Chapter 2: Q2.5-30Q (page 93)
30. Find a different factorization of the A in Exercise 29 i.e., \(A = \left( {\begin{aligned}{*{20}{c}}{{{\bf{4}} \mathord{\left/
{\vphantom {{\bf{4}} {\bf{3}}}} \right.
\kern-\nulldelimiterspace} {\bf{3}}}}&{ - {\bf{12}}}\\{{{ - {\bf{1}}} \mathord{\left/
{\vphantom {{ - {\bf{1}}} {\bf{4}}}} \right.
\kern-\nulldelimiterspace} {\bf{4}}}}&{\bf{3}}\end{aligned}} \right)\), and thereby design a different ladder network whose transfer matrix is A.
Short Answer
The different factorization of A in Exercise 29 is
\(A = \left( {\begin{aligned}{*{20}{c}}1&{ - {4 \mathord{\left/
{\vphantom {4 3}} \right.
\kern-\nulldelimiterspace} 3}}\\0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/
{\vphantom {1 4}} \right.
\kern-\nulldelimiterspace} 4}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&{ - 8}\\0&1\end{aligned}} \right)\).
Thus, the ladder network, whose transfer matrix is A, is
Step by step solution
Over 22 million students worldwide already upgrade their learning with Vaia!
Design a ladder network
Construct a ladder network that is different from the network in Exercise 29.
Write the transfer matrices
Here, the first circuit is the series circuit with resistance \({R_1}\) , the second circuit is the shunt circuit with resistance \({R_2}\), and the last circuit is the series circuit with resistance \({R_3}\) , whose transfer matrices are \(\left( {\begin{aligned}{*{20}{c}}1&{ - {R_1}}\\0&1\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/
{\vphantom {1 {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right)\), and \(\left( {\begin{aligned}{*{20}{c}}1&{ - {R_3}}\\0&1\end{aligned}} \right)\), respectively.
Find the transfer matrix for the above ladder network
The transfer matrix for the ladder network is given by in the reverse order.
\(\begin{aligned}{c}\left( {\begin{aligned}{*{20}{c}}1&{ - {R_3}}\\0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/
{\vphantom {1 {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&{ - {R_1}}\\0&1\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{1 + {{{R_3}} \mathord{\left/
{\vphantom {{{R_3}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_3}}\\{ - {1 \mathord{\left/
{\vphantom {1 {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&{ - {R_1}}\\0&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{1 + {{{R_3}} \mathord{\left/
{\vphantom {{{R_3}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_1} - {{{R_1}{R_3}} \mathord{\left/
{\vphantom {{{R_1}{R_3}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}} - {R_3}}\\{ - {1 \mathord{\left/
{\vphantom {1 {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&{{{{R_1}} \mathord{\left/
{\vphantom {{{R_1}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}} + 1}\end{aligned}} \right)\end{aligned}\)
Set matrix A, which is equal to the transfer matrix in step 2
\(\begin{aligned}{c}\left( {\begin{aligned}{*{20}{c}}{1 + {{{R_3}} \mathord{\left/
{\vphantom {{{R_3}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_1} - {{{R_1}{R_3}} \mathord{\left/
{\vphantom {{{R_1}{R_3}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}} - {R_3}}\\{ - {1 \mathord{\left/
{\vphantom {1 {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&{{{{R_1}} \mathord{\left/
{\vphantom {{{R_1}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}} + 1}\end{aligned}} \right) = A\\\left( {\begin{aligned}{*{20}{c}}{1 + {{{R_3}} \mathord{\left/
{\vphantom {{{R_3}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_1} - {{{R_1}{R_3}} \mathord{\left/
{\vphantom {{{R_1}{R_3}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}} - {R_3}}\\{ - {1 \mathord{\left/
{\vphantom {1 {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}}}&{{{{R_1}} \mathord{\left/
{\vphantom {{{R_1}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}} + 1}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{{4 \mathord{\left/
{\vphantom {4 3}} \right.
\kern-\nulldelimiterspace} 3}}&{ - 12}\\{{{ - 1} \mathord{\left/
{\vphantom {{ - 1} 4}} \right.
\kern-\nulldelimiterspace} 4}}&3\end{aligned}} \right)\end{aligned}\)
This implies that
\(\begin{aligned}{c}{{ - 1} \mathord{\left/
{\vphantom {{ - 1} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}} = - {1 \mathord{\left/
{\vphantom {1 4}} \right.
\kern-\nulldelimiterspace} 4}\\{R_2} = 4\end{aligned}\)
\(\begin{aligned}{c}{{{R_1}} \mathord{\left/
{\vphantom {{{R_1}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}} + 1 = 3\\{{{R_1}} \mathord{\left/
{\vphantom {{{R_1}} 4}} \right.
\kern-\nulldelimiterspace} 4} = 2\\{R_1} = 8\end{aligned}\)
\(\begin{aligned}{c}1 + {{{R_3}} \mathord{\left/
{\vphantom {{{R_3}} {{R_2}}}} \right.
\kern-\nulldelimiterspace} {{R_2}}} = {4 \mathord{\left/
{\vphantom {4 3}} \right.
\kern-\nulldelimiterspace} 3}\\{{{R_3}} \mathord{\left/
{\vphantom {{{R_3}} 4}} \right.
\kern-\nulldelimiterspace} 4} = {1 \mathord{\left/
{\vphantom {1 3}} \right.
\kern-\nulldelimiterspace} 3}\\{R_3} = {4 \mathord{\left/
{\vphantom {4 3}} \right.
\kern-\nulldelimiterspace} 3}\end{aligned}\)
Conclusion
The different factorization of A in Exercise 29 is:
\(A = \left( {\begin{aligned}{*{20}{c}}1&{ - {4 \mathord{\left/
{\vphantom {4 3}} \right.
\kern-\nulldelimiterspace} 3}}\\0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/
{\vphantom {1 4}} \right.
\kern-\nulldelimiterspace} 4}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&{ - 8}\\0&1\end{aligned}} \right)\)
Thus, the ladder network, whose transfer matrix is A, is
