(14 intermediate revisions by 2 users not shown)
Line 1: Line 1:
=Homework 4 Solutions=
+
=Homework 4 Solutions, [[2011 Spring ECE 301 Boutin|ECE301 Spring 2011 Prof. Boutin]]=
 +
Students should feel free to make comments/corrections or ask questions directly on this page. -pm
 
==Question 1==
 
==Question 1==
 
a)<u>Memory</u>
 
a)<u>Memory</u>
  
 
Since <math class="inline">h[n]=e^{j2\pi n}=1</math> for all <math>n</math>, then <math class="inline">h[n]\neq 0</math> for all <math class="inline">n\neq 0</math>. Hence, this system has '''memory'''.
 
Since <math class="inline">h[n]=e^{j2\pi n}=1</math> for all <math>n</math>, then <math class="inline">h[n]\neq 0</math> for all <math class="inline">n\neq 0</math>. Hence, this system has '''memory'''.
 +
:<span style="color:green">Warning: one has to be careful when negating the statement: "h[n]=0 for all non zero n's". The correct negated statement is: "there exists a non-zero n such that h[n] is not zero." So to show that a system has memory, it is sufficient to show that there exists a non-zero n for which h[n] is not zero.-pm </span>
  
 
<u>Causality</u>
 
<u>Causality</u>
  
 
<math class="inline">h[n]=e^{j2\pi n}=1</math> for all <math>n</math>, then <math class="inline">h[n]\neq 0</math> for all <math class="inline">n<0</math>. Hence, the system is '''not causal'''.
 
<math class="inline">h[n]=e^{j2\pi n}=1</math> for all <math>n</math>, then <math class="inline">h[n]\neq 0</math> for all <math class="inline">n<0</math>. Hence, the system is '''not causal'''.
 +
:<span style="color:green">Warning: one has to be careful when negating the statement: "h[n]=0 for all negative n's". The correct negated statement is: "there exists a non-negative n such that h[n] is not zero." So to show that a system is not causal, it is sufficient to show that there exists a non-negative n for which h[n] is not zero.-pm </span>
  
 
<u>Stability</u>
 
<u>Stability</u>
  
 
<math class="inline">\sum_{n=-\infty}^{\infty} |h[n]|=\sum_{n=-\infty}^{\infty} 1 = \infty</math>. Hence, the system is '''unstable'''.
 
<math class="inline">\sum_{n=-\infty}^{\infty} |h[n]|=\sum_{n=-\infty}^{\infty} 1 = \infty</math>. Hence, the system is '''unstable'''.
 +
 +
:<span style="color:green"> Note that you are summing the magnitude, not the magnitude squared. So this is not the same as computing the energy of the unit impulse response h[n]. -pm </span>
  
 
b)<u>Memory</u>
 
b)<u>Memory</u>
Line 17: Line 22:
 
Since <math class="inline">h(t)=e^{j2\pi t}\neq a\delta(t)</math>, where <math class="inline">a</math> is any number (can be complex). Hence, this system has '''memory'''.
 
Since <math class="inline">h(t)=e^{j2\pi t}\neq a\delta(t)</math>, where <math class="inline">a</math> is any number (can be complex). Hence, this system has '''memory'''.
  
 +
:<span style="color:green">To be more precise, one could say that if h(t) satisfied <math class="inline">h(t)= a\delta(t)</math>  for some constant a, then h(t) would be zero for all non-zero t's. Now in this case, we have <math class="inline"> h(1)=1 \neq 0 </math>, and so the system has memory. -pm </span>
 
<u>Causality</u>
 
<u>Causality</u>
  
Line 65: Line 71:
 
a) The period of <math>x(t)</math> is 1.  
 
a) The period of <math>x(t)</math> is 1.  
  
Now, <math>x(t)=\sin(2\pi t)=\frac{1}{2j}e^{j2\pi t}-\frac{1}{2j}e^{-j2\pi t}</math>.
+
Now, <math class="inline">x(t)=\sin(2\pi t)=\frac{1}{2j}e^{j2\pi t}-\frac{1}{2j}e^{-j2\pi t}</math>.
  
