(New page: ---- == Solution to Q3 of Week 10 Quiz Pool == ---- <work in progress> It is important to note that systems g1 and g2 are Length-2 FIR filters, these filters are of the form – [[Image...) |
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | g1[n] &= | + | g1[n] &= a_1*g1'[n] + b_1*g1'[n-1] \\ |
| − | g2[n] &= | + | g2[n] &= a_1*g2'[n] + b_1*g2'[n-1] \\ |
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | or, looking at it in terms of impulse responses, | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | g1[n] &= a_1*\delta[n] + b_1*\delta[n-1] \\ | ||
| + | g2[n] &= a_2*\delta[n] + b_2*\delta[n-1]\\ | ||
| + | |||
\\ | \\ | ||
\text{Given:} \\ | \text{Given:} \\ | ||
| − | g1[0] &= | + | g1[0] &= a_1 = (1/2) \\ |
\\ | \\ | ||
| − | \text{To Find:} \\ | + | \text{To Find:} \\ |
| − | g1[1] &= | + | g1[1] &= b_1 \\ |
| − | g2[0] &= | + | g2[0] &= a_2 \\ |
| − | g2[1] &= | + | g2[1] &= b_2 \\ |
\end{align} | \end{align} | ||
</math> | </math> | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | g1[n] = (1/2)\delta[n] + b_1\delta[n-1] \\ | ||
| + | g2[n] = a_2\delta[n] + b_2\delta[n-1] \\ | ||
| + | |||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | G1(\omega) &= \frac{1}{2} + b_1z^{-1} \\ | ||
| + | G2(\omega) &= a_2 + b_2z^{-1} \\ | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | It is also given to us that, <br/> | ||
| + | y[n] = x[n-1],<br/> | ||
| + | so feeding in <math>\delta[n]</math> as input (x[n]) would result in <math>\delta[n-1]</math>. <br/> | ||
| + | Thus we require:<br/> | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | h1[n] * g1[n] + h2[n] * g2[n] &= \delta[n-1] | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Taking Z transform, <br/> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | H1(z)G1(z) + H2(z)G2(z) &= z^{-1} \\ | ||
| + | \end{align} | ||
| + | </math><br/> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | \frac{1}{2}(1+z^{-1})(\frac{1}{2} + b_1z^{-1}) + \frac{1}{2} (1- z^{-1}) (a_2 + b_2z^{-1}) &= z^{-1} | ||
| + | |||
| + | \end{align} | ||
| + | </math><br/> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | (1+z^{-1})(\frac{1}{2} + b_1z^{-1}) + (1- z^{-1}) (a_2 + b_2z^{-1}) &= 2z^{-1} | ||
| + | |||
| + | \end{align} | ||
| + | </math><br/> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | \frac{1}{2} + b_1z^{-1} + \frac{1}{2}z^{-1} + b_1z^{-2} + a_2 + b_2z^{-1} - a_2z^{-1} - b_2z^{-2} = 2z^{-1} | ||
| + | |||
| + | \end{align} | ||
| + | </math><br/> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | (\frac{1}{2} + a_2) + (\frac{1}{2} + b_1 + b_2 - a_2)z^{-1} + (b_1 - b_2)z^{-2} &= 2z^{-1} | ||
| + | |||
| + | \end{align} | ||
| + | </math><br/> | ||
| + | |||
| + | Solve equation by equating coefficients of <math>z^0, z^{-1}, z^{-2}</math>, | ||
| + | |||
| + | <math>\frac{1}{2} + a_2 = 0, a_2 = -\frac{1}{2} </math><br/> | ||
| + | <math>\begin{align}b_1 - b_2 = 0, b_1 = b_2\end{align}</math><br/> | ||
