Signal processing

11. Input x(t) and output y(t) of an LTI system are related by the differential equation y”(t) – y'(t) – 6y(t) = x(t). If the system is neither causal nor stable, the impulse response h(t) of the system is

$${1 over 5}{e^{3t}}uleft( { - t} ight) + {1 over 5}{e^{ - 2t}}uleft( { - t} ight)$$
$$ - {1 over 5}{e^{3t}}uleft( { - t} ight) + {1 over 5}{e^{ - 2t}}uleft( { - t} ight)$$
$${1 over 5}{e^{3t}}uleft( { - t} ight) - {1 over 5}{e^{ - 2t}}uleft( t ight)$$
$$ - {1 over 5}{e^{3t}}uleft( { - t} ight) - {1 over 5}{e^{ - 2t}}uleft( t ight)$$

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equation y”(t) – y'(t) – 6y(t) = x(t). If the system is neither causal nor stable, the impulse response h(t) of the system is" class="read-more button" href="https://exam.pscnotes.com/mcq/input-xt-and-output-yt-of-an-lti-system-are-related-by-the-differential-equation-yt-yt-6yt-xt-if-the-system-is-neither-causal-nor-stable-the-impulse-response-ht-of-the-system/#more-43184">Detailed SolutionInput x(t) and output y(t) of an LTI system are related by the differential equation y”(t) – y'(t) – 6y(t) = x(t). If the system is neither causal nor stable, the impulse response h(t) of the system is

13. Consider the sequence x[n] = {-4 – j5, 1 + j2, 4} The conjugate antisymmetric part of the sequence is

{-4 - j2.5, j2, 4 - j2.5}
{-j2.5, 1, j2.5}
{-j5, j2, 0}
{-4, 1, 4}

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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
= {-4 – j5, 1 + j2, 4} The conjugate antisymmetric part of the sequence is" class="read-more button" href="https://exam.pscnotes.com/mcq/consider-the-sequence-xn-4-j5-1-j2-4-the-conjugate-antisymmetric-part-of-the-sequence-is/#more-42988">Detailed SolutionConsider the sequence x[n] = {-4 – j5, 1 + j2, 4} The conjugate antisymmetric part of the sequence is

14. Given that F(s) is the one-sided Laplace transform of f(t), the Laplace transform of $$\int\limits_0^t {f\left( \tau \right)} d\tau $$ is

”sF(s)
”$${1
”$$intlimits_0^s
”$${1
$$”
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124.1c-6.3-23.7-24.8-42.3-48.3-48.6C458.8 64 288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 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
correct=”option4″]

Detailed SolutionGiven that F(s) is the one-sided Laplace transform of f(t), the Laplace transform of $$\int\limits_0^t {f\left( \tau \right)} d\tau $$ is

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