Hence, <math>a_k=\left\{\begin{array}{ll}\frac{1}{2j},& \mbox{ for k=1, }\\ \frac{1}{2j},& \mbox{ for k=-1, }\\ 0,& \mbox{ otherwise.}\end{array}\right.</math>
+
Hence, <math class="inline">a_k=\left\{\begin{array}{ll}\frac{1}{2j},& \mbox{ for k=1, }\\ \frac{1}{2j},& \mbox{ for k=-1, }\\ 0,& \mbox{ otherwise.}\end{array}\right.</math>
  
 
b)Using Euler's properties, we get:
 
b)Using Euler's properties, we get:
Line 86: Line 92:
  
 
d)<math class="inline">x[n]=j^n=e^{j\frac{\pi}{2}  n}=e^{j\frac{2\cdot 2\pi}{4} n}</math>.
 
d)<math class="inline">x[n]=j^n=e^{j\frac{\pi}{2}  n}=e^{j\frac{2\cdot 2\pi}{4} n}</math>.
Hence, the period of the signal is 4, and <math class="inline">a_0=a_1=a_3=0</math> and <math class="inline">a_2=1</math>. Note that the DTFS is periodic with period 4 with respect to <math>k</math>.
+
Hence, the period of the signal is 4, and <math class="inline">a_0=a_2=a_3=0</math> and <math class="inline">a_1=1</math>. Note that the DTFS is periodic with period 4 with respect to <math>k</math>.
  
 
e)<math class="inline">x[n]=e^{j\frac{3}{5}\pi(n-\frac{1}{2})}=e^{-j\frac{3}{10}\pi}e^{j\frac{2\cdot 3}{10} \pi n}</math>.
 
e)<math class="inline">x[n]=e^{j\frac{3}{5}\pi(n-\frac{1}{2})}=e^{-j\frac{3}{10}\pi}e^{j\frac{2\cdot 3}{10} \pi n}</math>.
 
Hence, the period of the signal is 10, and <math class="inline">a_3=e^{-j\frac{3}{10}\pi}</math> and all other coefficients are zero in one period of the DTFS.
 
Hence, the period of the signal is 10, and <math class="inline">a_3=e^{-j\frac{3}{10}\pi}</math> and all other coefficients are zero in one period of the DTFS.
  
f)<math>\begin{align}
+
f)
 +
 
 +
<math>\begin{align}
 
x(t)&=\cos^2(t) \\
 
x(t)&=\cos^2(t) \\
&=\frac{1}{2} + \frac{1}{2}cos(2t) \\
+
&=\frac{1}{2} + \frac{1}{2}\cos(2t) \\
&=\frac{1}{2} + \frac{1}{4}e^{j2\frac{\pi}{\pi} t} + \frac{1}{4}e^{-j2\frac{\pi}{\pi} t}</math>
+
&=\frac{1}{2} + \frac{1}{4}e^{j2\frac{\pi}{\pi} t} + \frac{1}{4}e^{-j2\frac{\pi}{\pi} t}
 
\end{align}
 
\end{align}
 
</math>
 
</math>
Line 101: Line 109:
  
 
g)
 
g)
 +
 
<math>\begin{align}
 
<math>\begin{align}
x(t)&=1+e^{j\frac{4\pi n}{7}-e^{j\frac{2\pi n}{5} \\
+
x(t)&=1+e^{j\frac{4\pi n}{7}}-e^{j\frac{2\pi n}{5}} \\
&=1+e^{j\frac{2\pi \cdot 5 \cdot n}{35}-e^{j\frac{2\pi \cdot 7 \cdot n}{35} \\
+
&=1+e^{j\frac{2\pi \cdot 5 \cdot n}{35}}-e^{j\frac{2\pi \cdot 7 \cdot n}{35}} \\
 
\end{align}
 
\end{align}
 
</math>
 
</math>
  
Hence, the signal is periodic with period 35. The coefficients are <math class="inline">a_0=1</math>, <math class="inline">a_5=1</math>, and <math class="inline">a_7=-1</math>, and all other coefficients in one period of the DTFS are zero.
+
Hence, the signal is periodic with period 35. The coefficients are <math class="inline">a_0=1</math>, <math class="inline">a_7=-1</math>, and <math class="inline">a_{10}=1</math>, and all other coefficients in one period of the DTFS are zero.
  
 
h)
 
h)
For <math>k=0</math>, we can directly find <math>a_0=\frac{1}{2} \int_{-1}^{0} + \frac{1}{2} \int_{0}^{-1}=\frac{1}{2}</math>.
+
 
 +
For <math>k=0</math>, we can directly find <math class="inline">a_0=\frac{1}{2} \int_{-1}^{0} (t+1) dt + \frac{1}{2} \int_{0}^{1} (1-t) dt=\frac{1}{2}</math>.
 +
 
 +
 
  
 
Now,
 
Now,
 +
 +
 
<math>\begin{align}
 
<math>\begin{align}
 
a_k&= \frac{1}{2}\int_{-1}^{0} (t-1)e^{-j\pi k t} dt + \frac{1}{2}\int_{0}^{1} (1-t)e^{-j\pi k t} dt \\
 
a_k&= \frac{1}{2}\int_{-1}^{0} (t-1)e^{-j\pi k t} dt + \frac{1}{2}\int_{0}^{1} (1-t)e^{-j\pi k t} dt \\
 
