# Confusion with step response and laplace inverse

6 vues (au cours des 30 derniers jours)
Robert le 16 Jan 2017
Hey all
Im hoping someone can explain this
This is fairly embarrassing i don't remember why this is not working out
Simple RC circuit
We know the transfer function to be the following
Vout/Vin = (1)/(S*R*C+1)
Or if you want to rearrange it in the proper form you get.
(1/RC)/(S+1/RC)
Anyways. If you give this thing a step input you get a curve that looks like the following using matlab
Again no surprise. As the cap begins to block the voltage builds up as the current flow tends towards zero
No surprise here
So anyway here is what i can't remember.
Just for the hell of it, i was trying to take the laplace and do the inverse and plot it in math against time t without using the step command. Obviously i should get the same results
The inverse is easily found using the table of course
Now lets just take something simple
Say R = 1k and C = 1u
And plot the thing
You get something very different then what the step response should be
You clearly get a decay which is obviously not the case. Step response of RC is an exponential growth then settles which is very well known
So what am i missing? Why when put back into time does the function not behave properly? or more likely, what am i doing wrong here? Note the second plot is just an example not the exact plot
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens

Connectez-vous pour commenter.

### Réponse acceptée

Star Strider le 16 Jan 2017
Try this:
syms R C s t
Z = 1/(R*C)/(s + 1/(R*C));
V = 1/s;
H = Z*V;
h = ilaplace(H,s,t);
h = subs(h, {R,C}, {1, 1E-6});
figure(1)
fplot(h, [0, 5E-6])
grid
Since youâ€™re starting at t=0 (I assume), a=0, and:
h = 1 - exp(-1000000*t)
giving the expected output.
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens

Connectez-vous pour commenter.

### Catégories

En savoir plus sur MATLAB dans Help Center et File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by