Signal processing

71. If $$F\left( s \right) = L\left| {f\left( t \right)} \right| = {K \over {\left( {s + 1} \right)\left( {{s^2} + 4} \right)}},$$ then $$\mathop {\lim }\limits_{t \to \infty } f\left( t \right)$$ is given by

$${K over 4}$$
Zero
Infinite
Undefined

Detailed SolutionIf $$F\left( s \right) = L\left| {f\left( t \right)} \right| = {K \over {\left( {s + 1} \right)\left( {{s^2} + 4} \right)}},$$ then $$\mathop {\lim }\limits_{t \to \infty } f\left( t \right)$$ is given by

72. A system with input x[n] and output y[n] is given as $$y\left( n \right) = \left( {\sin {5 \over 6}\pi n} \right)x\left( n \right).$$ The system is

Linear, stable and invertible
Non-linear, stable and non-invertible
Linear, stable and non-invertible
Linear, unstable and invertible

Detailed SolutionA system with input x[n] and output y[n] is given as $$y\left( n \right) = \left( {\sin {5 \over 6}\pi n} \right)x\left( n \right).$$ The system is

76. A signal $$2\cos \left( {{{2\pi } \over 3}t} \right) – \cos \left( {\pi t} \right)$$ is the input to an LTI system with the transfer function $$H\left( s \right) = {e^s} + {e^{ – s}}$$ If Ck denote the kth coefficient in the exponential Fourier series of the output signal, then C3 is equal to

0
1
2
3

Detailed SolutionA signal $$2\cos \left( {{{2\pi } \over 3}t} \right) – \cos \left( {\pi t} \right)$$ is the input to an LTI system with the transfer function $$H\left( s \right) = {e^s} + {e^{ – s}}$$ If Ck denote the kth coefficient in the exponential Fourier series of the output signal, then C3 is equal to

77. If $$sL\left[ {f\left( t \right)} \right] = {\omega \over {\left( {{s^2} + {\omega ^2}} \right)}},$$ then the value of $$\mathop {\lim }\limits_{t \to \infty } f\left( t \right)$$

Cannot be determined
Is zero
Is unity
Is infinite

Detailed SolutionIf $$sL\left[ {f\left( t \right)} \right] = {\omega \over {\left( {{s^2} + {\omega ^2}} \right)}},$$ then the value of $$\mathop {\lim }\limits_{t \to \infty } f\left( t \right)$$

78. A signal containing only two frequency components (3 kHz and 6 kHz) is sampled at the rate of 8 kHz, and then passed through a low pass filter with a cut-off frequency of 8 kHz. The filter output

Is an undistorted version of the original signal
Contains only the 3 kHz component
Contains the 3 kHz component and a spurious component of 2 kHz
Contains both the components of the original signal and two spurious components of 2 kHz and 5 kHz

Detailed SolutionA signal containing only two frequency components (3 kHz and 6 kHz) is sampled at the rate of 8 kHz, and then passed through a low pass filter with a cut-off frequency of 8 kHz. The filter output

80. The transfer function of a discrete time LTI system is given by $$H\left( z \right) = {{2 – {3 \over 4}{z^{ – 1}}} \over {1 – {3 \over 4}{z^{ – 1}} + {1 \over 8}{z^{ – 2}}}}$$ Consider the following statements: S1 : The system is stable and causal for $$ROC:\left| z \right| > {1 \over 2}$$ S2 : The system is stable but not causal for $$ROC:\left| z \right| < {1 \over 4}$$ S3 : The system is neither stable nor causal for $$ROC:{1 \over 4} < \left| z \right| < {1 \over 2}$$ Which one of the following statements is valid?

Both S1 and S2 are true
Both S2 and S3 true
Both S1 and S3 are true
S1, S2 and S3 are all true

Detailed SolutionThe transfer function of a discrete time LTI system is given by $$H\left( z \right) = {{2 – {3 \over 4}{z^{ – 1}}} \over {1 – {3 \over 4}{z^{ – 1}} + {1 \over 8}{z^{ – 2}}}}$$ Consider the following statements: S1 : The system is stable and causal for $$ROC:\left| z \right| > {1 \over 2}$$ S2 : The system is stable but not causal for $$ROC:\left| z \right| < {1 \over 4}$$ S3 : The system is neither stable nor causal for $$ROC:{1 \over 4} < \left| z \right| < {1 \over 2}$$ Which one of the following statements is valid?

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