&=\frac{1}{2}\int_{-1}^{0}te^{-j\pi kt} dt + \frac{1}{2}\int_{-1}^{0} e^{-j\pi kt} dt + \frac{1}{2}\int_{0}^{1} e^{-j\pi kt} dt + \frac{1}{2}\int_{0}^{1}te^{-j\pi kt} dt \\
 
&=\frac{1}{2}\int_{-1}^{0}te^{-j\pi kt} dt + \frac{1}{2}\int_{-1}^{0} e^{-j\pi kt} dt + \frac{1}{2}\int_{0}^{1} e^{-j\pi kt} dt + \frac{1}{2}\int_{0}^{1}te^{-j\pi kt} dt \\
 
&=\frac{1}{2} \left[\frac{te^{-jk\pi t}}{-j\pi k}+\frac{e^{-j\pi kt}}{\pi^2 k^2}\right]^0_{-1} + \left[\frac{1}{2j\pi k} e^{-j\pi kt}\right]^{-1}_{1} - \frac{1}{2}\left[\frac{te^{-jk\pi t}}{-j\pi k}+\frac{e^{-j\pi kt}}{\pi^2 k^2}\right]^1_0 \\
 
&=\frac{1}{2} \left[\frac{te^{-jk\pi t}}{-j\pi k}+\frac{e^{-j\pi kt}}{\pi^2 k^2}\right]^0_{-1} + \left[\frac{1}{2j\pi k} e^{-j\pi kt}\right]^{-1}_{1} - \frac{1}{2}\left[\frac{te^{-jk\pi t}}{-j\pi k}+\frac{e^{-j\pi kt}}{\pi^2 k^2}\right]^1_0 \\
&=\frac{1}{2}\left[ \frac{1}{\pi^2 k^2} - \frac{e^{j\pi k}}{j\pi k} - \frac{e^{j\pi k}}{\pi^2 k^2}\right] + \frac{1}{2j\pi k}\left[e^{j\pi k}-e^{-j\pi k}\right]-\frac{1}{2}\left[\frac{e^{-j\pi k}}{-j\pi k}+\frac{e^{-j\pi k}}{\pi^2 k^2}-\frac{1}{\pi^2 k^2}\right] \\
+
&=\frac{1}{2}\left[\frac{1}{\pi^2 k^2} - \frac{e^{j\pi k}}{j\pi k}-\frac{e^{j\pi k}}{\pi^2 k^2}\right] + \frac{1}{2j\pi k}\left[e^{j\pi k}-e^{-j\pi k}\right]-\frac{1}{2}\left[\frac{e^{-j\pi k}}{-j\pi k}+\frac{e^{-j\pi k}}{\pi^2 k^2}-\frac{1}{\pi^2 k^2}\right] \\
&=\frac{1}{\pi^2 k^2}-\frac{1}{\pi}\left[\frac{e^{j\pi k}-e^{-j\pi k}}{2j}\right]-\frac{1}{\pi^2 k^2}\left[\frac{e^{j\pi k}+e^-j\pi k}}{2}\right]+\frac{1}{\pi k}\left[\frac{e^{j\pi k}-e^{-j\pi k}}{2j}\right] \\
+
&=\frac{1}{\pi^2 k^2}-\frac{1}{\pi k}\left[\frac{e^{j\pi k}-e^{-j\pi k}}{2j}\right]-\frac{1}{\pi^2 k^2}\left[\frac{e^{j\pi k}+e^{-j\pi k}}{2}\right]+\frac{1}{\pi k}\left[\frac{e^{j\pi k}-e^{-j\pi k}}{2j}\right] \\
&=\frac{1}{\pi^2 k^2}-\frac{1}{\pi^2 k^2}\cos(\pi k)  
+
&=\frac{1}{\pi^2 k^2}-\frac{1}{\pi^2 k^2}\cos(\pi k) \\
 +
&=\frac{1}{\pi^2 k^2}[1-(-1)^k]
 
