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jaydnul
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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?
First, where does it work? I don’t remember ever seeing that work in circuitsjaydnul said:why does that work?
So where is ##i \cos = \sin ## in that?jaydnul said:sin(wt) across a cap gives I = c*w*cos(wt). So V/I becomes 1/jwc. It works everywhere else too
Does that complex exponential equation show that i*sin = coshutchphd said:I think you are misinterpreting the fact that $$\frac d {dt}~e^{i\omega t}=i\omega e^{i\omega t}$$ as a new "magic rule" for circuit analysis. This is just a property of a circular function.
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 itjaydnul said:Does that complex exponential equation show that i*sin = cos
I am just wondering why it seems to always work, i cant find an example where it doesnt. But then also, i cant prove i*sin = cos.
Without every invoking complex exponentials, eulers formula, etc. Why can i simply replace all cos terms with isin, do the algebra, then replace all the isin terms back with cos and somehow magically always get the correct answer.
Ahh maybe thats it. But in circuits all the of the components are linearized. Whats an example of what you are talking about, maybe im confused?hutchphd said: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
Ok, so you seem to be mixing phasor concepts with time domain notation. So in phasor notation ##A \cos(\omega t+\phi)## is written as ##A\angle \phi## which is further interpreted as the complex number ##A e^{j\phi}##. So the phasor of ##\cos(\omega t)## would be ##1 e^{j0}=1## and the phasor of ##\sin (\omega t)=\cos (\omega t - \pi/2)## would be ##1 e^{-j\pi/2}=-j##. So ##phasor(\cos)=j \ phasor(\sin)## which does not mean ##\cos = j \ \sin##jaydnul said:should be i*sin = cos, sorry. V/I = sin/(cwcos). replace cos with jsin and it simplifies to 1/jwc
If you are using complex numbers in circuit analysis (like a complex impedance, or your transfer function), then you are using Euler's formula* to express quantities that have both an amplitude and phase. It won't do you much good in the long run to pretend that you aren't. It's not any easier your way, assuming correct answers are your goal.jaydnul said:another example:
i have a transfer function given by the complex number (A+jB). I multiply my input sin(wt) and i get Asin(wt)+jBsin(wt). Using the equality above, it becomes Asin(wt) + Bcos(wt), which is just standard form of any sinusoid with a magnitude of sqrt(A^2 + B^2). No need to convert into an exponential with eulers formula, then take the "real" part of that exponential.
Because that's just not true. EEs use math the way it's taught in math classes. Inventing something new without any rigorous basis will not work for you. The mathematical approaches to circuit analysis are not tricks that someone stumbled across or found through trial and error. There is a firm logically consistent reason for the analytical tools used.jaydnul said:i cant prove i*sin = cos.
Of course this example works, the way you did it anyway. All you did was change the symbols while still treating it as a 2-d vector (the original complex number). A & B terms didn't change and those were all you used to find the magnitude and phase. You could have substituted apples and monkeys in place of sin and cos, as long as apples and monkeys make an orthogonal basis. I don't really follow how this improves anything.jaydnul said:
That substitution doesn't work in the example that you gave in post 5.jaydnul said:Just asking for an example where that substitution doesnt work thats it.
If the voltage across a capacitor is ##v= \sin(\omega t)## then the current through it is ##i = C \dot v = C \omega \cos(\omega t)## so $$\frac{v}{i} = \frac{\sin(\omega t)}{C \omega \cos(\omega t)} \ne \frac{1}{j\omega C}$$ What does work, as I said, is to use phasor notation.jaydnul said:should be i*sin = cos, sorry. V/I = sin/(cwcos). replace cos with jsin and it simplifies to 1/jwc
Since you claim to understand phasors, why don’t you use them the same way as all other EEs I have ever worked with and gone to school with? Were you taught this method?jaydnul said:I get it, I understand phasors, i am an MSEE working in industry as an analog designer. Just asking for an example where that substitution doesnt work thats it. Phasors are already abstract, and its mysterious to me as to why that substitution seems to always work.
