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Power spectrum

I'm getting lost with the general method of finding the power spectrum. Here is an example:

"s1(t) - a square wave with a frequency of 1 MHz, a duty cycle of 50% and a peak-to-peak amplitude of 2 V across a load of 50 =D9."

To find the power spectrum, I first obtained P =3D VI =3D V^2/R. I'm wondering to obtain the power spectrum whether the Fourier series of s1(t) should be calculated, and then substitute that Fourier series into P equation that I have derived. I'm getting mixed up between the signal voltage and the Fourier series of the signal. Any clarifications would be much appreciated.

Reply to
ssylee
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I am having great trouble understanding the question. While the two introductory sentences will win no writing awards, at least I think that I understand them. Then comes the quote. THAT IS NOT A SENTENCE! There is no verb! Moreover, the terms are not defined.

While many engineers almost seem to take pride in their inability to communicate in English, which I presume is the poster's native language, it really does impede or completely stop transfer of thought.

Good Luck!

Bill

Reply to
Salmon Egg

My ESL isn't too good, but I'll try to translate.

^ Is this by any chance Mhos?

It' better to spell units out when posting on the Usenet. Not all special characters come across the way they were intended.

Small problem. The answer you will get here will differ depending on whether "peak-to-peak amplitude of 2 V" means +1V to -1V (no d.c. component) or +2V to 0V (1V d.c component).

The Fourier series will give you an answer in a different form than the P = VI formula above. What you will get is the power at each frequency (d.c component, fundamental, plus harmonics). When the Fourier coefficients are summed, they should add up to P.

signal. Any

Don't get confused by units. If you are looking for the 'power' spectrum, then the time series you need to put into the Fourier transform is the instantaneous values for power at each t. Not the instantaneous voltage, s1(t).

Reply to
Paul Hovnanian P.E.

Hi Paul,

Please f> ssylee wrote:

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 ^

=A0 =A0 =A0 =A0Is this by any chance Mhos?

Yes, it's in ohms, the special character didn't paste properly on Google Groups.

I'm not clear about that component either. This part is not very descriptive. I'll ask for further details.

I'm still confused on whether I substitute the fourier series coefficients into V so that I get the power spectrum coefficients =3D (sum of fourier series)^2/resistance.

of the signal. Any

Reply to
ssylee

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