Question 1 |
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 ________
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 ________
-18.05 | |
-32.45 | |
-27.00 | |
-20.25 |
Question 1 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
\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
Question 2 |
The direct form structure of an FIR (finite impulse response) filter is shown in the figure.
The filter can be used to approximate a
The filter can be used to approximate a
low-pass filter | |
high-pass filter | |
band-pass filter | |
band-stop filter |
Question 2 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.
\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.
Question 3 |
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 ________
0.01 | |
0.04 | |
0.1 | |
0.4 |
Question 3 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}
\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}
There are 3 questions to complete.