# How to calculate the average power for a square wave DC signal

1. Now, suppose that a square wave generator is used as the source. If the square wave signal has a peak-to-peak of 20 V and a zero average value, determine the average power supplied by the source connected to 1 k ohms resistor.

2. Next, if the square wave signal has a peak-to-peak of 20 V and a 10 V average value, determine the average power supplied by the source.

My attempt for Q 1. Half of peak to peak voltage which is 10. Then, average power is 0.1 W.

I am not sure what to do for question 2.

• Hint: draw the square wave. A 20V peak to peak with zero average is actually swinging between +10V and -10V. – Peter Smith Apr 14 '17 at 15:40
• Thanks. so for question 1 the average power is zero? – Crazy Apr 14 '17 at 15:42
• @Crazy no... let me splain – Trevor_G Apr 14 '17 at 15:42
• About question 2, putting average voltage into the average power equation? – Crazy Apr 14 '17 at 15:43
• @Trevor Can you help me with question 2 as well. – Crazy Apr 14 '17 at 15:45

In the first case your square waive is 20V biased around ground, so it is actually +-10V

The power for the upper half is $\frac{V^2/R}{2}$ =$\frac{10^2/1000}{2} = 0.05W$

By the same equation the power for the lower half is $\frac{-10^2/1000}{2} = 0.05W$

So the total power is $0.1W$

In the second case when biased to $10V$

The upper half becomes $\frac{20^2/1000}{2} = 0.2W$

The lower half becomes zero in this case.

The actual formula for any bias is...

$P$ = $\huge\frac{(V_{pk-pk}/2)^2 + V_{bias}^2}{R}$

• More complete than mine; I like the pictures. – Peter Smith Apr 14 '17 at 16:11

In case 1:

The signal is going from +10V to -10V; ignoring transition time, there is always 10V across the resistor, yielding $\frac {V^2} {R}$ = 100mW

For case 2:

The signal is switching from 0V to 20V with (assuming a proper square wave) a 50% duty cycle.

The formula for such a situation is $\frac {V^2} {R} \cdot DC$ which yields 200mW.

Note that this is only for a square wave under the circumstances shown.