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A function generator with an internal resistance of $$50\Omega$$ is set to produce a 1kHz square wave with $$20v_{p-p}$$ The duty cycle is 80%. The signal is run through a $$50\Omega$$ resistor.

I need to determine the peak power that is output.

Now, $$\frac{\Delta t}{T}= 0.8$$ so $$\Delta t=0.8T=0.8\left ( \frac{1}{f} \right )=0.0008s$$

But this is as far as I can go.

A useful hint is appreciated. Thanks in advance.

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  • \$\begingroup\$ Is the 50 ohm resistor connected to ground? Is the 20Vp-p into an open circuit? \$\endgroup\$ – Andy aka Apr 14 '16 at 8:39
  • \$\begingroup\$ The peak power doesn't depend on the duty cycle. It does depend on the dc offset, which you haven't specified. \$\endgroup\$ – The Photon Apr 14 '16 at 15:01
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Hints (mistake corrected)....

If the duty cycle were 100% the power delivered to the 50 ohm load would be 10^2/50 = 2 watts. Power is proportional to duty cycle of a signal so, if only 1 watt was consumed by the resistor the duty would be 50%.

The above is for average power and not peak power. Peak power, as always would be 2 watts for duty cycles of >0% to 100%.

For the above I'm assuming that the 20Vp-p sinewave is from 0V to +20V and that when the load is connected it will see 0V to +10V as a signal.

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