Signal processing

41. Consider a single input single output discrete-time system with x[n] as input and y[n] as output, where the two are related as $$y\left[ n \right] = \left\{ {\matrix{ {n\left| {x\left[ n \right]} \right|,} & {{\rm{for}}\,0 \le n \le 10} \cr {x\left[ n \right] – x\left[ {n – 1} \right],} & {{\rm{otherwise}}} \cr } } \right.$$ Which one of the following statements is true about the system?

It is causal and stable
It is causal but not stable
It is not causal but stable
It is neither causal nor stable

Detailed SolutionConsider a single input single output discrete-time system with x[n] as input and y[n] as output, where the two are related as $$y\left[ n \right] = \left\{ {\matrix{ {n\left| {x\left[ n \right]} \right|,} & {{\rm{for}}\,0 \le n \le 10} \cr {x\left[ n \right] – x\left[ {n – 1} \right],} & {{\rm{otherwise}}} \cr } } \right.$$ Which one of the following statements is true about the system?

42. A system with transfer function H(z) has impulse response h(n) defined as h(2) = 1, h(3) = -1 and h(k) = 0 otherwise. Consider the following statements. S1 : H(z) is a low-pass filter. S2 : H(z) is an FIR filter. Which of the following is correct?

Only S2 is true
Both S1 and S2 are false
Both S1 and S2 are true, and S2 is a reason for S1
Both S1 and S2 are true, but S2 is not a reason for S1

Detailed SolutionA system with transfer function H(z) has impulse response h(n) defined as h(2) = 1, h(3) = -1 and h(k) = 0 otherwise. Consider the following statements. S1 : H(z) is a low-pass filter. S2 : H(z) is an FIR filter. Which of the following is correct?

45. The input and output of a continuous time system are respectively denoted by x(t) and y(t). Which of the following descriptions corresponds to a casual system?

y(t) = x(t - 2) + x(t + 4)
y(t) = (t - 4)x(t + 1)
y(t) = (t + 4)x(t - 1)
y(t) = (t + 5)x(t + 5)

Detailed SolutionThe input and output of a continuous time system are respectively denoted by x(t) and y(t). Which of the following descriptions corresponds to a casual system?

48. The signal $$\cos \left( {10\pi t + \frac{\pi }{4}} \right)$$ is ideally sampled at a sampling frequency of 15 Hz. The sampled signal is passed through a filter with impulse response $$\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi \tau }}} \right)\cos \left( {40\pi t – \frac{\pi }{2}} \right).$$ The filter output is

$$rac{{15}}{2}cos left( {40pi t - rac{pi }{4}} ight)$$
$$rac{{15}}{2}left( {rac{{sin left( {pi t} ight)}}{{pi t}}} ight)cos left( {10pi t + rac{pi }{4}} ight)$$
$$rac{{15}}{2}cos left( {10pi t - rac{pi }{4}} ight)$$
$$rac{{15}}{2}left( {rac{{sin left( {pi t} ight)}}{{pi t}}} ight)cos left( {10pi t - rac{pi }{2}} ight)$$

Detailed SolutionThe signal $$\cos \left( {10\pi t + \frac{\pi }{4}} \right)$$ is ideally sampled at a sampling frequency of 15 Hz. The sampled signal is passed through a filter with impulse response $$\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi \tau }}} \right)\cos \left( {40\pi t – \frac{\pi }{2}} \right).$$ The filter output is

49. The Fourier series representation of an impulse train denoted by $$s\left( t \right) = \sum\limits_{n = – \infty }^\infty {\delta \left( {t – n{T_0}} \right)} \,{\rm{is}}\,{\rm{given}}\,{\rm{by}}$$

$${1 over {{T_0}}}sumlimits_{n = - infty }^infty {exp left( { - {{j2pi nt} over {{T_0}}}} ight)} $$
$${1 over {{T_0}}}sumlimits_{n = - infty }^infty {exp } left( { - {{jpi nt} over {{T_0}}}} ight)$$
$${1 over {{T_0}}}sumlimits_{n = - infty }^infty {exp } left( {{{jpi nt} over {{T_0}}}} ight)$$
$${1 over {{T_0}}}sumlimits_{n = - infty }^infty {exp } left( {{{j2pi nt} over {{T_0}}}} ight)$$

Detailed SolutionThe Fourier series representation of an impulse train denoted by $$s\left( t \right) = \sum\limits_{n = – \infty }^\infty {\delta \left( {t – n{T_0}} \right)} \,{\rm{is}}\,{\rm{given}}\,{\rm{by}}$$

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