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An Example

Home › Forums › temp › More on asymmetric peak limiting › An Example

January 25, 2011 at 4:51 am #20320
radio8z
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Total posts : 45366

In order not to complicate my previous post I left this out but here it is in case anyone is still wondering about average power.

Let’s keep it simple. Suppose we have a voltage which is 1 volt and it is on for 1 second and off for 1 second. The average voltage over time is (1+0)/(1+1) = 1/2 volt. Do the same for a 1 amp current waveform such as this and you get 1/2 amp. on average. Multiplying these together to get power will yield an average power of 1/4 watt.

But the correct way to get average power is to multiply volts and amps before averaging. If both volts and amps are on at the same time and off at the same time then the power will be 1 watt for 1 second and 0 watts for 1 second. The average is (1+0)/(1+1) which is 1/2 watt, not 1/4 watt as calculated by multiplying the average volts and amps.

So, to get average power it is necessary to multiply volts and amps on an instantaneous basis before dividing by the period to get the average.

This is why the power input to a final APPEARS constant with and without modulation but it is not.

Neil

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