Question 1 |

Let x_{1}(t) and x_{2}(t) be two band-limited signals having bandwidth B=4 \pi \times 10^{3} \mathrm{rad} / \mathrm{s} \mathrm{each}. In the figure below, the Nyquist sampling frequency, in rad/s, required to sample y(t), is

20 \pi \times 10^{3} | |

40 \pi \times 10^{3} | |

8 \pi \times 10^{3} | |

32 \pi \times 10^{3} |

Question 1 Explanation:

Given that, x_{1}(t) and x_{2}(t) are two bandlimited signals having bandwidth B=4 \pi \times 10^{3} \mathrm{rad} / \mathrm{sec}.

and y(t)=x_{2}(t) \cos \left(12 \pi \times 10^{3} t\right)+x_{1}(t) \cos \left(4 \pi \times 10^{3} t\right)

So, \text{Nyquist Rate }=2\omega _{max}=2[16 \pi \times 10^3]=32 \pi \times 10^3\; rad/sec

and y(t)=x_{2}(t) \cos \left(12 \pi \times 10^{3} t\right)+x_{1}(t) \cos \left(4 \pi \times 10^{3} t\right)

So, \text{Nyquist Rate }=2\omega _{max}=2[16 \pi \times 10^3]=32 \pi \times 10^3\; rad/sec

Question 2 |

Consider a real-valued base-band signal x(t), band limited to \text{10 kHz}. The Nyquist rate for the signal y\left ( t \right )=x\left ( t \right )x\left ( 1+\dfrac{t}{2} \right ) is

\text{15 kHz} | |

\text{30 kHz} | |

\text{60 kHz} | |

\text{20 kHz} |

Question 2 Explanation:

\mathrm{NR}=2 \times f_{\mathrm{max}}=2 \times 15=30 \mathrm{kHz}

Question 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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}

There are 5 questions to complete.