# Transfer function of high pass filter via impulse response function

I'm having a hard time computing the transfer function of a high pass RC filter, by taking the Fourier transform of its impulse response function:

$$H(\omega)=\int_{-\infty}^{\infty} h(t)e^{-i\omega t}\ = \frac{-1}{1+i\omega RC},$$ which is the transfer function corresponding to a low pass transfer function.

I obtained its impulse response function by taking the derivative of its step response function $$h(t) = \frac{d}{dt} e^{-t/RC} = \frac{-1}{RC} e^{-t/RC}\ .$$

The step function $$s(t) = e^{-t/RC}$$ is derived by solving the differential equation resulting from equating the expressions for the current through the resistor and capacitor, respectively.

• Your first equation is a low pass TF. Your s(t) is also wrong for a step response.
– Chu
May 27 '18 at 11:14
• $$\lim_{\omega \to \infty} |H(\omega)| = \lim_{\omega \to \infty} \frac{1}{\sqrt{1+(\omega RC)^2}} = 1.$$ Hence, $$H(\omega) = \frac{-1}{1+i \omega RC}$$ is a high pass TF. On the step response $$s(t)=e^{-t/RC}$$. At $$t=0$$ all the voltage is over the resistor, as the capacitor starts to load up. This complies with the step response function. If we were measuring the voltage over the capacitor, we would have $$s(t) = 1-e^{-t/RC}$$. May 27 '18 at 11:24
• @MusséRedi incorrect - that is a low pass TF. May 27 '18 at 11:27
• @Andyaka Never mind, of course, you're right; thanks for pointing it out! I don't know why I made that calculation mistake earlier. $$\lim_{\omega \to \infty} |H(\omega)| = 0$$ indeed. :) May 27 '18 at 11:32
• @Andyaka Could you elaborate on why the step response $$s(t) = e^{-t/RC}$$ is wrong? May 27 '18 at 11:33

If you consider the simple RC highpass:

Then you can write the two I/O equations:

$$v_i(t)=v_C(t)+v_R(t)$$ $$v_o(t)=v_R(t)$$

Consdering i(t) the current through the circuit (no load) and vC(t) the voltage across C:

$$v_i(t)=\frac{1}{C}\int{i(t)\text{d}t}+R i(t)$$ $$v_o(t)=R i(t)$$

Apply the Laplace transform to the first, I(s) being the Laplace transform of i(t):

$$V_i(s)=\frac{1}{sC}I(s)+R I(s)=I(s)\left(R+\frac{1}{sC}\right)=>$$ $$I(s)=\frac{V_i(s)}{R+\frac{1}{sC}}$$

The same Laplace transform for the output yields:

$$V_o(s)=R I(s)=\frac{V_i(s)R}{R+\frac{1}{sC}}=>$$ $$\frac{V_o(s)}{V_i(s)}=\frac{R}{R+\frac{1}{sC}}=\frac{R}{\frac{sRC+1}{sC}}=\frac{sRC}{sRC+1}=\frac{s}{s+\frac{1}{RC}}$$

This is why I said it should be something like s/(s+1). Now, if you do some inverse Laplace transforms, you'll end up with an interesting impulse response. First, arrange into strictly proper partial fractions:

$$\frac{s}{s+\frac{1}{RC}}=1-\frac{\frac{1}{RC}}{s+\frac{1}{RC}}$$

And now you see that 1 is the Laplace transform of the Dirac impulse, plus the rest of it, which is the lowpass RC with the impulse response $1-exp(-\frac{t}{RC})$, and you might be tempted to cancel the 1s, but the first one is the $\delta$(t), and the derivative of the step response is $\frac{exp(-\frac{t}{RC})}{RC}$, which results in the total impuse response (the point where you shoul have started from):

$$v_o(t)=\delta(t)-\frac{exp(-\frac{t}{RC})}{RC}$$

Here's the confirmation (the input impulse is pulse 0 1k 0 1n 1n 1m -- 1kV over 1ms):

and here's a zoom on the Y axis:

This is one reason why it didn't work out, you omitted the initial conditions and the influence of the Dirac impulse: at t<0 everything is zero (null conditions), at t=0 the (ideal) capacitor charges with the input, the derivative of the applied voltage, $\delta$(t), but the input voltage is not just a rise, it's also a fall, both in the same time (Dirac or, how friends called him, Chuck Norris), so the voltage across the capacitor goes back and then reaches its negative peak, after which the discharge happens.