Revision as of 07:04, 15 February 2011

Homework 4 Solutions

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 -pm

Stability

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

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.

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_1=a_3=0 $ and $ a_2=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 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 can not 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

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