# RC Transfer function

Discussion in 'Homework Help' started by alexmath, Apr 4, 2015.

1. ### alexmath Thread Starter New Member

May 2, 2014
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0
Suppose R = C = 1 then the transfer function from the input voltage to the voltage across the capacitor is 1/ (s+1). So Vc(S) / V(S) = 1/(s+1). Getting back to time-domain: Vc(t) = V(t) * e^-t. What's wrong here? Thank you!

Last edited: Apr 4, 2015
2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,504
512
C is complex, R isn't.

3. ### MrAl Distinguished Member

Jun 17, 2014
2,556
515
Hello there,

Constant voltage sources V(s) transform into V/s they do not stay the same when there is a step input. So V(s) => V/s so your Laplace is not V/(s+1) it is (V/s)/(s+1)=V/(s*(s+1)). If the voltage source is 1v then this is simply 1/(s^2+s).

Find the inverse of that and you'll be very happy

4. ### alexmath Thread Starter New Member

May 2, 2014
17
0
Thank you! I'm very happy right now! Have a good day.

5. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
The statement R=C=1 is nonsensical.

That's like saying that the height of a car and the weight of a building are both equal to 1. Totally meaningless.

6. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
If C is capacitance measured in farads, then it is a real number. The reactance of C, Xc, with units of ohms is also a real number. The impedance or admitaance of a capacitor C is a complex number. In the ideal case it would have a zero real part. For a real component there will be some equivalent series resistance.