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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

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.)

Let Abe an invertible \(n \times n\) matrix, and let B be an \(n \times p\) matrix. Show that the equation \(AX = B\) has a unique solution \({A^{ - 1}}B\).

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

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.

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Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?
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