Question 1 |
A signal x(t)=2 \cos (180 \pi t) \cos (60 \pi t) is sampled at 200 \mathrm{~Hz} and then passed through an ideal low pass filter having cut-off frequency of 100 \mathrm{~Hz}.
The maximum frequency present in the filtered signal in \mathrm{Hz} is ____ (Round off to the nearest integer)
The maximum frequency present in the filtered signal in \mathrm{Hz} is ____ (Round off to the nearest integer)
95 | |
80 | |
110 | |
115 |
Question 1 Explanation:
Given :
\begin{aligned} x(\mathrm{t})&=2 \cos (180 \pi \mathrm{t}) \cos (60 \pi \mathrm{t}) \\ &=\cos 240 \pi \mathrm{t}+\cos 120 \pi \mathrm{t} \end{aligned}
\mathrm{F}_{1}=120 \mathrm{~Hz} and \mathrm{F}_{2}=60 \mathrm{~Hz}
Now, present frequency component,
\begin{aligned} & =\mathrm{nf}_{\mathrm{s}} \pm \mathrm{f}_{1} \\ & =\mathrm{f}_{1}, \mathrm{f}_{\mathrm{s}} \pm \mathrm{f}_{1}, 2 \mathrm{f}_{\mathrm{s}} \pm \mathrm{f}_{1} \\ & =120 \mathrm{~Hz}, 200 \pm 120, \ldots . \\ & =120 \mathrm{~Hz}, 80 \mathrm{~Hz}, 320 \mathrm{~Hz}, \ldots . \end{aligned}
and other frequency component.
\begin{aligned} & =n f_{s} \pm f_{2} \\ & =f_{2}, f_{s} \pm f_{2}, 2 f_{s} \pm f_{2}, \ldots \\ & =60,200 \pm 60, \ldots \\ & =60 \mathrm{~Hz}, 140,260 \mathrm{~Hz}, \ldots \end{aligned}
Given cut-off frequency of LPF =100 \mathrm{~Hz}
\therefore It is pas only 60 \mathrm{~Hz} and 80 \mathrm{~Hz} frequency component.
\therefore Max. frequency =80 \mathrm{~Hz}.
\begin{aligned} x(\mathrm{t})&=2 \cos (180 \pi \mathrm{t}) \cos (60 \pi \mathrm{t}) \\ &=\cos 240 \pi \mathrm{t}+\cos 120 \pi \mathrm{t} \end{aligned}
\mathrm{F}_{1}=120 \mathrm{~Hz} and \mathrm{F}_{2}=60 \mathrm{~Hz}
Now, present frequency component,
\begin{aligned} & =\mathrm{nf}_{\mathrm{s}} \pm \mathrm{f}_{1} \\ & =\mathrm{f}_{1}, \mathrm{f}_{\mathrm{s}} \pm \mathrm{f}_{1}, 2 \mathrm{f}_{\mathrm{s}} \pm \mathrm{f}_{1} \\ & =120 \mathrm{~Hz}, 200 \pm 120, \ldots . \\ & =120 \mathrm{~Hz}, 80 \mathrm{~Hz}, 320 \mathrm{~Hz}, \ldots . \end{aligned}
and other frequency component.
\begin{aligned} & =n f_{s} \pm f_{2} \\ & =f_{2}, f_{s} \pm f_{2}, 2 f_{s} \pm f_{2}, \ldots \\ & =60,200 \pm 60, \ldots \\ & =60 \mathrm{~Hz}, 140,260 \mathrm{~Hz}, \ldots \end{aligned}
Given cut-off frequency of LPF =100 \mathrm{~Hz}
\therefore It is pas only 60 \mathrm{~Hz} and 80 \mathrm{~Hz} frequency component.
\therefore Max. frequency =80 \mathrm{~Hz}.
Question 2 |
Consider the two continuous-time signals defined below:
x_{1}(t)=\left\{\begin{matrix} |t|, & -1\leq t \leq 1 \\ 0 & otherwise \end{matrix}\right. ,
x_{2}(t)=\left\{\begin{matrix} 1-|t|, & -1\leq t \leq 1\\ 0, & otherwise \end{matrix}\right.
These signals are sampled with a sampling period of T=0.25 seconds to obtain discretetime signals x_{1}[n] \; and \; x_{2}[n], respectively. Which one of the following statements is true?
x_{1}(t)=\left\{\begin{matrix} |t|, & -1\leq t \leq 1 \\ 0 & otherwise \end{matrix}\right. ,
x_{2}(t)=\left\{\begin{matrix} 1-|t|, & -1\leq t \leq 1\\ 0, & otherwise \end{matrix}\right.
These signals are sampled with a sampling period of T=0.25 seconds to obtain discretetime signals x_{1}[n] \; and \; x_{2}[n], respectively. Which one of the following statements is true?
The energy of x_{1}[n] is greater than the energy of x_{2}[n] | |
The energy of x_{1}[n] is greater than the energy of x_{2}[n] | |
x_{1}[n] nor x_{2}[n] have equal energies. | |
Neither x_{1}[n] nor x_{2}[n] is a finite-energy signal |
Question 2 Explanation:
x_1(t)=\left\{\begin{matrix} |t|, & -1 \leq t \leq 1\\ 0, & \text{ otherwise} \end{matrix}\right.
