The frequency is not relevant. The average of a square wave with a 50% duty cycle that swings between 0V and +5V is V/2 or 2.5 volts. There is a more complicated derivation involving the integration of a pair of unit step functions which gives the same result. Another derivation is to integrate V(t) with respect to t divided by T, the length of a single period.Originally posted by Annu007@Jun 12 2006, 02:00 PM
I need to calculate an average value of a square wave (Frequency- 48MHz).
Amplitude- 5V.
50% duty cycle.
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Of course this is true for a square wave that goes from 0V to +5V. If it was symetrical around 0V ie. -2.5V to +2.5V then the average would be zero.Originally posted by Papabravo@Jun 13 2006, 08:41 AM
The frequency is not relevant. The average of a square wave with a 50% duty cycle that swings between 0V and +5V is V/2 or 2.5 volts. There is a more complicated derivation involving the integration of a pair of unit step functions which gives the same result. Another derivation is to integrate V(t) with respect to t divided by T, the length of a single period.
There is also the statistical approach which is to take the Expected Value of the Voltage. It is 1/2*5V + 1/2*0V = 2.5 Volts. This is the same case as a Bernoulli trial
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Yes. A DC multimeter will do that. It will measure the DC average of the signal.Originally posted by Annu007@Jun 13 2006, 09:18 PM
Hello,
Is there a circuit I can use (like a filter) with which I can measure the DC value of this wave?
Regards
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Assuming that it can follow the 48 MHz. signal. It would not surprise me that a typical multimeter would read something smaller than V/2.Originally posted by windoze killa@Jun 13 2006, 05:44 PM
Yes. A DC multimeter will do that. It will measure the DC average of the signal.
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ooooppppsss missed that little detail.Originally posted by Papabravo@Jun 14 2006, 11:48 AM
Assuming that it can follow the 48 MHz. signal. It would not surprise me that a typical multimeter would read something smaller than V/2.
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