Off Topic: Period/Frequency/Impedance of square wave vs sine wave

This is a very stupid question, but leave it too me to ask it.

I was wondering why is it IMPEDANCE of a inductor is

2PI(FL) for a square wave pulse train also? I will presume that it is because it is a periodic function.

With a sinusoidal function:

Z for an inductor is 2PI(FL) where 2PI(F) = w(omega)
I understand that v= Ldi/dt where i(t)=cos(wt+O) and that for a deveritive of di/dt brings some of the constants outside the function etc.

Doings some algrebra and arrangment for an inductor it is found that Z=2PIFL

My question is how how do we come to the same conclusion that Z=2PI(FL) for square-wave pulse trains also?

Thanks

plc_user1973 said:

My question is how how do we come to the same conclusion that Z=2PI(FL) for square-wave pulse trains also?

Thanks

Click to expand...

Who are the "we" coming to this conclusion - Can you quote a reference ?

If you are trying to calculate the current flow in a circuit, one method is to turn the square wave into the sum of lots of different frequency sine waves (using fourier). You calculate the current for each frequency and then sum them to give the total current flow.

plc_user1973 said:

This is a very stupid question, but leave it too me to ask it.

I was wondering why is it IMPEDANCE of a inductor is

2PI(FL) for a square wave pulse train also? I will presume that it is because it is a periodic function.

With a sinusoidal function:

Z for an inductor is 2PI(FL) where 2PI(F) = w(omega)
I understand that v= Ldi/dt where i(t)=cos(wt+O) and that for a deveritive of di/dt brings some of the constants outside the function etc.

Doings some algrebra and arrangment for an inductor it is found that Z=2PIFL

My question is how how do we come to the same conclusion that Z=2PI(FL) for square-wave pulse trains also?

Thanks

Click to expand...

Better a stupid question than a stupid mistake. (Not my line, one of the techs I know taught me that one.)

You can't come to that conclusion.

That formula is only valid for a pure sinusoidal wave. For a square wave, as stated, you have to add all the component sine waves together. You could do a quick plot to find out when you can stop, as the filtering on the inductor will act as a low-pass filter. If you're getting -20dB or more, it's getting close enough to diminishing returns.

Of course, that's going to take you all day if you do the calculations by hand. Drag out your oscilloscope and take a look at what the waves will do instead.

What you'll end up with is putting a square wave into the inductor and getting a sine wave out... or will you?

No, because your inductor will also integrate (I remember the nmemonic "If you're high, you feel different.") your signal, so you'll actually get a sawtooth wave coming out. (or a "jumpy" dual-ramp wave if you're using L as a high-pass filter) You'll also end up with a phasic difference between the voltage and current, with the voltage leading the current.

The exact amount of lag depends on the rest of your circuit. If there's nothing there but the inductor, you'll get about 90 degrees of lag. (You'll also end up with a fire, but hey, it's your shop.) If you have resistance about equivalent to the inductance at the interesting frequencies, it's closer to 45 degrees.

Even for a square wave the formula Z = 2*pi*f*L stands, but as mentioned by L D[AR2,P#0.0] a square wave is the sum of an infinite number of sinus waves. Therefor you'll find an infinite number of impedances, one for each sinusfrequency in the square wave. Do mind, it's not simply the sum of the various impedances. It all depends on the frequency you're looking at.

Kind regards,

jvdcande - For the square wave case, what is the "impedance" value used for ?

As already stated, for a square wave applied voltage (+/- V volts whose fundamental frequency is at w rad/s), the current waveform will be triangular and the rms current flow will be as follows:

I(rms) = V * pi / (2 * w * L * sqrt(3))

This differs from what you get using w * L for the impedance.

Can you point to any references for your source of info ?

Well, I accept the idea of the fourier series to compute the said impedance, but since T=period of a square wave f=1/T. Thus, 2PI*F*L should stand.

Please lets keep trying to figure this out.

plc_user1973 said:

Well, I accept the idea of the fourier series to compute the said impedance, but since T=period of a square wave f=1/T. Thus, 2PI*F*L should stand.

Please lets keep trying to figure this out.

Click to expand...

No, because the higher frequency waves that you add have a shorter period. Each square wave is the sum of the odd harmonics. Take a look at the wikipedia entry:

http://en.wikipedia.org/wiki/Square_wave

The third harmonic is going to be running at 3 times the original frequency. The fifth will run at five times.

Thanks, I think I understand it now. Would this same procedure apply the same why to only (Positive Pulses) ie., no negative polarity?

plc_user1973 said:

Thanks, I think I understand it now. Would this same procedure apply the same why to only (Positive Pulses) ie., no negative polarity?

Click to expand...

Yes, because it's the same wave with a DC offset. If it helps, you can think of DC as a 0Hz sin wave. (Don't do that for very long, because it's wrong. It's just DC.)

It really depends on what else is in your circuit. For example, you're not going to get a lot of reverse current (negative voltage) if you've got a diode in there, unless you've got a big inductor and blow past the reverse breakdown voltage, etc. As that wikipedia page said, you can't get all the harmonics for an ideal square wave due to bandwidth concerns, but you can probably get the 5th, and that will be close enough.

Really, analysis is a great way to spend your day, but you're much better off pulling out the scope and seeing exactly what's going on. It's even better if your scope has a spectrum analyzer on it. Math will only work on really trivial circuits. Once you get more than a 1/2 dozen nodes or active components, you're just as likely to make a mistake as not.

So the real question here is, "what exactly are you trying to ask?" I don't understand why you want to know the impedance of the inductor. What's going on?

If there is any DC component in the applied voltage waveform, the current will (theoretically) just keep on rising for ever.

V = L di/dt so i = Integral(V)dt/L

L D[AR2 said:

If there is any DC component in the applied voltage waveform, the current will (theoretically) just keep on rising for ever.

V = L di/dt so i = Integral(V)dt/L

Click to expand...

A billion billion years ago, someone put a 9V battery across an inductor. The current just kept rising until it got too hot to touch. There was nothing that could be done. The circuit was wired, and there was no way to disconnect it. The current JUST KEPT RISING!

Eventually the entire planet was engulfed. There were no survivors.

We now call this fireball "the sun".

So whatever you do, don't put DC across an inductor. Do you want to be the one responsible for burning our planet into its component hydrogen molecules?

MASEngr, I am attempting the Bedini Pulse Motor. Its a N POLE Mono Pole Motor created by John Bedini. If you look in to it, you will see N Pole Magnets interacting with an electromagnetic. The electro-magnet is bi-filar and one part of the coil is for sensing a magnetic field and the other part of the coil is for generating the pulse to drive the motor rotation. Sorry for the basic explanation. My thoughts on the matter are this, as the sensing coil picks up a "faster" rotation magnetic field which in turn causes faster firing of the POWER electromagnet which is based on the RPM of the motor. My thoughts are: Faster rotation, more pickups in the sensing coil which leads to more fires of the Power coil, which in turn raises the frequency in the coil which should lead to more impedance. More impedance, means less current.

Sorry for the basic explanation. I am just experimenting just wondering what to expect.

All of this is based on the basics of a square wave DC signal into an inductor, in my opinion.