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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

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

Suppose a linear transformation \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) has the property that \(T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\) for some pair of distinct vectors u and v in \({\mathbb{R}^n}\). Can Tmap \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\)? Why or why not?

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

1. \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\)

Suppose A, B,and Care invertible \(n \times n\) matrices. Show that ABCis also invertible by producing a matrix Dsuch that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).

Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]
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