Second order difference equations

2 Second order difference equations

You will learn in this section about solving second order linear constant coefficient difference equations. In this case two initial conditions are required, typically either $y_{0}$ and $y_{1}$ or $y_{- 1}$ and $y_{- 2}$ . In the first case we use the left shift property of the z-transform, in the second case we use the right shift property. The same three basic steps are involved as in the first order case.

Task!

By solving

$y_{n + 2} = y_{n + 1} + y_{n}$ (12)

$y_{0} = y_{1} = 1$

obtain the general term $y_{n}$ of the Fibonacci sequence.

Begin by taking the z-transform of (12), using the left shift property. Then insert the initial conditions and solve the resulting algebraic equation for $Y (z)$ , the z-transform of ${y_{n}}$ :

$z^{2} Y (z) - z^{2} y_{0} - z y_{1} = z Y (z) - z y_{0} + Y (z)$ (taking z-transforms )

$z^{2} Y (z) - z^{2} - z = z Y (z) - z + Y (z)$ (inserting initial conditions)

$(z^{2} - z - 1) Y (z) = z^{2}$

$Y (z) = \frac{z^{2}}{z^{2} - z - 1}$ (solving for $Y (z)$ ). Now solve the quadratic equation $z^{2} - z - 1 = 0$ and hence factorize the denominator of $Y (z)$ :

$z^{2} - z - 1 = 0$

$∴ z = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$

so if $a = \frac{1 + \sqrt{5}}{2}, b = \frac{1 - \sqrt{5}}{2}$

$Y (z) = \frac{z^{2}}{(z - a) (z - b)}$ This form for $Y (z)$ often arises in solving second order difference equations. Write it in partial fractions and find $y_{n}$ , leaving $a$ and $b$ as general at this stage:

$Y (z) = z (\frac{z}{(z - a) (z - b)}) = \frac{A z}{z - a} + \frac{B z}{(z - b)} in partial fractions$

where $A = \frac{a}{a - b} and B = \frac{b}{b - a}$

Hence, taking inverse z-transforms

$y_{n} = A a^{n} + B b^{n} = \frac{1}{(a - b)} (a^{n + 1} - b^{n + 1})$ (13) Now complete the Fibonacci problem:

With $a = \frac{1 + \sqrt{5}}{2} b = \frac{1 - \sqrt{5}}{2}$ so $a - b = \sqrt{5}$

we obtain, using (13)

$y_{n} = \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{n + 1} - {(\frac{1 - \sqrt{5}}{2})}^{n}) n = 2, 3, 4, \dots$

for the $n^{th}$ term of the Fibonacci sequence. With an appropriate computational aid you could (i) check that this formula does indeed give the familiar sequence

${1, 1, 2, 3, 5, 8, 13, \dots}$

and (ii) obtain, for example, $y_{50}$ and $y_{100}$ .

Key Point 14

The inverse z-transform of

$Y (z) = \frac{z^{2}}{(z - a) (z - b)} a \neq b$ is $y_{n} = \frac{1}{(a - b)} (a^{n + 1} - b^{n + 1})$

Task!

Use the right shift property of z-transforms to solve the second order difference equation

$y_{n} - 7 y_{n - 1} + 10 y_{n - 2} = 0$ with $y_{- 1} = 16$ and $y_{- 2} = 5$ .

[Hint: the steps involved are the same as in the previous Task]

\begin{array}{rcl} Y (z) - 7 (z^{- 1} Y (z) + 16) & + & 10 (z^{- 2} Y (z) + 16 z^{- 1} + 5) = 0 \\ Y (z) (1 - 7 z^{- 1} + 10 z^{- 2}) & - & 112 + 160 z^{- 1} + 50 = 0 \\ Y (z) (\frac{z^{2} - 7 z + 10}{z^{2}}) & = & 62 - 160 z^{- 1} \end{array}

\begin{array}{rcl} Y (z) & = & \frac{62 z^{2}}{z^{2} - 7 z + 10} - \frac{160 z}{z^{2} - 7 z + 10} \\ = & z \frac{(62 z - 160)}{(z - 2) (z - 5)} \\ = & \frac{12 z}{z - 2} + \frac{50 z}{z - 5} in partial fractions \\ so y_{n} & = & 12 \times 2^{n} + 50 \times 5^{n} n = 0, 1, 2, \dots \end{array}

We now give an Example where a quadratic equation with repeated solutions arises.

