(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...)
 
m
Line 11: Line 11:
 
<math>
 
<math>
 
\begin{align}
 
\begin{align}
g1[n] &= a1*\delta[n] + b1*\delta[n-1] \\
+
g1[n] &= a_1*g1'[n] + b_1*g1'[n-1] \\
g2[n] &= a2*\delta[n] + b2*\delta[n-1]\\
+
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] &= a1 = (1/2) \\
+
g1[0] &= a_1 = (1/2) \\
 
\\
 
\\
\text{To Find:} \\  
+
\text{To Find:} \\
g1[1] &= b1 \\
+
g1[1] &= b_1 \\
g2[0] &= a2 \\
+
g2[0] &= a_2 \\
g2[1] &= b2 \\
+
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 –

Qp10q3fir.jpg

$ \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.






Back to Lab Week 10 Quiz Pool

Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett