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The Functions e^x and e^jθ: Second-Order Differential and Difference Equations

Module by: Louis Scharf. E-mail the author

Note:

This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

With our understanding of the functions e x e x , e j Θ e j Θ , and the quadratic equation z 2 + b a z + c a = 0 z 2 + b a z+ c a =0, we can undertake a rudimentary study of differential and difference equations.

Differential Equations. In your study of circuits and systems you will encounter the homogeneous differential equation

d2dt2x(t)+a1ddtx(t)+a2=0.d2dt2x(t)+a1ddtx(t)+a2=0.
(1)

Because the function estest reproduces itself under differentiation, it is plausible to assume that x(t)=estx(t)=est is a solution to the differential equation. Let's try it:

d 2 d t 2 ( e s t ) + a 1 d d t ( e s t ) + a 2 ( e s t ) = 0 ( s 2 + a 1 s + a 2 ) e s t = 0 . d 2 d t 2 ( e s t ) + a 1 d d t ( e s t ) + a 2 ( e s t ) = 0 ( s 2 + a 1 s + a 2 ) e s t = 0 .
(2)

If this equation is to be satisfied for all t t, then the polynomial in s s must be zero. Therefore we require

s2+a1s+a2=0.s2+a1s+a2=0.
(3)

As we know from our study of this quadratic equation, the solutions are

s1,2=-a12±12a12-4a2.s1,2=-a12±12a12-4a2.
(4)

This means that our assumed solution works, provided s=s1s=s1 or s 2 s 2 . It is a fundamental result from the theory of differential equations that the most general solution for x(t)x(t) is a linear combination of these assumed solutions:

x(t)=A1es1t+A2es2t.x(t)=A1es1t+A2es2t.
(5)

If a12-4a2a12-4a2 is less than zero, then the roots s 1 s 1 and s 2 s 2 are complex:

s1,2=-a12±j124a2-a12.s1,2=-a12±j124a2-a12.
(6)

Let's rewrite this solution as

s 1 , 2 = σ ± j ω s 1 , 2 = σ ± j ω
(7)

where σ and ω are the constants

σ = - a 1 2 σ = - a 1 2
(8)
ω=124a2-a12.ω=124a2-a12.
(9)

With this notation, the solution for x(t)x(t) is

x(t)=A1eσtejωt+A2eσte-jtωt.x(t)=A1eσtejωt+A2eσte-jtωt.
(10)

If this solution is to be real, then the two terms on the right-hand side must be complex conjugates. This means that A2=A1*A2=A1* and the solution for x(t)x(t) is

x ( t ) = A 1 e σ t e j ω t + A 1 * e σ t e - J ω t = 2 Re { A 1 e σ t e j ω t } . x ( t ) = A 1 e σ t e j ω t + A 1 * e σ t e - J ω t = 2 Re { A 1 e σ t e j ω t } .
(11)

The constant A 1 A 1 may be written as A1=|A|ejφA1=|A|ejφ. Then the solution for x(t)x(t) is

x(t)=2|A|eσtcos(ωt+φ).x(t)=2|A|eσtcos(ωt+φ).
(12)

This “damped cosinusoidal solution” is illustrated in Figure 1.

Figure 1: The Solution to a Second-Order Differential Equation
Figure one is a graph with unlabeled axes, with the horizontal axis  ranging in value from 0 to 10 in increments of one, and the vertical axis ranging in value from -20 to 20 in increments of 5. There are three curves on this graph. The first begins at the upper-left corner of the graph, at (0, 20), and is drawn decreasing from left to right at a decreasing rate, until, by (8, 0) it is nearly indiscernible from the horizontal axis, where it continues and terminates at (10, 0). The second curve is a mirror image of the first across the horizontal axis, beginning in the lower-left corner of the graph and continuing to the right until it is horizontal along the axis. The third curve seems to follow a wave-like pattern that is bound by the two curves above and below. The curve begins at (0, 10) and starts in a positive direction, then moves right with five peaks and troughs to the right until it is nearly a horizontal line by the far right of the graph. Along these graphs is an expression that reads 2*abs(A)*exp(sigma*t), and the entire graph has a small label that reads |A|=10

