11. The Nyquist sampling rate for the signal $x\left( t \right) = \frac{{\sin \left( {500\pi t} \right)}}{{\pi t}}\: \times \frac{{\sin \left( {700\pi t} \right)}}{{\pi t}}$ is
12. Consider the sampling of the band-pass signal, whose spectrum is as shown below. What minimum sampling frequency would you use from the options given below, so as to avoid aliasing?
24 Hz
26.5 Hz
48 Hz
212 Hz
Answer.2. 26.5 Hz
Explanation
Bandpass sampling theorem:
The sampling frequency is given as:
fs = 2 × fH/k
fH → Highest frequency present in the signal
K = [fH/B]
B = fH − fL
Calculation:
Given:
fH = 106 Hz
fL = 94 Hz
B = (106 – 94) Hz = 12 Hz
K = [106/12] = 8
fs = (2 × 106)/8
fs = 26.5 Hz
13. A continuous-time sinusoid of frequency 33 Hz is multiplied with a periodic Dirac impulse train of frequency 46 Hz. The resulting signal is passed through an ideal analog low-pass filter with a cutoff frequency of 23 Hz. The fundamental frequency (in Hz) of the output is _________
10 Hz
15 Hz
13 Hz
20 Hz
Answer.3. 13 Hz
Explanation
sinusoid Frequency fm = 33 Hz
Impulse train Frequency fs = 46 Hz
fo/p = n. fs ± fm
where
n = 1, 2, 3, _ _ _ _ _ n
f (o/p) =46+ 33, 46–33
⇒ 79 Hz, 13 Hz
When this signal is passed through an LPF with a cut-off of 23 Hz, the fundamental frequency of output will be 13 Hz.
14. To satisfy the sampling theorem, a 100 Hz sine wave should be sampled at
10 Hz
100 Hz
200 Hz
50 Hz
Answer.3. 200 Hz
Explanation
Nyquist Sampling Theorem:
A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.
fs ≥ 2fm
Calculation:
With fm = 100 Hz, the minimum sampling frequency will be:
fs = 2fm = 2 × 100
fs = 200 Hz
15. A 4 GHz carrier is amplitude-modulated by a low-pass signal of the maximum cut-off frequency 1 MHz. If this signal is to be ideally sampled the minimum sampling frequency should be nearly
4 MHz
4 GHz
8 MHz
8 GHz
Answer.1. 4 MHz
Explanation
Given,
fc = 4GHz
fm = 1 MHz = 0.001GHz
For a baseband signa
Fs = 2fH/K
K = fH/(fH − fL)
= 4.001/(4.001 − 3.999) = 2000.5
We have to take K without any decimal value
K = 2000
Fs = 2fH/K
Fs = (2 × 4.001)/2000
= 4 MHz
16. When aliasing takes place
Sampling signals less than Nyquist rate
Sampling signals more than Nyquist rate
Sampling signals equal to Nyquist rate
Sampling signals at a rate which is twice of Nyquist rate
Answer.1. Sampling signals less than Nyquist rate
Explanation
The sampling theorem states that “a signal can be exactly reproduced if it is sampled at the rate fs which is greater than twice the maximum frequency W in the modulating signal.”
The minimum rate at which a signal can be sampled and still be reconstructed from its samples is known as the Nyquist rate.
If the sampled frequency is less than the Nyquist frequency, overlapping of lower and upper sidebands known as aliasing takes place.
17. A signal of 5kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are
2006 kHz, and 1895 kHz
2005 kHz, and 1995 kHz
2003 kHz, and 2995 kHz
2006 kHz, and 1395 kHz
Answer.2. 2005 kHz, and 1995 kHz
Explanation
fR = fC + fm = 2000kHz + 5kHz = 2005 kHz
fR = fC − fm = 2000kHz − 5kHz = 1995kHz
So, the frequency content of the resultant wave will have frequencies 1995kHz,2000kHz and 2005kHz
18. An AM wave is expressed as e = 10(1 + 0.6cos2000πt)cos 2×108πt volts, the maximum and minimum value of modulated carrier wave are
10V and 20V
4V and 8V
16V and 4V
8V and 20V
Answer.3. 16V and 4V
Explanation
Emax=(1+ma)Ec=(1 + 0.6) × 10=16V
Emin= (1−ma)Ec=(1− 0.6) × 10=4V.
19. An audio signal of 25 sin 2π (1400t) amplitude modulates to 80sin2π(105 )t. The two sideband frequencies are:
98.6 kHz, 101.4 kHz
92.5 kHz, 105.5 kHz
94 kHz, 102.5 kHz
96 kHz, 106 kHz
Answer.1. 98.6 kHz, 101.4 kHz
Explanation
Given
f1 =1400
f2 = 105
Sidebands are defined as f2 − f1 and f1 + f2
105 − 1400 = 98.6 kHz
105 + 1400 = 101.4 kHz
Hence, sidebands are 98600Hz,101400Hz, i.e., 98.6 KHz,101.4 KHz.
20. The limitation of amplitude modulation is
Clear reception
High efficiency
Small operating range
Good audio quality
Answer.3. Small operating range
Explanation
The most natural as well as man-made radio noise are of AM type. The AM receivers do not have any means to reject this kind of noise. Weak AM signals have a low magnitude compared to strong signals.
When it comes to power usage it is not efficient. It requires a very high bandwidth that is equivalent to that of the highest audio frequency. Amplitude Modulation has a very small operating range.
