Question 1 |

A band limited low-pass signal x(t) of bandwidth 5 kHz is sampled at a sampling rate f_{s} .
The signal x(t) is reconstructed using the reconstruction filter H(f) whose magnitude
response is shown below:

The minimum sampling rate f_{s} (in kHz) for perfect reconstruction of x(t) is _______.

The minimum sampling rate f_{s} (in kHz) for perfect reconstruction of x(t) is _______.

10 | |

11 | |

12 | |

13 |

Question 1 Explanation:

Let us assume an arbitrary spectrum for x(t) as
shown below:

The spectrum of the sampled signal can be given as,

For proper reconstruction of the signal,

\begin{aligned} f_{s}-5 & \geq 8 \\ f_{s} &\geq 8+5=13 \mathrm{kHz} \\ \text { So. } \quad f_{s(\mathrm{min})} &=13 \mathrm{kHz} \end{aligned}

The spectrum of the sampled signal can be given as,

For proper reconstruction of the signal,

\begin{aligned} f_{s}-5 & \geq 8 \\ f_{s} &\geq 8+5=13 \mathrm{kHz} \\ \text { So. } \quad f_{s(\mathrm{min})} &=13 \mathrm{kHz} \end{aligned}

Question 2 |

The signal x(t)=sin(14000\pi t) , where t is in seconds is sampled at a rate of 9000 samples per second. The sampled signal is the input to an ideal low pass filter with frequency response H(f) as follows :

H(f)=\left\{\begin{matrix} 1, & |f|\leq 12 kHz\\ 0, & |f|\geq 12 kHz \end{matrix}\right.

What is the number of sinusoids in the output and their frequencies in kHz?

H(f)=\left\{\begin{matrix} 1, & |f|\leq 12 kHz\\ 0, & |f|\geq 12 kHz \end{matrix}\right.

What is the number of sinusoids in the output and their frequencies in kHz?

Number = 1, frequency = 7 | |

Number = 3, frequencies= 2,7,11 | |

Number = 2, frequencies = 2, 7 | |

Number = 2, frequencies = 2, 11 |

Question 2 Explanation:

\begin{aligned} x(t) &=\sin (14000 \pi t) ; f_{m}=7 \mathrm{kHz} \\ f_{s} &=9000 \text { samples per second } \\ &=9 \mathrm{kHz} \end{aligned}

The spectrum of the sampled signal can be given as shown below:

So, three sinusoids will be there at the output of the LPF and the frequencies of those sinusoids are 2 kHz, 7 kHz and 11 kHz.

The spectrum of the sampled signal can be given as shown below:

So, three sinusoids will be there at the output of the LPF and the frequencies of those sinusoids are 2 kHz, 7 kHz and 11 kHz.

Question 3 |

Consider the signal x(t)=cos(6\pi t)+sin(8\pi t), where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t)=x(2t+ 5) is

8 | |

12 | |

16 | |

32 |

Question 3 Explanation:

\begin{aligned} X(t) & =\cos (6 \pi t)+\sin (8 \pi t) \\ Y(t) & =x(2 t+5) \\ Y(t) & =\cos [6 \pi(2 t+5)+\sin (8 \pi(2 t+5)] \\ & =\cos (12 \pi t+30 \pi)+\sin (16 \pi t+40 \pi) \\ f_{m_{1}} & =6 \mathrm{Hz}_{i} \qquad f_{m 2}=8 \mathrm{Hz}\\ \text{Nyquist }&\text{sampling rate},\\ f_{s} &=2 f_{\max } \\ &=16 \text { samples/second } \end{aligned}

Question 4 |

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 _________

5 | |

12 | |

18 | |

24 |

Question 4 Explanation:

If x(t) is a message signal and y(t) is a sampled
signal, then y(t) is related to x(t) as

y(t)=x(t) \sum_{n=-\infty}^{\infty} \delta\left(t-n T_{s}\right)

r(f)=f_{s} \sum_{n=--\infty}^{\infty} X\left(f-n f_{s}\right)

Spectrum of X(f) and Y(f) are as shown

Cut off frequency of LPF = 23 Hz

Hence, frequency at the output is 13 Hz

y(t)=x(t) \sum_{n=-\infty}^{\infty} \delta\left(t-n T_{s}\right)

r(f)=f_{s} \sum_{n=--\infty}^{\infty} X\left(f-n f_{s}\right)

