Q1GATE 2019NAT

Let h[n] be a length-7 discrete-time finite impulse response filter, given by h[0]=4, h[1]=3, h[2]=2, h[3]=1, h[-1]=-3, h[-2]=-2, h[-3]=-1, and h[n] is zero for $|n|\geq 4$ . A length-3 finite impulse response approximation g[n] of g[n] has to be obtained such that $E(h,g)=\int_{-\pi}^{\pi}|H(e^{j\omega })-G(e^{j\omega })|^2d\omega$ is minimized, where $H(e^{j\omega })$ and $G(e^{j\omega })$ are the discrete-time Fourier transforms of h[n] and g[n], respectively. For the filter that minimizes E(h,g), the value of 10g[-1]+g[1], rounded off to 2 decimal places, is ________

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📖 Explanation

From Parseval's theorem, $\sum_{n=-\infty }^{\infty }|x[n]|^{2}=\frac{1}{2 \pi} \int_{-\pi}^{\pi}\left|X\left(e^{j \omega}\right)\right|^{2} d \omega$ So, $\int_{-\pi}^{\pi}\left|H\left(e^{j \omega}\right)-G\left(e^{j \omega}\right)\right|^{2} d \omega=2 \pi \sum_{n=-3}^{3}|h(n)-g(n)|^{2}$ The solution of g(n) that minimizes E(h, g) also minimizes $\sum_{n=-3}^{3}|h(n)-g(n)|^{2}$ $\sum_{n=-3}^{3}|h(n)-g(n)|^{2}$ $=|4-g(0)|^{2}+|3-g(1)|^{2}+|-3-g(-1)|^{2}+10$ The solution of g(n) that minimizes the above equation is $g(n)=(-3,4,3\}$ $So, 10 g(-1)+g(1)=10(-3)+3=-27$

Q2GATE 2016MCQ

The direct form structure of an FIR (finite impulse response) filter is shown in the figure. The filter can be used to approximate a

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📖 Explanation

\begin{aligned} y(n] &=5[x[n]-x[n-2]] \\ H(z) &=\frac{Y(z)}{X(z)}=5\left[1-z^{-2}\right]\$ \\ \text{But,}\quad z &=e^{i \omega} \\ H\left(e^{j \omega}\right) &=5\left[1-e^{-j2w}\right]\$\\ \end{aligned}

\begin{aligned} \omega&=0 \left|H\left(e^{j \omega}\right)\right|=0 \\ \omega&=\frac{\pi}{2} \left|H\left(e^{j \omega}\right)\right|=10 \\ \omega&=\pi \left|H\left(e^{j \omega)} \mid=0\right.\right. \end{aligned}

$\therefore \quad$ The given one is bandpass filter.

Q3GATE 2016NAT

A continuous-time filter with transfer function $H(S)=\frac{2s+6}{s^{2}+6s+8}$ is converted to a discretetime filter with transfer function $G(z)=\frac{2z^{2}-0.5032z}{z^{2}-0.5032z+k}$ so that the impulse response of the continuous-time filter, sampled at 2 Hz, is identical at the sampling instants to the impulse response of the discrete time filter. The value of k is ________

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📖 Explanation

\begin{aligned} \text{Given, }H(s)&=\frac{2 s+6}{s^{2}+6 s+8}=\frac{1}{s+2}+\frac{1}{s+4} \\ h(t) &=e^{-2 t} u(t)+e^{-4 t} u(t) \\ \text{Given, }f_{s} &=2 \mathrm{Hz} \end{aligned}

\begin{aligned} \text { For discrete time }, t &=n T_{s}=\frac{n}{2} \\ h[n] &=\left(e^{-n}+e^{-2 n}\right) \cup(n] \\ H(z) &=\frac{1}{1-e^{-1} Z^{-1}}+\frac{1}{1-e^{-2} Z^{-1}} \\ &=\frac{Z}{Z-e^{-1}}+\frac{Z}{Z-e^{-2}} \\ &=\frac{2 Z^{2}-0.5032 Z}{Z^{2}-0.5032 Z+0.049} \\ K &=0.049 \end{aligned}

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