V(AVG) of a sine, again

Discussion in 'Homework Help' started by Agonche, Aug 21, 2012.

Aug 26, 2011
30
0
Hey, it's a similar question again.
This time, I think my teacher got a wrong answer.
Here's the problem:
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Draw the output voltage waveform, and find $V_{AVG}$.
The diode is considered ideal.

$V_i=30sin\omega t$
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Diode is ON when $V_i>-11.3V$, in this case $V_o=-11.3V$
Diode is OFF when $V_i<-11.3V$, in this case $V_o=V_i$

$V_o$ waveform:

$\theta_1=\pi + arcsin \frac{-11.3}{-30}$
$\theta_2=2\pi - arcsin \frac{-11.3}{-30}$

OK, here's what I've done. I pushed the waveform up +11.3.

$V_{AVG}=\frac {1}{2\pi}[\int_{\theta_1}^{\theta_2}18.7sin\theta \cdot d\theta]-11.3$
$V_{AVG}=-16.8V$

The teachers answer is -15.8. I don't know how he got there but can anyone confirm that I'm right.

2. The Electrician AAC Fanatic!

Oct 9, 2007
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If the diode is considered ideal doesn't that mean that the forward voltage drop is zero? That's what you assumed in your previous thread, wasn't it?

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Aug 26, 2011
30
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Im sorry, the diode has the .7V barrier. It seems I was looking somewhere else when I wrote ideal.
Anyway, thank you for the very clear explanation. I didn't really catch the easy way, with the MIN function. (Im not that good at math :/).

btw, can you tell me what software are you using to display the function graph, and these integrals.

4. The Electrician AAC Fanatic!

Oct 9, 2007
2,347
347
I'm using Mathematica. There are several similar programs that you have to pay for such as Maple, Mathcad, Matlab and there are some free ones that can also solve integrals. I think t_n_k can tell you what they are.

Oct 2, 2009
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Oct 9, 2007
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