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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

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

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\)as\(n \times 1\)matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is a \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

28. If u and v are in \({\mathbb{R}^n}\), how are \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \) related? How are \({{\mathop{\rm uv}\nolimits} ^T}\) and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) related?

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 \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

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 \({\mathbb{R}^n}\).
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