T_s= sampling time-period = 0.25 sec
x_1(n)=\{ 1,0.75,0.5,0.25,0,0.25,0.5,0.75,1\}
x_2(t)=\left\{\begin{matrix} 1-|t|, & -1 \leq t \leq 1\\ 0, & \text{ otherwise} \end{matrix}\right.
x_2(n)=\{ 0,0.25,0.5,0.75,1,0.75,0.5,0.25,0\}
Since, x_1(n) is having one more non-zero sample of amplitude '1' as compared to x_2(n). Therefore, energy of x_1(n) greater than energy of x_2(n).
T_s= sampling time-period = 0.25 sec
x_1(n)=\{ 1,0.75,0.5,0.25,0,0.25,0.5,0.75,1\}
x_2(t)=\left\{\begin{matrix} 1-|t|, & -1 \leq t \leq 1\\ 0, & \text{ otherwise} \end{matrix}\right.
x_2(n)=\{ 0,0.25,0.5,0.75,1,0.75,0.5,0.25,0\}
Since, x_1(n) is having one more non-zero sample of amplitude '1' as compared to x_2(n). Therefore, energy of x_1(n) greater than energy of x_2(n).
Question 3 |
The output y(t) of the following system is to be sampled, so as to reconstruct it from its samples
uniquely. The required minimum sampling rate is
1000 samples/s | |
1500 samples/s | |
2000 samples/s | |
3000 samples/s |
Question 3 Explanation:
From the above block diagram,
Z(t)=x(t) \cos 1000 \pi t
By using modulation property of fourier transform,
Z(\omega )=\frac{1}{2}[X(\omega + 1000 \pi t)+X(\omega - 1000 \pi t)]
Now, h(t)=\frac{\sin 1500 \pi t}{\pi t}=1500 Sa(1500 \pi t)
Thus, H(\omega ) is a low pass filter and it will pass frequency, component of Z(\omega ) upto 1500 \pi rad/sec.
Therefore, maximum frequency component of y(t) is
\omega _m =1500 \pi rad/sec or f_m =750 Hz
So, the minimum sampling rate for y(t) is
f_{s\; min}=2f_m=1500Hz=1500 samples/sec
Question 4 |
Let x_{1}(t)\leftrightarrow X_{1}(\omega ) \; and \; x_{2}(t) \leftrightarrow X_{2}(\omega \omega ) be two signals whose Fourier Transforms are as shown in the figure below. In the figure, h(t)=e^{-2|t|} denotes the impulse response.
For the system shown above, the minimum sampling rate required to sample y(t), so that y(t) can be uniquely reconstructed from its samples, is
For the system shown above, the minimum sampling rate required to sample y(t), so that y(t) can be uniquely reconstructed from its samples, is
2B_{1} | |
2(B_{1}+B_{2}) | |
4(B_{1}+B_{2}) | |
\infty |
Question 4 Explanation:
Given that,
Bandwidth of X_1(\omega )=B_1
Bandwidth of X_2(\omega )=B_2
system has h(t)=e^{-2|t|} and input to the system is x_1(t) \cdot x_2(t)
The bandwidth of x_1(t) \cdot x_2(t) is B_1+B_2.
The bandwidth of output will be B_1+B_2.
So, sampling rate will be 2(B_1+B_2).
Bandwidth of X_1(\omega )=B_1
Bandwidth of X_2(\omega )=B_2
system has h(t)=e^{-2|t|} and input to the system is x_1(t) \cdot x_2(t)
The bandwidth of x_1(t) \cdot x_2(t) is B_1+B_2.
The bandwidth of output will be B_1+B_2.
So, sampling rate will be 2(B_1+B_2).
Question 5 |
A sinusoid x(t) of unknown frequency is sampled by an impulse train of period
20 ms. The resulting sample train is next applied to an ideal lowpass filter with
cutoff at 25 Hz. The filter output is seen to be a sinusoid of frequency 20 Hz. This
means that x(t)
10Hz | |
60 Hz | |
30 Hz | |
90 Hz |
Question 5 Explanation:
Given, impulse train of period 20 ms.
Then, sampling frequency =\frac{1}{20 \times 10^{-3}}=50Hz
If the input signal x(t)= \cos \omega _m (t) having spectrum
The filtered out sinusoidal signal has 20 Hz frequency, the sampling must be under sampling. The output signal which is an under sampled signal with sampling frequency 50 Hz is
and 50-f_m=20Hz\;\Rightarrow \;f_m=30Hz
Then, sampling frequency =\frac{1}{20 \times 10^{-3}}=50Hz
If the input signal x(t)= \cos \omega _m (t) having spectrum
The filtered out sinusoidal signal has 20 Hz frequency, the sampling must be under sampling. The output signal which is an under sampled signal with sampling frequency 50 Hz is
and 50-f_m=20Hz\;\Rightarrow \;f_m=30Hz
There are 5 questions to complete.