Signal processing

31. The Fourier transform F{e-t u(t)} is equal to $${1 \over {1 + j2\pi f}}.$$ Therefore, $$F\left\{ {{1 \over {1 + j2\pi f}}} \right\}$$ is

ef u(f)
e-f u(f)
ef u(-f)
e-f u(-f)

Detailed

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SolutionThe Fourier transform F{e-t u(t)} is equal to $${1 \over {1 + j2\pi f}}.$$ Therefore, $$F\left\{ {{1 \over {1 + j2\pi f}}} \right\}$$ is

32. Let $$g\left( t \right) = {e^{ – \pi {t^2}}}$$ and h(t) is a filter matched to g(t). If g(t) is applied as input to h(t), then the Fourier transform of the output is

$${e^{ - pi {t^2}}}$$
$${e^{ - {{pi {t^2}} over 2}}}$$
$${e^{ - pi left| f ight|}}$$
$${e^{ - 2pi {t^2}}}$$

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81.2-142.7 81.2z"/> Subscribe on YouTube
is" class="read-more button" href="https://exam.pscnotes.com/mcq/let-gleft-t-right-e-pi-t2-and-ht-is-a-filter-matched-to-gt-if-gt-is-applied-as-input-to-ht-then-the-fourier-transform-of-the-output-is/#more-54947">Detailed SolutionLet $$g\left( t \right) = {e^{ – \pi {t^2}}}$$ and h(t) is a filter matched to g(t). If g(t) is applied as input to h(t), then the Fourier transform of the output is

33. Consider the following statements for continuous-time linear time invariant (LTI) systems. 1. There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane. 2. There is no causal and BIBO stable system with a pole in the right half of the complex plane. Which one among the following is correct?

Both 1 and 2 are true
Both 1 and 2 are not true
Only 1 is true
Only 2 is true

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There is no causal and BIBO stable system with a pole in the right half of the complex plane. Which one among the following is correct?" class="read-more button" href="https://exam.pscnotes.com/mcq/consider-the-following-statements-for-continuous-time-linear-time-invariant-lti-systems-1-there-is-no-bounded-input-bounded-output-bibo-stable-system-with-a-pole-in-the-right-half-of-the-complex/#more-54876">Detailed SolutionConsider the following statements for continuous-time linear time invariant (LTI) systems. 1. There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane. 2. There is no causal and BIBO stable system with a pole in the right half of the complex plane. Which one among the following is correct?

34. Let x(t) and y(t) (with Fourier transforms X(f) and Y(f) respectively) be related as shown in the figure. Then Y(f) is

$$ - rac{1}{2}Xleft( { rac{f}{2}} ight){e^{ - j2pi f}}$$
$$ - rac{1}{2}Xleft( { rac{f}{2}} ight){e^{j2pi f}}$$
$$ - Xleft( { rac{f}{2}} ight){e^{j2pi f}}$$
$$ - Xleft( { rac{f}{2}} ight){e^{ - j2pi f}}$$

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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 figure. Then Y(f) is" class="read-more button" href="https://exam.pscnotes.com/mcq/let-xt-and-yt-with-fourier-transforms-xf-and-yf-respectively-be-related-as-shown-in-the-figure-then-yf-is/#more-54340">Detailed SolutionLet x(t) and y(t) (with Fourier transforms X(f) and Y(f) respectively) be related as shown in the figure. Then Y(f) is

35. Two sequences [a, b, c] and [A, B, C] are related as, \[\left[ {\begin{array}{*{20}{c}} A \\ B \\ C \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{W_3^{ – 1}}&{W_3^{ – 2}} \\ 1&{W_3^{ – 2}}&{W_3^{ – 4}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a \\ b \\ c \end{array}} \right]\] where, $${W_3} = {e^{i\frac{{2\pi }}{3}}}.$$ If another sequence [p, q, r] is derived as, \[\left[ {\begin{array}{*{20}{c}} a \\ b \\ c \end{array}} \right] = \] \[\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{W_3^1}&{W_3^2} \\ 1&{W_3^2}&{W_3^4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{W_3^2}&0 \\ 0&0&{W_3^4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {A/3} \\ {B/3} \\ {C/3} \end{array}} \right]\] then the relationship between the sequences [p, q, r] and [a, b, c] is

”[p,
= [b, a, c]” option2=”[p, q, r] = [b, c, a]” option3=”[p, q, r] = [c, a, b]” option4=”[p, q, r] = [c, b, a]” correct=”option1″]

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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 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{W_3^2}&0 \\ 0&0&{W_3^4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {A/3} \\ {B/3} \\ {C/3} \end{array}} \right]\] then the relationship between the sequences [p, q, r] and [a, b, c] is" class="read-more button" href="https://exam.pscnotes.com/mcq/two-sequences-a-b-c-and-a-b-c-are-related-as-left-beginarray20c-a-b-c-endarray-right-left-beginarray20c-111-1w_3-1w_3/#more-54045">Detailed SolutionTwo sequences [a, b, c] and [A, B, C] are related as, \[\left[ {\begin{array}{*{20}{c}} A \\ B \\ C \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{W_3^{ – 1}}&{W_3^{ – 2}} \\ 1&{W_3^{ – 2}}&{W_3^{ – 4}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a \\ b \\ c \end{array}} \right]\] where, $${W_3} = {e^{i\frac{{2\pi }}{3}}}.$$ If another sequence [p, q, r] is derived as, \[\left[ {\begin{array}{*{20}{c}} a \\ b \\ c \end{array}} \right] = \] \[\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{W_3^1}&{W_3^2} \\ 1&{W_3^2}&{W_3^4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{W_3^2}&0 \\ 0&0&{W_3^4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {A/3} \\ {B/3} \\ {C/3} \end{array}} \right]\] then the relationship between the sequences [p, q, r] and [a, b, c] is

