Three phase phasor proof or simplification

[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

and I couldn't figure out how that could be so (still can't, please help)

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!

 

[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).

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

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

 

[tim9000]

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.

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.

What about:

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

no idea how it reduces to that?

 

[milesyoung]

Yeah as far as I remember a space vector is the addition of phasors ...

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.

... through feeding a sineusoidally current to a sineusoidally distributed winding in space.

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

Yeah I tried that thanks.

I can help you with the algebra if you show your work.

no idea how it reduces to that?

Try doing the complex algebra first.

 

[tim9000]

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.

Could please you elaborate on that?

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

 

[milesyoung]

Could please you elaborate on that?

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

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

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

... I(cos(wt)<0 + cos(wt - 120)<120 + cos(wt - 240)<240 ) ...

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.

 

[milesyoung]

$$ \mathbf{i_s} = i_a(t) + i_b(t) \mathbf{a} + i_c(t) \mathbf{a}^2, \mathbf{a} = e^{2\pi/3}$$

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

 

[tim9000]

I can help you with the algebra if you show your work.

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

ok thanks, found the paper:

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

=[e^i(wt)+e^-i(wt)]/2 + [e^{i(wt +120)}+e^{i(wt + 120)}]/2 * (-1/2 + i√3/2) + [e^{i(wt -120)}+e^{i(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!

 

[tim9000]

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.

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?

 

[milesyoung]

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?

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.

 

[tim9000]

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.

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

 

[milesyoung]

What did you think of the track I went down with my working for the initial two expression derivations?

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

 

[tim9000]

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

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 e^j2pi/3

Cos(wt) + e^j2pi/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

 

[milesyoung]

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

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

Sorry, this is embarrassing

No it's not .

 

[tim9000]

No it's not .

I appreciate your tact.

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

Thank you

P.S also what about:

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

 

[milesyoung]

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

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.

P.S also what about:
A = CosΘ.Cos(wt) + Cos(Θ - 120).Cos(wt - 120) + Cos(Θ + 120).Cos(wt + 120)
= 3/2 * Cos(Θ - wt)

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.

 

[tim9000]

Convert the complex exponentials to their rectangular form (like you did), i.e. ej2π/3=−12+j3√2,ej4π/3=−12−j3√2e^{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.

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

 

[milesyoung]

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

Did you have a look at the angle sum identities?

 

[tim9000]

Did you have a look at the angle sum identities?

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

 

[tim9000]

Did you have a look at the angle sum identities?

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

do you just take the real part?

 

[milesyoung]

do you just take the real part?

No, what's the reasoning?

This:

... cos(wt - 120)<120 + cos(wt - 240)<240 ...

should reduce to this:
$$
\frac{1}{2}\cos(\omega t) + j\frac{3}{2}\sin(\omega t)
$$
using the angle sum identities. Check your algebra carefully.

 

[tim9000]

what's the reasoning?

Fair call.

should reduce to this:

12cos(ωt)+j32sin(ωt)​

Right you are:

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) + (-0.5 + j√3/2)*(-0.5cos(wt) - √3/2*sin(wt)) + (-0.5 - j√3/2)*(- 0.5*cos(wt) + √3/2*sin(wt))]

= I[cos(wt) + -0.5*(-0.5cos(wt) - √3/2*sin(wt) - 0.5*cos(wt) + √3/2*sin(wt)) - j3/4*sin(wt) - j3/4*sin(wt)]

= I[cos(wt) + 0.5*-cos(wt) - j3/2*sin(wt)]

So I [ 0.5*cos(wt) - j3/2*sin(wt)] is 3/2<wt?

Thanks very much

 

[tim9000]

No, what's the reasoning?

This:

should reduce to this:
$$
\frac{1}{2}\cos(\omega t) + j\frac{3}{2}\sin(\omega t)
$$
using the angle sum identities. Check your algebra carefully.

Hey,
As in post #22, So getting from [ 0.5*cos(wt) - j3/2*sin(wt)]
to 3/2<wt
Is that a conceptual leap rather than a mathematical one?

Thanks

 

[milesyoung]

As in post #22, So getting from [ 0.5*cos(wt) - j3/2*sin(wt)]
to 3/2<wt
Is that a conceptual leap rather than a mathematical one?

It's true that:
$$
\cos\left(\omega t - \frac{2\pi}{3}\right)\mathbf{a} + \cos\left(\omega t - \frac{4\pi}{3}\right)\mathbf{a}^2 = \frac{1}{2}\cos(\omega t) + j\frac{3}{2}\sin(\omega t)
$$
but you still need to add the $\cos(\omega t)$ term. Then apply Euler's formula.

 

[tim9000]

It's true that:
$$
\cos\left(\omega t - \frac{2\pi}{3}\right)\mathbf{a} + \cos\left(\omega t - \frac{4\pi}{3}\right)\mathbf{a}^2 = \frac{1}{2}\cos(\omega t) + j\frac{3}{2}\sin(\omega t)
$$
but you still need to add the $\cos(\omega t)$ term. Then apply Euler's formula.

I'm really sorry, but what do you mean?

 

[milesyoung]

I'm really sorry, but what do you mean?

There's nothing to be sorry about.

This is what you posted originally:

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

Notice that besides:

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

You also have:

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

 

[tim9000]

There's nothing to be sorry about.

This is what you posted originally:

Notice that besides:

You also have:

Ah, I see what you're saying, but didn't I have:

cos(wt) -0.5*cos(wt) in the final expession, so wasn't that cos(wt) accounted for?

