Driving vibrations with sawtooth waves

The previous post looked at driving a vibrating system with square waves rather than the more customary sine waves.

You could think of a square wave as a crude approximation to a sine wave. A sawtooth wave is another crude approximation to a sine wave, and so it would be interesting to see how systems driven by a sawtooth forcing function differ from those driven by a square or sinusoidal forcing function.

Sine wave, sawtooth wave, and square wave

With a square wave, the differences were most pronounced at low frequencies, i.e. when the driving frequency was low relative to the natural frequency. The same is true for sawtooth waves, but with some interesting differences.

As before we will look at the equation

u'' + u = sin(ωt)

and with sine replaced with a variation, this time a sawtooth wave rather than a square. We will use initial conditions u(0) = u‘(0) = 1.

We start with ω = 1, i.e. driving the system at its natural frequency.

Resonance with sawtooth forcing

We see the resonance we’d expect when driving a system at its natural frequency. The amplitudes are lower for the sawtooth forcing function than the sinusoidal forcing function.

When we back away a little from the resonant frequency, setting ω = 0.9, we see beats.

Sawtooth wave beats

When we reduce ω to 0.5, we get resonance again for the sawtooth forcing solutions.

Beats for sawtooth-driven solution

At ω = 0.4 we see something like beats again.

Sawtooth driven beats

And at ω = 1/3 it looks like we have resonance again for the sawtooth-driven system.

Sawtooth driven resonance

We also see beats at ω = 1/n for all integer n. I’ve tried this for n up to 12. But at other frequencies that are not reciprocals of integers I see periodic solutions.

I have only investigated this numerically. Do any of you know of analytical results that apply here?

Update: There’s a simple reason why driving frequencies of 1/n do indeed create resonance: The Fourier series for the driving function has a component at the resonant frequency. Thanks to gmvh for pointing this out.

The square wave has a similar resonance at driving frequency 1/n, but only when n is an odd number. I missed that because I didn’t happen to try any frequencies of that form. Here’s an example with n = 5.

2 thoughts on “Driving vibrations with sawtooth waves

  1. I must be missing something. The system appears to me to be linear. If so, the output will be a weighted sum of the inputs (represented as sum of sinusoidal functions). More accurately, as a co-worker observed a long ago, the eigenfunctions of linear operators are complex exponentials.

    For the sawtooth (IIR) at w=1/2 and w=1/3, there’s a component (overtone, harmonic) at w=1, and resonance would occur.

    So, analytically, you should be able to get answers from Laplace xforms, Green’s function, impulse responses, etc.

    Bob76

  2. The used sawtooth function has lower energy than sine. It would be interesting to see the camparison where the driving functions have equal RMS instead of equal amplitude. I think the difference would be less then.

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