Dirac Delta Impulse Response of series RC circuit (Voltage across the capacitor)

Question

I have evaluated the transfer function of an \$RC\$ circuit, getting the following transfer function:

$$H(s)= \frac{1}{1+sCR}$$

the impulse response (i.e. output response to a Dirac's Delta) would thus be the inverse laplace of the same expression and that would be a decaying exponential function with a peak value of \$1/(CR)\$.

How can we explain the sudden rise of capacitor voltage to \$1/(CR)\$ (remember, a capacitor needs a finite interval of time before completely charging)?

Moreover, is this the maximum limit of voltage across the capacitor in this circuit ? (considering that we have applied an impulse at the input and thus applied a very high value of voltage at the input)?

Answer 1

The instantaneous rise on the capacitor is not possible in the real world, but then again neither is the impulse it takes to cause that. The rate of voltage rise on the capacitor is proportional to the input voltage. During the impulse, that input voltage is infinite, so the capacitor voltage can rise infinitely fast.

This is sortof a word problem illustrating the mathematical concept of limit. As the input voltage goes towards infinity the capacitor voltage rise time goes towards 0. In the limit this works out to a finite voltage left on the capacitor, which got there instantaneously.

Answer 2

It's important to keep in mind that $$\frac{1}{1+sRC}$$ is the transfer function you still have to multiply by a step function to obtain the voltage on the output of the capacitor. A step function has a Laplace transform of \$1/s\$ (prove this for yourself by calculating $$\int_{0}^{\infty} 1 \times e^{-st}dt$$

That means the output has a Laplace transform of $$\frac{1}{s(1+sRC)}$$ The inverse is \$1-e^{-t/RC}\$ which starts from zero as you would expect.

The impulse is just a infinitesimal mathematical construct and so shouldn't be expected to adhere to physical laws.

Answer 3

If the capacitor's voltage changes instantaneously it was because a impulse current occurred, raising the voltage by \$1/C\$. It's not true that you have to divide by \$s\$ to get the answer. In fact, if the input is the impulse, its Laplace transform is \$1\$. So, \$Y(s)=H(s)X(s)=H(s)\$.