Implementation of a harmonic oscillator¶

Laboratory Class 07 - 14 October 2021¶

Topics¶

Theoretical and experimental aspects linked to the development of a harmonic oscillator:
- general discussion on the difficulty of implementing an oscillator;
- from the differential equation of a forced oscillator to the z–transform of the simulator response function V (z);
- derivation of the difference equation;
- setting of the boundary conditions for the cosine operation;
- dependency of the working frequency \(f_0 = \frac{ω_0}{2\pi}\) on the parameter k, provided that \(ω_0T ≪ 1\): and \(f_0 = f_s/\left( 2^{\frac{k}{2}+1}\pi\right)\), with \(f_s = 1/T\).

Problems¶

implementation and characterization of a harmonic oscillator.

Harmonic Oscillator¶

Differential Equation¶

Let’s start by writing the differential equation into the “Laplace” space or “s-space”. In this case the Fourier transform is not enough because it can not handle initial conditions.

\[\begin{align*} \mathcal{L} \left\{ \ddot{x}+\omega_0 x\right\} &= s^2 \tilde{x}(s)-s x_0 - \dot{x}_0 + \omega_0^2\tilde{x}(s) = 0 \\ &\\ \tilde{x}(s)&=\frac{sx_0 + \dot{x}_0}{s^2+\omega_0^2} \end{align*}\]

Bilinear transform¶

It is possible to obtain the approximate transfer function by applyng the bilinear transform

\[\begin{align*} \omega = \frac{2i}{T}\frac{z-1}{z+1} + \mathcal{o}\left(\omega^3 T^3\right) \qquad s=-i\omega = + \frac{2}{T}\frac{z-1}{z+1} \end{align*}\]

After the sobstitution we obtain:

\[\begin{align*} V(z) &= \frac{Y(z)}{X(z)}= \frac{ (z+1)^2 T/4}{(z-1)^2 + (z+1)^2 \left(\frac{\omega T}{2}\right)}\\ \implies&Y(z)\left( 1-\frac{2c}{z}-\frac{1}{z^2}\right) = \frac{N(z)}{z^2} \\ \text{with}\qquad& c=(1-\frac{\omega_0^2 T^2}{4})/(1+\frac{\omega_0^2 T^2}{4}) &N(z)=\frac{(z+1)^2 T/4}{1+\frac{\omega_0^2 T^2}{4}} \end{align*}\]

By inverting the z-transform we get:

\[\begin{align*} y[n]-2cy[n-1]+y[n-2]=\text{boundary conditions}[n] \end{align*}\]

As an fpga works best with powers of two, \(c = 1-2^{k+1}\) is assumed.

Boundary conditions¶

Instead of actually computing the boundary condition, the fact that \(sin(x)\) is symmetric with respect to its maximum is used:

\[\begin{align*} y[-1] &= y[1] \\ y[1] &= (1-2^{-k+1}) y[0] = y[-1]\\ y[0] &= \text{boundary conditions}[n]= 0 \end{align*}\]

The value of \(y[1]\) was computed by setting \(\text{boundary conditions}[n]= 0\) and assuming an infinite wave.

Characterization of the oscillator¶

By inverting the relation

\[\begin{align*} c&=(1-\frac{\omega_0^2 T^2}{4})/(1+\frac{\omega_0^2 T^2}{4}) \\ \implies& \omega_0 = \frac{2}{T}\sqrt{\frac{1-c}{1+c}} \end{align*}\]

In terms of the k parameter:

\[\begin{align*} \omega_0 = \frac{2}{T}\sqrt{\frac{2^{-(k+1)}}{2+2^{-(k+1)}}} = \frac{2}{T}\sqrt{ \frac{1}{1+2^{k+2}} } \end{align*}\]

Cosine¶

The quadrature (cosine) WF can be obtained by making a second sinusoid starting when the other wf crosses zero.

Notes of Advanced Electronics