# Nonsinusoidal waveforms + Fourier series

Discussion in 'Homework Help' started by notoriusjt2, Sep 21, 2010.

1. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
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0

allright where did I go wrong here....

Io = Vo/R = 6/16 = .375A

I1 = $\frac{V1}{R+jwL}$ = $\frac{5}{16+j(2pi60)(.025)}$ = $\frac{5}{16 + j9.42}$ = .269@-30.48deg

I2 = $\frac{V2}{R+jwL}$ = $\frac{3}{16+j(6pi60)(.025)}$ = $\frac{3}{16 + j28.27}$ = .092@-60.49deg

therefore

i(t) = .375 + .269cos(2pi60t - 30.48deg) + .092cos(6pi60t - 60.49deg)

P0 = 6*.375 = 2.25w
P1 = $\frac{(5)(.269)}{.375}$cos(30.48) = 3.09w
P2 = $\frac{(3)(.092)}{.375}$cos(60.49) = .363w
Ptotal = 2.25 + 3.09 + .363 = 5.7w

2. ### Ghar Active Member

Mar 8, 2010
655
72
Your approach is correct as far as I can tell, but I don't know how you got the equation for P1 and P2.

You're taking power provided by the sinusoidal source divided by DC current and multiplying by power factor? I don't follow.

I think the simpler way to do this is by forgetting about the voltage source and just looking at the resistor, it is asking for the power in the load after all.

Power in the resistor is I^2*R for RMS values.
Equivalently it's 0.5*Ipeak^2 * R for sinusoids. Frequency and phase shift don't matter in a resistor.

So, the power should be $R(I_0^2 + 0.5I_1^2 + 0.5I_2^2)$

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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P1 & P2 are incorrect.

Since the circuit is linear you can simply add the individual DC and AC power contributions algebraically.

I would first convert the AC current components to their rms values.

Then

P1=I1_rms^2*R=(0.269/√2)^2*16=0.579W

P2=I2_rms^2*R etc.

4. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
P0=6*.375=2.25W
P1=I1_rms^2*R=(0.269/√2)^2*16=0.579W
P2=I2 rms^2*R=(0.092/1.414)^2*16=0.068W

P0+P1+P2=2.9W

This forum is easily teaching me more than my online classes. Thank you

5. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,255
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It's not the linearity of the circuit that allows the individual power components to be added; it's the fact that the 3 current components are orthogonal.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Thanks Electrician - point taken. I was thinking vaguely about the likelihood of non-linear elements giving rise to other inter-modulation (?) terms. I'm not sure how that could happen in fact with a passive (albeit non-linear) circuit.

7. ### Georacer Moderator

Nov 25, 2009
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The power value we found here is the Apparent Power, not the Real Power, right?

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Real Power was found.

9. ### Georacer Moderator

Nov 25, 2009
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How can it be real power? First of all, the load is non-linear (it contains an inductor, hence making the load not a pure resistance but an impedance). Consequently, as the AC source is concerned, the voltage and the current will be out of phase. As a result the $I^2 \cdot R$ will give the apparent power not the real one.

The way I see it is the following:
For an AC voltage source with frequency f and amplitude Vs over a load comprised of a resistance and an inductor, the power is
$I=\frac{V}{Z}=\frac{V}{R+sL}=I_0 \angle \phi\\
S=I_0^2 \cdot |Z|\\
P=S \cdot cos(\phi)\ and \\
Q=S \cdot sin(\phi)$

Do I have a misconseption?

10. ### The Electrician AAC Fanatic!

Oct 9, 2007
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An inductor is not non-linear unless it contains an iron core, for example. Academic problems like this are assumed to involve an inductor which is linear unless specifically stated to be non-linear.

If the RMS current through a resistor is I, then the quantity $I^2R$ is always real power, and that will be the only real power absorbed by the series circuit of resistor and inductor since an inductor doesn't absorb real power..

In fact, it's not possible to have a reactive power component in a pure resistance.

