the only option i can see would be to pull 1 s out of the first 2 terms of each polynomial. if the K weren't there i could factor the polynomial using quadratic.
chester that too is what i got, only i had plugged in some ICs along the way, making it harder for me to investigate different values of drag and what influence they have on other parameters. this makes it a little easier.
@rude man: can you explain that a little further? i know what separation of variables is but I'm having a hard time seeing it here.
edit: i contacted my professor and he said Laplace was acceptable. thank god, i wouldn't have figured conventional int out!
Laplace Transform expressions for y'', y', and vd(t).y''(t):
s2Y(s) - sy(0) - y'(0)
+
(D/M)y'(t):
(D/M)(sY(s) - y(0))
+
(K/M)y(t):
(K/M)Y(s)
=
(K/M)vd(t):
(K/M)(Vo/s)then I factor out Y(s) and then solve for Y(s). that's when i get that ugly fraction up there that i don't know what to...
Homework Statement
Solve the DE for y(t) with the IC's
y(0)=20.8m/s and y'(0)=0
if the input is a step function scaled by the desired velocity Vo.
vd(t)=Vou(t).
Assume the desired velocity Vo=27.8m/s
Homework Equations
y''(t) + (D/M)y'(t) + (K/M)y(t) = (K/M)vd(t)
M = 1,000kg
D = 100kg/s
K...
Alright thanks so much guys, it's 5am so I'm going to go sleep on it. Laplace, laplace, laplace... As soon as I get over this speed bump the rest is a breeze, MATLAB plots, etc.
Thanks again!
yes Fourier and laplace. he told us to use the time domain. the block diagram is in laplace format i.e. S instead of jw. I've also tried:
(1/m) / (s + D/m)
D/m = .1
.001 / (s + .1)
1 / (s + .1) ---> e^(-.1t)u(t)
i'm not sure if I'm doing the right thing with the m in the numerator's...