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

[jaydnul]

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?

 

[Dale]

why does that work?

First, where does it work? I don’t remember ever seeing that work in circuits

 

[jaydnul]

sin(wt) across a cap gives I = c*w*cos(wt). So V/I becomes 1/jwc. It works everywhere else too

 

[Dale]

sin(wt) across a cap gives I = c*w*cos(wt). So V/I becomes 1/jwc. It works everywhere else too

So where is $i \cos = \sin$ in that?

 

[jaydnul]

should be i*sin = cos, sorry. V/I = sin/(cwcos). replace cos with jsin and it simplifies to 1/jwc

 

[jaydnul]

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.

 

[hutchphd]

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.

 

[jaydnul]

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.

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.

 

[hutchphd]

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.

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

 

[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

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?

 

[Dale]

should be i*sin = cos, sorry. V/I = sin/(cwcos). replace cos with jsin and it simplifies to 1/jwc

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]

 

[DaveE]

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.

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.

i cant prove i*sin = cos.

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.

I think it's time to take a big step back and look not just at how these problems are solved, but also why those tools are used; why they work.

* Alternately you can think of everything as a 2-D vector instead of complex numbers, it's really the same thing.

 

[jaydnul]

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.

 

[DaveE]

View attachment 333813

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.

 

[Dale]

Just asking for an example where that substitution doesnt work thats it.

That substitution doesn't work in the example that you gave in post 5.

should be i*sin = cos, sorry. V/I = sin/(cwcos). replace cos with jsin and it simplifies to 1/jwc

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.

 

[jasonRF]

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.

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.

 

[jaydnul]

You could have substituted apples and monkeys in place of sin and cos, as long as apples and monkeys make an orthogonal basis.

But you need the derivative of apples to be monkeys, as well as orthogonal. So i dont see the point you are making.

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.

Yes it does? Substitute cos(wt) with j*sin(wt) and thats exactly what you get, and kinda my entire point here.

 

[Dale]

Yes it does? Substitute cos(wt) with j*sin(wt) and thats exactly what you get, and kinda my entire point here.

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]

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.

I really appreciate all the posts but this is the only one that is attempting to answer my question. Is this expandable?

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.

It is interesting to think about and i enjoy it. Why do people post things like this lol. The only reasonable responses should be:

i dont know

it is unknowable

this is why that happens to work but here is the context where it doesn't work/falls apart

 

[jaydnul]

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,

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]

the two sin(wt) just cancel eachothers impacts

That is exactly the problem. $\cos/\sin=\tan\ne 1/j$

 

[jaydnul]

Im wondering why IT WORKS when you substitute jsin. Thats it

 

[Dale]

Im wondering why IT WORKS when you substitute jsin. Thats it

And I am telling you that the substitution doesn’t work.

What does work is to transform to phasors.

 

[jaydnul]

And I am telling you that the substitution doesn’t work.

What does work is to transform to phasors.

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

 

[hutchphd]

It might be easier if you give an explicit (excruciatingly detailed) example of what you mean. I think you will see it is not very interestng when looked at in detail. Phasors and complex notation give the same results when applied appropriately. So please give one concrete example of your substition (in gory detail)

 

[DaveE]

Because of the trig identity $A sin(\omega t) + B cos(\omega t) = \sqrt{A^2+B^2}~ sin(\omega t + atan(\frac{B}{A}))$

And
$|(A+jB) sin(\omega t)| = |(A+jB)| ⋅ |sin(\omega t)| = \sqrt{A^2+B^2} ~|sin(\omega t)|$

$arg((A+jB) sin(\omega t)) = arg(A+jB) + arg(sin(\omega t)) = atan(\frac{B}{A}) + arg(sin(\omega t))$

Then when you jump to phasor notation and drop the $sin(\omega t)$ terms they are the same. It's the same solution algorithm, that is all. Nothing Euler didn't know 300 years ago. If you aren't working with phasors, it's just wrong.

 

[hutchphd]

I am now officially lost.

 

[haushofer]

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?

 

[Dale]

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

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.

 

[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.