\end{align}
 
\end{align}
 
</math>
 
</math>
 +
 +
==Question 3==
 +
a)
 +
 +
<math>x(t)=\frac{1}{2j}e^{j2\pi t}-\frac{1}{2j}e^{-j2\pi t}</math>.
 +
 +
b)
 +
 +
<math>x(t)=\frac{e^{j\frac{5\pi}{2} t}}{4j} + \frac{e^{j\frac{3\pi}{2} t}}{4j} - \frac{e^{-j\frac{3\pi}{2} t}}{4j} - \frac{e^{-j\frac{5\pi}{2} t}}{4j}</math>.
 +
 +
c)
 +
 +
<math>x[n]=e^{j\pi  n}</math>.
 +
 +
d)
 +
 +
<math>x[n]=e^{j\frac{\pi}{2}  n}</math>.
 +
 +
e)
 +
 +
<math>x[n]=e^{-j\frac{3}{10}\pi}e^{j\frac{3}{5} \pi n}</math>.
 +
 +
f)
 +
 +
<math>x(t)=\frac{1}{2} + \frac{1}{4}e^{2jt} + \frac{1}{4}e^{-2jt}</math>.
 +
 +
g)
 +
 +
<math>x[n]=1+e^{j\frac{4\pi n}{7}}-e^{j\frac{2\pi n}{5}}</math>.
 +
 +
h)
 +
 +
<math>\begin{align}
 +
x(t)&= \frac{1}{2}+\sum_{k=-\infty}^{-1}a_k e^{j\pi kt} + \sum_{k=1}^{\infty}a_k e^{j\pi kt} \\
 +
&= \frac{1}{2}+\sum_{k=1}^{\infty}a_{-k} e^{-j\pi kt} + \sum_{k=1}^{\infty}a_k e^{j\pi kt} \\
 +
&= \frac{1}{2}+\sum_{k=1}^{\infty}a_k(e^{j\pi kt} + e^{-j\pi kt}) \\
 +
&= \frac{1}{2}+\sum_{k=1}^{\infty} \frac{2}{\pi^2 k^2}[1-(-1)^k]\cos(\pi kt) \\
 +
&= \frac{1}{2}+\frac{4}{\pi^2} \cos(\pi t)+\frac{4}{9\pi^2}\cos(3\pi t)+\frac{4}{25\pi^2}\cos(5\pi t)+\dots
 +
\end{align}
 +
</math>
 +
 +
Note that we have used the fact that <math>a_{-k}=a_k</math>.
 +
==Question 4==
 +
a)
 +
 +
<math>\begin{align}
 +
\mathcal{H}(\omega)=H(j\omega)&=\int_{-\infty}^{\infty}e^t[u(t-100)-u(t)]e^{-j\omega t}dt \\
 +
&=-\int_{0}^{100}e^t\cdot e^{-j\omega t}dt \\
 +
&=-\int{0}^{100}e^{(-j\omega+1)t}dt \\
 +
&=\frac{1}{j\omega-1}\left[e^{(-j\omega+1)t}\right]^{100}_0 \\
 +
&=\frac{1}{j\omega-1}e^{(1-j\omega)100}-\frac{1}{j\omega -1} \\
 +
&=\frac{e^{100}e^{-j\omega 100}}{j\omega -1}-\frac{1}{j\omega -1}
 +
\end{align}
 +
</math>
 +
 +
b)
 +
 +
For <math class="inline">x(t)=\frac{1}{2j}e^{j2\pi t}-\frac{1}{2j}e^{-j2\pi t}</math>  :
 +
 +
 +
<math>\begin{align}
 +
\mathcal{H}(2\pi)&=\frac{e^{100}e^{-j200\pi}}{2j\pi-1}-\frac{1}{2j\pi-1} \\
 +
&=\frac{e^{100}-1}{2j\pi-1} \\
 +
\end{align}
 +
</math>
 +
 +
<math>\mathcal{H}(-2\pi)=\frac{1-e^{100}}{1+2j\pi}</math>
 +
 +
Then,
 +
 +
<math>\begin{align}
 +
y(t)&=\frac{1}{2j}\mathcal{H}(2\pi)e^{j2\pi t}-\frac{1}{2j}\mathcal{H}(-2\pi)e^{-j2\pi t} \\
 +
&=\frac{1-e^{100}}{4\pi+2j}e^{j2\pi t}-\frac{1-e^{100}}{2j-4\pi}e^{-j2\pi t}
 +
\end{align}
 +
</math>
 +
 +
==Question 5==
 +
a)
 +
 +
<math>\begin{align}
 +
\mathcal{H}(\omega)=H(e^{j\omega})&=\sum_{n=-\infty}^{\infty} h[n] e^{-j\omega n} \\
 +
&=\sum_{n=-\infty}^\infty\left(\sum_{k=-7}^8 \delta[n-k]\right)e^{-j\omega n} \\
 +
&=\sum_{k=-7}^8 \sum_{n=-\infty}^\infty e^{-j\omega n}\delta[n-k] \\
 +
&=\sum_{k=-7}^8 e^{-j\omega k} \\
 +
&=\frac{e^{7j\omega}-e^{-9j\omega}}{1-e^{-j\omega}}
 +
\end{align}
 +
</math>
 +
 +
 +
<math>\begin{align}
 +
\mathcal{H}(0)&=H(e^{j0})&=\sum_{n=-\infty}^{\infty} h[n] \\
 +
&=16
 +
\end{align}
 +
</math>
 +
 +
b)
 +
 +
For <math class="inline">x[n]=1+e^{j\frac{4\pi n}{7}}-e^{j\frac{2\pi n}{5}}</math>  :
 +
 +
 +
<math> \mathcal{H}(0)= 16</math>
 +
 +
<math> \mathcal{H}\left(\frac{4\pi}{7}\right)=\frac{1+e^{-j\frac{\pi}{7}}}{1-e^{-j\frac{4\pi}{7}}}</math>
 +
 +
<math> \mathcal{H}\left(\frac{2\pi}{5}\right)=\frac{e^{j\frac{4\pi}{5}}+e^{-j\frac{3\pi}{5}}}{1-e^{-j\frac{2\pi}{5}}}</math>
 +
 +
Then,
 +
 +
<math>\begin{align}
 +
y[n]&=\mathcal{H}(0)+\mathcal{H}\left(\frac{4\pi}{7}\right)e^{j\frac{4\pi n}{7}}-\mathcal{H}\left(\frac{2\pi}{5}\right)e^{j\frac{2\pi n}{5}} \\
 +
&=16+\frac{1+e^{-j\frac{\pi}{7}}}{1-e^{-j\frac{4\pi}{7}}}e^{j\frac{4\pi n}{7}}-\frac{e^{j\frac{4\pi}{5}}+e^{-j\frac{3\pi}{5}}}{1-e^{-j\frac{2\pi}{5}}}e^{j\frac{2\pi n}{5}}
 +
\end{align}
 +
</math>
 +
 +
 +
==Question 6==
 +
a) No. The system cannot be LTI, since <math>y(t)</math> is a sin wave with different frequency than <math>x(t)</math>.
 +
 +
b) Yes. The system could be LTI since the transfer function might be zero at this frequency and thus it will suppress it.
 +
 +
c) Yes. The system could be LTI. That is because the output has the same frequency as the input. In this case, the system has suppressed the negative frequency of the input.
 +
 +
d) No. The system cannot be LTI, since the output and the input have different frequencies.
 +
 +
e)Yes. The system could be LTI. The output has the same frequency as the input. We notice that the system has suppressed the positive frequency this time.
 +
 +
==Question 7==
 +
a)Yes. The system could be LTI, since the output is just a scalar multiple of the input.
 +
 +
b)Yes. <math class="inline">x[n]=j^n=e^{j\frac{\pi}{2}n}=e^{j\frac{\pi}{2}n}</math> and thus the output and the input are the same signal.
 +
 +
c)No. Since the output is a DC signal while the input has a frequency of 1/4.
 +
 +
d)No. Since the output signal has a frequency of 1/2 and the input signal has a frequency of 1/4.
 +
 +
----
 +
[[HW4 ECE301 Spring2011 Prof Boutin| HW4]]
 +
 +
[[2011 Spring ECE 301 Boutin|Back to 2011 Spring ECE 301 Boutin]]
 +
 +
[[Category:2011_Spring_ECE_301_Boutin]]

