Phasors and steady-state solutions

  • Thread starter Thread starter ABoul
  • Start date Start date
  • Tags Tags
    Phasors
ABoul
Messages
26
Reaction score
0

Homework Statement


use phasors to evaluate the steady-state solution to the equation f'' + 1.5f' + f = Ce^2tj


Homework Equations





The Attempt at a Solution


let f = Ce^2tj (this is a 'trial solution', i think)

f' = 2tCe^2tj
f'' = 4(t^2)Ce^2tj

therefore 4(t^2)Ce^2tj + 3tCe^2jt + Ce^2tj = Ce^2tj
4t^2 + 3t = 0

therefore t = 0 and t = -3/4
therefore f = C and f = Ce^-1.5tj.

^ the above is NOTHING like the answer. what am i doing wrong?
 
Physics news on Phys.org
attempt #2:

f = Ae^2tj
f' = 2tAe^2tj
f'' = 4(t^2)Ae^2tj

so...

4(t^2)Ae^2tj + 3Ate^2tf + Ae^2tj = Ce^2tj

dividing through by e^2tj gives

A(4t^2 + 3t + 1) = C

so A = C/(4t^2 + 3t + 1)? that's still not the answer
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top