Last edited: Sep 22, 2010
11. ### Ghar Active Member

Mar 8, 2010
655
72
He found the phasor current through the inductor in series with the resistor using the impedance calculation.
Once you have that the resistor power is 0.5*Ipk^2*R

For sinusoids all you need for resistor power is current or voltage magnitude.
The voltage and current in a resistor are in phase, it doesn't matter what angle it has relative to the source or what frequency.

Power factor only applies to sources and complex impedances (meaning resistive + reactive). A resistor cannot absorb reactive power and an inductor cannot absorb real power.

Yes, exactly, but what does R have to do with the AC source?

12. ### Georacer Moderator

Nov 25, 2009
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1,266
Right! So instead of going for the apparent power and then bringing it down to the rel power, we go for the real power directly, knowing that all the real power will dissipate on the resistance, independently of its phase. I think I got it.
So what I did in my calculations whasn't wrong, but a waste of time, yes?

God, I hate power electronics!

13. ### The Electrician AAC Fanatic!

Oct 9, 2007
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But you were wrong when you said:

"First of all, the load is non-linear..."

14. ### The Electrician AAC Fanatic!

Oct 9, 2007
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For any waveform, all you need is current or voltage magnitude, assuming magnitude means RMS value, as is conventional in Electrical Engineering.

15. ### Ghar Active Member

Mar 8, 2010
655
72
I meant magnitude as vector magnitude (peak) which is the only definition I'm aware of, which is why I made sure to refer to sinusoids.
Personally I won't ever equate RMS and magnitude but at least I'll be on the look out for potential miscommunication about it, thanks.

16. ### Georacer Moderator

Nov 25, 2009
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I was so overwhelmed that I figured out what was going on, that I forgot to acknowledge that misconception. I stand corrected.

17. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Since phasors aren't vectors, we might not be surprised that vector magnitude isn't conventionally used with them.

See http://www.tutorvista.com/ks/what-is-rms for an example of use of the word magnitude to mean RMS value: "The root mean square ( RMS or rams ) is a measure of the magnitude of a varying quantity that can be a number, voltage, current or power."

Also see http://en.wikipedia.org/wiki/Root_mean_square: "...root mean square...is a statistical measure of the magnitude of a varying quantity."

One can find alternative uses, e.g.: http://www.allaboutcircuits.com/vol_2/chpt_1/3.html where the words "intensity", "magnitude" and "amplitude" appear to be used synonymously.

The way to avoid all ambiguity is to use an adjective in front of words like "magnitude" or "value", such as "peak", "average", "RMS", etc. For example, "effective value", which I've never seen used to mean anything other than "RMS value".

For example, here: http://www.bcae1.com/voltages.htm, the word "magnitude" is used for 3 different things, distinguished only by qualifying adjectives.

The author of a power electronics text often will specify early on that the unqualified word "magnitude" will be taken to mean "RMS value".

I consulted several classic texts dealing with power and their convention is to use the word "magnitude" to mean "RMS value".

A good reason for this use is that for any waveform, not just sinusoids, the RMS value of a current or voltage waveform plugs right into the various formulae such as $P=I^2R$, or $P=IE$ and gives the expected result of real power dissipated in a resistive load.

18. ### Ghar Active Member

Mar 8, 2010
655
72
You're right, I guess I haven't been paying attention. I flipped through some of my books and they specify 'peak magnitude' or 'rms magnitude'.
Looks like I'll just avoid the word magnitude in general

I guess I leaned more towards the vector one because by far my most common usage is for transfer functions where magnitude is indeed vector magnitude. I don't do much phasor analysis.

It does make sense, rms is really the only meaningful magnitude for complicated waveforms.

Oct 9, 2007
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20. ### Georacer Moderator

Nov 25, 2009
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Phasors are vectors. Vectors with standard angular velocity that rotate in the cartesian plane.

All my classes confirmed that and wikipedia seems to agree: http://en.wikipedia.org/wiki/Phasor
In fact, it reads "In physics and engineering, a phase vector ("phasor") is a representation of a sine wave whose amplitude (A), phase (θ), and frequency (ω) are time-invariant."