2.6: Second-Order Differential and Difference Equations

With our understanding of the functions \(e^x\), \(e^{jΘ}\), and the quadratic equation \(z^2 + \frac b a z + \frac c a =0\), 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

\[\frac {d^2} {dt^2} x(t)+a_1\frac d {dt} x(t)+a_2=0 \nonumber \]

Because the function \(e^{st}\) reproduces itself under differentiation, it is plausible to assume that x(t)=est is a solution to the differential equation. Let's try it:

\[\frac {d^2} {dt^2}(e^{st})+a_1\frac d {dt}(e^{st})+a_2(e^{st})=0 \nonumber \]

\[(s^2+a_1s+a_2)e^{st}=0 \nonumber \]

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

\[s^2+a_1s+a_2=0 \nonumber \]

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

\[s_{1,2}=−\frac {a_1} 2 ± \frac 1 2 \sqrt{a^2_1−4a_2} \nonumber \]

This means that our assumed solution works, provided \(s=s_1\) or \(s_2\). It is a fundamental result from the theory of differential equations that the most general solution for x(t) is a linear combination of these assumed solutions:

\[x(t)=A_1e^{s_1t}+A_2e^{s_2t} \nonumber \]

If \(a^2_1−4a_2\) is less than zero, then the roots \(s_1\) and \(s_2\) are complex:

\[s_{1,2}=−\frac {a_1} 2 ± j\frac 1 2 \sqrt{4a_2−a^2_1} \nonumber \]

Let's rewrite this solution as

\[s_{1,2}=σ±jω \nonumber \]

where σ and ω are the constants

\[σ=−\frac {a_1} 2 \nonumber \]

\[ω=\frac 1 2 \sqrt{4a_2−a^2_1} \nonumber \]

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

\[x(t)=A_1e^{σt}e^{jωt}+A_2e^{σt}e^{−jωt} \nonumber \]

If this solution is to be real, then the two terms on the right-hand side must be complex conjugates. This means that \(A_2=A^∗_1\) and the solution for x(t) is

\[x(t)=A_1e^{σt}e^{jωt}+A^∗_1e^{σt}e^{−Jωt} = 2\mathrm{Re} \{A_1e^{σt} e^{jωt}\} \nonumber \]

The constant \(A_1\) may be written as \(A_1=|A|e^{jφ}\). Then the solution for x(t) is

\[x(t)=2|A|e^{σt}\cos(ωt+φ) \nonumber \]

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

The Solution to a Second-Order Differential Equation

Exercise \(\PageIndex{1}\)

Find the general solutions to the following differential equations:

a. \(\frac {d^2} {dt^2} X(t)+2\frac d {dt} x(t)+2=0\)
b. \(\frac {d^2} {dt^2} x(t)+2\frac d {dt} x(t)−2=0\)
c. \(\frac {d^2} {dt^2} x(t)+2=0\)

Difference Equations

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

\[x_n+a_1x_n−1+a_2x_{n−2}=0 \nonumber \]

What this means is that the sequence \(\{x_n\}\) obeys a homogeneous recursion:

\[x_n=−a_1x_{n−1}−a_2x_{n−2} \nonumber \]

A plausible guess at a solution is the geometric sequence \(x_n=z^n\). With this guess, the difference equation produces the result

\[z^n+a_1z^n−1+a_2z^{n−2}=0 \nonumber \]

\[(1+a_1z^{−1}+a_2z^{−2})z^n=0 \nonumber \]

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

\[1+a_1z^{−1}+a_2z^{−2}=0 \nonumber \]

\[z^2+a_1z+a_2=0 \nonumber \]

The solutions are

\[z_{1,2}=−\frac {a_1} 2 ± j\frac 1 2 \sqrt{4a_2−a^2_1} = re^{jθ} \nonumber \]

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

\[x_n=A_1z^n_1+A_2(z^∗_1)^n \nonumber \]

\[=A_1z^n_1+A^∗_1(z^∗_1) \nonumber \]

\[=2\mathrm{Re}{A_1z^n_1} \nonumber \]

\[=2|A|r^n\cos(θn+φ) \nonumber \]

This general solution is illustrated in the Figure.

The Solution to a Second-Order Difference Equation

Exercise \(\PageIndex{2}\)

Find the general solutions to the following difference equations:

a. x_n+2x_{n−1}+2=0
b. x_n−2x_{n−1}+2=0
c. x_n+2x_{n−2}=0