Question about Ce^at

Discussion in 'Homework Help' started by boulwabd, Oct 12, 2016.

  1. boulwabd

    Thread Starter New Member

    Oct 12, 2016
    I am reading my textbook and have been trying to figure this out all night and day, why is a = r +jw0 and what is r? The book described this as "a in rectangular form" but I was under the impression rectangular form specified a real x component and an imaginary y component similar to cartesian fashion. I understand that r is the real part and jw0 is the imaginary part but can't quite grasp why we used the fundamental angular velocity instead of a "y" value or click on the context and how it relates to the periodic complex exponential e^jw0t. Any help is appreciated, I've moved on but would like to fundamentally understand how this is derived.
  2. ci139


    Jul 11, 2016
    withveryvaryingfrequency it jω₀t likely won't describe much anything - as well as - Xc XL are statistically derived averages for steady frequency time window being examined

    ▼. . . the imaginary unit being another statistical variable - that just enables to describe things more comfortably -- considering the θ=(ω₀-jr)t . . . shows i haven't read some 30 chapters of something that points out extremely limited use of such . . .

    . . . ? phasors - never heard - i'm the differential cat
    Last edited: Oct 12, 2016
  3. WBahn


    Mar 31, 2012
    Just expand out the exponential term:

    <br />
e^{at} \; = \; e^{\(r \, + \, j \omega_0 \) t} \; = \; e^{rt} e^{j \omega_0 t}<br />

    In the first factor, r is almost always negative, so e^(rt) represents a magnitude that is decaying exponentially with time. If r = 0; then the magnitude is a constant. If r is positive, then the magnitude is growing exponentially with time.

    The second factor represents a signal that is varying sinusoidally with time at frequency w_0. Depending on how this were set up, the actual signal of interest is usually either just the real part or just the imaginary part. In either case we have a sinusoid.
    boulwabd likes this.