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In Exercises 25-28, show that the given signal is a solution of the difference equation. Then find the general solution of that difference equation.

\({y_k} = {\bf{2}}k - {\bf{4}}\); \({y_{k + {\bf{2}}}} + \frac{{\bf{3}}}{{\bf{2}}}{y_{k + {\bf{1}}}} - {y_k} = {\bf{1}} + {\bf{3}}k\)

Short Answer

Expert verified

The given signal is the solution of difference equation, and the general solution of the difference equation is \(y = {c_1}{\left( {\frac{1}{2}} \right)^k} + {c_2}{\left( { - 2} \right)^k} + \left( {2k - 4} \right)\).

Step by step solution

01

Substitute \({\bf{1}} + k\) for \({y_k}\) in the difference equation

Use \({y_k} = {k^2}\) in the difference equation.

\(\begin{aligned} {y_{k + 2}} + \frac{3}{2}{y_{k + 1}} - {y_k} &= \left( {2\left( {k + 2} \right) - 4} \right) + \frac{3}{2}\left( {2\left( {k + 1} \right) - 4} \right) - \left( {2k - 4} \right)\\ &= 2k + 3k - 3 - 2k + 4\\ &= 1 + 3k\end{aligned}\)

So, the signal \({y_k} = 2k - 4\) is the solution of the given the difference equation.

02

Solve the auxiliary equation for the difference equation

The auxiliary equation for the difference equation \({y_{k + 2}} + \frac{3}{2}{y_{k + 1}} - {y_k} = 1 + 3k\) is obtained below:

\(\begin{aligned} {r^{k + 2}} + \frac{3}{2}{r^{k + 1}} - {r^k} &= 0\\{r^k}\left( {{r^2} + \frac{3}{2}r - 1} \right) &= 0\\2{r^2} + 3r - 2 &= 0\\2{r^2} + 4r - r - 2 &= 0\\\left( {r + 2} \right)\left( {2r - 1} \right) &= 0\\r &= - 2,\,\frac{1}{2}\end{aligned}\)

So, the general solutions of the auxiliary set are \( - {2^k}\) and \({\left( {\frac{1}{2}} \right)^k}\).

03

Find the general solution

The general solution of the difference equation is

\(y = {c_1}{\left( {\frac{1}{2}} \right)^k} + {c_2}{\left( { - 2} \right)^k} + \left( {2k - 4} \right)\).

So, the given signal is the solution of the difference equation, and the general solution of the difference equation is \(y = {c_1}{\left( {\frac{1}{2}} \right)^k} + {c_2}{\left( { - 2} \right)^k} + \left( {2k - 4} \right)\).

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Most popular questions from this chapter

Use Exercise 28 to explain why the equation\(Ax = b\)has a solution for all\({\rm{b}}\)in\({\mathbb{R}^m}\)if and only if the equation\({A^T}x = 0\)has only the trivial solution.

Let V be a vector space that contains a linearly independent set \(\left\{ {{u_{\bf{1}}},{u_{\bf{2}}},{u_{\bf{3}}},{u_{\bf{4}}}} \right\}\). Describe how to construct a set of vectors \(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\) in V such that \(\left\{ {{v_{\bf{1}}},{v_{\bf{3}}}} \right\}\) is a basis for Span\(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\).

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

20. \(A = \left( {\begin{array}{*{20}{c}}{.8}&{ - .3}&0\\{.2}&{.5}&1\\0&0&{ - .5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right)\).

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

13. Show that if \(P\) is an invertible \(m \times m\) matrix, then rank\(PA\)=rank\(A\).(Hint: Apply Exercise12 to \(PA\) and \({P^{ - 1}}\left( {PA} \right)\).)

In Exercise 3, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

3. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\\{\bf{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{2}}\\{ - {\bf{2}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{0}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{0}}\\{ - {\bf{1}}}\end{array}} \right)\)

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