Impedance, square waves, and duty cycle

Question

Will a coil present a different impedance to a 10 kHz sine wave as opposed to a 10 kHz square wave with a 10% duty cycle?

I could not find a way to calculate it, are there any formulas available?

Answer 1

Will a coil present a different impedance to a 10 kHz sine wave as opposed to a 10 kHz square wave with a 10% duty cycle?

Yes. You can calculate equivalent impedance. There can be a few different definitions of it, with different interpretations and applications.

Perhaps the most useful one would be the value of an equivalent resistor that could replace the inductor's reactance. I.e., you want equivalent reactance for the composite waveform. Such equivalent reactance is the reciprocal of the weighted average of real conductances of the load (here: inductor) for each of the harmonics of the input waveform, with weights set per the Fourier amplitudes of those harmonics.

Another useful equivalent impedance may be one that dissipates the same complex power - in W and VAR - as the original impedance, but when fed with only the fundamental harmonic of the composite waveform. This would be useful for thermal considerations, i.e. to know how much real power the load will dissipate, and how much circulating current there will be due to the load, when driving with a sine wave as a stand-in for the composite waveform. I could see some utility of such an equivalent back in the times of slide rules, but with today's ability to get numerical results almost instantly, there's no reason to worry about it.

Answer 2

Technically, yes, because a square wave is a sum of multiple sine waves, in theory infinite amount of sine waves, so since it is not a single sine wave you compare it to, it will present a different impedance.

Answer 3

A square wave is a sum of sine waves at harmonics of the fundamental frequency. The higher harmonics will see higher impedance than the lower ones, so generally a square wave of a given frequency would see a higher impedance overall than a sine wave at the same frequency.

You could determine the impedance by decomposing your pulse into its harmonics using a Fourier transform, calculating the impedance of each harmonic, then combining them back together to get the total impedance.

But probably it’s easier to just simulate it with Spice or similar and determine it empirically.

Answer 4

Will a coil present different impedance in a 10 khz sine wave as opposed to a 10 kHz square wave

Impedance, by definition, is the electrical opposition that a component presents when a sinewave is applied to its terminals. Because a square wave is not a sine wave, there is no meaningful impedance value or concept. Here's a quotation from wikipedia: -

Quantitatively, the impedance of a two-terminal circuit element is the ratio of the complex representation of the \$\color{red}{\text{sinusoidal voltage}}\$ between its terminals, to the complex representation of the current flowing through it. In general, it depends upon the frequency of the sinusoidal voltage.

I could not find a way to calculate it, are there any formulas available?

My answer explains why that is the case. If you have a problem to solve, maybe there's a better approach? For instance, it's much easier to solve the current in the inductor when a square wave is applied using the time domain (rather than the impedance/frequency domain)

Answer 5

Impedance is a function of frequency, not a single value. Your coil presumably has a similar impedance to an ideal inductor, which is \$Z = i\omega L\$, where \$L\$ is the inductance of the coil and \$\omega\ = 2\pi f\$ is the angular frequency. You can see that a coil has a large impedance at high frequencies, and at d.c. (\$\omega = 0\$) it has zero impedance. (In fact there is also some resistance, so the impedance at d.c. is finite.)

So now you need to think about the frequency components of your square wave (https://en.wikipedia.org/wiki/Square_wave#Fourier_analysis). Briefly, different frequency components of the square wave will see different impedances. The output waveform will then have different weightings of these components to the input, and so will no longer be a square wave. Your coil will have acted as a low-pass filter.

I think the really crucial thing to get your head round here is thinking of both the input wave and the impedance of the coil in frequency space.