Exercise 1

Find the general solutions to the following differential equations:

  1. d 2 d t 2 X ( t ) + 2 d d t x ( t ) + 2 = 0 d 2 d t 2 X ( t ) + 2 d d t x ( t ) + 2 = 0 ;
  2. d 2 d t 2 x ( t ) + 2 d d t x ( t ) - 2 = 0 d 2 d t 2 x ( t ) + 2 d d t x ( t ) - 2 = 0 ;
  3. d 2 d t 2 x ( t ) + 2 = 0 d 2 d t 2 x ( t ) + 2 = 0 .

Difference Equations. In your study of digital filters you will encounter homogeneous difference equations of the form

xn+a1xn-1+a2xn-2=0.xn+a1xn-1+a2xn-2=0.
(13)

What this means is that the sequence{xn}{xn} obeys a homogeneous recursion:

x n = - a 1 x n - 1 - a 2 x n - 2 . x n = - a 1 x n - 1 - a 2 x n - 2 .
(14)

A plausible guess at a solution is the geometric sequence xn=znxn=zn. With this guess, the difference equation produces the result

z n + a 1 z n - 1 + a 2 z n - 2 = 0 ( 1 + a 1 z - 1 + a 2 z - 2 ) z n = 0 . z n + a 1 z n - 1 + a 2 z n - 2 = 0 ( 1 + a 1 z - 1 + a 2 z - 2 ) z n = 0 .
(15)

If this guess is to work, then the second-order polynomial on the left-hand side must equal zero:

1 + a 1 z - 1 + a 2 z - 2 = 0 1 + a 1 z - 1 + a 2 z - 2 = 0

1 + a 1 z - 1 + a 2 z - 2 = 0 z 2 + a 1 z + a 2 = 0 . 1 + a 1 z - 1 + a 2 z - 2 = 0 z 2 + a 1 z + a 2 = 0 .
(16)

The solutions are

z 1 , 2 = - a 1 2 ± j 1 2 4 a 2 - a 1 2 = r e j θ . z 1 , 2 = - a 1 2 ± j 1 2 4 a 2 - a 1 2 = r e j θ .
(17)

The general solution to the difference equation is a linear combination of the assumed solutions:

x n = A 1 z 1 n + A 2 ( z 1 * ) n = A 1 z 1 n + A 1 * ( z 1 * ) n = 2 Re { A 1 z 1 n } = 2 | A | r n cos ( θ n + φ ) . x n = A 1 z 1 n + A 2 ( z 1 * ) n = A 1 z 1 n + A 1 * ( z 1 * ) n = 2 Re { A 1 z 1 n } = 2 | A | r n cos ( θ n + φ ) .
(18)

This general solution is illustrated in Figure 2.

Figure 2: The Solution to a Second-Order Difference Equation
Figure two is a graph with unlabeled axes, with the horizontal axis  ranging in value from 0 to 10 in increments of one, and the vertical axis ranging in value from -20 to 20 in increments of 5. There are two curves on this graph. The first begins at the upper-left corner of the graph, at (0, 20), and is drawn decreasing from left to right at a decreasing rate, until, by (8, 0) it is nearly indiscernible from the horizontal axis, where it continues and terminates at (10, 0). The second curve is a mirror image of the first across the horizontal axis, beginning in the lower-left corner of the graph and continuing to the right until it is horizontal along the axis. Inside these two curves are various vertical lines beginning from the horizontal axis that follow a wave-like pattern, with five positive groups of lines and five negative groups of lines. The figure has an expression near the curves that reads, 2*abs(A)*r.^n). A second expression above the curves reads |A|=10.

Exercise 2

Find the general solutions to the following difference equations:

  1. xn+2xn-1+2=0xn+2xn-1+2=0;
  2. xn-2xn-1+2=0xn-2xn-1+2=0;
  3. xn+2xn-2=0xn+2xn-2=0.

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