Signal processing

11. The impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by $${h_1}\left( t \right) = 1,$$ $${h_2}\left( t \right) = u\left( t \right),$$ $${h_3}\left( t \right) = \frac{{u\left( t \right)}}{{t + 1}},$$ $${h_4}\left( t \right) = {e^{ – 3t}}u\left( t \right)$$ Where u(t) is the unit step function. Which of these systems is time invariant, causal, and stable?

S1
S2
S3
S4

Detailed SolutionThe impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by $${h_1}\left( t \right) = 1,$$ $${h_2}\left( t \right) = u\left( t \right),$$ $${h_3}\left( t \right) = \frac{{u\left( t \right)}}{{t + 1}},$$ $${h_4}\left( t \right) = {e^{ – 3t}}u\left( t \right)$$ Where u(t) is the unit step function. Which of these systems is time invariant, causal, and stable?

12. The amplitude spectrum of a Gaussian pulse is

Uniform
A sine function
Gaussian
An impulse function

Detailed SolutionThe amplitude spectrum of a Gaussian pulse is

13. A discrete-time all-pass system has two of its poles at 0.25<0° and 2<30°. Which one of the following statements about the system is TRUE?

It has two more poles at 0.5<30° and 4<0°
It is stable only when the impulse response is two-sided
It has constant phase response over all frequencies
It has constant phase response over the entire z-plane

Detailed SolutionA discrete-time all-pass system has two of its poles at 0.25<0° and 2<30°. Which one of the following statements about the system is TRUE?

14. A stable linear time invariant (LTI) system has a transfer function $$H\left( s \right) = {1 \over {{s^2} + s – 6}}.$$ To make this system causal it needs to be cascaded with another LTI system having a transfer function H1(s). A correct choice for H1(s) among the following options is

s + 3
s - 2
s - 6
s + 1

Detailed SolutionA stable linear time invariant (LTI) system has a transfer function $$H\left( s \right) = {1 \over {{s^2} + s – 6}}.$$ To make this system causal it needs to be cascaded with another LTI system having a transfer function H1(s). A correct choice for H1(s) among the following options is

15. Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is \[\xrightarrow{{U\left( s \right)}}\boxed{\frac{1}{s}}\xrightarrow{{Y\left( s \right)}}\]

u(t)
tu(t)
$$ rac{{{t^2}}}{2}uleft( t ight)$$
e-tu(t)

Detailed SolutionAssuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is \[\xrightarrow{{U\left( s \right)}}\boxed{\frac{1}{s}}\xrightarrow{{Y\left( s \right)}}\]

16. The function x(t) is shown in the figure. Even and odd parts of a unit-step function u(t) are respectively,

$$ rac{1}{2}, rac{1}{2}xleft( t ight)$$
$$ - rac{1}{2}, rac{1}{2}xleft( t ight)$$
$$ rac{1}{2}, - rac{1}{2}xleft( t ight)$$
$$ - rac{1}{2}, - rac{1}{2}xleft( t ight)$$

Detailed SolutionThe function x(t) is shown in the figure. Even and odd parts of a unit-step function u(t) are respectively,

17. Laplace transforms of the functions tu(t) and u(t)sin(t) are respectively

$${1 over {{s^2}}},{s over {{s^2} + 1}}$$
$${1 over s},{1 over {{s^2} + 1}}$$
$${1 over {{s^2}}},{1 over {{s^2} + 1}}$$
$$s,{s over {{s^2} + 1}}$$

Detailed SolutionLaplace transforms of the functions tu(t) and u(t)sin(t) are respectively

18. It is desired to find three-tap causal filter which gives zero signal as an output to and input of the form \[x\left[ n \right] = {c_1}\exp \left( { – \frac{{j\pi n}}{2}} \right) + {c_2}\exp \left( {\frac{{j\pi n}}{2}} \right),\] Where c1 and c2 are arbitrary real numbers. The desired three-tap filter is given by h[0] = 1, h[1] = a, h[2] = b and h[n] = 0 for n < 0 or n > 2. What are the values of the filter taps a and b if the output is y[n] = 0 for all n, when x[n] is as given above? \[\xrightarrow{{x\left[ n \right]}}\boxed{\begin{array}{*{20}{c}} {n = 0} \\ \downarrow \\ {h\left[ n \right] = \left\{ {1,a,b} \right\}} \end{array}}\xrightarrow{{y\left[ n \right] = 0}}\]

a = -1, b = 1
a = 0, b = 1
a = 1, b = 1
a = 0, b = -1

Detailed SolutionIt is desired to find three-tap causal filter which gives zero signal as an output to and input of the form \[x\left[ n \right] = {c_1}\exp \left( { – \frac{{j\pi n}}{2}} \right) + {c_2}\exp \left( {\frac{{j\pi n}}{2}} \right),\] Where c1 and c2 are arbitrary real numbers. The desired three-tap filter is given by h[0] = 1, h[1] = a, h[2] = b and h[n] = 0 for n < 0 or n > 2. What are the values of the filter taps a and b if the output is y[n] = 0 for all n, when x[n] is as given above? \[\xrightarrow{{x\left[ n \right]}}\boxed{\begin{array}{*{20}{c}} {n = 0} \\ \downarrow \\ {h\left[ n \right] = \left\{ {1,a,b} \right\}} \end{array}}\xrightarrow{{y\left[ n \right] = 0}}\]

19. Let x[n] = x[-n]. Let X(z) be the z-transform of x[n]. If 0.5 + j0.25 is a zero of X(z), which one of the following must also be a zero of X(z).

0.5 - j0.25
$${1 over {left( {0.5 + j0.25} ight)}}$$
$${1 over {left( {0.5 - j0.25} ight)}}$$
2 + j4

Detailed SolutionLet x[n] = x[-n]. Let X(z) be the z-transform of x[n]. If 0.5 + j0.25 is a zero of X(z), which one of the following must also be a zero of X(z).

20. The bilateral Laplace transform of a function $$f\left( t \right) = \left\{ {\matrix{ {1,} & {{\rm{if}}\,a \le t \le b} \cr 0 & {{\rm{otherwise}}} \cr } } \right.$$ is

$${{a - b} over s}$$
$${{{e^z}left( {a - b} ight)} over s}$$
$${{{e^{ - as}} - {e^{ - bs}}} over s}$$
$${{{e^{ - left( {a - b} ight)}}} over s}$$

Detailed SolutionThe bilateral Laplace transform of a function $$f\left( t \right) = \left\{ {\matrix{ {1,} & {{\rm{if}}\,a \le t \le b} \cr 0 & {{\rm{otherwise}}} \cr } } \right.$$ is

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