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Chapter 4: Q1E (page 191)

Verify that the signals in Exercises 1 and 2 are solutions of the accompanying difference equation.

\({{\bf{2}}^k}\),\({\left( { - {\bf{4}}} \right)^k}\);\({y_{k + {\bf{2}}}} + {\bf{2}}{y_{k + {\bf{1}}}} - {\bf{8}}{y_k} = {\bf{0}}\)

Short Answer

Expert verified

\({2^k}\), \({\left( { - 4} \right)^k}\) are the solutions of the difference equation \({y_{k + 2}} + 2{y_{k + 1}} - 8{y_k} = 0\).

Step by step solution

01

Check the given difference equation for \({{\bf{2}}^k}\)

If\({2^k}\)is the solution,

\({y_{k + 2}} = {2^{k + 2}}\),\({y_{k + 1}} = {2^{k + 1}}\)and\({y_k} = {2^k}\).

By the difference equation, you get

\(\begin{aligned}{c}{2^{k + 2}} + 2\left( {{2^{k + 1}}} \right) - 8\left( {{2^k}} \right) &= 0\\{2^k}\left( {{2^2} + {2^2} - 8} \right) &= 0\\{2^k}\left( {8 - 8} \right) &= 0.\end{aligned}\)

So,\({2^k}\)is the solution of the given difference equation.

02

Check the given difference equation for \({\left( { - {\bf{4}}} \right)^k}\)

If\({\left( { - 4} \right)^k}\)is the solution,

\({y_{k + 2}} = {\left( { - 4} \right)^{k + 2}}\), \({y_{k + 1}} = {\left( { - 4} \right)^{k + 1}}\) and \({y_k} = {\left( { - 4} \right)^k}\).

By the difference equation, you get

\(\begin{aligned} {\left( { - 4} \right)^{k + 2}} + 2\left[ {{{\left( { - 4} \right)}^{k + 1}}} \right] - 8\left[ {{{\left( { - 4} \right)}^k}} \right] &= 0\\{\left( { - 4} \right)^k}\left( {{{\left( { - 4} \right)}^2} + 2\left( { - 4} \right) - 8} \right) &= 0\\{\left( { - 4} \right)^k}\left( {16 + 8 - 8} \right) &= 0.\end{aligned}\)

So, \({\left( { - 4} \right)^k}\) is the solution of the difference equation.

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Most popular questions from this chapter

Consider the polynomials \({{\bf{p}}_{\bf{1}}}\left( t \right) = {\bf{1}} + t\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = {\bf{1}} - t\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{4}}\), \({{\bf{p}}_{\bf{4}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), and \({{\bf{p}}_{\bf{5}}}\left( t \right) = {\bf{1}} + {\bf{2}}t + {t^{\bf{2}}}\), and let H be the subspace of \({P_{\bf{5}}}\) spanned by the set \(S = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}},\,{{\bf{p}}_{\bf{4}}},\,{{\bf{p}}_{\bf{5}}}} \right\}\). Use the method described in the proof of the Spanning Set Theorem (Section 4.3) to produce a basis for H. (Explain how to select appropriate members of S.)

Let \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Find \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) such that \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right] = {{\mathop{\rm uv}\nolimits} ^T}\) .

Question 18: Suppose A is a \(4 \times 4\) matrix and B is a \(4 \times 2\) matrix, and let \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) represent a sequence of input vectors in \({\mathbb{R}^2}\).

  1. Set \({{\mathop{\rm x}\nolimits} _0} = 0\), compute \({{\mathop{\rm x}\nolimits} _1},...,{{\mathop{\rm x}\nolimits} _4}\) from equation (1), and write a formula for \({{\mathop{\rm x}\nolimits} _4}\) involving the controllability matrix \(M\) appearing in equation (2). (Note: The matrix \(M\) is constructed as a partitioned matrix. Its overall size here is \(4 \times 8\).)
  2. Suppose \(\left( {A,B} \right)\) is controllable and v is any vector in \({\mathbb{R}^4}\). Explain why there exists a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) in \({\mathbb{R}^2}\) such that \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation.

a. What is the dimension of range of T if T is one-to-one mapping? Explain.

b. What is the dimension of the kernel of T (see section 4.2) if T maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^m}\)? Explain.

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).
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