But you need the derivative of apples to be monkeys, as well as orthogonal. So i dont see the point you are making.DaveE said:You could have substituted apples and monkeys in place of sin and cos, as long as apples and monkeys make an orthogonal basis.
Yes it does? Substitute cos(wt) with j*sin(wt) and thats exactly what you get, and kinda my entire point here.Dale said:That substitution doesn't work in the example that you gave in post 5. If the voltage across a capacitor is ##v= \sin(\omega t)## then the current through it is ##i = C \dot v = C \omega \cos(\omega t)## so $$\frac{v}{i} = \frac{\sin(\omega t)}{C \omega \cos(\omega t)} \ne \frac{1}{j\omega C}$$ What does work, as I said, is to use phasor notation.
If you make the substitution then you do get the right hand side of the inequality, but it doesn’t equal the left hand side of the inequality. In other words, once you make the substitution, the remaining quantity is no longer equal to ##v/i## where ##v## and ##i## are functions of time, not phasors.jaydnul said:Yes it does? Substitute cos(wt) with j*sin(wt) and thats exactly what you get, and kinda my entire point here.
hutchphd said: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.
It is interesting to think about and i enjoy it. Why do people post things like this lol. The only reasonable responses should be:jasonRF said:Since you claim to understand phasors, why don’t you use them the same way as all other EEs I have ever worked with and gone to school with? Were you taught this method?
Of course, if ##\omega t## is real then ##\cos(\omega t)## is real and ##i \, \sin(\omega t)## is imaginary, so they are never equal. Writing ##i \, \sin(\omega t)=\cos(\omega t)## makes as much sense as writing ##1=2##. They are both false.
the units for 1/jwc is the same as V/I. I mean it is a function of t, the two sin(wt) just cancel eachothers impacts…Dale said:If you make the substitution then you do get the right hand side of the inequality, but it doesn’t equal the left hand side of the inequality. In other words, once you make the substitution, the remaining quantity is no longer equal to ##v/i## where ##v## and ##i## are functions of time,
That is exactly the problem. ##\cos/\sin=\tan\ne 1/j##jaydnul said:the two sin(wt) just cancel eachothers impacts
And I am telling you that the substitution doesn’t work.jaydnul said:Im wondering why IT WORKS when you substitute jsin. Thats it
Using phasors to derive the magnitude of the transfer function of a circuit gives THE EXACT SAME results as substituting jsin=cos. Thats what i mean by work. Input output. If you want to argue what word to use for that instead of WORKS thats fineDale said:And I am telling you that the substitution doesn’t work.
What does work is to transform to phasors.
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##.jaydnul said:Using phasors to derive the magnitude of the transfer function of a circuit gives THE EXACT SAME results as substituting jsin=cos. Thats what i mean by work. Input output. If you want to argue what word to use for that instead of WORKS thats fine
Hutchphd had what seemed to be the only helpful post so far. Can anyone expand on that
Hey haushofer, ya i tried that. if you do that then it just spits it back out. jsinx = jsinx.haushofer said: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?
I understand all of this, ill stop using the word "works" since that seems to be what we are hung up on.Dale said: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.
DaveE said:If you aren't working with phasors, it's just wrong.
This is an interesting comment. I'm not sure how to take it further, but maybe it is an explaination?hutchphd said: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 step is a sloppy transform to the phasor representation. The purpose of the phasor representation is indeed to simplify the math.jaydnul said:substitute jsin for cos terms to further simplify the algebra
This step is a sloppy transform back from the phasor representation.jaydnul said:substitute cos back in for all the jsin terms
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?jaydnul said: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 am actually hung up on saying ##\cos = j \sin## when what you really mean is that in the phasor representation ## 1 = j (-j)##.jaydnul said:ill stop using the word "works" since that seems to be what we are hung up on.
Isnt that curious? Idk i find it strange.hutchphd said: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.