Exercise solution (Digital Communications)
1.
p (t )
1 cos
2(
2 t
T
t 2
)
T
2 t
2
2 t
2 t
sin
sin
T lim T
T lim
T 1
p (0) lim
t 0
t 0
t 0
t 2
t
t
2( )
4 ( )
2( )
T
T T
T
1 cos 2n
p (nT )
0,
n0
2(n )2
1 cos
The zero ISI condition can be satisfied.
2. Consider the binary sequence 0100101. Draw the waveform for the following signaling formats.
(a) Unipolar NRZ signaling format;
(b) Bipolar RZ signaling format
Solution
3. The waveform as shown is transmitting data at a rate of 100,000 bits/s. State (a) the modulation
technique; (b) the carrier frequency; (c) the transmitted data.
Solution
(a) PSK;
(b) T = 5 s fc = 200 kHz;
(c) Data = 110100
4. (a) The overall transfer function of a transmission medium is given as H a(w). The Fourier transform of
this function is
2 sin 2 10 6 t
.
10 6 tthrough the medium given above, discuss
If the sequence of pulses, shown below, is to be transmitted
ha (t )
whether there will be any possibility of error due to intersymbol interference at the receiver. Explain.
(Hint: there is an inverse time-bandwidth relationship, i.e. B ~ 1 /, whereis pulse duration)
(b) If we now send the same sequence of pulses through another transmission medium which has the
transfer function Hb(w) shown below, will there be intersymbol interference? Explain.
Solution
(a) Impulse response is:
ha (t )
2 sin(2 106 t )
4 Sa(2 106 t )
6
10 t
The first zero point of the sampling function curve is at 0.5 10-6s.
The system bandwidth can be estimated from the first zero crossing of the impulse response, i.e.
System bandwidth 1 / (0.5 10-6 2) = 1 MHz
Whereas the signal bandwidth can be estimated from the inverse of the duration of the PCM pulses, i.e.
Signal bandwidth 1 / = 2 / Ts = 2 /(0.4 10-6) = 5 MHz
Since the signal bandwidth is larger than the transmission system bandwidth, distortion occurs.
Therefore, intersymbol interference occurs and may lead to detection error.
(b) The system bandwidth is
108 / 2 > 5 MHz
So there will be no intersymbol interference.