sin kπ kπ d
d
cos
kω
st
⎤ ⎥
⎦
v1 (t ) = ws (t ) cos nωst
∑ =
w(t
)[d
cos
nωst
+
2d
∞ k =1
sin kπ kπ d
d
cos
kωst
cos
nωst
n≠k
+
2d
sin nπ nπ d
d
cos2
nωst]
cos2
nωst
=
1 2
+
1 2
cos
2nωst
Solution:
( ) ∑ ∑ s t
=
∞ k =−∞
Π
⎡t ⎢⎣
− kTs τ
⎤ ⎥⎦
=
∞ k =−∞
ck e−
jkωst
∑ ∑ =
∞
d
k =−∞
sin kπ d kπ d
e− jkωst
=
d
+ 2d
∞ k =1
sin kπ d kπ d
cos kωst
∑ ws
(t)
=
w(t)
⎡⎢⎣d
+
2d
∞ k =1
is obtained with this demodulation technique , where w(t) is the oriqinal waveform and Cw(t) is the recovered waveform. Note that C is a function of n ,where the oscillator frequency is nfs.
⎞ ⎟⎠
⎛ ⎜⎝
fs
words sec
⎞ ⎟⎠
=
(9)200
=
1.8
K bits sec
(d) Determine the minimum absolute channel
bandwidth required for transmission of this
PCM signal.
For binary PCM D=R
number of bits in each PCM word needs to be
n ≥ [log2
10]
⎡ ⎢log10 ⎣
3.32
log10
⎛ ⎜ ⎝
50 P
⎞ ⎟ ⎠
(Hint: Look at Fig. 3-8c.)
Solution:
Binary PCM
M= 2n levels
Solution: (a) Determine the minimum sampling rate needed.
f s = 2B = 2(100) = 200 samples /sec
(b) Determine the number of bits needed in each PCM word. Using the results given in prob. 3-7.
⇒ B = D = 900Hz 2
3-9 An 850-Mbyte hard disk is used to store
3-4 (a)Show that an analog output waveform (which is proportional to the original input analog waveform) may be recovered from a naturally sampled PAM waveform by using the demodulation technique showed in Fig.3-4. (b) Find the constant of proportionality C, that
±δ 2 = ±0.1% →
δ = 0.1%
2V
4V
M = 2V = 500 and 29 = 512 > 500 δ
n = 9 bits → a PCM word
(c) Determine the minimum bit rate required in
the PCM signal.
R
=
⎛ ⎜⎝
n bits word
(c) Determine the minimum bit rate required in
the PCM signal.
(d) Determine the minimum absolute channel bandwidth required for transmission of this PCM signal.
3-2 (a) Sketch the naturally sampled PAM waveform that results from sampling a 1-kHz sine wave at a 4-kHz rate.
(b) Repeat part (a) for the case of a flat-topped PAM waveform. Solution:
=
loga (b) logb
(x)
3-8 The information in an analog voltage
waveform is to be transmitted over a PCM
system with a ± 0.1% accuracy (full scale).
The analog waveform has an absolute
for
P | nq |≤ 100 VPP
We need
step size
=δ
= VPP M
2P ≤ 100 VPP
1≤P M 50
2n = M ≥ 50 P
n
≥
log
2
⎛ ⎜ ⎝
50 P
⎞ ⎟ ⎠
50 ≥ log 2 (10) log10 ( p )
log a
(x)
=
logb (x) logb (a)
bandwidth of 100 Hz and an amplitude range
of –10 to +10V.
(a) Determine the minimum sampling rate
needed.
(b) Determine the number of bits needed in
each PCM word.
after LPF:
wo
(t
)
=
d
sin nπ nπ d
d
w(t
)
=
cw ( t
)
∴c = d sin nπ d nπ d
3-7 In a binary PCM system, if the
quantizing noise is not to exceed ±P percent
of the peak-to-peak analog level, show that the