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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

Expert verified

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

01

Design a ladder network

Construct a ladder network that is different from the network in Exercise 29.

02

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.

03

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}\)

04

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}\)

05

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

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Most popular questions from this chapter

In exercise 11 and 12, mark each statement True or False.Justify each answer.

a. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}&{{A_{\bf{2}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), with \({A_{\bf{1}}}\) and \({A_{\bf{2}}}\) the same sizes as \({B_{\bf{1}}}\) and \({B_{\bf{2}}}\), respectively then \(A + B = \left[ {\begin{array}{*{20}{c}}{{A_1} + {B_1}}&{{A_{\bf{2}}} + {B_{\bf{2}}}}\end{array}} \right]\).

b. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_{\bf{2}}}}\end{array}} \right]\), then the partitions of \(A\) and \(B\) are comfortable for block multiplication.

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. [Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)].

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.

3. \[\left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\I&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}W&X\\Y&Z\end{array}} \right]\]

Suppose Ais an \(m \times n\) matrix and there exist \(n \times m\) matrices C and D such that \(CA = {I_n}\) and \(AD = {I_m}\). Prove that \(m = n\) and \(C = D\). (Hint: Think about the product CAD.)
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