Spectrum of X(f) and Y(f) are as shown

Cut off frequency of LPF = 23 Hz

Hence, frequency at the output is 13 Hz

Question 5 |

Consider a continuous-time signal defined as

x(t)=(\frac{sin(\pi t/2)}{(\pi t/2)})*\sum_{n=-\infty }^{\infty }\delta (t-10n)

where '*' denotes the convolution operation and t is in seconds. The Nyquist sampling rate (in samples/sec) for x(t) is _____.

x(t)=(\frac{sin(\pi t/2)}{(\pi t/2)})*\sum_{n=-\infty }^{\infty }\delta (t-10n)

where '*' denotes the convolution operation and t is in seconds. The Nyquist sampling rate (in samples/sec) for x(t) is _____.

0.4 | |

0.8 | |

0.2 | |

1 |

Question 5 Explanation:

x(t)=\frac{\sin \left(\frac{\pi t}{2}\right)}{\left(\frac{\pi t}{2}\right)} * \sum_{n=-\infty}^{\infty} \delta[t-10 n]

\begin{aligned} f_{\max } &=0.2 \mathrm{Hz} \\ \Rightarrow \quad f_{s} &=2 f_{\max }=0.4 \mathrm{Hz} \end{aligned}

\begin{aligned} f_{\max } &=0.2 \mathrm{Hz} \\ \Rightarrow \quad f_{s} &=2 f_{\max }=0.4 \mathrm{Hz} \end{aligned}

Question 6 |

The signal cos(10\pi t+\frac{\pi }{4}) is ideally sampled at a sampling frequency of 15 Hz. The sampled signal is passed through a filter with impulse response (\frac{sin(\pi t)}{\pi t})cos(40\pi t-\frac{\pi }{2}). The filter output is

\frac{15}{2}cos(40\pi t-\frac{\pi }{4}) | |

\frac{15}{2}(\frac{sin(\pi t)}{\pi t})cos(10\pi t+\frac{\pi }{4}) | |

\frac{15}{2}cos(10\pi t-\frac{\pi }{4}) | |

\frac{15}{2}(\frac{sin(\pi t)}{\pi t})cos(40\pi t-\frac{\pi }{4}) |

Question 6 Explanation:

\begin{array}{l} x(t)=\cos \left(10 \pi t+\frac{\pi}{4}\right) \\ X(f)=\frac{1}{2}[\delta(f-5)+\delta(f+5)] e^{j \pi / 4} \end{array}

After sampling x(t) by 15 Hz, we get

X_{s}(f)=15 \sum_{n=-\infty}^{\infty} X(f-15 n)

Given, \quad h(t)=\left[\frac{\sin (\pi t)}{\pi t}\right] \cos \left(40 \pi t-\frac{\pi}{2}\right)

H(f)=\frac{1}{2}\left[\text{rect} \frac{(f-20)}{2 \pi}+\text{rect} \frac{(f+20)}{2 \pi}\right] e^{-j \frac{\pi}{2}}

\begin{aligned} \text{Output }\quad Y(t)&=X_{s}(f) \times H(f) \\ y(t)&=\frac{15}{2} \cos \left(40 \pi t-\frac{\pi}{4}\right) \end{aligned}

After sampling x(t) by 15 Hz, we get

X_{s}(f)=15 \sum_{n=-\infty}^{\infty} X(f-15 n)

Given, \quad h(t)=\left[\frac{\sin (\pi t)}{\pi t}\right] \cos \left(40 \pi t-\frac{\pi}{2}\right)

H(f)=\frac{1}{2}\left[\text{rect} \frac{(f-20)}{2 \pi}+\text{rect} \frac{(f+20)}{2 \pi}\right] e^{-j \frac{\pi}{2}}

\begin{aligned} \text{Output }\quad Y(t)&=X_{s}(f) \times H(f) \\ y(t)&=\frac{15}{2} \cos \left(40 \pi t-\frac{\pi}{4}\right) \end{aligned}

Question 7 |

For a given sample-and-hold circuit, if the value of the hold capacitor is increased, then

droop rate decreases and acquisition time decreases | |

droop rate decreases and acquisition time increases | |

droop rate increases and acquisition time decreases | |

droop rate increases and acquisition time increases |

Question 7 Explanation:

We know that

\begin{aligned} Q&=C V=i \cdot t &\ldots(i)\\ t&=\frac{C V}{i} &\ldots(ii) \end{aligned}

