Difference between revisions of "ECE438 Week9 Quiz Q1sol" - Rhea

Revision as of 11:56, 19 October 2010

a. The difference equation for this system is

\begin{align} & H(z) = 1-2\cos\theta z^{-1}+z^{-2} = \frac{Y(z)}{X(z)} \\ & Y(z) = X(z)-2\cos\theta X(z)z^{-1} + X(z)z^{-2} \\ \end{align}\,\!

Appling the inverse DTFT,

\quad y[n]=x[n]-2\cos\theta x[n-1]+x[n-2]\,\!

.

b. Find the frequency response $$ H(w) $$ from the difference equation:

i. In an LTI system decribed by a difference equation, if we input

e^{jwn}

, then the output is always

H(w)e^{jwn}

.

\begin{align} H(w)e^{jwn} & =e^{jwn}-2\cos\theta e^{jw(n-1)}+e^{jw(n-2)} \\ & =e^{jwn}(1-2\cos\theta e^{-jw} + e^{-j2w}) \\ \end{align}

Hence,

H(w)=1-2\cos\theta e^{-jw} + e^{-j2w}\,\!

ii. It is easily driven that

h[n]=\delta[n]-2\cos\theta\delta[n-1]+\delta[n-2]\,\!

.

Appling DTFT,

H(w)=1-2\cos\theta e^{-jw} + e^{-j2w}\,\!

c. Find the response of this system to the input x[n]:

x[n]=\delta[n+1]+\delta[n]\,\!

X(w)=e^{jw}+1\,\!

\begin{align} Y(w) & = X(w)H(w) \\ & = (e^{jw}+1)(1-2\cos\theta e^{-jw} + e^{-j2w}) \\ & = e^{jw} - 2\cos\theta + e^{-jw} + 1 - 2\cos\theta e^{-jw} + e^{-j2w} \\ \end{align}

therefore,

y[n]=\delta[n+1] + (1-2\cos\theta)\delta[n] + (1-2\cos\theta)\delta[n-1] + \delta[n-2]\,\!

d. If we further look at the frequency response of this filter,

\begin{align} H(w)&=1-2\cos\theta e^{-jw} + e^{-j2w} \\ &= e^{-jw}(e^{jw}+e^{-jw}-2\cos\theta) \\ &= e^{-jw}(2\cos w -2\cos\theta) \\ \end{align}

The magnitude reponse is

|H(w)|=2\cos w -2\cos\theta\,\!

, which becomes zero at

w=\pm \theta

.

therefore, when

\theta=\pi/2

, it is a bandstop filter.

e. The signal modulated at $$ f $$ Hz and sampled at $$ F_s $$ is mapped to the DTFT by the following rule.

w=2\pi\frac{f}{F_s}\,\!

.

Thus,

w=2\pi\frac{2000}{8000}=\frac{\pi}{4}\,\!

Since this filter bandstops at

w=\pm \theta

,

\theta

must be

\pi/4

.