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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

Suppose \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), \({{\bf{p}}_{\bf{3}}}\), and \({{\bf{p}}_{\bf{4}}}\) are specific polynomials that span a two-dimensional subspace H of \({P_{\bf{5}}}\). Describe how one can find a basis for H by examining the four polynomials and making almost no computations.

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

17. A submatrix of a matrix A is any matrix that results from deleting some (or no) rows and/or columns of A. It can be shown that A has rank \(r\) if and only if A contains an invertible \(r \times r\) submatrix and no longer square submatrix is invertible. Demonstrate part of this statement by explaining (a) why an \(m \times n\) matrix A of rank \(r\) has an \(m \times r\) submatrix \({A_1}\) of rank \(r\), and (b) why \({A_1}\) has an invertible \(r \times r\) submatrix \({A_2}\).

The concept of rank plays an important role in the design of engineering control systems, such as the space shuttle system mentioned in this chapter’s introductory example. A state-space model of a control system includes a difference equation of the form

\({{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}\)for \(k = 0,1,....\) (1)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(\left\{ {{{\mathop{\rm x}\nolimits} _k}} \right\}\) is a sequence of “state vectors” in \({\mathbb{R}^n}\) that describe the state of the system at discrete times, and \(\left\{ {{{\mathop{\rm u}\nolimits} _k}} \right\}\) is a control, or input, sequence. The pair \(\left( {A,B} \right)\) is said to be controllable if

\({\mathop{\rm rank}\nolimits} \left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right) = n\) (2)

The matrix that appears in (2) is called the controllability matrix for the system. If \(\left( {A,B} \right)\) is controllable, then the system can be controlled, or driven from the state 0 to any specified state \({\mathop{\rm v}\nolimits} \) (in \({\mathbb{R}^n}\)) in at most \(n\) steps, simply by choosing an appropriate control sequence in \({\mathbb{R}^m}\). This fact is illustrated in Exercise 18 for \(n = 4\) and \(m = 2\). For a further discussion of controllability, see this text’s website (Case study for Chapter 4).

Suppose a nonhomogeneous system of nine linear equations in ten unknowns has a solution for all possible constants on the right sides of the equations. Is it possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other? Discuss.

If A is a \({\bf{7}} \times {\bf{5}}\) matrix, what is the largest possible rank of A? If Ais a \({\bf{5}} \times {\bf{7}}\) matrix, what is the largest possible rank of A? Explain your answer.

In Exercise 6, find the coordinate vector of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

6. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{6}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{0}}\end{array}} \right)\)

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