Faculty of Engineering Sciences
Electronics and Communications Department
Fall 2012
ELE 466
Communication Systems
Sheet(4)
Dr. Amr Abouel Magd Ahmed
Eng : Walid Raslan
2013 – 2014
Sheet ( 4 )
(1)
Five telemetry signals, each of bandwidth 1 kHz, are to be transmitted simultaneously by
binary PCM. The maximum tolerable error in sample amplitudes is 0.2% of the peak
signal amplitude. The signals must be sampled at least 20% above the Nyquist rate.
Framing and synchronizing requires an additional 0.5% extra bits.
Determine the minimum possible data rate (bits per second) that must be transmitted, and
the minimum bandwidth required to transmit this signal.
Sol.
(2)
Specify the Nyquist rate and the Nyquist interval for each of the following signals:
(a) g(t) = sinc(200t)
(b) g(t) = sinc2(200t)
(c) g(t) = sinc(200t) + sinc2(200t)
Sol.
(3)
A certain continuous-time signal is 2 minutes long. The spectrum of the signal ranges
from near dc to 100 Hz. It is to be sampled and converted to digital form for computer
processing.
(a) What is the theoretical minimum number of samples that must be taken?
(b) Assume that each sample is represented as a 12-bit binary number. What is the
minimum storage in bits required to handle this signal?
Sol.
a) fs ≥ 2 fm  fs min = 2 fm = 200 sample/sec
b) bit rate = fs*n = 200*12 =2400 bps  min. storage = 2400*120 = 288 kbits
(4)
Figure 1 shows Fourier spectra of signals g1(t) and g2(t). Determine the Nyquist
interval for signals g1(t), g2(t), g12(t), g23(t) and g1(t)g2(t).
Figure 1
Sol.
(5)
A signal g(t) = sinc(10πt) is sampled (using uniformly spaced impulses) at a rate of: (i) 5
Hz; (ii) 10 Hz; (iii) 20 Hz. For each of the three cases:
(a) Sketch the sampled signal.
(b) Sketch the spectrum of the sampled signal.
(c) Explain weather you can recover the signal g(t) from sampled signal.
(d) If the sampled signal is passed through an ideal low-pass filter of bandwidth 5 Hz,
sketch the spectrum of the output signal.
(6)
Signals g1(t) = 104rect(104t) and g2(t) = δ(t) are applied at the inputs of ideal low-pass
filters H1(w) = rect (w/40000π) and H2(w) = rect(w/20000π) (fig.2). The outputs y1(t) and
y2(t) of these filters are multiplied to obtain the signal y(t) = y1(t)y2(t). Find the Nyquist
rate of y1(t), y2(t) and y(t).
Figure 2
Sol.
The B.W. of y1(t), y2(t) and y(t) are 10kHz, 5kHz and 15kHz respectively. Hence the
nyquist rate for three signals equal is 20 kHz, 10 kHz and 30 kHz respectively.
(7)
The signal shown in the following figure was sampled and quantized. The variance of the
quantization error is 2.0833. Compute the following:
Smax=30
Δ
0
Smin=-10
(a) The quantization step (Δ).
(b) Maximum quantization error.
(c) Minimum quantization error.
(d) Total number of levels.
(e) The number of bits required to represent each sample.
Sol.
a) σ2 = △2/12 = 2.0833  △ = 5
b) max. quantization error = △/2 = 2.5
c) min. quantization error = 0
d) L = (mmax - mmin)/△ = (30+10)/5 = 8 levels
e) n ≥ log2 L  n = 3 bits/sample