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The following are quotes from a textbook,

If S(t) is represented as a rotating phasor, the angular frequency of the phasor can be thought of as velocity at the end of the phasor. In particular the velocity \$\omega\$ is always at right angles to the phasor. $$\mathbf S(t) = Ae^{j\omega t}$$ enter image description here

However when we consider the general case when the velocity vector is inclined at an arbitrary angle \$\psi\$. In this case is a velocity is given by the symbol \$s\$ which now is composed of a component \$\omega\$ at right angle to the phasor as well as a component sigma which is parallel to \$s\$ enter image description here

Question1: I'm not able think about a complex number whose derivative is not \$i\$ times something, since any complex number can be represented by \$e^{jt}\$, their derivative is given by \$j e^{jt}\$ which means the velocity vector is perpendicular to the current position. But in this "general case" the velocity vector is pointing at an arbitrary angle, how?

Question 2: If the velocity vector is pointing in an arbitrary direction, then it's stated that the amplitude increases or decays "exponentially" Why is this true? Why not some other rate of decrease or increase?

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  • \$\begingroup\$ $e^{-t/\tau + i \omega t}$ is a sin/cosine oscillation that exponentially decays. It’s not just $e^{it}$ \$\endgroup\$ – Bob Jacobsen Aug 17 at 3:26
  • \$\begingroup\$ I wrote something, some of which may help a little. At the bottom, there are some potentially helpful videos included, as well. An error you make is assuming that "any complex number can be represented by \$e^{i\,t}\$." That assumption fails, so the rest of the reasoning isn't sound. At the beginning of this, I write another short intro at the top of it that may help a small bit. (Or not. I can't say.) Multiplication in complex number group theory combines both rotation and scaling. \$\endgroup\$ – jonk Aug 17 at 7:18
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In the figure a the function being represented must be of the type \${\bf S}(t)=Ae^{jωt}\$, as it is described. But in the figure b the function must be \${\bf S}(t)=Ae^{(jω+\sigma)t}\$ to have the trajectory depicted.

Question1: I'm not able think about a complex number whose derivative is not i times something, since any complex number can be represented by \$e^{jt}\$, their derivative is given by \$je^{jt}\$ which means the velocity vector is perpendicular to the current position. But in this "general case" the velocity vector is pointing at an arbitrary angle, how?

Think of \$e^{(j\omega +a)t}\$, the derivative is \$(j\omega +a)e^{(j\omega +a)t}\$, do observe that the \$j\omega e^{(j\omega +a)t}\$ is the part perpendicular to \$e^{(j\omega +a)t}\$, the angular velocity, and \$ae^{(j\omega +a)t}\$ is parallel to \$e^{(j\omega +a)t}\$, the velocity going away from the center. Hence you can an arbitrary angle.

Question 2: If the velocity vector is pointing in an arbitrary direction, then it's stated that the amplitude increases or decays "exponentially" Why is this true? Why not some other rate of decrease or increase?

It has the behavior you describe if the function is of the form \$e^{(j\omega +a)t}\$, the \$a\$ will determine the exponential growth/decay. But you could have a function such as \$t \cdot e^{j\omega t}\$, that will not grow exponentially. The reason complex functions of the form \$e^{(j\omega +a)t}\$ are more used/studied is that they are solutions to many ODEs, being used a lot in engineering.

obs: if the function is of the type \$2^{at} \cdot e^{jωt} = e^{ln(2)at} \cdot e^{jωt}\$, and \$ln(2)a\$ can be sen as a new number, like, call it \$c=ln(2)a\$, so

\$2^{at} \cdot e^{jωt} = e^{(j\omega +c)t}\$

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  • \$\begingroup\$ Why should the function be \$e^{at}\times e^{j\omega t}\$ and not \$ 2^{at} \times e^{j\omega t}\$ or something else, that is my question 2. But now I understand the answer to question 1, the derivative contains only imaginary component if amplitude is constant, if the amplitude isn't constant then the velocity vector contains two components. \$\endgroup\$ – Aravindh Vasu Aug 17 at 4:06
  • \$\begingroup\$ If the answer is good, please upvote. I will add that part, but it is surprisingly simple \$\endgroup\$ – jDAQ Aug 17 at 4:08
  • \$\begingroup\$ Please add that part as to why the amplitude decays or increases exponentially and also, please modify the answer to question 1, without using \$e^{at}\$ \$\endgroup\$ – Aravindh Vasu Aug 17 at 4:10
  • \$\begingroup\$ was that it? the term \$e^{at}\$ will remain, it is the most common form of a complex function growing/decay that appears as a solution to an ODE, you could have other functions of course but it is beyond the question. Does the part that answers the second question not shows "why the amplitude decays or increases exponentially ..." ? \$\endgroup\$ – jDAQ Aug 17 at 4:20
  • \$\begingroup\$ My Question 2 is rather, why e^{(at+j\omega t)} is the most common form of a complex function? \$\endgroup\$ – Aravindh Vasu Aug 17 at 4:25

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