unilateral Laplace transform of g(t) = t.f(t)?" class="read-more button" href="https://exam.pscnotes.com/mcq/the-unilateral-laplace-transform-of-ft-is-1-over-s2-s-1-which-one-of-the-following-is-the-unilateral-laplace-transform-of-gt-t-ft/#more-49296">Detailed 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)?
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
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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
\over {dt}} = 10 – 0.2x$$ with initial conduction x(0) = 1. The response x(t) for t > 0 is" class="read-more button" href="https://exam.pscnotes.com/mcq/consider-the-differential-equation-dx-over-dt-10-0-2x-with-initial-conduction-x0-1-the-response-xt-for-t-0-is/#more-48164">Detailed SolutionConsider the differential equation $${{dx} \over {dt}} = 10 – 0.2x$$ with initial conduction x(0) = 1. The response x(t) for t > 0 is
class="read-more button" href="https://exam.pscnotes.com/mcq/for-a-periodic-signal-vleft-t-right-30sin-100t-10cos-300t-6sin-left-500t-pi-over-4-right-the-fundamental-frequency-in-rad-s-is/#more-47963">Detailed SolutionFor a periodic signal $$v\left( t \right) = 30\sin 100t + 10\cos 300t + 6\sin \left( {500t + {\pi \over 4}} \right),$$ the fundamental frequency in rad/s is
= $${\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},$$ then g[1] equals" class="read-more button" href="https://exam.pscnotes.com/mcq/let-yn-denote-the-convolution-of-hn-and-gn-where-hn-left-frac12-rightn-un-and-gn-is-a-causal-sequence-if-y0-1-and-y1-frac12-then-g1-equals/#more-47823">Detailed SolutionLet y[n] denote the convolution of h[n] and g[n], where h[n] = $${\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
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the sampling theorem, the sampling frequency which is not valid is" class="read-more 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