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sin over cos equals -j ?

abaz

New member
Feb 23, 2013
9
Hi all

I am trying to go through and understand the derivation of
\[X_{c} = \frac{1}{j\omega C}\]
Where $X_{c}$ is capacitive reactance in ohms, $\omega$ is the angular velocity or $2\pi frequency$ and $C$ is capacitance in Farads


To start with we already have $I=C\frac{dv}{dt}$, $V=V_{pk}\sin\left (\omega \right )t$ and that $X_{c}=\frac{V}{I}$
Then
\[I=C\frac{d}{vt}V_{pk}\sin \left ( \omega \right )t = C\cdot V_{pk}\omega \cos \left ( \omega \right )t\]
Therefore
\[X_{c}=\frac{V_{pk}\sin\left (\omega \right )t}{C\cdot V_{pk}\omega \cos \left ( \omega \right )t}\]
$V_{pk}$ and $t$ cancel out and we can be left with
\[X_{c}=\frac{1}{\omega C}\cdot \frac{\sin \left (\omega \right )}{\cos \left (\omega \right )}\]

Now I am up to the part I don't understand. The only non-Euler derivation of this formula that I have been able to find was on youtube and unfortunately it gets a bit vague here. He says that because we know that for capacitance current leads the voltage that any real number multiplied by $90^{\circ}$, or $pi$ on the Argand plane, it swings up to the imaginary axis and thus it becomes $-j$.

I'd like to see how the second term can be mathematically turned into $-j$ or $\frac{1}{j}$ for that matter.

Any help or guidance would be greatly appreciated
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hi abaz! :)

It seems you are stuck with the Euler derivation after all.
The vagueness you are referring to, relates directly to a derivation using Euler's formula.
When you swing an angle by 90 degrees, a sine becomes a cosine.
In Euler's representation of a complex number that corresponds to a swing from the real numbers into the imaginary numbers (j or -j).

As it is, the capacitance current leads by 90 degrees, meaning we get an expression for Xc that contains both a sine and a cosine.
Basically you can leave out all sines and cosines to get the expression for the impedance.

I do not know how to explain this better without going into complex numbers and Euler's formula.
 

zzephod

Well-known member
Feb 3, 2013
134
Hi all

I am trying to go through and understand the derivation of
\[X_{c} = \frac{1}{j\omega C}\]
Where $X_{c}$ is capacitive reactance in ohms, $\omega$ is the angular velocity or $2\pi frequency$ and $C$ is capacitance in Farads


To start with we already have $I=C\frac{dv}{dt}$, $V=V_{pk}\sin\left (\omega \right )t$ and that $X_{c}=\frac{V}{I}$
Then
\[I=C\frac{d}{vt}V_{pk}\sin \left ( \omega \right )t = C\cdot V_{pk}\omega \cos \left ( \omega \right )t\]
Therefore
\[X_{c}=\frac{V_{pk}\sin\left (\omega \right )t}{C\cdot V_{pk}\omega \cos \left ( \omega \right )t}\]
$V_{pk}$ and $t$ cancel out and we can be left with
\[X_{c}=\frac{1}{\omega C}\cdot \frac{\sin \left (\omega \right )}{\cos \left (\omega \right )}\]

Now I am up to the part I don't understand. The only non-Euler derivation of this formula that I have been able to find was on youtube and unfortunately it gets a bit vague here. He says that because we know that for capacitance current leads the voltage that any real number multiplied by $90^{\circ}$, or $pi$ on the Argand plane, it swings up to the imaginary axis and thus it becomes $-j$.

I'd like to see how the second term can be mathematically turned into $-j$ or $\frac{1}{j}$ for that matter.

Any help or guidance would be greatly appreciated
Your input should be \( \displaystyle V_{pk} e^{j\omega t} \) not \(V_{pk} \sin(\omega t)\)

.
 

abaz

New member
Feb 23, 2013
9
Thank you I like Serena (your avatar had me in stitches!)
It would seem that you and zzephod are both steering me toward Euler

I may be over thinking this (or being to simplistic) but if Euler's formula yields $X_{c}=\frac{-j}{\omega C}$
and that assuming the partial derivation that I showed is correct does that mean
$\frac{\sin \left ( \omega \right )}{\cos \left ( \omega \right )}$
can somehow (Eulers?) evaluate to $-j$?

Any way thank you both for the input I'm going to try and get my head around Euler's formula. It may show that I'm being silly :)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If you use the input suggest by zzephod, you find that:

$\displaystyle \frac{dV}{dt}=j\omega V$

and so:

$\displaystyle I=Cj\omega V$

therefore:

$\displaystyle X_C=\frac{V}{I}=\frac{1}{j\omega C}$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Thank you I like Serena (your avatar had me in stitches!)
It would seem that you and zzephod are both steering me toward Euler
Thanks!
And yes.

I may be over thinking this (or being to simplistic) but if Euler's formula yields $X_{c}=\frac{-j}{\omega C}$
and that assuming the partial derivation that I showed is correct does that mean
$\frac{\sin \left ( \omega \right )}{\cos \left ( \omega \right )}$
can somehow (Eulers?) evaluate to $-j$?

