Q21GATE 2008MCQ

The impulse response h(t) of a linear time invariant continuous time system is described by h(t) = exp( $\alpha$ t)u(t) + exp( $\beta$ t)u(-t), where u(-t) denotes the unit step function, and $\alpha$ and $\beta$ are real constants. This system is stable if

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📖 Explanation

$h(t)=e^{\alpha t} u(t)+e^{\beta t} u(-t)$ For the system to be stable, $\int_{-\infty }^{\infty } h(t) d t \lt \infty$ For the above condition, h(t) should be as shown below.

Q23GATE 2005MCQ

Which of the following can be impulse response of a causal system ?

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📖 Explanation

For a causal system $h(t)=0 \text{ for }t \lt 0$

Q24GATE 2004MCQ

The impulse response h[n] of a linear time-invariant system is given by h[n]=u[n+3] + u[n-2) - 2n[n-7] where u[n] is the unit step sequence. The above system is

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📖 Explanation

\begin{aligned} \sum_{k=-\infty }^{\infty } h(k)=& \sum_{k=-3}^{\infty } u(k+3)+\sum_{k=2}^{\infty } u(k-2) \\ &-2 \sum_{k=7}^{\infty } u(k-7) \\ =& \sum_{k=-3}^{6} 1+\sum_{k=2}^{6} 1 \\ =& 10+5=15 \lt \infty \end{aligned}

For bounded input, bounded output. So system is stable. Response depends on future value of input signal i.e. u(n + 3). So, system is not causal

Q25GATE 2002MCQ

Convolution of x(t+5) with impulse function $\delta$ (t-7) is equal to

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📖 Explanation

\begin{aligned} x(t+5) * \delta(t-7) &=x(t+5-7) \\ &=x(t-2) \end{aligned}

Q26GATE 2001MCQ

The impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by $h_{1}(t)=1$ $h_{2}(t)=U(t)$ $h_{3}(t)=\frac{U(t)}{t+1}$ $h_{4}(t)=e^{-3t}U(t)$ where U(t) is the step function. Which of these systems is time invariant, causal and stable?

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📖 Explanation

$h_{1}(t) \neq 0 \text { for } t\lt 0$ Therefores, is non causal

\begin{aligned} h_{2}(t)&=u(t)=0 \text { for } t \lt 0 \\ \int_{-\infty }^{\infty } n_{2}(t) d t&=\int_{-\infty }^{\infty } u(t) d t=\int_{0}^{\infty } d t=\infty \end{aligned}

Therefore $s_{2}$ is unstable and causal

\begin{aligned} h_{3}(t)&=\frac{u(t)}{t+1} \\ \text { at } t&=-1 \\ h_{f}(t)&=\infty , h_{3}(t)=0 \text { for } t \lt 0 \end{aligned}

Therefore $s_{3}$ is unstable and causal $h_{4}(t)=e^{-3 t} u(t)$ $s_{4}$ is time invariant, causal and stable