Revision as of 19:03, 14 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.

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.

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)$ x(t)=\cos^2(t)= $