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 $ ?
Best Answer
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 $\text{a}>0$