Q21GATE 2005MCQ

The power in the signal $s(t)=8cos(20\pi t -\frac{\pi }{2})+4sin(15\pi t)$ is

📖 Explanation

\begin{aligned} s(t) &=8 \cos \left(\frac{\pi}{2}-20 \pi t\right)+4 \sin 15 \pi t \\ &=8 \sin 20 \pi t+4 \sin 15 \pi t \\ P &=\frac{8^{2}}{2}+\frac{4^{2}}{2}=32+8=40 \end{aligned}

Q22GATE 2004MCQ

Consider the sequence x[n]=[-4-j51+j25]. The conjugate anti-symmetric part of the sequence is

📖 Explanation

\begin{aligned} x(n) &=[-4-j 5,1+2 j, 4] \\ x^{*}(-n) &=[4,1-2 j,-4+j 5] \\ x_{\mathrm{CAS}}(n) &=\frac{x(n)-x^{*}(-n)}{2} \\ &=[-4-j2.5 , \underset{\uparrow}{j2} , 4-j2.5]\\ \end{aligned}

Q23GATE 2003MCQ

Let P be linearity, Q be time-invariance, R be causality and S be stability. A discrete time system has the input-output relationship,

y(n)=\left\{\begin{matrix} x(n) &n\geq 1 \\ 0, &n=0 \\ x(n+1)&n\leq -1 \end{matrix}\right.

where x(n) is the input and y(n) is the output. The above system has the properties

📖 Explanation

\begin{aligned} y\left(n-n_{0}\right)&=x\left(n-n_{0}+1\right)(\text { time varying })\\ y(n)&=x(n+1) \text{(depends on future)} \\ \text{i.e.}\quad y(1)&=x(2) \quad \text{(non causal) } \end{aligned}

For bounded input, system has bounded output. So it is stable.

\begin{aligned} y(n) &=x(n), n \geq 1 \\ &=0, n=0 \\ &=x(n+1), n \leq-1 \end{aligned}

So, system is linear.

Q24GATE 2001MCQ

Let $\delta(t)$ denote the delta function. The value of the integral $\int_{-\infty }^{\infty }\delta (t)\cos \left ( \frac{3t}{2} \right )dt$ is

📖 Explanation

$\int_{-\infty }^{\infty } \delta(t) \cos \left(\frac{3 t}{2}\right) d t=f(0)=\cos \left(\frac{3 \times 0}{2}\right)=\cos 0=1$

Q25GATE 2001MCQ

If a signal f(t) has energy E, the energy of the signal f(2t) is equal to

📖 Explanation

\begin{aligned} E &=\int_{-\infty }^{\infty } f(t)^{2} d t \\ E^{\prime} &=\int_{-\infty }^{\infty } f(2 t)^{2} d t \\ &=\int_{-\infty }^{\infty } f(p)^{2}-\frac{d p}{2}\left(2 t=p ; d t=\frac{d p}{2}\right) \\ E^{\prime} &=\frac{E}{2} \end{aligned}