# Three phase phasor proof or simplification

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1. Sep 20, 2015

### tim9000

So I'm really rusty on phasors, I was reading that a space vector current was

i(t) = I(cos(wt)<0 + cos(wt - 120)<120 + cos(wt - 240)<240 ) = 3/2 * I < wt

So I tried to go back to basics and I went back and read:

A = CosΘ.Cos(wt) + Cos(Θ - 120).Cos(wt - 120) + Cos(Θ + 120).Cos(wt + 120)
= 3/2 * Cos(Θ - wt)

so I went back to Trig identities.....and pages of scribbling later still couldn't figure it out.
Does anyone have a proof for either of these?

Thanks!

2. Sep 20, 2015

### milesyoung

A note: phasors and space vectors are two very different concepts.
- A phasor is a representation of a sinusoidal function with parameters that are time-invariant.
- A space vector is a transformation that maps a set of real-valued functions to a complex-valued function, which usually has some type of spatial interpretation (used a lot in electric machine theory).

Have you tried converting all the terms to their complex representation? Then do a bit of algebra.

3. Sep 20, 2015

### tim9000

Yeah as far as I remember a space vector is the addition of phasors through feeding a sineusoidally current to a sineusoidally distributed winding in space.
Yeah I tried that thanks.
As you can see in what you quoted the angle term becomes the frequency * time. It should be simple, I'm just not doing something right.

no idea how it reduces to that?

4. Sep 21, 2015

### milesyoung

No, phasors aren't involved at all when you're working with space vectors, which is a good thing, since you'd otherwise be limited to steady-state analysis.

If you have a function $i(t) = A\sin(\omega t + \theta)$, where $A, \omega, \theta$ are constants, then its phasor representation is $I = Ae^{j\theta}$.

If this:
$$\mathbf{i_s} = i_a(t) + i_b(t) \mathbf{a} + i_c(t) \mathbf{a}^2, \mathbf{a} = e^{2\pi/3}$$
is the space-vector transformation, then, for instance, the term $i_b(t) \mathbf{a}$ is not a phasor, even though it is complex. $i_b$ is just an arbitrary real-valued function of $t$, and $\mathbf{a}$ isn't a phasor representation of a sinusoidal function.

$i_a, i_b, i_c$ aren't restricted to sinusoidal functions, and the space-vector transformation is just as useful for windings that aren't sinusoidally distributed.

Try doing the complex algebra first.

5. Sep 21, 2015

### tim9000

Could please you elaborate on that?

Anyway isn't what I've posted just phasors?

6. Sep 21, 2015

### milesyoung

That was the purpose of the rest of my post. Can you be a bit more specific?

No, a phasor is a complex constant, i.e. they have no time dependency.

None of these terms are phasors.

Edit:
Just to reiterate: a phasor is a complex constant, but not all complex constants are phasors. As I wrote, a phasor encodes the time-invariant amplitude and phase of a sinusoidal function.

You can have voltage and current phasors, but their complex ratio, the impedance, isn't a phasor, since it doesn't encode the parameters of a sinusoidal function.

Last edited: Sep 21, 2015
7. Sep 21, 2015

### milesyoung

Oups, that should be $\mathbf{a} = e^{j2\pi/3}$.

8. Sep 21, 2015

### tim9000

ok thanks, found the paper:

i(t) = I(cos(wt)<0 + cos(wt - 120)<120 + cos(wt - 240)<240 ) = 3/2 * I < wt

=[ei(wt)+e-i(wt)]/2 + [ei(wt +120)+ei(wt + 120)]/2 * (-1/2 + i√3/2) + [ei(wt -120)+ei(wt - 120)]/2 * (-1/2 + i√3/2) = ...

And

A = CosΘ.Cos(wt) + Cos(Θ - 120).Cos(wt - 120) + Cos(Θ + 120).Cos(wt + 120)
= 3/2 * Cos(Θ - wt)

using: sin(wt).sin(Θ) = [Cos(Θ - wt) - Cos(Θ + wt)] / 2
and Cos(a).Cos(b) = Cos(a - b) + sin(a).Sin(b)

cos(Θ - wt) + sin(wt).sin(Θ) + cos(Θ - 120 -(wt - 120)) + sin(Θ - 120).sin(wt - 120) + cos(Θ + 120 - (wt + 120) )+ sin(Θ + 120).sin(wt + 120)

cos(Θ - wt) + [Cos(Θ - wt) - Cos(Θ + wt)] / 2 + cos(Θ - wt) + [cos(Θ - wt) - cos(Θ + wt + 240)] / 2 + cos(Θ - wt) + [cos(Θ - wt) - cos(Θ + wt + 240)/2 = ...

cheers!

