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Chapter 4: Q32E (page 191)

What is the order of the following difference equation? Explain your answer\({y_{k + 3}} + {a_1}{y_{k + 2}} + {a_2}{y_{k + 1}} + {a_3}{y_k} = 0\).

Short Answer

Expert verified

If \({a_3} = 0\) and \({a_2} \ne 0\), the order is 2.

If \({a_3} \ne 0\), the order is 3.

If \({a_3} = {a_2} = 0\) and \({a_1} \ne 0\), the order is 1.

If \({a_3} = {a_2} = {a_1} = 0\), the order is 0.

Step by step solution

01

Discuss the attributes of a difference equation

The given difference equation is \({y_{k + 3}} + {a_1}{y_{k + 2}} + {a_2}{y_{k + 1}} + {a_3}{y_k} = 0\). The order of a difference equation depends on the values of the coefficients \({a_1}\), \({a_2}\), and \({a_3}\).

02

Discuss the cases corresponding to the values of the coefficients

If \({a_3} = 0\) and \({a_2} \ne 0\), then the difference equation can be reduced to \({y_{k + 2}} + {a_1}{y_{k + !}} + {a_2}{y_k} = 0\), which implies that its order is 2.

If \({a_3} \ne 0\), then the difference equation cannot be reduced further, which implies that its order is 3.

If \({a_3} = {a_2} = 0\) and \({a_1} \ne 0\), then the difference equation can be reduced to \({y_{k + 1}} + {a_1}{y_k} = 0\), which implies that its order is 1.

If all the coefficients are zero, that is, \({a_3} = {a_2} = {a_1} = 0\), then the difference equation does not possess any signal value, which implies that its order is 0.

03

Draw a conclusion

If \({a_3} = 0\) and \({a_2} \ne 0\), the order is 2.

If \({a_3} \ne 0\), the order is 3.

If \({a_3} = {a_2} = 0\) and \({a_1} \ne 0\), the order is 1.

If \({a_3} = {a_2} = {a_1} = 0\), the order is 0.

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Most popular questions from this chapter

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 \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

Find a basis for the set of vectors in\({\mathbb{R}^{\bf{2}}}\)on the line\(y = {\bf{5}}x\).

Define by \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 1 \right)}\end{array}} \right)\). For instance, if \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + 5t + 7{t^2}\), then \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}3\\{15}\end{array}} \right)\).

  1. Show that \(T\) is a linear transformation. (Hint: For arbitrary polynomials p, q in \({{\mathop{\rm P}\nolimits} _2}\), compute \(T\left( {{\mathop{\rm p}\nolimits} + {\mathop{\rm q}\nolimits} } \right)\) and \(T\left( {c{\mathop{\rm p}\nolimits} } \right)\).)
  2. Find a polynomial p in \({{\mathop{\rm P}\nolimits} _2}\) that spans the kernel of \(T\), and describe the range of \(T\).

Question 11: Let \(S\) be a finite minimal spanning set of a vector space \(V\). That is, \(S\) has the property that if a vector is removed from \(S\), then the new set will no longer span \(V\). Prove that \(S\) must be a basis for \(V\).

In Exercise 5, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

5. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\end{array}} \right)\)
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