Confusion with step response and laplace inverse

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

Answer 1

[Star Strider]

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 you’re starting at t=0 (I assume), a=0, and:

h = 1 - exp(-1000000*t)

giving the expected output.