Confusion with step response and laplace inverse

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Robert
Robert 2017 年 1 月 16 日
回答済み: Star Strider 2017 年 1 月 16 日
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

採用された回答

Star Strider
Star Strider 2017 年 1 月 16 日
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.

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