B Understanding the Relationship Between i*cos and sin in Circuit Analysis

[jaydnul]

Maybe I missed it (sorry if I did), but why don't you write sin(x) and cos(x) in terms of complex exponentials and check it explicitly?

Hey haushofer, ya i tried that. if you do that then it just spits it back out. jsinx = jsinx.

So the analytic representation of $A \cos(\omega t+\phi)$ is $A e^{j\phi}e^{j\omega t}$ and the phasor representation is the non-time-dependent part $A e^{j\phi}$. So the phasor representation of $\cos(\omega t)$ is $e^{j0}=1$.

Since $\sin(\theta)=\cos(\theta-\pi/2)$ the phasor representation of $\sin(\omega t)$ is $e^{-j\pi/2}=-j$.

So $$phasor(\cos(\omega t))=j \ phasor(\sin(\omega t))$$ This is where the relationship you are seeing comes in. But this in no way implies that $$\cos(\omega t)=j \ \sin(\omega t)$$ That substitution does not work for the reasons I showed above.

You are just being sloppy. You need to clearly identify when you are working with functions of time and when you are working with their phasor representations. A function is not the same as the representation of that function in another domain.

Similarly, when you are working with Fourier transforms you need to be clear about when a function is in the time domain and when it is in the frequency domain. If $X(\omega)=\mathcal F [x(t)]$ then $$\mathcal F[ \dot x]=j\omega X$$ is correct whereas $$\dot x=j\omega x$$ is simply wrong. This is what you are doing and claiming that it works. Sloppy mixing of domains doesn't work.

I understand all of this, ill stop using the word "works" since that seems to be what we are hung up on.

Mathematically, if you input a sin signal into a circuit, calculate the transfer characteristics, substitute jsin for cos terms to further simplify the algebra, then substitute cos back in for all the jsin terms, you end up with a sin wave with exactly the same magnitude as when you do it with phasor representation. I can plug in x values into the answer i arrived at, it will give me the right values for y. plain and simple.

Intellectually it is an interesting exercise to find out why that happens to spit out the same answer as the phasor method, and why it wont work in some instances.

Just a bunch of dogmatic explanations so far like just listen to us, thats not how you are supposed to do it

If you aren't working with phasors, it's just wrong.

 

[jaydnul]

because the problem is linear and all the i does is identify the imaginary part. So switching and then switching back works OK. If your problem scrambled them (by squaring for instance) the result would not work out. Try it

This is an interesting comment. I'm not sure how to take it further, but maybe it is an explaination?

Something else I was thinking about (which may be complete nonsense) is that -sin^2 is equal to cos^2 when looking at phase and magnitude ONLY. The dc bias separates them, but working with j seems to remove the DC bias from the calculations? idk attacking the problem from an EE perspective haha probably just nonsense

 

[Dale]

substitute jsin for cos terms to further simplify the algebra

This step is a sloppy transform to the phasor representation. The purpose of the phasor representation is indeed to simplify the math.

substitute cos back in for all the jsin terms

This step is a sloppy transform back from the phasor representation.

Intellectually it is an interesting exercise to find out why that happens to spit out the same answer as the phasor method, and why it wont work in some instances

I don't think that figuring out when it is ok to be sloppy is a good idea. I think it is better to not be sloppy in the first place. What possible disadvantage is there to being clear and careful?

Personally, I think transforming to and from phasors is so easy that there is no reason to make it sloppy. I would rather have one clear rule that I know works and can apply systematically instead of a collection of ad hoc rules that are unclear and their limits are unknown. Particularly when the clear and systematic rule is as easy as phasors.

It seems likely that the sloppy approach will not work any time that phasors will not work. Are there cases where phasors will work and the sloppy approach will not? I don't know.

ill stop using the word "works" since that seems to be what we are hung up on.

I am actually hung up on saying $\cos = j \sin$ when what you really mean is that in the phasor representation $1 = j (-j)$.

$\cos = j \sin$ is simply false. $1=j(-j)$ is true. The fact that the second is the phasor representation of the first doesn't make the first true.

 

[hutchphd]

In phasor world what you are doing is equivalent to taking your phasor graph, rotating it 90deg on the table, doing some stuff and then rotating it back. As long as tyou work consistently, this should work out fine. Only the relative phases matter.

 

[jaydnul]

In phasor world what you are doing is equivalent to taking your phasor graph, rotating it 90deg on the table, doing some stuff and then rotating it back. As long as tyou work consistently, this should work out fine. Only the relative phases matter.

