Signal processing

41. Consider a single input single output discrete-time system with x[n] as input and y[n] as output, where the two are related as $$y\left[ n \right] = \left\{ {\matrix{ {n\left| {x\left[ n \right]} \right|,} & {{\rm{for}}\,0 \le n \le 10} \cr {x\left[ n \right] – x\left[ {n – 1} \right],} & {{\rm{otherwise}}} \cr } } \right.$$ Which one of the following statements is true about the system?

It is causal and stable
It is causal but not stable
It is not causal but stable
It is neither causal nor stable

Detailed SolutionConsider a single input single output discrete-time system with x[n] as input and y[n]

Join Our Telegram Channel
Subscribe on YouTube
as output, where the two are related as $$y\left[ n \right] = \left\{ {\matrix{ {n\left| {x\left[ n \right]} \right|,} & {{\rm{for}}\,0 \le n \le 10} \cr {x\left[ n \right] – x\left[ {n – 1} \right],} & {{\rm{otherwise}}} \cr } } \right.$$ Which one of the following statements is true about the system?

42. A system with transfer function H(z) has impulse response h(n) defined as h(2) = 1, h(3) = -1 and h(k) = 0 otherwise. Consider the following statements. S1 : H(z) is a low-pass filter. S2 : H(z) is an FIR filter. Which of the following is correct?

Only S2 is true
Both S1 and S2 are false
Both S1 and S2 are true, and S2 is a reason for S1
Both S1 and S2 are true, but S2 is not a reason for S1

Join Our Telegram Channel
Subscribe on YouTube
the following statements. S1 : H(z) is a low-pass filter. S2 : H(z) is an FIR filter. Which of the following is correct?" class="read-more button" href="https://exam.pscnotes.com/mcq/a-system-with-transfer-function-hz-has-impulse-response-hn-defined-as-h2-1-h3-1-and-hk-0-otherwise-consider-the-following-statements-s1-hz-is-a-low-pass-filter-s2-hz-is-an/#more-52834">Detailed SolutionA system with transfer function H(z) has impulse response h(n) defined as h(2) = 1, h(3) = -1 and h(k) = 0 otherwise. Consider the following statements. S1 : H(z) is a low-pass filter. S2 : H(z) is an FIR filter. Which of the following is correct?

where A and B are constants" class="read-more button" href="https://exam.pscnotes.com/mcq/the-fourier-transform-of-the-signal-xleft-t-right-e-3t2-is-of-the-following-form-where-a-and-b-are-constants/#more-52777">Detailed SolutionThe Fourier Transform of the signal $$x\left( t \right) = {e^{ – 3{t^2}}}$$ is of the following form, where A and B are constants

44. The Fourier transform of a signal h(t) is $$H\left( {j\omega } \right) = {{\left( {2\cos \omega } \right)\left( {\sin \omega } \right)} \over \omega }$$ The value of h(0) is

$${1 over 4}$$
$${1 over 2}$$
1
2

Detailed SolutionThe Fourier transform of a signal h(t) is $$H\left( {j\omega } \right) = {{\left( {2\cos \omega }

Join Our Telegram Channel
Subscribe on YouTube
\right)\left( {\sin \omega } \right)} \over \omega }$$ The value of h(0) is

45. The input and output of a continuous time system are respectively denoted by x(t) and y(t). Which of the following descriptions corresponds to a casual system?

y(t) = x(t - 2) + x(t + 4)
y(t) = (t - 4)x(t + 1)
y(t) = (t + 4)x(t - 1)

Detailed SolutionThe input and output of a continuous time system are respectively denoted by x(t) and y(t). Which of the following descriptions corresponds to a casual system?

46. The first five points of the 8-point DFT of a real valued sequence are 5, 1 – j3, 0, 3 – j4 and 3 + j4. The last two points of the DFT are respectively

0, 1 - j3
0, 1 + j3
1 + j3, 5
1 - j3, 5

Detailed Solution

Join Our Telegram Channel
Subscribe on YouTube
class="screen-reader-text">The first five points of the 8-point DFT of a real valued sequence are 5, 1 – j3, 0, 3 – j4 and 3 + j4. The last two points of the DFT are respectively

47. The trigonometric Fourier series of an even function of time does not have

The dc term
Cosine terms
Sine terms
Odd harmonic terms

Detailed SolutionThe

Join Our Telegram Channel
Subscribe on YouTube
trigonometric Fourier series of an even function of time does not have

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

$$ rac{{15}}{2}cos left( {40pi t - rac{pi }{4}} ight)$$
$$ rac{{15}}{2}left( { rac{{sin left( {pi t} ight)}}{{pi t}}} ight)cos left( {10pi t + rac{pi }{4}} ight)$$
$$ rac{{15}}{2}cos left( {10pi t - rac{pi }{4}} ight)$$
$$ rac{{15}}{2}left( { rac{{sin left( {pi t} ight)}}{{pi t}}} ight)cos left( {10pi t - rac{pi }{2}} ight)$$

Detailed SolutionThe signal $$\cos \left( {10\pi t + \frac{\pi }{4}} \right)$$ is ideally sampled at a

Join Our Telegram Channel
sampling frequency of 15 Hz. The sampled signal is
Subscribe on YouTube
passed through a filter with impulse response $$\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi \tau }}} \right)\cos \left( {40\pi t – \frac{\pi }{2}} \right).$$ The filter output is

49. The Fourier series representation of an impulse train denoted by $$s\left( t \right) = \sum\limits_{n = – \infty }^\infty {\delta \left( {t – n{T_0}} \right)} \,{\rm{is}}\,{\rm{given}}\,{\rm{by}}$$

$${1 over {{T_0}}}sumlimits_{n = - infty }^infty {exp left( { - {{j2pi nt} over {{T_0}}}} ight)} $$
$${1 over {{T_0}}}sumlimits_{n = - infty }^infty {exp } left( {{{jpi nt} over {{T_0}}}} ight)$$
$${1 over {{T_0}}}sumlimits_{n = - infty }^infty {exp } left( {{{j2pi nt} over {{T_0}}}} ight)$$

Detailed SolutionThe Fourier series representation of an impulse train denoted by $$s\left( t \right) = \sum\limits_{n = – \infty }^\infty {\delta \left( {t – n{T_0}} \right)} \,{\rm{is}}\,{\rm{given}}\,{\rm{by}}$$

50. If the Laplace transform of a signal y(t) is $$Y\left( s \right) = {1 \over {s\left( {s – 1} \right)}},$$ then its final value is

-1
0
1
Unbounded

Detailed SolutionIf the Laplace transform of a signal

Join Our Telegram Channel
23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTube
y(t) is $$Y\left( s \right) = {1 \over {s\left( {s – 1} \right)}},$$ then its final value is

Page 1Page 2Page 3Page 4Page 5
Test 1Test 2Test 3