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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)

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f}}.$$ Therefore, $$F\left\{ {{1 \over {1 + j2\pi f}}} \right\}$$ is" class="read-more button" href="https://exam.pscnotes.com/mcq/the-fourier-transform-fe-t-ut-is-equal-to-1-over-1-j2pi-f-therefore-fleft-1-over-1-j2pi-f-right-is/#more-55054">Detailed Solution

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

$${e^{ - pi {t^2}}}$$

$${e^{ - {{pi {t^2}} over 2}}}$$

$${e^{ - pi left| f ight|}}$$

$${e^{ - 2pi {t^2}}}$$

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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systems. 1. There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane.

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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?" 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 Solution

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

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

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

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

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

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″]

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

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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

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

Detailed SolutionThe pole-zero pattern of a certain filter is

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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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H(s) is given by $$H\left( s \right) = {{K\left( {{s^2} + 1\omega _0^2} \right)} \over {{s^2} + \left( {{{{\omega _0}} \over Q}} \right)s + \omega _0^2}}$$" 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 Solution

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

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

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

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

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

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

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

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

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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