2403 General Case

It is actually just as easy to write out the formula for the Fourier series expansion of the steady-periodic solution $x_{sp}(t)$ to the general secondorder LTI DE $p(D)x = f(t)$ with $f(t)$ periodic as it was to work out the previous example - the only difference is that now we use letters instead of numbers. We will choose the letters used for the spring-mass-dashpot system, but clearly the derivation and formulas will work with any three parameters. For simplicity we will take the case of $f(t)$ even (i.e. cosine series).

Problem: Solve $mx''+bx'+kx=f(t)$, for the steady-periodic response $x_{sp}(t)$, where $f(t)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos(n\frac{\pi}{L}t)dt$

Solution
Characteristic polynomial: $p(s) = ms^2 + bs + k$.
Solving for the component pieces: $$mx_n''+bx_n'+kx_n=\cos(n\frac{\pi}{L}t)$$ For $n=0$ we get $x_{0,p}=\frac{1}{k}$.
For $n\geq1$:
Complex replacement: $$mz_n''+bz_n'+kz_n=e^{in\frac{\pi}{L}t}, x_n=Re(z_n)$$ Exponential Response formula: $$z_{n,p}(t)=\frac{e^{in\frac{\pi}{L}t}}{p(in\frac{\pi}{L})}$$ Polar coords: $$p(in\frac{\pi}{L})=(k-m(n\frac{\pi}{L})^2)+ibn\frac{\pi}{L}=|p(in\frac{\pi}{L})|e^{i\phi_n}$$ where $$|p(in\frac{\pi}{L})|=\sqrt{(k-m(n\frac{\pi}{L})^2)^2+b^2(n\frac{\pi}{L})^2}$$ and $$\phi_n=Arg(p(in\frac{\pi}{L}))=\tan^{-1}(\frac{bn\frac{\pi}{L}}{k-m(n\frac{\pi}{L})^2})\tag{phase lag}$$ Thus, $$z_{n,p}(t)=g_ne^{i(n\frac{\pi}{L}t-\phi_n)}$$ with $$g_n=\frac{1}{|p(in\frac{\pi}{L})|}\tag{gain}$$ Taking the real part of $x_{n,p}$ we get $$x_{n,p}(t)=g_n\cos(n\frac{\pi}{L}t-\phi_n)$$ Now using superposition and putting back in the coefficients $a_n$ we get: $$x_{sp}(t)=\frac{a_0}{2}x_{0,p}+\sum_{n=1}^\infty a_n x_{n,p}(t)=\frac{a_0}{2k}+\sum_{n=1}^\infty a_n g_n \cos(n\frac{\pi}{L}t-\phi_n)$$ This is the general formula for the steady periodic response of a secondorder LTI DE to an even periodic driver $f(t)$.