Isnt that curious? Idk i find it strange.

Maybe it works in circuits because everything is linearized as you said, but in quantum mechanics for instance would this fall apart? Its been so long since i took qm so i probably dont remember well enough to generate an example like that

 

[hutchphd]

You have not supplied any exact example of the process so I no longer know what you are talking about. This is why exact referrals to published works is usually required by PF. You need to be more definitive about what "it" is.......

 

[jaydnul]

You have not supplied any exact example of the process so I no longer know what you are talking about. This is why exact referrals to published works is usually required by PF. You need to be more definitive about what "it" is.......

That's a fair point. I guess in electronics you can derive a transfer equation when you input a sinusoid into the circuit, but quantum mechanics isn't exactly similar in that way

 

[DaveE]

In circuit analysis, everything seems to work out when you set i*cos = sin. But thats not a legitimate equation, so why does that work? Is there a proof that this is a real equation?

Just a bunch of dogmatic explanations so far like just listen to us, thats not how you are supposed to do it

What do you want from us?

Sometimes education is dogmatic, sometimes that's the best way to get people on the right track. You have several people that know a lot about this subject trying to help. Do you want to listen or are you seeking irrational affirmation?

 

[hutchphd]

Ask a particular question please. One with well-defined terms. If you wish to be bemused, please do it privately.
Comparison similar to electronic I/O in QM is often called scattering theory. The arithmetic is similar.

 

[vanhees71]

In AC analysis, it's simplifying things tremendously, when you use the fact that you deal with linea differential equations with real coefficients for real functions $U(t)$ and $I(t)$ ("voltage" and currents). Further usual household voltage is harmonic.

Take for example the series of a coil and a resistande. The corresponding equation reads
$$L \dot{I} + R I=U_0 \cos(\omega t).$$
The idea is that to find the solution for this equation it's much simpler to introduce complex quantities $i(t)$ and $u(t)$ and define the physical real quantities as the real part of these complex quantities, i.e., $U=\mathrm{Re} u$, $I=\mathrm{Re} i$. The external AC voltage is given by $U_0 \exp(\mathrm{i} \omega t)$ (using the engineers' sign convention in the exponential, because then $\mathrm{Re} [U_0 \exp(\mathrm{i} \omega t)=U_0 \cos (\omega t).$
Then we can solve the equation for the complex quantities,
$$L \dot{i} + R i=U_0 \exp(\mathrm{i} \omega t).$$
Obviously the real part of the solution for $i$ then gives you $I$, because $L$, $R$, and $U_0$ are all real.

To find the "stationary state" you can simply make the ansatz
$$i(t)=i_0 \exp(\mathrm{i} \omega t),$$
and plugging this into the equation you only have to solve an algebraic equation for the complex amplitude $i_0$:
$$(\mathrm{i} \omega L + R) i_0 = U_0 \; \Rightarrow \; i_0 = \frac{U_0}{R+\mathrm{i} \omega L} = \frac{R-\mathrm{i} \omega L}{R^2+(\omega L)^2}.$$
The physical current thus is
$$I(t)=\mathrm{Re} i(t)=\frac{U_0}{|R^2+(\omega L)^2} [R \cos(\omega t) + \omega L \sin(\omega t)].$$
This becomes a bit more convenient if you write the complex "inverse resistance" (impedance) in "polar form"
$$\frac{R-\mathrm{i} \omega L}{R^2+(\omega L)^2} = \frac{1}{\sqrt{R^2+(\omega L)^2}} \exp(\mathrm{i} \phi), \quad \varphi=-\arccos \left (\frac{R}{\sqrt{R^2+(\omega L)^2}} \right).$$
Then you get
$$I(t)=\frac{U}{\sqrt{R^2+(\omega L)^2}} \cos(\omega t+ \phi).$$
This gives you immediately the amplitude of the current,
$$I_0=\frac{U}{\sqrt{R^2+(\omega L)^2}},$$
and the phase shift $\phi$, given above. It's negative, which means that the current's phase is always behind the external voltages phase, as expected for an inductance.

With this complex treatment you can calculate any AC circuit's stationary state as if it were a DC circuit, using Kirchhoff's two rules but use for resistors, capacitors, and inductances the corresponding complex impedances,
$$Z_R=R, \quad Z_L=\mathrm{i} \omega L, \quad Z_C=-\frac{\mathrm{i}}{\omega C}.$$
At the end of the cacluation you get all real physical currents and voltages by taking the real part of the corresponding complex quantities.