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  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.

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

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}\$

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

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