Latest revision as of 09:13, 3 March 2011

Homework 4 Solutions, ECE301 Spring 2011 Prof. Boutin

Students should feel free to make comments/corrections or ask questions directly on this page. -pm

Question 1

a)Memory

Since $ h[n]=e^{j2\pi n}=1 $ for all $ n $, then $ h[n]\neq 0 $ for all $ n\neq 0 $. Hence, this system has memory.

Warning: one has to be careful when negating the statement: "h[n]=0 for all non zero n's". The correct negated statement is: "there exists a non-zero n such that h[n] is not zero." So to show that a system has memory, it is sufficient to show that there exists a non-zero n for which h[n] is not zero.-pm

Causality

$ h[n]=e^{j2\pi n}=1 $ for all $ n $, then $ h[n]\neq 0 $ for all $ n<0 $. Hence, the system is not causal.

Warning: one has to be careful when negating the statement: "h[n]=0 for all negative n's". The correct negated statement is: "there exists a non-negative n such that h[n] is not zero." So to show that a system is not causal, it is sufficient to show that there exists a non-negative n for which h[n] is not zero.-pm

Stability

$ \sum_{n=-\infty}^{\infty} |h[n]|=\sum_{n=-\infty}^{\infty} 1 = \infty $. Hence, the system is unstable.

Note that you are summing the magnitude, not the magnitude squared. So this is not the same as computing the energy of the unit impulse response h[n]. -pm

b)Memory

Since $ h(t)=e^{j2\pi t}\neq a\delta(t) $, where $ a $ is any number (can be complex). Hence, this system has memory.

To be more precise, one could say that if h(t) satisfied $ h(t)= a\delta(t) $ for some constant a, then h(t) would be zero for all non-zero t's. Now in this case, we have $ h(1)=1 \neq 0 $, and so the system has memory. -pm

Causality

$ h(t)=e^{j2\pi t}\neq 0 $ for $ t<0 $. Hence, the system is not causal.

Stability

$ \int_{-\infty}^{\infty} |h(t)|dt=\int_{-\infty}^{\infty} 1dt = \infty $. Hence, the system is unstable.

c)Memory

Since $ h(t)=e^{j2\pi t}u(-t)\neq a\delta(t) $, where $ a $ is any number (can be complex). Hence, this system has memory.

