Principles of Communication Systems-I
Assignment #7 - Solutions
1. From the Nyquist sampling theorem, the minimum required sampling rate to avoid aliasing
is 2fm , where fm is the maximum frequency of the message signal.
Ans b
∞
X
2. The spectrum of the resulting sampled signal is fs M (f − nfs ), as described in the
n=−∞
lectures.
Ans a
3. Consider 5sinc (2t). Its spectrum corresponds to a pulse of height 25 from -1 Hz to 1 Hz.
Hence, the spectrum 5sinc (2t) cos(10πt) has two pulses of height 54 each from 4 Hz to 6 Hz
and -6 Hz to -4 Hz. As can be seen, fm is 6 Hz. Hence the minimum sampling frequency
required to avoid distortion using conventional low pass sampling Nyquist criterion is 2×6 =
12 Hz.
Ans a
4. Given the signal m(t) = sinc2 (5t). Consider sinc (5t). The corresponding spectrum is 51 p5 (f ).
2 1 1
Hence the spectrum
corresponding to sinc (5t) is given by the convolution 5 p5 (f ) ∗ 5 p5 (f )
1 |f |
which is 1− , |f | ≤ 5 and 0 otherwise. The maximum frequency component is seen
5 5
to be 5 Hz. Hence the minimum sampling frequency required is 10 Hz.
Ans d
5. From the previous 2
problem it is seen that the spectrum of sinc (5t) is given by the triangular
1 |f |
pulse M (f ) = 1− , |f | ≤ 5 and 0 otherwise. If this is sampled at fs =5 Hz, the
5 5
X ∞
resulting spectrum is 5 M (f − 5n) which can be seen to be equal to 1 for all f . Thus,
n=−∞
when it is sampled with an LPF with cutoff frequency 10 Hz, the resulting spectrum is a
pulse of height 1 between -10 Hz to 10 Hz i.e. p20 (f ). The corresponding time domain signal
is 20sinc (20t).
Ans a
6. Given that a wagon wheel is rotating clockwise at a speed of 2400 rpm (40 Hz) and is
viewed under a fluorescent light in India. The AC line frequency in India is 50 Hz at which
the fluorescent lamp flickers. This is equivalent to sampling a 40 Hz signal at a sampling
frequency of 50 Hz. Hence, the resulting sampled signal contains components at 40-n50 Hz,
1
Principles of Communication Systems-I
where n is any integer. Therefore, the frequencies present are 40 Hz, -10 Hz, -60 Hz etc.
Since the eye is usually sensitive to the lowest frequency, the wagon wheel will be observed
to rotate at -10 Hz i.e. 10 Hz anti-clockwise. This is termed as the wagon wheel effect.
Ans d
7. The pulse amplitude modulated signal m(t), with pulse p(t) and sampling duration Ts can
P∞
be expressed as m(nTs )p(t − nTs ).
n=−∞
Ans c
|t|
8. Given triangular pulse p(t) = 1 − for |t| ≤ τ and 0 otherwise, with τ << Ts . This
τ
corresponds to a triangle of height 1 from −τ to τ . This can be obtained as the convolution
1 1 1
to two rectangular pulses of height √ from − τ2 to τ2 i.e. √ pτ (t) ∗ √ pτ (t). The spectrum
τ τ τ
1 √ √ 2
of √ pτ (t) is τ sinc (τ f ). Hence, spectrum of given triangular pulse is ( τ sinc (τ f )) =
τ
τ sinc2 (τ f ). Since there is no aliasing, original signal can be reconstructed by filtering with
1 1
system with overall response 2
for |f | ≤ and 0 otherwise.
τ sinc (τ f ) 2Ts
Ans d
9. Uniform quantizers with odd and even number of quantization levels respectively are known
as mid-tread and mid-rise quantizers respectively.
Ans a
10. As described in class lectures, the resulting quantization noise power is well approximated by
the formula −10 log10 3 + 20 log10 ν − 6R.
Ans b
