I know the impedance of the inductance is: 'jwl'; but this is correct with sinusoidal input; now my question is:
how to calculate the impedance when the input is the periodic square wave(like a digital clock)?
any help will be appreciated.
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\$\begingroup\$ Is your aim to measure inductance with your secondary (and probably not-well-thought-out) aim to use a square wave? \$\endgroup\$– Andy akaCommented May 25, 2018 at 8:47
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5 Answers
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You could use the fundamental to get a very approximate stab, but because a square wave has significant energy at higher frequencies, there will be a fair degree of error.
However, given that we know that the square wave is treatable as a summed series of sine waves with a well known equation - you could iterate towards a better solution by applying the sine wave equation for impedance at each of these harmonics in turn, and summing the current flowing at each. You then get a series of sinusoidal currents which you can sum to get total current, and then get impedance back.
As you include more and more of the upper frequencies you will iterate towards a more accurate solution. This could be done easily enough in a spreadsheet or with a simple script if you know a little programming.
This gives a decent solution without delving into higher mathematics - or a useful sanity check on results that you might obtain by other methods.
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Form the context of your question, it sounds like you are asking for an impedance magnitude in ohms. The problem is that this is only defined at a single frequency.
As a square wave is composed of a fundamental frequency, plus 0.33 3rd harmonic, plus 0.2 5th harmonic etc, you can make a good stab at an effective impedance (depending on how you want to define effective) by simply using the fundamental frequency.
If you want to define an effective impedance as (perhaps) rms voltage divided by rms current, then there are several ways to arrive at that figure, from analytical, to simulation with SPICE. The latter might be easiest, and also illustrate to you what happens to the waveforms (hint, the current wave is not square).
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\$\begingroup\$ in this case; because of the small effect of harmonics(amplitude of harmonics are small) this approximation can be accepted; but if the amplitude of the harmonics were not small; what should I do? Is there any formula for that? \$\endgroup\$– hojjatCommented May 25, 2018 at 7:50
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\$\begingroup\$ It's more complicated than that, because the current response of an inductor is decreasing even as the higher harmonic energy decreases, so the fundamental approximation is quite good. With a capacitor, the opposite applies, and there's increasing response to harmonics, so the current is infinite with a square wave voltage excitation. Back to the inductor, you have to choose what 'effective' means. You may want a peak sum, you may want a power sum, depending on what's important. Analyse for all frequencies, and see whether one captures your application, or if you need to combine them somehow. \$\endgroup\$– Neil_UKCommented May 25, 2018 at 8:38
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assuming your usecase allows it, you can build a simple LC filter with a known value capacitor.
From there, you can use your square wave as an input, Vc as an output, an work out your impedance from the step response of your system.
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Impedance is defined as the ratio of the Laplace transforms of the Input voltage and the current flowing through the said input voltage source.
You can use the method suggested by Neil_UK for an approximate analysis, but if you are looking for an exact answer, then you must keep in mind that impedance comes out in the jwL form only for a purely sinusoidal input.
One of the methods to handle a general input voltage is to first solve the problem by writing the differential equations, for example in this case,
$$L\frac{di(t)}{dt} = v_i(t)$$
Then taking a Laplace transform like so
$$sL\mathscr{L}(i(t)) = \mathscr{L}(v_i(t))$$
And then the current established in the circuit as a function of time may be found by expressing the Laplace transform of the current in terms of known quantities (including the Laplace of v_i) and taking the inverse Laplace transform using a tool like Matlab.
If the OP is familiar with these ideas then they may lookup the iLaplace operator in Matlab.
A useful link to learn more about generalised impedance as an s domain (Laplace Domain) proportionality factor is this.
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1\$\begingroup\$ I'm always willing to learn. Would you provide a link so I can read up some more on the concept of impedance over multiple frequencies. A public web sort of link, I don't have Matlab. Thanks. \$\endgroup\$– Neil_UKCommented May 25, 2018 at 8:34
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1\$\begingroup\$ That is true for an impedance over a range of frequencies, but OP wanted to know at a specific frequency made up of a square wave which, in turn, is comprised of (higher) harmonics. IMHO, it's this that should be addressed in the first place (since OP seems to think the impedance is only based on the square wave's frequency), to show OP the invalidity of OP's question, then answer based on the meaning of OP's question. Your answer only takes care of the 2nd part, could you also add the 1st part? \$\endgroup\$ Commented May 25, 2018 at 8:45
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\$\begingroup\$ @Neil_UK It is just a matter of terminology. I have added a lecture from UCSD that discusses the general idea of impedance as an s domain proportionality factor. Rather than impedance 'over multiple frequencies' it is better to think of it as generalised impedance for any input voltage waveform (Periodic or Aperiodic) \$\endgroup\$– ijunejaCommented May 25, 2018 at 9:11
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1\$\begingroup\$ for instance example 10-3 T&Rp461 - gives a Z as a function of s. While Z(s) is the ratio of V(s) to I(s), I'm not sure that's what the OP had in mind. I think that the fact he asked the question automatically puts his level of understand somewhat below being able to use your answer. Myself, I'm always forming the Laplace expressions and then substituting s by \$j\omega\$ to actually evaluate them. \$\endgroup\$– Neil_UKCommented May 25, 2018 at 13:30
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1\$\begingroup\$ @IshankJuneja Sure, it is a valid question, but, somehow, it seems to me that OP meant to equivalate jw (the particular frequency) to the square wave frequency, without being able to discern the differences, that's all the reason I meant. So, from this point of view, I thought a minor clarification to address this might not be a bad idea. After all, the word itself, impedance, suggests a frequency behaviour, and not everybody considers all the frequencies. Of course, it's just a tiny comment with an even tinier opinion. :-) \$\endgroup\$ Commented May 25, 2018 at 13:44
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Another answer makes mention of laplace, but since your signal is periodic, it's easier to think in terms of fourier.
Your square wave can be regarded as a series of harmonics. Since the inductor is linear we can use the superposition principle to analyse each harmonic seperately. I'm going to ignore the issue of phase and just think about the magnitudes of the components.
In a square wave the fundament is the strongest component, the 3rd harmonic is \$\frac{1}{3}\$ the strength of the fundemental, the 5th harmonic is \$\frac{1}{5}\$ the strength of the fundemental and so-on.
But the impedance of the inductor is proportional to frequency. So in our current waveform, the 3rd harmonic is \$\frac{1}{9}\$ the strength of the fundemental, the 5th harmonic is \$\frac{1}{25}\$ the strength of the fundemental and so-on. The RMS current through the inductor will be dominated by the fundamental.
If you had used a capacitor instead then the impedance would be inversely proportional to frequency. All the harmonics in the current waveform would have the same strength as the fundamental and the RMS current would be infinite.
answered May 25, 2018 at 14:27