Chapter 8: Applications of Discrete Time Signals and Systems
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Test 8.1
The impulse response $h[n]$ having the magnitude response shown
below is:
Solution: The plot has zeros
$\{e^{j0}=1,\ e^{\pm j\pi/2}=\pm j,\ e^{j\pi}=-1\}$.
$$
{\bf H} ({\bf z})=({\bf z}-1)({\bf z}+1)({\bf z}-j)({\bf z}+j)={\bf z}^4-1.
$$ Then $h[n]=\{1,0,0,0,-1\}$.
Test 8.2
The impulse response $h[n]$ having the magnitude response shown
below is:
Solution: The plot has zeros
$\{e^{\pm j\pi/3}=\frac1{2}\pm j\frac{\sqrt{3}}{2},\ e^{\pm j\pi/2}=\pm j\}$.
\begin{eqnarray*}
{\bf H} ({\bf z})&=&\left({\bf z}-\frac1{2}-j\frac{\sqrt{3}}{2}\right)
\left({\bf z}-\frac1{2}+j\frac{\sqrt{3}}{2}\right)
({\bf z}-j)({\bf z}+j)\\
&=&({\bf z}^2-{\bf z}+1)({\bf z}^2+1).
\end{eqnarray*}
Then $h[n]=\{1,-1,1\}*\{1,0,1\}=\{1,-1,2,-1,1\}$
(2 notch filters
in series).
Test 8.3
An out-of-tune trumpet plays note A, but with a fundamental
frequency of 441 Hz. The trumpet signal is sampled at 44100
samples/s. Which filter can eliminate the out-of-tune signal?
Solution: The signal has period
$N_0=\frac{44100}{441}=100$. Use a comb filter (Eq. (8.31)).
Test 8.4
Which filter rejects a 1000 Hz sinusoid sampled at 3000
samples/s?
Solution: The discrete-time frequency to be
rejected is $\Omega_0=2\pi\frac{1000}{3000}=\frac{2\pi}{3}$.
The FIR notch filter that rejects $\Omega_0$ is
$h[n]=\{1,-2\cos(\Omega_0),1\}$ (see Eq. (8.14)).
$-2\cos(\Omega_0)=-2\cos(\frac{2\pi}{3})=1$, so $h[n]=\{1,1,1\}$.
Test 8.5
Which filter rejects signal $x[n]=5\cos(2\pi 250t)+7\cos(2\pi
1250t)$ sampled at 3000 samples/s?
Solution: The frequencies to be rejected are
$\Omega_1=2\pi\frac{250}{3000}=\frac{\pi}{6}$ and
$\Omega_2=2\pi\frac{1250}{3000}=\frac{5\pi}{6}$. Use two FIR notch filters
connected in series: $-2\cos(\frac{\pi}{6})=-\sqrt{3}$ and
$-2\cos(\frac{5\pi}{6})=\sqrt{3}$.
$h_1[n]=\{1,-\sqrt{3},1\}$ and
$h_2[n]=\{1,\sqrt{3},1\}$. Then $h_1[n]*h_2[n]=\{1,0,-1,0,1\}$.
Solution: $\{1,4,5\}*\{2,3,7\}=\{2,11,29,43,35\}$. Alias to
$\{2+43,\ 11+35,\ 29\}$.
Test 8.7
A signal is sampled at 80 samples/s and stored in MATLAB vector
X. length(X) gives 16. fft(X) gives [0,0,0,7,0,0,0,0,0,0,0,0,0,7,0,0].
What is the frequency of the sinusoid in Hz?
Solution: Use f=(K-1)*F/N $=(4-1)80/16=15$ Hz. See Example 8-22.
Test 8.8
What is the inverse DFT of $\{0,3+j4,0,0,0,0,0,3-j4\}$?
Solution: See Example 8-22.
$$
\frac{5}{8}\;e^{(j2\pi 1n/8+j53^\circ )}
+\frac{5}{8}\;e^{(j2\pi 7n/8-j53^\circ)}
=\frac{10}{8}\cos\left(\frac{\pi}{4}n+53^\circ \right).
$$
Test 8.9
A minimum phase discrete-time system:
Solution: See Sections 8-4.1 and 8-4.2.
Test 8.10
An LTI system has impulse response
$h[n]=3(-3)^n\;u[n]-2(-2)^n\;u[n]$. Its inverse system has impulse
response $g[n]=$?
