# Homework Help: Calculating power of a pulse

1. Jul 18, 2006

### axawire

Hi,

Im trying to calculate the power of a pulse. I know the duration of the pulse t in seconds and I know the peak current Ipeak in Amps. If need be I can also calculate the resistance R of the conductor. I know of P=RI^2 but am not sure if this holds for pulses.

Help/Tips/Suggestions

Thanks.

2. Jul 18, 2006

### vanesch

Staff Emeritus
Given that power is an instantaneous concept, there's no need to involve time, and it doesn't matter if it is a pulse or a continuous current. Yes, the power is given by I^2.R.

But maybe you're wondering what is the ENERGY of the pulse ? Then you will have to integrate the power over time. Assuming a constant current during time T, the energy becomes then I^2.R.T.

3. Jul 19, 2006

### desA

Current is the rate of moving a bunch of charge down a wire... Would it be acceptable to assume that a pulse represents a constant current?

Perhaps you could consider a charge-time envelope, then integrate this over time, then work in charge-energy relationships to obtain a final form.

4. Jul 19, 2006

### vanesch

Staff Emeritus
Depends. If it is a square-pulse generator feeding a resistor, why not ?
However, if it is some kind of discharge, it will certainly not be constant.

Well, you'd need to take the derivative of the charge-time curve to find back the current, SQUARE IT, and integrate it back over time.

5. Jul 19, 2006

### desA

You're finding the RMS value for current, I assume.

You could probably also perform a simple integral average of the current (charge-time envelope differentiated). Depends how you wanted to define the final energy form.

6. Jul 19, 2006

### vanesch

Staff Emeritus
Well, yes. If you integrate the square of the root of the average of the square, you find, eh, the integral of the square.

The integral of the differentiated charge-time envelope is simply the difference between the initial and final value of the charge-time curve ; in other words, the total amount of charge displaced.
However, depending on how this is delivered to a resistance, the dissipated energy is different! In the case of a true delta-function, the dissipated energy is infinite. In the case of a square pulse, the energy is indeed <I>^2 R T. For an intermediate pulse form, the energy dissipated in the resistor will be higher than <I>^2 R T: it will be <I^2> R T.
Now, the difference, <I^2> - <I>^2 is nothing else but the variance of the current (that's why for a square pulse, both are equal: the current doesn't change and has variance 0 during the time it flows).
So with <I> alone, you can only estimate a lower bound on the dissipated energy and all variation will increase it.

7. Jul 19, 2006

### axawire

Well... I do not know the exact form of the wave, but I am assuming its sinusoidal in shape (but on the positive part of the wave). Now can I actually use the RMS for current when its a pulse and not an alternating current?

Also Im not so sure about using P=R*I^2 as if I use a super conductor does this equation still apply?

My goal here is I have a rough idea of the shape of this pulse I need, but I am trying to get a ball park figure for how much energy a pulse generater would have to use to generate such pulses on a continous basis.

Thanks.