36. The pole-zero diagram of a causal and stable discrete-time system is shown in the figure. The zero at the origin has multiplicity 4. The impulse response of the system is h[n]. If h[0] = 1, we can conclude

”h[n
is real for all n” option2=”h[n] is purely imaginary for all n” option3=”h[n] is real for only even n” option4=”h[n] is purely imaginary for only odd n” correct=”option1″]

Detailed SolutionThe pole-zero diagram of a

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causal and stable discrete-time system is shown in the figure. The zero at the origin has multiplicity 4. The impulse response of the system is h[n]. If h[0] = 1, we can conclude

37. The pole-zero pattern of a certain filter is shown in figure. The filter must be of the following type

Low-pass
High-pass
All-pass
Band-pass

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must be of the following type" class="read-more button" href="https://exam.pscnotes.com/mcq/the-pole-zero-pattern-of-a-certain-filter-is-shown-in-figure-the-filter-must-be-of-the-following-type/#more-53149">Detailed SolutionThe pole-zero pattern of a certain filter is shown in figure. The filter must be of the following type

38. Specify the filter type if its voltage transfer function H(s) is given by $$H\left( s \right) = {{K\left( {{s^2} + 1\omega _0^2} \right)} \over {{s^2} + \left( {{{{\omega _0}} \over Q}} \right)s + \omega _0^2}}$$

All pass filter
Low pass filter
Band pass filter
Notch filter

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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 {{{{\omega _0}} \over Q}} \right)s + \omega _0^2}}$$" class="read-more button" href="https://exam.pscnotes.com/mcq/specify-the-filter-type-if-its-voltage-transfer-function-hs-is-given-by-hleft-s-right-kleft-s2-1omega-_02-right-over-s2-left-omega-_0-over-q-rig/#more-53148">Detailed SolutionSpecify the filter type if its voltage transfer function H(s) is given by $$H\left( s \right) = {{K\left( {{s^2} + 1\omega _0^2} \right)} \over {{s^2} + \left( {{{{\omega _0}} \over Q}} \right)s + \omega _0^2}}$$

39. Consider the system shown in the figure below. The transfer function $$\frac{{Y\left( z \right)}}{{X\left( z \right)}}$$ of the system is

$$ rac{{1 + a{z^{ - 1}}}}{{1 + b{z^{ - 1}}}}$$
$$ rac{{1 + b{z^{ - 1}}}}{{1 + a{z^{ - 1}}}}$$
$$ rac{{1 + a{z^{ - 1}}}}{{1 - b{z^{ - 1}}}}$$
$$ rac{{1 - b{z^{ - 1}}}}{{1 - a{z^{ - 1}}}}$$

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z \right)}}$$
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of the system is" class="read-more button" href="https://exam.pscnotes.com/mcq/consider-the-system-shown-in-the-figure-below-the-transfer-function-fracyleft-z-rightxleft-z-right-of-the-system-is/#more-52902">Detailed SolutionConsider the system shown in the figure below. The transfer function $$\frac{{Y\left( z \right)}}{{X\left( z \right)}}$$ of the system is

40. The complex envelope of the bandpass signal $$x\left( t \right) = \sqrt 2 \left( {{{\sin \left( {{{\pi t} \over 5}} \right)} \over {{{\pi t} \over 5}}}} \right)\sin \left( {\pi t – {\pi \over 4}} \right),$$ centered about $$f = {1 \over 2}Hz,$$ is

$$left( {{{sin left( {{{pi t} over 5}} ight)} over {{{pi t} over 5}}}{e^{j{pi over 4}}}} ight)$$
$$left( {{{sin left( {{{pi t} over 5}} ight)} over {{{pi t} over 5}}}{e^{ - j{pi over 4}}}} ight)$$
$$sqrt 2 left( {{{sin left( {{{pi t} over 5}} ight)} over {{{pi t} over 5}}}{e^{j{pi over 4}}}} ight)$$
$$sqrt 2 left( {{{sin left( {{{pi t} over 5}} ight)} over {{{pi t} over 5}}}{e^{ - j{pi over 4}}}} ight)$$

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= {1 \over 2}Hz,$$ is" class="read-more button" href="https://exam.pscnotes.com/mcq/the-complex-envelope-of-the-bandpass-signal-xleft-t-right-sqrt-2-left-sin-left-pi-t-over-5-right-over-pi-t-over-5-rightsin-left-pi-t/#more-52869">Detailed SolutionThe complex envelope of the bandpass signal $$x\left( t \right) = \sqrt 2 \left( {{{\sin \left( {{{\pi t} \over 5}} \right)} \over {{{\pi t} \over 5}}}} \right)\sin \left( {\pi t – {\pi \over 4}} \right),$$ centered about $$f = {1 \over 2}Hz,$$ is

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