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4.6: Second-Order Differential and Difference Equations

  • Page ID
    9959
  • 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

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

    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:

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

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

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

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

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

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

    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:

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

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

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

    Let's rewrite this solution as

    \[s_{1,2}=σ±jω\]

    where σ and ω are the constants

    \[σ=−\frac {a_1} 2\]

    \[ω=\frac 1 2 \sqrt{4a_2−a^2_1}\]

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

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

    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

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

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

    \[x(t)=2|A|e^{σt}\cos(ωt+φ)\]

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

    DifEqSolution.PNG

    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

    \[x_n+a_1x_n−1+a_2x_{n−2}=0\]

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

    \[x_n=−a_1x_{n−1}−a_2x_{n−2}\]

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

    \[z^n+a_1z^n−1+a_2z^{n−2}=0\]

    \[(1+a_1z^{−1}+a_2z^{−2})z^n=0\]

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

    \[1+a_1z^{−1}+a_2z^{−2}=0\]

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

    The solutions are

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

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

    \[x_n=A_1z^n_1+A_2(z^∗_1)^n\]

    \[=A_1z^n_1+A^∗_1(z^∗_1)\]

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

    \[=2|A|r^n\cos(θn+φ)\]

    This general solution is illustrated in the Figure.

    DifEqSolution2.PNG
    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