Causality

$ h(t)=e^{j2\pi t}u(-t)\neq 0 $ for $ t<0 $. Hence, the system is not causal.

Stability

$ \int_{-\infty}^{\infty} |h(t)|dt=\int_{-\infty}^{0} 1dt = \infty $. Hence, the system is unstable.

d)Memory

Since $ h[n]=e^{j2\pi n}\delta[n]=\delta[n] $, then this system is memoryless.

Causality

$ h[n]=\delta[n] $, then $ h[n]=0 $ for all $ n<0 $. Hence, the system is causal. We can also directly say that it is a causal system since we know that it is memoryless.

Stability

$ \sum_{n=-\infty}^{\infty} |h[n]|=\sum_{n=-\infty}^{\infty} \delta[n] = 1 <\infty $. Hence, the system is stable.

e)Memory

Since $ h[n]=e^{j2\pi n}(u[n+7]-u[n])=u[n+7]-u[n] $, then $ h[n]\neq 0 $ for all $ n\neq 0 $. Hence, this system has memory.

Causality

$ h[n]=u[n+7]-u[n]\neq 0 $ for all $ n<0 $. Hence, the system is not causal.

Stability

$ \sum_{n=-\infty}^{\infty} |h[n]|=\sum_{n=-7}^{-1} 1 = 7< \infty $. Hence, the system is stable.

Question 2

a) The period of $ x(t) $ is 1.

Now, $ x(t)=\sin(2\pi t)=\frac{1}{2j}e^{j2\pi t}-\frac{1}{2j}e^{-j2\pi t} $.

Hence, $ a_k=\left\{\begin{array}{ll}\frac{1}{2j},& \mbox{ for k=1, }\\ \frac{1}{2j},& \mbox{ for k=-1, }\\ 0,& \mbox{ otherwise.}\end{array}\right. $

b)Using Euler's properties, we get:

$ \begin{align} x(t)&= \left[\frac{e^{j2\pi t}}{2j}-\frac{e^{-j2\pi t}}{2j}\right] \left[\frac{e^{j\frac{\pi}{2} t}}{2}+\frac{e^{-j\frac{\pi}{2} t}}{2}\right] \\ &= \frac{e^{j\frac{5\pi}{2} t}}{4j} + \frac{e^{j\frac{3\pi}{2} t}}{4j} - \frac{e^{-j\frac{3\pi}{2} t}}{4j} - \frac{e^{-j\frac{5\pi}{2} t}}{4j} \\ &= \frac{e^{j\frac{2(5)\pi}{4} t}}{4j} + \frac{e^{j\frac{2(3)\pi}{4} t}}{4j} - \frac{e^{-j\frac{2(3)\pi}{4} t}}{4j} - \frac{e^{-j\frac{2(5)\pi}{4} t}}{4j} \end{align} $

Hence, $ a_5=\frac{1}{4j} $, $ a_3=\frac{1}{4j} $, $ a_{-3}=-\frac{1}{4j} $, $ a_{-5}=-\frac{1}{4j} $,

and all other coefficients are zero.

c)$ x[n]=(-1)^n=e^{j\pi n}=e^{j\frac{2\pi}{2} n} $. Hence, the period of the signal is 2, and $ a_0=0 $ and $ a_1=1 $. Note that the DTFS is periodic with period 2 with respect to $ k $.

d)$ x[n]=j^n=e^{j\frac{\pi}{2} n}=e^{j\frac{2\cdot 2\pi}{4} n} $. Hence, the period of the signal is 4, and $ a_0=a_2=a_3=0 $ and $ a_1=1 $. Note that the DTFS is periodic with period 4 with respect to $ k $.

e)$ x[n]=e^{j\frac{3}{5}\pi(n-\frac{1}{2})}=e^{-j\frac{3}{10}\pi}e^{j\frac{2\cdot 3}{10} \pi n} $. Hence, the period of the signal is 10, and $ a_3=e^{-j\frac{3}{10}\pi} $ and all other coefficients are zero in one period of the DTFS.

f)

$ \begin{align} x(t)&=\cos^2(t) \\ &=\frac{1}{2} + \frac{1}{2}\cos(2t) \\ &=\frac{1}{2} + \frac{1}{4}e^{j2\frac{\pi}{\pi} t} + \frac{1}{4}e^{-j2\frac{\pi}{\pi} t} \end{align} $

Hence, the signal is periodic with period $ \pi $ and has coefficients $ a_0=\frac{1}{2} $,$ a_1=a_{-1}=\frac{1}{4} $.

g)

$ \begin{align} x(t)&=1+e^{j\frac{4\pi n}{7}}-e^{j\frac{2\pi n}{5}} \\ &=1+e^{j\frac{2\pi \cdot 5 \cdot n}{35}}-e^{j\frac{2\pi \cdot 7 \cdot n}{35}} \\ \end{align} $

Hence, the signal is periodic with period 35. The coefficients are $ a_0=1 $, $ a_7=-1 $, and $ a_{10}=1 $, and all other coefficients in one period of the DTFS are zero.

h)

For $ k=0 $, we can directly find $ a_0=\frac{1}{2} \int_{-1}^{0} (t+1) dt + \frac{1}{2} \int_{0}^{1} (1-t) dt=\frac{1}{2} $.