From equation (ii)

t \propto C and it is clear that if value of capacitor increases then the acquisition time increases. For capacitor the drop rate is given as \frac{d v}{d t}

\begin{aligned} i &=C \frac{d v}{d t} \\ \frac{d v}{d t} &=\frac{i}{C} \\ \frac{d v}{d t} & \propto \frac{1}{C} \end{aligned}

From above relation it is clear that if capacitor value increases drop rate decreases.

\begin{aligned} Q&=C V=i \cdot t &\ldots(i)\\ t&=\frac{C V}{i} &\ldots(ii) \end{aligned}

From equation (ii)

t \propto C and it is clear that if value of capacitor increases then the acquisition time increases. For capacitor the drop rate is given as \frac{d v}{d t}

\begin{aligned} i &=C \frac{d v}{d t} \\ \frac{d v}{d t} &=\frac{i}{C} \\ \frac{d v}{d t} & \propto \frac{1}{C} \end{aligned}

From above relation it is clear that if capacitor value increases drop rate decreases.

Question 8 |

Let x(t)=cos(10\pi t)+cos(30\pi t) be sampled at 20 Hz and reconstructed
using an ideal low-pass filter with cut-off frequency of 20 Hz. The frequency/
frequencies present in the reconstructed signal is/are

5 Hz and 15 Hz only | |

10 Hz and 15 Hz only | |

5 Hz, 10 Hz and 15 Hz only | |

5 Hz only |

Question 8 Explanation:

x(t)=\cos (10 \pi t)+\cos (30 \pi t)

Given sampling frequency, f_{s}=20 \mathrm{Hz}

\omega_{s}=40 \pi \mathrm{rad} / \mathrm{sec}

Y(\omega)=\sum_{K=-\infty}^{\infty} X\left(\omega-K \omega_{s}\right)

After sampling waveform will be

Applying an ideal low-pass filter of cut-off frequency of 20 \mathrm{Hz} or 40 \pi \; \mathrm{ rad} / \mathrm{sec}, we get the frequencies in reconstructed signal as 10 \pi \text{ and } 30 \pi \; \mathrm{rad} / \mathrm{sec}.

or 5 \mathrm{Hz} and 15 \mathrm{Hz}

Given sampling frequency, f_{s}=20 \mathrm{Hz}

\omega_{s}=40 \pi \mathrm{rad} / \mathrm{sec}

Y(\omega)=\sum_{K=-\infty}^{\infty} X\left(\omega-K \omega_{s}\right)

After sampling waveform will be

Applying an ideal low-pass filter of cut-off frequency of 20 \mathrm{Hz} or 40 \pi \; \mathrm{ rad} / \mathrm{sec}, we get the frequencies in reconstructed signal as 10 \pi \text{ and } 30 \pi \; \mathrm{rad} / \mathrm{sec}.

or 5 \mathrm{Hz} and 15 \mathrm{Hz}

Question 9 |

Consider two real valued signals, x(t) band-limited to [-500 Hz, 500 Hz] and
y(t) band-limited to [-1 kHz, 1 kHz]. For z(t) = x(t)\cdot y(t), the Nyquist sampling
frequency (in kHz) is ____.

1 | |

2 | |

3 | |

4 |

Question 9 Explanation:

Multiplication in time domain = convolution in frequency domain.

x_{1}(t) \cdot x_{2}(t) \leftrightarrow X_{1}(\omega) * X_{2}(\omega)

So, highest frequency component contained by the convolved signal z(t)=1500 \mathrm{Hz}

\therefore Nyquist rate =2 \times 1500

=3000 \mathrm{Hz}=3 \mathrm{kHz}

x_{1}(t) \cdot x_{2}(t) \leftrightarrow X_{1}(\omega) * X_{2}(\omega)

So, highest frequency component contained by the convolved signal z(t)=1500 \mathrm{Hz}

\therefore Nyquist rate =2 \times 1500

=3000 \mathrm{Hz}=3 \mathrm{kHz}

Question 10 |

A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency which is not valid is

5 kHz | |

12 kHz | |

15 kHz | |

20 kHz |

Question 10 Explanation:

\begin{aligned} \left(f_{s}\right)_{\min } &=2 f_{m} \\ \left(f_{s}\right)_{\min } &=2 \times 5=10 \mathrm{kHz} \\ \mathrm{So}, \quad f_{s} & \geq 10 \mathrm{kHz} \end{aligned}

There are 10 questions to complete.