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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

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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

Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

Suppose Ais an \(n \times n\) matrix with the property that the equation \(Ax = 0\)has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation \(Ax = b\) must have a solution for each b in \({\mathbb{R}^n}\).

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.]

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

Assume \(A - s{I_n}\) is invertible and view (8) as a system of two matrix equations. Solve the top equation for \({\bf{x}}\) and substitute into the bottom equation. The result is an equation of the form \(W\left( s \right){\bf{u}} = {\bf{y}}\), where \(W\left( s \right)\) is a matrix that depends upon \(s\). \(W\left( s \right)\) is called the transfer function of the system because it transforms the input \({\bf{u}}\) into the output \({\bf{y}}\). Find \(W\left( s \right)\) and describe how it is related to the partitioned system matrix on the left side of (8). See Exercise 15.

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]
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