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Chapter 2: Q2.5-29Q (page 93)

  1. Compute the transfer matrix of the network in the figure.
  2. Let \[A = \left[ {\begin{array}{*{20}{c}}{{{\bf{4}} \mathord{\left/
  3. {\vphantom {{\bf{4}} {\bf{3}}}} \right.
  4. \kern-\nulldelimiterspace} {\bf{3}}}}&{ - {\bf{12}}}\\{{{ - {\bf{1}}} \mathord{\left/
  5. {\vphantom {{ - {\bf{1}}} {\bf{4}}}} \right.
  6. \kern-\nulldelimiterspace} {\bf{4}}}}&{\bf{3}}\end{array}} \right]\]. Design a ladder network whose transfer matrix is Aby finding a suitable matrix factorization of A.

Short Answer

Expert verified
  1. The transfer matrix for the given network is \[\left[ {\begin{array}{*{20}{c}}{1 + {{{R_2}} \mathord{\left/
  2. {\vphantom {{{R_2}} {{R_1}}}} \right.
  3. \kern-\nulldelimiterspace} {{R_1}}}}&{ - {R_2}}\\{ - {1 \mathord{\left/
  4. {\vphantom {1 {{R_3} - {{{R_2}} \mathord{\left/
  5. {\vphantom {{{R_2}} {\left( {{R_1}{R_3}} \right)}}} \right.
  6. \kern-\nulldelimiterspace} {\left( {{R_1}{R_3}} \right)}} - {1 \mathord{\left/
  7. {\vphantom {1 {{R_1}}}} \right.
  8. \kern-\nulldelimiterspace} {{R_1}}}}}} \right.
  9. \kern-\nulldelimiterspace} {{R_3} - {{{R_2}} \mathord{\left/
  10. {\vphantom {{{R_2}} {\left( {{R_1}{R_3}} \right)}}} \right.
  11. \kern-\nulldelimiterspace} {\left( {{R_1}{R_3}} \right)}} - {1 \mathord{\left/
  12. {\vphantom {1 {{R_1}}}} \right.
  13. \kern-\nulldelimiterspace} {{R_1}}}}}}&{{{{R_2}} \mathord{\left/
  14. {\vphantom {{{R_2}} {{R_3}}}} \right.
  15. \kern-\nulldelimiterspace} {{R_3}}} + 1}\end{array}} \right]\].
  16. The suitable matrix factorization of A is

\[A = \left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 6}} \right.

\kern-\nulldelimiterspace} 6}}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 12}\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {36}}} \right.

\kern-\nulldelimiterspace} {36}}}&1\end{array}} \right]\].

Thus, the ladder network is

Step by step solution

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01

Write the transfer matrices

(a)

The first circuit is the shunt circuit with resistance \[{R_1}\] , the second circuit is the series circuit with resistance \[{R_2}\], and the last circuit is the shunt circuit with resistance \[{R_3}\] , whose transfer matrices are \[\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&1\end{array}} \right],\left[ {\begin{array}{*{20}{c}}1&{ - {R_2}}\\0&1\end{array}} \right]\], and \[\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}}}&1\end{array}} \right]\], respectively.

02

Find the transfer matrix for the given network

The transfer matrix for the given network is given by in the reverse order.

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}}}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - {R_2}}\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - {R_2}}\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}}}&{{{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}} + 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 + {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&{ - {R_2}}\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3} - {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {\left( {{R_1}{R_3}} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {{R_1}{R_3}} \right)}} - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}}} \right.

\kern-\nulldelimiterspace} {{R_3} - {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {\left( {{R_1}{R_3}} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {{R_1}{R_3}} \right)}} - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}}}&{{{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}} + 1}\end{array}} \right]\end{array}\]

03

Design a ladder network whose transfer matrix is A

(b)

To find a ladder network having a structure resembling the one in part (a) with the transfer matrix A, the resistances \[{R_1},{R_2},\] and \[{R_3}\] must satisfy the equation.

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{1 + {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&{ - {R_2}}\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3} - {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {\left( {{R_1}{R_3}} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {{R_1}{R_3}} \right)}} - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}}} \right.

\kern-\nulldelimiterspace} {{R_3} - {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {\left( {{R_1}{R_3}} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {{R_1}{R_3}} \right)}} - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}}}&{{{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}} + 1}\end{array}} \right] = A\\\left[ {\begin{array}{*{20}{c}}{1 + {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}&{ - {R_2}}\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_3} - {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {\left( {{R_1}{R_3}} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {{R_1}{R_3}} \right)}} - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}}} \right.

\kern-\nulldelimiterspace} {{R_3} - {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {\left( {{R_1}{R_3}} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {{R_1}{R_3}} \right)}} - {1 \mathord{\left/

{\vphantom {1 {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}}}}}&{{{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}} + 1}\end{array}} \right] = \left[ {\begin{array}{*{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{array}} \right]\end{array}\]

This implies that

\[\begin{array}{c} - {R_2} = - 12\\{R_2} = 12\end{array}\]

\[\begin{array}{c}{{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}} + 1 = 3\\{{12} \mathord{\left/

{\vphantom {{12} {{R_3}}}} \right.

\kern-\nulldelimiterspace} {{R_3}}} = 2\\{R_3} = 6\end{array}\]

\[\begin{array}{c}1 + {{{R_2}} \mathord{\left/

{\vphantom {{{R_2}} {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}} = {4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}\\{{12} \mathord{\left/

{\vphantom {{12} {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}} = {4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3} - 1\\{{12} \mathord{\left/

{\vphantom {{12} {{R_1}}}} \right.

\kern-\nulldelimiterspace} {{R_1}}} = {1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}\\{R_1} = 36\end{array}\]

Hence, the suitable matrix factorization of A is:

\[A = \left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 6}} \right.

\kern-\nulldelimiterspace} 6}}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 12}\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {36}}} \right.

\kern-\nulldelimiterspace} {36}}}&1\end{array}} \right]\]

Thus, theladder network is:

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Most popular questions from this chapter

Prove the Theorem 3(d) i.e., \({\left( {AB} \right)^T} = {B^T}{A^T}\).

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

5. \[\left[ {\begin{array}{*{20}{c}}A&B\\C&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&Y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\Z&{\bf{0}}\end{array}} \right]\]

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

a. The definition of the matrix-vector product \(A{\bf{x}}\) is a special case of block multiplication.

b. If \({A_{\bf{1}}}\), \({A_{\bf{2}}}\), \({B_{\bf{1}}}\), and \({B_{\bf{2}}}\) are \(n \times n\) matrices, \[A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}\\{{A_{\bf{2}}}}\end{array}} \right]\] and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), then the product \(BA\) is defined, but \(AB\) is not.

Give a formula for \({\left( {ABx} \right)^T}\), where \({\bf{x}}\) is a vector and \(A\) and \(B\) are matrices of appropriate sizes.

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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