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Chapter 2: Q2.5-27Q (page 93)

27. Design two different ladder networks that each output 9 volts and 4 amps when the input is 12 volts and 6 amps.

Short Answer

Expert verified

The first ladder network is

Here, \[{v_1} = 12,{i_1} = 6,\] and \[{v_3} = 9,{i_3} = 4\].

The second ladder network is

Here, \[{v_1} = 12,{i_1} = 6,\] and \[{v_3} = 9,{i_3} = 4\].

Step by step solution

01

Write the transfer matrices

The transfer matrices of the series circuit with resistance \[{R_1}\] and shunt circuit with resistance \[{R_2}\] are \[\left[ {\begin{array}{*{20}{c}}1&{ - {R_1}}\\0&1\end{array}} \right]\], and \[\left[ {\begin{array}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{array}} \right]\], respectively.

02

Design the first ladder network

Consider the first ladder network as aseries circuitwith resistance \[{R_1}\], followed by a shunt circuitwith resistance \[{R_2}\]. Therefore, the transfer matrix for this network is

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

{\vphantom {1 {{R_2}}}} \right.

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

{\vphantom {1 {{R_2}}}} \right.

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

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

\kern-\nulldelimiterspace} {{R_2}}}}\end{array}} \right]\].

To produce an output of 9 volts and 4 amps for an input of 12 volts and 6 amps, the transfer matrix must satisfy the equation:

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}1&{ - {R_1}}\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

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

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

\kern-\nulldelimiterspace} {{R_2}}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{12}\\6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\4\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{12 - 6{R_1}}\\{ - {{12} \mathord{\left/

{\vphantom {{12} {{R_2} + 6 + {{6{R_1}} \mathord{\left/

{\vphantom {{6{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}}} \right.

\kern-\nulldelimiterspace} {{R_2} + 6 + {{6{R_1}} \mathord{\left/

{\vphantom {{6{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\4\end{array}} \right]\end{array}\]

This implies that

\[\begin{array}{c}12 - 6{R_1} = 9\\6{R_1} = 3\\{R_1} = {1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}\end{array}\]

And

\[\begin{array}{c} - {{12} \mathord{\left/

{\vphantom {{12} {{R_2} + 6 + {{6{R_1}} \mathord{\left/

{\vphantom {{6{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}}} \right.

\kern-\nulldelimiterspace} {{R_2} + 6 + {{6{R_1}} \mathord{\left/

{\vphantom {{6{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}} = 4\\ - 12 + 6{R_2} + 6\left( {{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}} \right) = 4{R_2}\\2{R_2} = 12 - 3\\{R_2} = {9 \mathord{\left/

{\vphantom {9 2}} \right.

\kern-\nulldelimiterspace} 2}\end{array}\]

Thus, the firstladder network is

Here, \[{v_1} = 12,{i_1} = 6,\] and \[{v_3} = 9,{i_3} = 4\].

03

Design the second ladder network

Consider the second ladder network as a shunt circuit with resistance \[{R_2}\], followed by a series circuit with resistance \[{R_1}\]. Therefore, the transfer matrix for this network is

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

{\vphantom {1 {{R_2}}}} \right.

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

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

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

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{array}} \right]\].

To produce an output of 9 volts and 4 amps for an input of 12 volts and 6 amps, the transfer matrix must satisfy the equation:

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{1 + {{{R_1}} \mathord{\left/

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

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

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{12}\\6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\4\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{12 + {{12{R_1}} \mathord{\left/

{\vphantom {{12{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} - 6{R_1}}\\{ - {{12} \mathord{\left/

{\vphantom {{12} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9\\4\end{array}} \right]\end{array}\]

This implies that

\[\begin{array}{c} - {{12} \mathord{\left/

{\vphantom {{12} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + 6 = 4\\{{12} \mathord{\left/

{\vphantom {{12} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} = 2\\{R_2} = 6\end{array}\]

And

\[\begin{array}{c}12 + {{12{R_1}} \mathord{\left/

{\vphantom {{12{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} - 6{R_1} = 9\\3 + {{12{R_1}} \mathord{\left/

{\vphantom {{12{R_1}} 6}} \right.

\kern-\nulldelimiterspace} 6} - 6{R_1} = 0\\3 + 2{R_1} - 6{R_1} = 0\\{R_1} = {3 \mathord{\left/

{\vphantom {3 4}} \right.

\kern-\nulldelimiterspace} 4}\end{array}\]

Thus, the second ladder network is

Here, \[{v_1} = 12,{i_1} = 6,\] and \[{v_3} = 9,{i_3} = 4\].

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Most popular questions from this chapter

If a matrix \(A\) is \({\bf{5}} \times {\bf{3}}\) and the product \(AB\)is \({\bf{5}} \times {\bf{7}}\), what is the size of \(B\)?

Suppose \(\left( {B - C} \right)D = 0\), where Band Care \(m \times n\) matrices and \(D\) is invertible. Show that B = C.

2. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{2}}\\{\bf{7}}&{\bf{4}}\end{aligned}} \right)\).

a. Verify that \({A^2} = I\) when \(A = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\).

b. Use partitioned matrices to show that \({M^2} = I\) when\(M = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\).

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

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