = so that 1 = => :7 = 0.0785 54 , _ r723 _ (0.0785)2(3500) __ 4
(b) A __ 3?: _ _————(3)(64000) _. 1.12 x 10 (c) Here 50 = = 0.5. and
so 0.5 .1
No 112 x 10m: = 4.46 X10 (d) For uniform distribution W 2
so = 1.12m .—. ___ So 0.333 N_o = 1.12 x 104
8 kHz. For a bipolar case. we need a so that = 2.94 x 103 .1.
3 (e) For onoﬂ’ signaling with a bit. rate 64 kHz, we need a bandwidth of 12
bandwidth of 64 kHz Q1
C‘ ...
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 Spring '08
 Dandakar

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