9. Sep 21, 2015

### tim9000

hmm,
so are you saying the thing that makes it a phasor is that there is not 't', the 'wt' is implicit, you have to keep it in the back of your mind? Like a phasor has an invisible frequency, but the phasor itself just a complex value?

10. Sep 22, 2015

### milesyoung

Not exactly. What makes a complex constant a phasor is the mapping it represents. It's just a compact way of representing a sinusoidal function with time-invariant parameters, which, in the framework of steady-state AC analysis, is very convenient mathematically to work with. But this is really getting away from the point I was trying to make:

Space vectors have nothing to do with phasors. There are no complex constants, like $Ae^{j\theta}$, which maps to a sinusoidal function, present in this expression:
$$i_a(t) + i_b(t) \mathbf{a} + i_c(t) \mathbf{a}^2, \mathbf{a} = e^{j2\pi/3}$$
You're just working with complex numbers here. Phasors belong in a completely different framework.

There are similarities between space vectors and symmetrical components, in which you actually do work with phasors, but the theory of symmetrical components only deal with steady-state analysis, which is the crucial difference.

11. Sep 22, 2015

### tim9000

I sort of hear what you're saying, I'll endevour to think about it more, I think some specific maths would help. What did you think of the track I went down with my working for the initial two expression derivations?
Cheers

12. Sep 22, 2015

### milesyoung

Focusing on the first bit, I'd start with polar form. It'll make the algebra lot easier:
$$\cos(\omega t) + \cos(\omega t + 2\pi/3)e^{j2\pi/3} + \cos(\omega t - 2\pi/3)e^{-j2\pi/3}$$

13. Sep 22, 2015

### tim9000

I can see in your script the last negative is there, but for some reason it didn't display on my browser.

Right, well, I suppose I could factor out the ej2pi/3

Cos(wt) + ej2pi/3[cos(wt + 2pi/3) + cos(wt - 2pi/3).e^-1]
...actually that didn't help
So I suppose those trig relations were of no relevance?

Sorry, this is embarrassing

14. Sep 22, 2015

### milesyoung

You were on the right track before by converting the cosines to their complex form. I just wanted you to start out with:
$$\cos(\omega t) + \cos(\omega t + 2\pi/3)e^{j2\pi/3} + \cos(\omega t - 2\pi/3)e^{-j2\pi/3}$$

No it's not .

15. Sep 22, 2015

### tim9000

So If I leave the Cosines in their current form, how can I simplify them?

Thank you

Last edited: Sep 22, 2015
16. Sep 22, 2015

### milesyoung

Convert the complex exponentials to their rectangular form (like you did), i.e. $e^{j2\pi/3} = -\frac{1}{2} + j\frac{\sqrt3}{2}, e^{j4\pi/3} = -\frac{1}{2} - j\frac{\sqrt3}{2}$, expand terms, and see what you get.

There's probably a more elegant way to show it, but I'd just use the angle sum identities, so you have, for instance:
$$\cos\left(\theta - \frac{2\pi}{3}\right) = \cos(\theta)\cos\left(\frac{2\pi}{3}\right) + \sin(\theta)\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\cos(\theta) + \frac{\sqrt{3}}{2}\sin(\theta)$$
and expand terms.

17. Sep 22, 2015

### tim9000

Very well, here goes:
i(t) = I(cos(wt)<0 + cos(wt - 120)<120 + cos(wt - 240)<240 ) = I[ cos(wt) + cos(wt + 120)*(-0.5 + j√3/2) + cos(wt + 240)*(-0.5 - j√3/2)]
= .... what am I not seeing?
= 3/2 * I < wt

18. Sep 22, 2015

### milesyoung

Did you have a look at the angle sum identities?

19. Sep 22, 2015

### tim9000

Ah, I see:
i(t) = I(cos(wt)<0 + cos(wt - 120)<120 + cos(wt - 240)<240 ) = I[ cos(wt) + cos(wt + 120)*(-0.5 + j√3/2) + cos(wt + 240)*(-0.5 - j√3/2)]
= I[ cos(wt) + (j√3/2-0.5)*(-0.5*cos(wt) + √3/2*sin(wt) - 0.5*cos(wt) - √3/2*sin(wt))]
=I[ 1.5*cos(wt) - j√3/2*cos(wt)] ....hmm I don't have a calculator on me atm, but if we just took the real part that'd be right?
= 3/2 * I < wt

20. Sep 26, 2015

### tim9000

do you just take the real part?