The bilateral $\displaystyle{\bf z}$-transform of
$x[n]=2(\frac1{2})^n\;u[n]+2(3)^n\;u[-n-1]$ is:
Solution: Recall that ${\cal Z}\{-{\bf
a}^n\;u[-n-1]\}=\frac{\bf z}{{\bf z}-{\bf a}}$ (see Eqs. (8.76) and (8.81)).
$$
{\bf X} ({\bf z})=\frac{2{\bf z}}{{\bf z}-\frac1{2}}\frac{{\bf z}-3}{{\bf z}-3}-
\frac{2{\bf z}}{{\bf z}-3}\frac{{\bf z}-\frac1{2}}{{\bf z}-\frac1{2}}=
\frac{-5{\bf z}}{({\bf z}-\frac1{2})({\bf z}-3)}.
$$
Note the minus sign.
Test 8.12
What is the ROC of the $\displaystyle{\bf z}$-transform of $2({1\over
2})^n\;u[n]+2(3)^n\;u[-n-1]$? (Note that you need not compute the
function ${\bf H} ({\bf z})$ to solve this problem.)
Solution: ROC of ${\cal Z}\{-{\bf
a}^n\;u[-n-1]\}=\frac{\bf z}{{\bf z}-{\bf a}}$ is $\{|{\bf z}|<|{\bf a}|\}$
(Eq. (8.76)). ROC of ${\cal Z}\{{\bf a}^n\;u[n]\}=\frac{\bf z}{{\bf z}-{\bf a}}$
is $\{|{\bf z}|>|{\bf a}|\}$ (Eq. (8.81)).
$$
\left\{|{\bf z}|>\frac1{2}\right\}\cap\{|{\bf z}|<3\}=\left\{\frac1{2}<|{\bf z}|<3\right\}.
$$
Test 8.13
A stable LTI system can have an ROC of the form (for some
constants ${\bf a}_1,{\bf a}_2$):
Solution: See Section 8-7.
Test 8.14
A system's ROC has inner radius $=2$ and outer radius
$=\infty$. The inverse $\displaystyle{\bf z}$-transform is:
Solution: Unit circle $\{|z|=1\}\notin{\rm
ROC}$, so the system is unstable. $\infty\in {\rm ROC}$, so the
system is causal.
Test 8.15
An ROC has inner radius $=\frac1{2}$ and outer radius
$=3$. The inverse $\displaystyle{\bf z}$-transform is:
Solution: $\{|z|=1\}\in$ ROC, so the system
is stable. ROC form: $\{|a_1|<|{\bf z}|<|a_2|\}$, so the system is
2-sided.
Test 8.16
An ROC has inner radius $=2$ and outer radius $=3$. The inverse
$\displaystyle{\bf z}$-transform is:
Solution: $\{|z|=1\}\notin$ ROC, so the
system is unstable. ROC form: $\{|a_1|<|{\bf z}|<|a_2|\}$, so the system
is 2-sided.
Test 8.17
Any rational function has an inverse bilateral $\displaystyle{\bf z}$-transform
that is:
Solution: See Sections 8-5 and 8-6.
Test 8.18
Impulse response $g[n]$ of the inverse system to
$h[n]=\delta[n]-(2)^{n-1}\;u[n-1]$ is:
Which system rejects $3\cos(2\pi 125t)+4\cos(2\pi 250t)$ sampled
at 1000 samples/s?
Solution: The frequencies to be rejected are:
$\Omega_1=2\pi\frac{125}{1000}=\frac{\pi}{4}$ and
$\Omega_2=2\pi\frac{250}{1000}=\frac{\pi}{2}$.
Use two FIR notch filters connected in series:
$-2\cos(\frac{\pi}{4})=-\sqrt{2}$ and $-2\cos(\frac{\pi}{2})=0$.
$h_1[n]=\{1,-\sqrt{2},1\}$ and $h_2[n]=\{1,0,1\}$.
Then
$h_1[n]*h_2[n]=\{1,-\sqrt{2},2,-\sqrt{2},1\}$.
OR: The periods of the two sinusoids are:
$$
T_1=\frac{2\pi}{\Omega_1}=\frac{2\pi}{\pi/4}=8
$$ and
$$
T_2=\frac{2\pi}{\Omega_2}=\frac{2\pi}{\pi/2}=4.
$$
So the FIR comb filter $h[n]=\{1,0,0,0,0,0,0,-1\}$ also works.