Cheers

 

[milesyoung]

Ah, I see what you're saying, but didn't I have:

cos(wt) -0.5*cos(wt) in the final expession, so wasn't that cos(wt) accounted for?

$$
\begin{align*}
\cos\!\left(\omega t - \frac{2\pi}{3}\right)\mathbf{a} &= \left(-\frac{1}{2}\cos(\omega t) + \frac{\sqrt{3}}{2}\sin(\omega t)\right)\left(-\frac{1}{2} + j\frac{\sqrt{3}}{2}\right)\\
&= \frac{1}{4}\cos(\omega t) - \frac{\sqrt{3}}{4}\sin(\omega t) - j\frac{\sqrt{3}}{4}\cos(\omega t) + j\frac{3}{4}\sin(\omega t)\\\\

\cos\!\left(\omega t - \frac{4\pi}{3}\right)\mathbf{a}^2 &= \left(-\frac{1}{2}\cos(\omega t) - \frac{\sqrt{3}}{2}\sin(\omega t)\right)\left(-\frac{1}{2} - j\frac{\sqrt{3}}{2}\right)\\
&= \frac{1}{4}\cos(\omega t) + \frac{\sqrt{3}}{4}\sin(\omega t) + j\frac{\sqrt{3}}{4}\cos(\omega t) + j\frac{3}{4}\sin(\omega t)
\end{align*}
$$
And then you have:
$$
\begin{align*}
&\frac{1}{4}\cos(\omega t) - \frac{\sqrt{3}}{4}\sin(\omega t) - j\frac{\sqrt{3}}{4}\cos(\omega t) + j\frac{3}{4}\sin(\omega t)\\
+ &\frac{1}{4}\cos(\omega t) + \frac{\sqrt{3}}{4}\sin(\omega t) + j\frac{\sqrt{3}}{4}\cos(\omega t) + j\frac{3}{4}\sin(\omega t)\\
= &\frac{1}{2}\cos(\omega t) + j\frac{3}{2}\sin(\omega t)
\end{align*}
$$
so check your algebra again carefully.

 

[tim9000]

So was what you were saying before, if the 'x' in:

has a frequency, I can see that the magnitude would rotate around the zero between the real and imaginary planes, but are you saying that's true for...
Hmm I'd better start again, that ...<wt on the magnetising current, what is that actually indicating?

so check your algebra again carefully.

I'm going to take your advice, so I better go over it in the morning...or afternoon, it's 4:20am here.

Cheers!

 

[milesyoung]

Hmm I'd better start again, that ...<wt on the magnetising current, what is that actually indicating?

Say you have some complex number $z = re^{j\phi}$. The $re^{j\phi}$ bit is its polar form, which you can think of as a vector in the complex plane with magnitude $r$ and angle (phase) $\phi$.

The complex number $\frac{3}{2}e^{j\omega t}$ then represents a vector of constant magnitude $\frac{3}{2}$, which is rotating counterclockwise in the complex plane with angular frequency $\omega$. See here for an illustration.

That's exactly the result you would expect in the framework of space vectors as they relate to electrical machines.

The vector shows you what happens if you apply a balanced set of three-phase currents to the stator of a symmetrical machine: it produces a stator current space vector, which rotates CCW in the plane normal to the rotor axis of the machine. There's no actual current with any spatial direction, that's nonsense, but the stator current space vector is aligned with the magnetic axis of the resulting stator field, which makes it a very useful abstraction (and one of the many great properties of space vectors).

 

[tim9000]

Say you have some complex number $z = re^{j\phi}$. The $re^{j\phi}$ bit is its polar form, which you can think of as a vector in the complex plane with magnitude $r$ and angle (phase) $\phi$.

The complex number $\frac{3}{2}e^{j\omega t}$ then represents a vector of constant magnitude $\frac{3}{2}$, which is rotating counterclockwise in the complex plane with angular frequency $\omega$. See here for an illustration.

That's exactly the result you would expect in the framework of space vectors as they relate to electrical machines.

The vector shows you what happens if you apply a balanced set of three-phase currents to the stator of a symmetrical machine: it produces a stator current space vector, which rotates CCW in the plane normal to the rotor axis of the machine. There's no actual current with any spatial direction, that's nonsense, but the stator current space vector is aligned with the magnetic axis of the resulting stator field, which makes it a very useful abstraction (and one of the many great properties of space vectors).

Hey I'm back, sorry about the delay.
Ok this is really dumb question but:
I see how the phasee shifted cos-es are adding up to 3/2 with the frequency omega. But the way w're representing it with e^jwt
with the circle on the left hand side, does that imply that the y-axis is actually the imaginary axis?
Cheers

 

[milesyoung]

I see how the phasee shifted cos-es are adding up to 3/2 with the frequency omega. But the way w're representing it with e^jwt
with the circle on the left hand side, does that imply that the y-axis is actually the imaginary axis?

Yes, did you have a look at the Wikipedia page?

 

[tim9000]

Yes, did you have a look at the Wikipedia page?

I did, it's what I based my assertion. It just seems odd to me that we would model this circular path that takes place in an imaginary and real set of axies, to represent the actual vector of the peak mmf in real life. I can accept it though.

 

[milesyoung]

It just seems odd to me that we would model this circular path that takes place in an imaginary and real set of axies, to represent the actual vector of the peak mmf in real life.

There's nothing in the framework of space vectors that you can't also express with vectors and matrix algebra, see, for example, the equivalent Alpha–beta transformation, but the algebra of complex numbers is often preferred by electrical engineers. You can just think of the real and imaginary components as coordinates in a plane (the complex plane).