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I am struggling to find the magnitude and phase of complex exponential signals ($z$) of the form:

$z=(a+bi) e^{(A+Bi)t}$ where $a, b, A, B$ are all real numbers.

I am aware of the following equations, but these do not seem to work in this context:

$|z| = \sqrt {Re(z)^2+Im(z)^2} $, $\angle z = \tan^{-1}\left(\frac{Im(z)}{Re(z)}\right) $

What is a method that works for all complex exponential signals of the aforementioned form? Please present it as applied to the following example:

$z=e^{(-1-i)t}$ where $a = 1, b=0, A=-1, B=-1 $

Thus, what is the value of $|z|$ and $\angle z $ ?

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Assuming that all variables are real:

$$\left(\text{a}+\text{b}i\right)e^{\left(\text{A}+\text{B}i\right)t}=\left(\text{a}+\text{b}i\right)e^{\text{A}t}e^{\text{B}ti}$$

So:

$$\left|\left(\text{a}+\text{b}i\right)e^{\left(\text{A}+\text{B}i\right)t}\right|=\left|\text{a}+\text{b}i\right|\left|e^{\text{A}t}\right|\left|e^{\text{B}ti}\right|=\sqrt{\text{a}^2+\text{b}^2}\cdot e^{\text{A}t}\cdot1=\sqrt{\text{a}^2+\text{b}^2}\cdot e^{\text{A}t}$$

And:

$$\arg\left(\left(\text{a}+\text{b}i\right)e^{\left(\text{A}+\text{B}i\right)t}\right)=\arg\left(\text{a}+\text{b}i\right)+\arg\left(e^{\text{A}t}\right)+\arg\left(e^{\text{B}ti}\right)=$$ $$\arg\left(\text{a}+\text{b}i\right)+0+\text{B}t=\arg\left(\text{a}+\text{b}i\right)+\text{B}t$$


Notice that:

$$\arg\left(\text{a}+\text{b}i\right)=\arctan\frac{\text{b}}{\text{a}}$$

ONLY works when $\text{a}>0$

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    Fantastically clear! Thank you!2017-01-08
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    You're welcome, I'm glad that I could help!2017-01-09
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Your formulas are correct, it's just not as obvious what the real and imaginary parts are since you have an exponential that does not have a purely imaginary argument. You can expand $z$ using Euler's formula (on only the purely imaginary part of the exponential):

$$z=(a+bi)e^{At}e^{iBt}=(a+bi)e^{At}[\cos(Bt)+i\sin(Bt)]=[ae^{At}\cos(Bt)-be^{At}\sin(Bt)]+i[be^{At}\cos(Bt)+ae^{At}\sin(Bt)]$$ and so

$Re(z)=e^{At}[a\cos(Bt)-b\sin(Bt)]$ and

$Im(z)=e^{At}[b\cos(Bt)+a\sin(Bt)]$

You can then plug these into your formulas for $|z|$ and $\angle z$ for any real numbers $a,b,A,B$. Note that when you're calculating the phase you must check which quadrant of the complex plane your number lies in as the inverse tangent function only returns values in $(-\frac \pi 2,\frac \pi 2)$ and $\tan (\theta)=\tan (\theta+n\pi)$.

For the case $a=1$, $b=0$, $A=-1$ and $B=-1$ I get that $|z|=e^{-t}$ and $\angle z=-t$ (since $Re(z)<0$ when $t>0$).

Note that if you're comfortable with writing complex numbers in polar form, it is much quicker to write $z$ in the form $z=re^{i\theta}$ where $r,\theta \in \mathbb R$, then $|z|=r$ and $\angle z=\theta$ so for the case $z=e^{(-1-i)t}=e^{-t}e^{i(-t)}$ we immediately get that $|z|=e^{-t}$ and $\angle z=-t$ as before.

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    A different and slightly longer method but useful nonetheless, thanks!2017-01-08