Periodic Driving Forces

In the last section, we investigated the response of a one-dimensional dynamical system, close to a stable equilibrium point, to an external force which varies as $\cos(\omega\,t)$. Let us now examine the response of the same system to a more complicated external force.

Consider a general external force which is periodic in time, with period $T$. By analogy with Eq. (182), we can write such a force as

$$f_{ext}(t) = m\,\omega_0^{\,2}\,X(t),$$ (190)

where

$$X(t+T) = X(t)$$ (191)

for all $t$.

Now, we can represent $X(t)$ as a Fourier series in time. In other words, we can write

$$X(t) = \sum_{n=0}^{\infty} X_{n}\,\cos(n\,\omega\,t),$$ (192)

where $\omega = 2\pi/T$. By writing $X(t)$ in this form, we automatically satisfy the periodicity constraint (191). [Note that by choosing a cosine Fourier series we are limited to even functions in $t$: i.e., $X(-t)=X(t)$. Odd functions in $t$ can be represented by sine Fourier series, and mixed functions require a combination of cosine and sine Fourier series.] The constant coefficients $X_n$ are known as Fourier coefficients. But, how do we determine these coefficients for a given functional form, $X(t)$?

Well, it follows from the periodicity of the cosine function that

$$\frac{1}{T} \int_0^T \cos (n\,\omega\,t)\,dt = \delta_{n\,0},$$ (193)

where $\delta_{n\,n'}$ is unity if $n=n'$, and zero otherwise, and is known as the Kronecker delta function. Thus, integrating Eq. (192) over $t$ from $t=0$ to $t=T$, and making use of Eq. (193), we obtain

$$X_0 = \frac{1}{T}\int_0^T X(t) \,dt.$$ (194)

It is also easily demonstrated that

$$\frac{2}{T} \int_0^T \cos(n\,\omega\,t)\,\cos(n'\,\omega\,t) \,dt= \delta_{n\,n'},$$ (195)

provided $n, n'>0$. Thus, multiplying Eq. (192) by $\cos(n\,\omega\,t)$, integrating over $t$ from $t=0$ to $t=T$, and making use of Eqs. (193) and (195), we obtain

$$X_n = \frac{2}{T} \int_0^T X(t)\,\cos(n\,\omega\,t)\,dt$$ (196)

for $n>0$. Hence, we have now determined the Fourier coefficients of the general periodic function $X(t)$.

We can incorporate the periodic external force (190) into our perturbed equation of motion by writing

$$\frac{d^2 x}{dt^2} + 2\,\nu\,\frac{dx}{dt} + \omega_0^{\,2}\... ...\,2}\,\sum_{n=0}^\infty X_n\,{\rm e}^{-{\rm i}\,n\,\omega\,t},$$ (197)

where we are again using the convention that the physical solution corresponds to the real part of the complex solution. Note that the above differential equation is linear. This means that if $x_a(t)$ and $x_b(t)$ represent two independent solutions to this equation then any linear combination of $x_a(t)$ and $x_b(t)$ is also a solution. We can exploit the linearity of the above equation to write the solution in the form

$$x(t) = \sum_{n=0}^\infty X_n\,a_n\,{\rm e}^{-{\rm i}\,n\,\omega\,t},$$ (198)

where the $a_n$ are the complex amplitudes of the solutions to

$$\frac{d^2 x}{dt^2} + 2\,\nu\,\frac{dx}{dt} + \omega_0^{\,2}\,x = \omega_0^{\,2}\,{\rm e}^{-{\rm i}\,n\,\omega\,t}.$$ (199)

In other words, $a_n$ is obtained by substituting $x = a_n\,\exp(-{\rm i}\,n\,\omega\,t)$ into the above equation. Hence, it follows that

$$a_n = \frac{\omega_0^{\,2}}{\omega_0^{\,2} - n^2\,\omega^2 - {\rm i}\,2\,\nu\,n\,\omega}.$$ (200)

Thus, the physical solution takes the form

$$x(t) = \sum_{n=0}^{\infty} X_n\,x_n\,\cos(n\,\omega\,t-\phi_n),$$ (201)

where

$$a_n = x_n \,{\rm e}^{\,{\rm i}\,\phi_n},$$ (202)

and $x_n$ and $\phi_n$ are real parameters. It follows from Eq. (200) that

$$x_n = \frac{\omega_0^{\,2}}{\left[(\omega_0^{\,2}-n^2\,\omega^2)^2 + 4\,\nu^2\,n^2\,\omega^2\right]^{1/2}},$$ (203)

and

$$\phi_n = \tan^{-1}\left(\frac{2\,\nu\,n\,\omega}{\omega_0^{\,2}-n^2\,\omega^2}\right).$$ (204)

We have now fully determined the response of our dynamical system to a general periodic driving force.

As an example, suppose that the external force periodically delivers a brief kick to the system. For instance, let $X(t) = A$ for $0\leq t\leq T/10$ and $9\,T/10<t<T$, and $X(t)=0$ otherwise (in the period $0\leq t\leq T$). It follows from Eq. (194) and (196) that, in this case,

$$X_0 = 0.2\,A,$$ (205)

and

$$X_n = \frac{2\,\sin(n\,\pi/5)\,A}{n\,\pi},$$ (206)

for $n>0$. Obviously, to obtain an exact solution, we would have to include every Fourier harmonic in Eq. (201), which is impractical. However, we can obtain a fairly accurate approximate solution by truncating the Fourier series (i.e., by neglecting all the terms with $n>N$, where $N\gg 1$).

Figure 22: Periodic forcing.

Figure 22 shows an example calculation in which the Fourier series is truncated after 100 terms. The parameters used in this calculation are $\omega = 1.2\,\omega_0$ and $\nu= 0.8\,\omega_0$. The left panel shows the Fourier reconstruction of the driving force, $X(t)$. The glitches at the rising and falling edges of the pulses are called Gibbs phenomena, and are an inevitable consequence of attempting to represent a discontinuous periodic function as a Fourier series. The right panel shows the Fourier reconstruction of the response, $x(t)$, of the dynamical system to the applied force.