Example 1

Obtain the z-transform of ${f_{n}} = {n a^{n}}$ .
Solve
$\begin{matrix} y_{n} - 6 y_{n - 1} + 9 y_{n - 2} = 0 & n = 0, 1, 2, \dots \\ y_{- 1} = 1 y_{- 2} = 0 \end{matrix}$

[Hint: use the result from (1) at the inversion stage.]

Solution

$Z {n} = \frac{z}{{(z - 1)}^{2}}$ $∴ Z {n a^{n}} = \frac{z ∕ a}{{(z ∕ a - 1)}^{2}} = \frac{a z}{{(z - a)}^{2}}$ where we have used the property $Z {f_{n} a^{n}} = F (\frac{z}{a})$
Taking the z-transform of the difference equation and inserting the initial conditions:
$Y (z) - 6 (z^{- 1} Y (z) + 1) + 9 (z^{- 2} Y (z) + z^{- 1}) = 0$

$Y (z) (1 - 6 z^{- 1} + 9 z^{- 2}) = 6 - 9 z^{- 1}$

$Y (z) (z^{2} - 6 z + 9) = 6 z^{2} - 9 z$

$Y (z) = \frac{6 z^{2} - 9 z}{{(z - 3)}^{2}} = z (\frac{6 z - 9}{{(z - 3)}^{2}}) = z (\frac{6}{z - 3} + \frac{9}{{(z - 3)}^{2}}) in partial fractions$

from which, using the result (1) on the second term,

$y_{n} = 6 \times 3^{n} + 3 n \times 3^{n} = (6 + 3 n) 3^{n}$

We shall re-do this inversion by an alternative method shortly.

Task!

Solve the difference equation

$y_{n + 2} + y_{n} = 0$ with $y_{0}, y_{1}$ arbitrary. (14)

Start by obtaining $Y (z)$ using the left shift theorem:

\begin{array}{rcl} z^{2} Y (z) - z^{2} y_{0} - z y_{1} + Y (z) & = & 0 \\ (z^{2} + 1) Y (z) & = & z^{2} y_{0} + z y_{1} \\ Y (z) & = & \frac{z^{2}}{z^{2} + 1} y_{0} + \frac{z}{z^{2} + 1} y_{1} \end{array}

To find the inverse z-transforms recall the results for $Z {cos ω n}$ and $Z {sin ω n}$ from Key Point 6 (page 21) and some of the particular cases discussed in Section 21.2. Hence find $y_{n}$ here:

Taking $Z {cos ω n}$ and $Z {sin ω n}$ with $ω = \frac{π}{2}$

\begin{array}{rcl} Z \{cos (\frac{n π}{2})\} & = & \frac{z^{2}}{z^{2} + 1} \\ Z \{sin (\frac{n π}{2})\} & = & \frac{z}{z^{2} + 1} \end{array}

Hence $y_{n} = y_{0} ℤ^{- 1} {\frac{z^{2}}{z^{2} + 1}} + y_{1} ℤ^{- 1} {\frac{z}{z^{2} + 1}} = y_{0} cos (\frac{n π}{2}) + y_{1} sin (\frac{n π}{2})$ (15) Those of you who are familiar with differential equations may know that

$\frac{d^{2} y}{d t^{2}} + y = 0 y (0) = y_{0}, y^{'} (0) = y_{0}^{'}$ (16)

has solutions $y_{1} = cos t$ and $y_{2} = sin t$ and a general solution

$y = c_{1} cos t + c_{2} sin t$ (17)

where $c_{1} = y_{0}$ and $c_{2} = y_{0}^{'}$ .

This differential equation is a model for simple harmonic oscillations. The difference equation (14) and its solution (15) are the discrete counterparts of (16) and (17).

2 Second order difference equations

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Key Point 14

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Example 1

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