I’m self-studying this problem, and it seems that have a problem solving this.

The only thing I didn’t understand is this solution is how to calculate the average power dissipation. It’s never mentioned in class and I can’t find any formula even to proceed.

I know that:

pavg=Vm⋅Im2p_{avg} = \dfrac{V_m \cdot I_m}{2}

And I don’t see how I can apply this formula to the problem. And I don’t think it is helpful anyway since it gives the average power in Watts.

So, how can I calculate the average power dissipation?

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Hint: JOULES measures energy. Power is energy/time. (1Joule/sec = 1 WATT)

– JIm Dearden

Sep 14 ’14 at 18:30

1

Remember Cal 2? Yeah, you’ll need that.

– Ignacio Vazquez-Abrams

Sep 14 ’14 at 18:30

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2 Answers

2

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The J is probably a typo, unit of power is watt and not joule. To calculate signal power: P=1TT∫0u(t)i(t)dt=1TT∫0u2(t)Rdt=4TRT/4∫0(10t)2dt=4⋅1004⋅1001∫0t2dt=[t33]10=13−0=13WP=\frac{1}{T}\int\limits_0^T{u(t)i(t)\,dt}=\frac{1}{T}\int\limits_0^T{\frac{u^2(t)}{R}dt}=\frac{4}{TR}\int\limits_0^{T/4}{(10t)^2dt}=\frac{4\cdot 100}{4\cdot 100}\int\limits_0^{1}{{t^2}\,dt}=\left[\frac{t^3}{3}\right]_0^1=\frac{1}{3}-0=\frac{1}{3}\,\mathrm{W}

In part (b) you solved for the instantaneous power p(t)p(t).

You can calculate the average power as

1T∫t+Ttp(t)dt\dfrac{1}{T}\int_t^{t+T}p(t)\mathrm{d}t

Where taking T equal to one cycle of the periodic function will give the average over all time. If you are clever you can work out from symmetry how to get the answer using just the period from t = 0 to t = 1 min.

The units of power are watts, not joules, so the answer given in your text is incorrect.