Any way thank you both for the input I'm going to try and get my head around Euler's formula. It may show that I'm being silly :)
I wouldn't say that exactly, although it does look like it.

What happens is that instead of $\sin\omega t$ you have $e^{j\omega t}$ which includes an imaginary component that sine does not.
And instead of $\cos \omega t$ you have $j e^{j\omega t}$, where the $j$ comes out due to the differentiation and we had already isolated $\omega$.

So instead of $\dfrac{\sin \omega t}{\cos \omega t}$ you get $\dfrac{e^{j\omega t}}{j e^{j\omega t}} = \dfrac 1 j = -j$.

As Mark showed, the complete derivation is pretty elegant.
 

abaz

New member
Feb 23, 2013
9
Thank you very much Mark and I Like Serena (is just ILS or Serena ok?)

I have read a little on Euler's number $e$ since this post and it's significance is impressive to say the least, so I will spend a while learning more...

In my original post I cancelled out the $t$'s when I shouldn't have?
Should it have been $\sin ( \omega t)$...

This would make more sense otherwise $X_{c} = 0$

I have only ever used degrees and only recently started using radians
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Thank you very much Mark and I Like Serena (is just ILS or Serena ok?)
I'm used to ILS by now but either is fine.

I have read a little on Euler's number $e$ since this post and it's significance is impressive to say the least, so I will spend a while learning more...
Good!
Its simplicity and usefulness is impressive.

In my original post I cancelled out the $t$'s when I shouldn't have?
Should it have been $\sin ( \omega t)$...
Yep, it should have been $\sin(\omega t)$.
Note how dividing by $\cos(\omega t)$ will give problems when the cosine gets to be zero...

This would make more sense otherwise $X_{c} = 0$

I have only ever used degrees and only recently started using radians
There's a first time for everything. ;)
It's like switching from feet or yards to metric meters.
Once you get used to it, it really makes more sense.
 

abaz

New member
Feb 23, 2013
9
Yep, it should have been $\sin(\omega t)$.
Note how dividing by $\cos(\omega t)$ will give problems when the cosine gets to be zero...
I actually plotted $\frac{\sin \left ( \theta\right )}{\cos \left (\theta\right )}$ and said Hey thats tan!. Substituted sin and cos with (opp/hyp)/(adj/hyp) and ended up with opp/adj Thought I was so :cool: LOL ... forgotten trig id's
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I actually plotted $\frac{\sin \left ( \theta\right )}{\cos \left (\theta\right )}$ and said Hey thats tan!. Substituted sin and cos with (opp/hyp)/(adj/hyp) and ended up with opp/adj Thought I was so :cool: LOL ... forgotten trig id's
Yep. ;-)
Btw, $X_c$ is a so called complex impedance.
It only works if you use complex numbers.
 

abaz

New member
Feb 23, 2013
9
Funny thing that.
I'm an electrician, did 3yr trade course and I'm almost certain nothing complex was ever mentioned! Yet here we were, day in, day out, drawing phasor's on the x/y plane. Probably why I find j terms so elusive
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Funny thing that.
I'm an electrician, did 3yr trade course and I'm almost certain nothing complex was ever mentioned! Yet here we were, day in, day out, drawing phasor's on the x/y plane. Probably why I find j terms so elusive
Phasor's... isn't that a way to visualize complex numbers in practice?

Or as wiki puts it:
Electrical engineers, electronics engineers, electronic engineering technicians and aircraft engineers all use phasor diagrams to visualize complex constants and variables (phasors).
 

abaz

New member
Feb 23, 2013
9
Phasor's... isn't that a way to visualize complex numbers in practice?

Or as wiki puts it:
Electrical engineers, electronics engineers, electronic engineering technicians and aircraft engineers all use phasor diagrams to visualize complex constants and variables (phasors).
AFAIK They're just vectors. Since the angles between them in the electrical game is usually the phase difference they call them phasors.

As for them being complex, I don't know. It's probably a question for another thread.

This is one of the reasons I'm looking at $X_C$, and $X_L$ for that matter using complex notation.

Typically, as far as I can tell, if values are all in phase then you can just add, subtract etc. If not you have to do phasor addition, the results are very real so I don't know where $\sqrt{-1}$ comes into it

It's all beyond me at the moment, but from the bits and pieces I have picked up, in a linear system design, if your calc's end up with a j term in it it shows the system will be unstable.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
AFAIK They're just vectors. Since the angles between them in the electrical game is usually the phase difference they call them phasors.

As for them being complex, I don't know. It's probably a question for another thread.

This is one of the reasons I'm looking at $X_C$, and $X_L$ for that matter using complex notation.

Typically, as far as I can tell, if values are all in phase then you can just add, subtract etc. If not you have to do phasor addition, the results are very real so I don't know where $\sqrt{-1}$ comes into it

It's all beyond me at the moment, but from the bits and pieces I have picked up, in a linear system design, if your calc's end up with a j term in it it shows the system will be unstable.
Well, phasors are not just vectors - they represent complex numbers.
Complex numbers are represented as vectors with the x-axis as the real axis, and the y-axis as the imaginary axis (the $j$-axis).

In a phasor diagram a resistance is a horizontal vector which is along the real axis.
A capacitor is represented by a vertical downward vector which is in the direction of the $-j$ that you just found.