# I Sinusoids as Phasors, Complex Exp, I&Q and Polar form

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1. May 10, 2018

### Natalie Johnson

Hi,

I am going around in circles, excuse the pun, with phasors, complex exponentials, I&Q and polar form...

1. A cos (ωt+Φ) = Acos(Φ) cos(ωt) - Asin(Φ)sin(ωt)
Right hand side is polar form .... left hand side is in cartesian (rectangular) form via a trignometric identity?

2. But then sometimes I read...
A cos (ωt+Φ) in polar form has corresponding cartesian form of Bcos(ωt)+Csin(ωt), which is fine to understand because this cartesian form gives X and Y coordinates on a cartesian coordinate axes of a vector in that axes.

3. But point 1 and 2 are different, how can Acos (ωt+Φ) in polar represent Bcos(ωt)+Csin(ωt) in cartesian but also be equal to Acos(Φ) cos(ωt) - Asin(Φ)sin(ωt) via a trignometric identity ---> Is it because Acos(Φ) and Asin(Φ) are constants and therefore also B and C? Might be obvious but I need to ask for my own sanity of seeing so much different ways its written.

What about if B and C are not constants due to the phase changing with time Φ(t)?

I am further purplexed by notation used for complex sinusoids.

3. Acos (ωt+Φ) can be represented as the real part of Aei(ωt+Φ)= Acos(ωt+Φ) + iAsin(ωt+Φ)
but from point 1, the right hand side of this equation can be then re-written with the trigometric identity in point 1, expanding it into 4 terms which removes the phase from the argument and giving constants, like in point 3. So why cannot it not be written without the Φ in the argument on the right hand side and use different constants
Aei(ωt+Φ)=Bcos(ωt)+iCsin(ωt)

2. May 10, 2018

### kuruman

Start with $e^{\pm i\omega t}=\cos\omega t\pm i\sin\omega t$,
Now,
\begin{align} e^{i (\omega t+\phi)}& =e^{i\omega t}e^{i\phi}=(\cos\omega t+i\sin\omega t)(\cos\phi +i\sin\phi)\nonumber \\ & =(\cos\omega t~cos\phi-\sin\omega t~sin\phi)+i(\sin\omega t~\cos\phi+\cos\omega t~sin\phi)\nonumber \end{align}Equating real and imaginary parts, gives you two trig identities
$$\cos(\omega t+\phi)=\cos\omega t~cos\phi-\sin\omega t~sin\phi~;~~~\sin(\omega t+\phi)=\sin\omega t~\cos\phi+\cos\omega t~sin\phi$$
You can easily show that $$\cos\omega t= \frac{e^{i\omega t}+e^{-i\omega t}}{2}~;~~~\sin\omega t= \frac{e^{i\omega t}-e^{-i\omega t}}{2i}=-i\frac{e^{i\omega t}-e^{-i\omega t}}{2}.$$
Then,
\begin{align} A \cos\omega t +B\sin \omega t &=A\frac{e^{i\omega t}+e^{-i\omega t}}{2}-iB\frac{e^{i\omega t}-e^{-i\omega t}}{2}\nonumber \\ & = \frac{(A-iB)e^{i\omega t}}{2} +\frac{(A+iB)e^{-i\omega t}}{2}\nonumber \end{align}
Note that the last expression is real because it is the sum of a number plus its complex conjugate. You can write the complex coefficient in polar form, $A-iB = re^{i\phi},~r=\sqrt{A^2+B^2}$. In that case,
\begin{align} A \cos\omega t +B\sin \omega t &= \frac{re^{i(\omega t +\phi)}}{2} +\frac{re^{-i(\omega t +\phi)}}{2}\nonumber\\ & =\sqrt{A^2+B^2}\cos(\omega t+\phi)\nonumber \end{align}.
Note: In the above expression $\phi = \arctan(B/A)$.
I hope this clarifies things for you and convinces you that the complex exponential plays a pivotal role in trigonometric transformations.

3. May 11, 2018

### sophiecentaur

Can that be right? I would have thought that the magnitude of the left hand side is constant but the magnitude of the right hand side is not (it's an ellipse) unless B=C.

4. May 11, 2018

### kuruman

I don't think that $B=C$ will do it, but $A=\sqrt{B^2+C^2}$ will.
Suppose $z = Ae^{i \omega t}=B\cos(\omega t)+iC\sin(\omega t)$, where $A$, $B$ and $C$ are assumed real.
Then
$|z|^2=(Ae^{i \omega t})(Ae^{-i \omega t})=A^2$
But also
$|z|^2=[B\cos(\omega t)+iC\sin(\omega t)][B\cos(\omega t)-iC\sin(\omega t)]=B^2+C^2$
The two expressions will match if $A^2=B^2+C^2$ as also shown in post #2.

5. May 12, 2018

### sophiecentaur

But the phase has been lost. How can varying B and C coefficients achieve an arbitrary phase for the resultant?