Signal processing

11. The unit impulse response of a linear time invariant system is the unit step function u(t). For t > 0, the response of the system to an excitation e-at u(t), a > 0 will be

ae-at
$$left( {{1 over a}} ight)left( {1 - {e^{ - at}}} ight)$$
a(1 - e-at)
1 - e-at

Detailed SolutionThe unit impulse response of a linear time invariant system is the unit step function

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u(t). For t > 0, the response of the system to an excitation e-at u(t), a > 0 will be

12. The unilateral Laplace transform of f(t) is $${1 \over {{s^2} + s + 1}}.$$ Which one of the following is the unilateral Laplace transform of g(t) = t.f(t)?

$${{ - s} over {{{left( {{s^2} + s + 1} ight)}^2}}}$$
$${{ - left( {2s + 1} ight)} over {{{left( {{s^2} + s + 1} ight)}^2}}}$$
$${s over {{{left( {{s^2} + s + 1} ight)}^2}}}$$
$${{2s + 1} over {{{left( {{s^2} + s + 1} ight)}^2}}}$$

Detailed

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SolutionThe unilateral Laplace transform of f(t) is $${1 \over {{s^2} + s + 1}}.$$ Which one of the following is the unilateral Laplace transform of g(t) = t.f(t)?

13. A 1 kHz sinusoidal signal is ideally sampled at 1500 samples/sec and the sampled signal is passed through an ideal low-pass filter with cutoff frequency 800 Hz. The output signal has the frequency

Zero Hz
0.75 kHz
0.5 kHz
0.25 kHz

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the sampled signal is passed through an ideal low-pass filter with cutoff frequency 800 Hz. The output signal has the frequency" class="read-more button" href="https://exam.pscnotes.com/mcq/a-1-khz-sinusoidal-signal-is-ideally-sampled-at-1500-samples-sec-and-the-sampled-signal-is-passed-through-an-ideal-low-pass-filter-with-cutoff-frequency-800-hz-the-output-signal-has-the-frequency/#more-48771">Detailed SolutionA 1 kHz sinusoidal signal is ideally sampled at 1500 samples/sec and the sampled signal is passed through an ideal low-pass filter with cutoff frequency 800 Hz. The output signal has the frequency

14. Flat top sampling of low pass signals

Gives rise to aperture effect
Implies oversampling
Leads to aliasing
Introduces delay distortion

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signals" class="read-more button" href="https://exam.pscnotes.com/mcq/flat-top-sampling-of-low-pass-signals/#more-48698">Detailed SolutionFlat top sampling of low pass signals

15. The Fourier series of an odd periodic function, contains only

Odd harmonics
Even harmonics
Cosine terms
Sine terms

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href="https://exam.pscnotes.com/mcq/the-fourier-series-of-an-odd-periodic-function-contains-only/#more-48280">Detailed SolutionThe Fourier series of an odd periodic function, contains only

16. Consider the differential equation $${{dx} \over {dt}} = 10 – 0.2x$$ with initial conduction x(0) = 1. The response x(t) for t > 0 is

2 - e-0.2t
2 - e-0.2t
50 - 49e-0.2t
50 - 49e0.2t

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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 button" href="https://exam.pscnotes.com/mcq/consider-the-differential-equation-dx-over-dt-10-0-2x-with-initial-conduction-x0-1-the-response-xt-for-t-0-is/#more-48164">Detailed SolutionConsider the differential equation $${{dx} \over {dt}} = 10 – 0.2x$$ with initial conduction x(0) = 1. The response x(t) for t > 0 is

17. For a periodic signal $$v\left( t \right) = 30\sin 100t + 10\cos 300t + 6\sin \left( {500t + {\pi \over 4}} \right),$$ the fundamental frequency in rad/s is

100
300
500
1500

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href="https://www.youtube.com/channel/UCNHT8lW-JmLC68rjBfZhdkg?sub_confirmation=1" target="_blank" class="youtube-subscribe-button"> Subscribe on YouTube {\pi \over 4}} \right),$$ the fundamental frequency in rad/s is" class="read-more button" href="https://exam.pscnotes.com/mcq/for-a-periodic-signal-vleft-t-right-30sin-100t-10cos-300t-6sin-left-500t-pi-over-4-right-the-fundamental-frequency-in-rad-s-is/#more-47963">Detailed SolutionFor a periodic signal $$v\left( t \right) = 30\sin 100t + 10\cos 300t + 6\sin \left( {500t + {\pi \over 4}} \right),$$ the fundamental frequency in rad/s is

18. Let y[n] denote the convolution of h[n] and g[n], where h[n] = $${\left( {\frac{1}{2}} \right)^n}$$ u[n] and g[n] is a causal sequence. If y[0] = 1 and y[1] = $$\frac{1}{2},$$ then g[1] equals

0
$${1 over 2}$$
1
$${3 over 2}$$

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causal sequence. If y[0] = 1 and y[1] = $$\frac{1}{2},$$ then g[1] equals" class="read-more button" href="https://exam.pscnotes.com/mcq/let-yn-denote-the-convolution-of-hn-and-gn-where-hn-left-frac12-rightn-un-and-gn-is-a-causal-sequence-if-y0-1-and-y1-frac12-then-g1-equals/#more-47823">Detailed SolutionLet y[n] denote the convolution of h[n] and g[n], where h[n] = $${\left( {\frac{1}{2}} \right)^n}$$ u[n] and g[n] is a causal sequence. If y[0] = 1 and y[1] = $$\frac{1}{2},$$ then g[1] equals

19. The impulse response h[n] of a linear time-invariant system is given by h[n] = u[n + 3] + u[n – 2] – 2u[n – 7], where u[n] is the unit step sequence. The above system is

Stable but not causal
Stable and causal
Causal but unstable
Unstable and not causal

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where u[n] is the unit step sequence. The above system is" class="read-more button" href="https://exam.pscnotes.com/mcq/the-impulse-response-hn-of-a-linear-time-invariant-system-is-given-by-hn-un-3-un-2-2un-7-where-un-is-the-unit-step-sequence-the-above-system-is/#more-47800">Detailed SolutionThe impulse response h[n] of a linear time-invariant system is given by h[n] = u[n + 3] + u[n – 2] – 2u[n – 7], where u[n] is the unit step sequence. The above system is

20. 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

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button" href="https://exam.pscnotes.com/mcq/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/#more-47793">Detailed SolutionA 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

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