average voltage square wave

Thread Starter


Joined Jan 5, 2005
Sedra and Smith 3.13,14
A square wave of 10-V peak-to-peak amplitude and zero average is applied to a circuit employing a 100 ohm resistor and a diode (simple circuit).

What is the average output voltage (across the resistor)? The book gives a answer of 2.5 volts, which suggest that you simply take the peak voltage and divide by 2.

However in 3.14 They change the average voltage of the square wave to 2volts while keeping the peak-to-peak at 10volts. What is the average output voltage (across the resistor)?

Incidently how can you have a 10V p-p square wave that averages to 2 volts?

I also looked up root-mean-squared to see if it applied to this problem. But usually that is defined for sine-wave and its application to square waves is not that clear.

Please help me sort this out. Thanks


Joined Jan 19, 2004
the 10 V p-p waveform with zero avg value is symmetric wrt to the time axis.
however the 2V (avg value) waveform is not.
refer to the wave form diagrams drawn

avg value voltage over the resistance in the 1st case over the entire cycle must 2.5V...

now that we know how the avg value can be 2 volts ...the avg value across the reistor i think must be 3.5V


Joined Dec 12, 2004
The average is 2,5V on the resistor, after the diode. If the duration of the positive voltage is: T1, then in general, the average on the resistor, after the diode would be: Vm*T1/T . If T1=T/2 then Vaver=5/2=2,5V. T is the period of the square wave.
In the case 2, Vaver=7*1/2=3,5V