GATE EC LTI Systems questions

A low-pass filter having a frequency response $\mathrm{H}(\mathrm{j} \omega)=\mathrm{A}(\omega) \;e^{j{\phi(\omega)}}$ does not produce any phase distortion, if

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

Common Data for Questions 74, 75: Let $g(t)=p(t)^{*} p(t)$, where * denotes convolution and $p(t)=u(t)-u(t-1)$ with $u(t)$ being the unit step function The impulse response of filter matched to the signal $s(t)=g(t)-\delta(t-2)^{*} g(t)$ is given as

A signal $x(n)=\sin \left(\omega_{0} n+\phi\right)$ is the input to a linear time-invariant system having a frequency response $\text{H}(e^{j \omega})$. If the output of the system $A x\left(n-n_{0}\right)$, then the most general form of $\angle \mathrm{H}\left(\mathrm{e}^{j \omega}\right)$ will be

A signal as shown in the figure is applied to a matched filter. Which of the following does represent the output of this matched filter?

The circuit in the given figure is a

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

A rectangular pulse train $s(t)$ as shown in the figure is convolved with the signal $\cos ^{2}\left(4 p \times 10^{3} t\right)$. The convolved signal will be a

A causal $\text{LTI}$ system is described by the difference equation $$2 y[n]=\alpha y[n-2]-2 x[n]+\beta x[n-1]$$ The system is stable only if

The impulse response $h|n|$ of a linear time invariant system is given as $$h[n]=\left\{\begin{array}{cc} -2 \sqrt{2} & n=1,-1 \\ 4 \sqrt{2} & n=2,-2 \\ 0, & \text{ otherwise } \end{array}\right.$$ If the input to the above system is the sequence $\mathrm{e}^{j \pi n / 4}$, then the output is

Consider the signal $x(t)$ shown in the figure. Let $h(t)$ denote the impulse response of the filter matched to $x(t)$, with $h(t)$ being non-zero only in the interval 0 to $4 \; \mathrm{sec}$. The slope of $h(t)$ in the interval $3 < t < \text{sec}.$ is

Let $x(t)$ be the input to a linear, time-invariant system. The required output is $4 x(t-2)$. The transfer function of the system should be

Let $\mathrm{P}$ be linearity, $\mathrm{Q}$ be time-invariance, $\mathrm{R}$ be causality and $\mathrm{S}$ be stability. A discrete-time system has the input-output relationship, $$y(n)=\left\{\begin{array}{ll} x(n) & n \geq 1 \\ 0, & n=0 \\ x(n+1), & n \leq-1 \end{array}\right.$$ where $x(n)$ is the input and $y(n)$ is the output. …

Data for Q. 63-64 are given below. Solve the problems and choose the correct answers. The system under consideration is an RC low-pass filter (RC-LPF) with $R=1.0 \mathrm{k} \; \Omega \mathrm{k}$ and $\mathrm{C}=1.0 \; \mu \mathrm{F}$ Let $\mathrm{H}(f)$ denote the frequency response of the RC-LPF. Let $f_{1}$ be the highest frequency such that …

Data for Q. 63-64 are given below. Solve the problems and choose the correct answers. The system under consideration is an $R C$ low-pass filter (RC-LPF) with $R=1.0 \; \mathrm{k} \Omega \mathrm{k}$ and $\mathrm{C}=1.0 \; \mu \mathrm{F}$ Let $t_{g}(f)$ be the group delay function of the given RC-LPF and $f_{2}=100 \mathrm{~Hz}$. Then …

A linear phase channel with phase delay $T_p$ and group delay $T_g$ must have ($f$ denotes frequency)

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

If the variance $\sigma_{x}^{2}$ of $d(n)=x(n)-x(n-1)$ is one-tenth the variance $\sigma_{x}^{2}$ of a stationary zero-mean discrete-time signal $x(n)$, then the normalized autocorrelation function $R_{x x}(k) / \sigma_{x}^{2}$ at $k=1$ is

The impulse response functions of four linear systems $\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}, \mathrm{~S}_{4}$ are given respectively by $\begin{array}{ll} h_{1}(t)=1 & h_{2}(t)=U(t) \\ h_{3}(t)=\frac{U(t)}{t+1} & h_{4}(t)=e^{3 t} U(t) \end{array}$ Where $\mathrm{U}(t)$ is the unit step function. Which of these systems is time …

A linear time invariant system has an impulse response $e^{2 t}, t>0$. If the initial conditions are zero and the input is $e^{3 t}$, then output for $t>0$ is