(New page: Category:2010 Fall ECE 438 Boutin ---- == Solution to Q1 of Week 9 Quiz Pool == ---- a. The difference equation for this system is :<math>\begin{align} & H(z) = 1-2\cos\theta z^{-1}+...) |
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e. The signal modulated at <math>f</math> Hz and sampled at <math>F_s</math> is mapped to the DTFT by the following rule. | e. The signal modulated at <math>f</math> Hz and sampled at <math>F_s</math> is mapped to the DTFT by the following rule. | ||
:<math>w=2\pi\frac{f}{F_s}\,\!</math>. | :<math>w=2\pi\frac{f}{F_s}\,\!</math>. | ||
| − | :Thus, <math>w=2\pi\frac{2000}{8000}=\frac{\pi}{ | + | :Thus, <math>w=2\pi\frac{2000}{8000}=\frac{\pi}{2}\,\!</math> |
| − | :Since this filter bandstops at <math>w=\pm \theta</math>, <math>\theta</math> must be <math>\pi/ | + | :Since this filter bandstops at <math>w=\pm \theta</math>, <math>\theta</math> must be <math>\pi/2</math>. |
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Latest revision as of 11:00, 19 October 2010
Solution to Q1 of Week 9 Quiz Pool
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}{2}\,\! $
- Since this filter bandstops at $ w=\pm \theta $, $ \theta $ must be $ \pi/2 $.
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