Now,


$ \begin{align} a_k&= \frac{1}{2}\int_{-1}^{0} (t-1)e^{-j\pi k t} dt + \frac{1}{2}\int_{0}^{1} (1-t)e^{-j\pi k t} dt \\ &=\frac{1}{2}\int_{-1}^{0}te^{-j\pi kt} dt + \frac{1}{2}\int_{-1}^{0} e^{-j\pi kt} dt + \frac{1}{2}\int_{0}^{1} e^{-j\pi kt} dt + \frac{1}{2}\int_{0}^{1}te^{-j\pi kt} dt \\ &=\frac{1}{2} \left[\frac{te^{-jk\pi t}}{-j\pi k}+\frac{e^{-j\pi kt}}{\pi^2 k^2}\right]^0_{-1} + \left[\frac{1}{2j\pi k} e^{-j\pi kt}\right]^{-1}_{1} - \frac{1}{2}\left[\frac{te^{-jk\pi t}}{-j\pi k}+\frac{e^{-j\pi kt}}{\pi^2 k^2}\right]^1_0 \\ &=\frac{1}{2}\left[\frac{1}{\pi^2 k^2} - \frac{e^{j\pi k}}{j\pi k}-\frac{e^{j\pi k}}{\pi^2 k^2}\right] + \frac{1}{2j\pi k}\left[e^{j\pi k}-e^{-j\pi k}\right]-\frac{1}{2}\left[\frac{e^{-j\pi k}}{-j\pi k}+\frac{e^{-j\pi k}}{\pi^2 k^2}-\frac{1}{\pi^2 k^2}\right] \\ &=\frac{1}{\pi^2 k^2}-\frac{1}{\pi k}\left[\frac{e^{j\pi k}-e^{-j\pi k}}{2j}\right]-\frac{1}{\pi^2 k^2}\left[\frac{e^{j\pi k}+e^{-j\pi k}}{2}\right]+\frac{1}{\pi k}\left[\frac{e^{j\pi k}-e^{-j\pi k}}{2j}\right] \\ &=\frac{1}{\pi^2 k^2}-\frac{1}{\pi^2 k^2}\cos(\pi k) \\ &=\frac{1}{\pi^2 k^2}[1-(-1)^k] \end{align} $

Question 3

a)

$ x(t)=\frac{1}{2j}e^{j2\pi t}-\frac{1}{2j}e^{-j2\pi t} $.

b)

$ x(t)=\frac{e^{j\frac{5\pi}{2} t}}{4j} + \frac{e^{j\frac{3\pi}{2} t}}{4j} - \frac{e^{-j\frac{3\pi}{2} t}}{4j} - \frac{e^{-j\frac{5\pi}{2} t}}{4j} $.

c)

$ x[n]=e^{j\pi n} $.

d)

$ x[n]=e^{j\frac{\pi}{2} n} $.

e)

$ x[n]=e^{-j\frac{3}{10}\pi}e^{j\frac{3}{5} \pi n} $.

f)

$ x(t)=\frac{1}{2} + \frac{1}{4}e^{2jt} + \frac{1}{4}e^{-2jt} $.

g)

$ x[n]=1+e^{j\frac{4\pi n}{7}}-e^{j\frac{2\pi n}{5}} $.

h)

$ \begin{align} x(t)&= \frac{1}{2}+\sum_{k=-\infty}^{-1}a_k e^{j\pi kt} + \sum_{k=1}^{\infty}a_k e^{j\pi kt} \\ &= \frac{1}{2}+\sum_{k=1}^{\infty}a_{-k} e^{-j\pi kt} + \sum_{k=1}^{\infty}a_k e^{j\pi kt} \\ &= \frac{1}{2}+\sum_{k=1}^{\infty}a_k(e^{j\pi kt} + e^{-j\pi kt}) \\ &= \frac{1}{2}+\sum_{k=1}^{\infty} \frac{2}{\pi^2 k^2}[1-(-1)^k]\cos(\pi kt) \\ &= \frac{1}{2}+\frac{4}{\pi^2} \cos(\pi t)+\frac{4}{9\pi^2}\cos(3\pi t)+\frac{4}{25\pi^2}\cos(5\pi t)+\dots \end{align} $

Note that we have used the fact that $ a_{-k}=a_k $.

