# V(AVG) of a sine

Discussion in 'Homework Help' started by Agonche, Aug 18, 2012.

1. ### Agonche Thread Starter Member

Aug 26, 2011
30
0
Hello guys.
Im having a problem finding the average value of a specific V(o) function.
Heres the circuit:

When the diode is considered ideal, V(o)=V(i) (for the positive values of Vi) and V(o)=0V (for the negative values of V(i)).

then: $V_{AVG}=\frac{1}{2\pi }(\int_{0}^{\pi}20\cdot sinwt\cdot d(wt))$
which can be simplified: $V_{AVG}=\frac{20}{\pi}$

But when the diode is considered real, with a 0,7V barrier...
V(i)=20sinwt-0.7

$V_{AVG}=\frac{1}{2\pi }(\int_{\theta_1}^{\theta_2}20sinwt-0.7\cdot d(wt))$

How to solve this integral !

Last edited: Aug 18, 2012
2. ### MrChips Moderator

Oct 2, 2009
12,625
3,451
$\int_{\theta_1}^{\theta_2}\!a \, \mathrm{d} \theta= a(\theta_2 - \theta_1)$

3. ### ramancini8 Member

Jul 18, 2012
447
119
Maybe this is incorrect, but I would subtract the diode drop from 20 because that is what the circuit sees.

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
Try this:

$V_{AVG}=\frac{1}{2\pi }(\int_{\arcsin (\frac{7}{200})}^{\pi-\arcsin (\frac{7}{200})}20 sin(wt)-0.7\cdot d(wt))$

5. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
What these people are trying to say to you is that you have to first find at which angle your sine reaches the 0.7V threshold and then use that angle value $\theta$ to integrate from $\theta$ to $\pi-\theta$.

6. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
While this seems reasonable at first glance, notice that for the first part, and last part, of the integration you have a negative voltage whereas you should have zero.

So, as the others pointed out, you have to narrow the integration limits as well as substract off the voltage drop.

7. ### Agonche Thread Starter Member

Aug 26, 2011
30
0
I hate those irregular sine waves.

I know how to find $\theta _1$ and $\theta _2$
$\theta _1=arcsin(\frac{0.7}{20})$
$\theta _2=\pi -arcsin(\frac{0.7}{20})$

My question is:
How to solve the integral, what to do with the -0.7 part.
you know, a step by step solve.
I know that first we have to 'divide' the integral in two parts.

8. ### MrChips Moderator

Oct 2, 2009
12,625
3,451
The $-0.7$ part is a constant.
The integral is $0.7 (\theta_2 - \theta_1)$

9. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
$V_{AVG}=\frac{1}{2\pi } \int_{\theta_1}^{\theta_2}(20Vsin(\theta)-0.7V)\cdot d\theta$

What is it about this integral that you don't know how to solve?

10. ### Agonche Thread Starter Member

Aug 26, 2011
30
0
I don't know what am I doing wrong.
if $V_{i}=20sin\theta$ then $V_{AVG}$ should be a positive number, because the diode is ON only for positive values of Vi (Diode ON if Vi>0.7V)

but I get a negative result. How is this possible ?

$V_{AVG}=\frac{1}{2\pi } [\int_{\theta_1}^{\theta_2}(20Vsin(\theta)-0.7V)\cdot d\theta]$
$V_{AVG}=\frac{1}{2\pi }[ \int_{\theta_1}^{\theta_2}(20Vsin(\theta)\cdot d\theta)-\int_{\theta_1}^{\theta_2}(0.7V\cdot d\theta)]$
$V_{AVG}=\frac{1}{2\pi }[-20(cos(\theta_2)-cos(\theta_1))-(0.7V\cdot (\theta_2-\theta_1))]$
$V_{AVG}=\frac{1}{2\pi }[39.975-123.2]$
$V_{AVG}=-13.24V$

Do I have to convert $\theta_1,\theta_2$ to radian on the second part, when they multiply with 0.7 ?

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Yes the angles θ1 & θ2 must be in radians rather than degrees.

12. ### Agonche Thread Starter Member

Aug 26, 2011
30
0
$V_{AVG}=\frac{1}{2\pi }[39.975-2.15]$

$V_{AVG}\approx 6V$

cool, got it.

13. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
How about an exact value?

See the attachment.

• ###### Avg.png
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Agonche likes this.
14. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Yes, you must have theta in radians. If you don't, the units don't work out.

$V_{AVG}=\frac{1}{2\pi }[-20V(cos(\theta_2)-cos(\theta_1)\,)\,-\,0.7V(\theta_2-\theta_1)]$

While putting something with units of degrees into the cosine function is well defined, the return value is still dimensionless making the units on that term volts. But your second term will have units of volt-degrees, which can't be added to volts. Hence, you KNOW the answer is wrong.

But, like so many people, you won't track your units and so you can't catch a common error like this. Had your wrong answer come out positive, you very likely would have blindly reported it as your answer (and at least you did ask if the answer made sense and thereby caught the error that way -- good for you).

Most mistakes that you make will mess up the units. If you are tracking your units, you catch the vast majority of these mistakes. As an engineer, as far as I am concerned, it is bordering on gross professional negligence not to track units and thereby fail to use one of the most powerful error detection devices around. As a student, failing to do so lowers your grade. As an engineer, failing to do so can kill a planeload of people. Never forget that doctors kill people one at a time while engineers do it in job lots.

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
This begs the question as to whether the OP would have been any the wiser had they noted the units as volts-radians.

16. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Maybe, maybe not. Since radians are dimensionless, they can come and go as need be. I tend to carry them explicitly where it makes sense to flag the quantity as involving an angular measure and drop them elsewhere. But whether you keep them or not, degrees can't be dropped willy-nilly and the inconsistency is still evident if units are tracked.

So whether he would have or not, hopefully he can see that value that tracking units has and will choose to track them in the future.

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
No doubt there is genuine advantage with tracking units and I admire your resolve to convince others of the merits of the approach.

I would have also taken the OP to task had they blithely passed over the blatant inconsistency that their calculations showed [post #10] the integration of a 20V peak rectified sinusoid component having less value than the integration of a 0.7V DC value over the same interval. Clearly the OP was concerned at the strange result and proposed the notional reason as to the source of the error.

An appreciation of the consistent magnitudes of various terms in a solution also provides some feedback for the observant student / professional as they work through a problem.

At the moment I'd rather take the OP to task for that annoying avatar.

18. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Absolutely agree. Units are a big part of error checking, but not the only thing. That's why you'll see me tout the dynamic duo of always track the units and always ask if the answer makes sense. The OP here is well ahead of so many people because he is asking if the answer makes sense, and just asked the question is the lion's share of the battle.

Another extremely valuable tool is asking if the answer makes sense in the limits; in other words, take the answer (including at points along the way) and see if it reduces to something that is obviously right under simplifying assumptions, such as the frequency going to DC, or the capacitance and inductance going away, or the power factor going to unity, or one resistor becoming arbitrarily larger than another.