| + | <math>\begin{align}\frac{1}{2} + b_1 + b_2 - a_2 = 2\end{align}</math><br/> | ||
| + | <math>\begin{align}\frac{1}{2} + 2b_1 + \frac{1}{2} = 2\end{align}</math><br/> | ||
| + | <math>\begin{align}2b_1 = 1, b_1 = \frac{1}{2}\end{align}</math><br/> | ||
| + | <math>\begin{align}b_2 = \frac{1}{2}\end{align}</math><br/> | ||
| + | |||
| + | |||
| + | Therefore our two systems are - <br/> | ||
| + | <math>\begin{align}g1[n] = \frac{1}{2}g1'[n] + \frac{1}{2}g1'[n-1]\end{align}</math><br/> | ||
| + | <math>\begin{align}g2[n] = -\frac{1}{2}g2'[n] + \frac{1}{2}g2'[n-1]\end{align}</math><br/> | ||
| + | |||
| + | |||
| + | b. | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
Revision as of 17:40, 27 October 2010
Solution to Q3 of Week 10 Quiz Pool
<work in progress>
It is important to note that systems g1 and g2 are Length-2 FIR filters, these filters are of the form –
$ \begin{align} g1[n] &= a_1*g1'[n] + b_1*g1'[n-1] \\ g2[n] &= a_1*g2'[n] + b_1*g2'[n-1] \\ \end{align} $
or, looking at it in terms of impulse responses,
$ \begin{align} g1[n] &= a_1*\delta[n] + b_1*\delta[n-1] \\ g2[n] &= a_2*\delta[n] + b_2*\delta[n-1]\\ \\ \text{Given:} \\ g1[0] &= a_1 = (1/2) \\ \\ \text{To Find:} \\ g1[1] &= b_1 \\ g2[0] &= a_2 \\ g2[1] &= b_2 \\ \end{align} $
$ \begin{align} g1[n] = (1/2)\delta[n] + b_1\delta[n-1] \\ g2[n] = a_2\delta[n] + b_2\delta[n-1] \\ \end{align} $
$ \begin{align} G1(\omega) &= \frac{1}{2} + b_1z^{-1} \\ G2(\omega) &= a_2 + b_2z^{-1} \\ \end{align} $
It is also given to us that,
y[n] = x[n-1],
so feeding in $ \delta[n] $ as input (x[n]) would result in $ \delta[n-1] $.
Thus we require:
$ \begin{align} h1[n] * g1[n] + h2[n] * g2[n] &= \delta[n-1] \end{align} $
Taking Z transform,
$ \begin{align} H1(z)G1(z) + H2(z)G2(z) &= z^{-1} \\ \end{align} $
$ \begin{align} \frac{1}{2}(1+z^{-1})(\frac{1}{2} + b_1z^{-1}) + \frac{1}{2} (1- z^{-1}) (a_2 + b_2z^{-1}) &= z^{-1} \end{align} $
$ \begin{align} (1+z^{-1})(\frac{1}{2} + b_1z^{-1}) + (1- z^{-1}) (a_2 + b_2z^{-1}) &= 2z^{-1} \end{align} $
$ \begin{align} \frac{1}{2} + b_1z^{-1} + \frac{1}{2}z^{-1} + b_1z^{-2} + a_2 + b_2z^{-1} - a_2z^{-1} - b_2z^{-2} = 2z^{-1} \end{align} $
$ \begin{align} (\frac{1}{2} + a_2) + (\frac{1}{2} + b_1 + b_2 - a_2)z^{-1} + (b_1 - b_2)z^{-2} &= 2z^{-1} \end{align} $
Solve equation by equating coefficients of $ z^0, z^{-1}, z^{-2} $,
$ \frac{1}{2} + a_2 = 0, a_2 = -\frac{1}{2} $
$ \begin{align}b_1 - b_2 = 0, b_1 = b_2\end{align} $
$ \begin{align}\frac{1}{2} + b_1 + b_2 - a_2 = 2\end{align} $
$ \begin{align}\frac{1}{2} + 2b_1 + \frac{1}{2} = 2\end{align} $
$ \begin{align}2b_1 = 1, b_1 = \frac{1}{2}\end{align} $
$ \begin{align}b_2 = \frac{1}{2}\end{align} $
Therefore our two systems are -
$ \begin{align}g1[n] = \frac{1}{2}g1'[n] + \frac{1}{2}g1'[n-1]\end{align} $
$ \begin{align}g2[n] = -\frac{1}{2}g2'[n] + \frac{1}{2}g2'[n-1]\end{align} $
b.
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