Ex Solution Pulse
Ex Solution Pulse
Ex Solution Pulse
1.
Solution
31
t cos 1 t
2
2
F ( ) = [ ( 1 ) + ( + 1 ) + ( 21 ) + ( + 21 )]
f (t ) = cos 1t + cos 21t = 2 cos
2.
The digital audio compact optical disc (CD) system uses 16 bit quantization. Assuming the audio
signal has a peak to mean power ratio of 13 dB,
mp2
m2 ( t )
SNRQ =
3L2 m 2 (t ) 3 232
=
= 6.44 108 = 88.1dB
2
mp
20
3. Suppose the signal as shown is sampled and each sample is quantized using 4 bits. Determine the
maximum quantization error.
Solution
2mp = 10V, n = 4 L = 24 = 16,
= 2mp /L = 10/16 = 0.625 V
The maximum quantization error = /2 = 0.3125V
4. A TDM system consists of 24 transmission channels and a synchronization channel, a sampling rate of
8kHz is used. The bandwidth of the signal for each channel is below 3.3kHz. Determine the minimum
channel bandwidth required to transmit TDM signal in the system.
Solution
fs = 8 kHz = 3 fm, thus fm = 4 kHz
The minimum channel bandwidth required to transmit TDM signal:
(24 + 1)fm = 25 4 = 100 kHz
5. Five signals are combined to be transmitted in a TDM system, the combined signals will pass through a
low-pass filter. Three channels are used to transmit the signals of frequency range between 300 to 3300 Hz
and the rest two channels transmit the signals of 50 Hz to 10 kHz range.
(a) What is the minimum sampling rate required?
(b) What is the minimum bandwidth of the low-pass filter required corresponding to the minimum
sampling rate?
Solution
(a) For 10 kHz signal, fs = 20 kHz;
(b) The bandwidth of the low-pass filter
B = 5 10 = 50 kHz
6. The information in an analog voltage waveform is to be transmitted over a PCM system with a 0.5%
accuracy (full scale). The analog waveform has an absolute bandwidth of 100 Hz and an amplitude range
of 5 to 5 V.
(a) Determine the minimum sampling frequency needed to recover the analog signal;
(b) Determine the number of bits needed in each PCM word;
(c) Determine the minimum bit rate required in PCM signal;
(d) Determine the minimum channel bandwidth required for transmission of this PCM signal.
Solution
mp = 5 V
(a)
(b)
(c)
(d)
fs = 2B = 200 Hz
= 0.5% 2 2mp = 0.02mp, L = 2mp / = 1/0.01 = 100 < 128 = 27
thus,
n=7
nfs = 7 200 = 1400 bits/s
BT = nB = 7 100 = 700 Hz
7. For a signal f(t) = 9 + Amcosmt, with Am 10, to be quantized into exactly 41 binary levels, with one
level set at the smallest value of f(t).
(a) Determine the number of bits needed in each PCM word;
(b) What are the values of extreme quantum levels Vmax and Vmin if the quantized levels are centered
to [f(t)max + f(t)min] / 2;
(c) If Am = 10 V, find the signal to quantized noise ratio.
Solution
(a)
(b)
(c)
Vmax = 9 + 32 / 2 = 9 + 16 = 25,
N q=
Vmin = 9 16 = -7
( ) 2 0.25
=
12
12