Question 4

a)

$ \begin{align} \mathcal{H}(\omega)=H(j\omega)&=\int_{-\infty}^{\infty}e^t[u(t-100)-u(t)]e^{-j\omega t}dt \\ &=-\int_{0}^{100}e^t\cdot e^{-j\omega t}dt \\ &=-\int{0}^{100}e^{(-j\omega+1)t}dt \\ &=\frac{1}{j\omega-1}\left[e^{(-j\omega+1)t}\right]^{100}_0 \\ &=\frac{1}{j\omega-1}e^{(1-j\omega)100}-\frac{1}{j\omega -1} \\ &=\frac{e^{100}e^{-j\omega 100}}{j\omega -1}-\frac{1}{j\omega -1} \end{align} $

b)

For $ x(t)=\frac{1}{2j}e^{j2\pi t}-\frac{1}{2j}e^{-j2\pi t} $  :


$ \begin{align} \mathcal{H}(2\pi)&=\frac{e^{100}e^{-j200\pi}}{2j\pi-1}-\frac{1}{2j\pi-1} \\ &=\frac{e^{100}-1}{2j\pi-1} \\ \end{align} $

$ \mathcal{H}(-2\pi)=\frac{1-e^{100}}{1+2j\pi} $

Then,

$ \begin{align} y(t)&=\frac{1}{2j}\mathcal{H}(2\pi)e^{j2\pi t}-\frac{1}{2j}\mathcal{H}(-2\pi)e^{-j2\pi t} \\ &=\frac{1-e^{100}}{4\pi+2j}e^{j2\pi t}-\frac{1-e^{100}}{2j-4\pi}e^{-j2\pi t} \end{align} $

Question 5

a)

$ \begin{align} \mathcal{H}(\omega)=H(e^{j\omega})&=\sum_{n=-\infty}^{\infty} h[n] e^{-j\omega n} \\ &=\sum_{n=-\infty}^\infty\left(\sum_{k=-7}^8 \delta[n-k]\right)e^{-j\omega n} \\ &=\sum_{k=-7}^8 \sum_{n=-\infty}^\infty e^{-j\omega n}\delta[n-k] \\ &=\sum_{k=-7}^8 e^{-j\omega k} \\ &=\frac{e^{7j\omega}-e^{-9j\omega}}{1-e^{-j\omega}} \end{align} $


$ \begin{align} \mathcal{H}(0)&=H(e^{j0})&=\sum_{n=-\infty}^{\infty} h[n] \\ &=16 \end{align} $

b)

For $ x[n]=1+e^{j\frac{4\pi n}{7}}-e^{j\frac{2\pi n}{5}} $  :


$ \mathcal{H}(0)= 16 $

$ \mathcal{H}\left(\frac{4\pi}{7}\right)=\frac{1+e^{-j\frac{\pi}{7}}}{1-e^{-j\frac{4\pi}{7}}} $

$ \mathcal{H}\left(\frac{2\pi}{5}\right)=\frac{e^{j\frac{4\pi}{5}}+e^{-j\frac{3\pi}{5}}}{1-e^{-j\frac{2\pi}{5}}} $

Then,

$ \begin{align} y[n]&=\mathcal{H}(0)+\mathcal{H}\left(\frac{4\pi}{7}\right)e^{j\frac{4\pi n}{7}}-\mathcal{H}\left(\frac{2\pi}{5}\right)e^{j\frac{2\pi n}{5}} \\ &=16+\frac{1+e^{-j\frac{\pi}{7}}}{1-e^{-j\frac{4\pi}{7}}}e^{j\frac{4\pi n}{7}}-\frac{e^{j\frac{4\pi}{5}}+e^{-j\frac{3\pi}{5}}}{1-e^{-j\frac{2\pi}{5}}}e^{j\frac{2\pi n}{5}} \end{align} $


Question 6

a) No. The system cannot be LTI, since $ y(t) $ is a sin wave with different frequency than $ x(t) $.

b) Yes. The system could be LTI since the transfer function might be zero at this frequency and thus it will suppress it.

c) Yes. The system could be LTI. That is because the output has the same frequency as the input. In this case, the system has suppressed the negative frequency of the input.

d) No. The system cannot be LTI, since the output and the input have different frequencies.

e)Yes. The system could be LTI. The output has the same frequency as the input. We notice that the system has suppressed the positive frequency this time.

Question 7

a)Yes. The system could be LTI, since the output is just a scalar multiple of the input.

b)Yes. $ x[n]=j^n=e^{j\frac{\pi}{2}n}=e^{j\frac{\pi}{2}n} $ and thus the output and the input are the same signal.

c)No. Since the output is a DC signal while the input has a frequency of 1/4.

d)No. Since the output signal has a frequency of 1/2 and the input signal has a frequency of 1/4.


HW4

Back to 2011 Spring ECE 301 Boutin

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood