Don't understand high pass filters?

Consider a RC circuit functioning as a high pass filter. The transfer function is given by:

$T(j\omega)=\frac{K}{1-j(RC/\omega)}$

For a frequency of 0, the transfer function is 0. Which means for DC input signals, the output response is 0.

However, by modelling the circuit with a differential equation and then solving it shows that the voltage across the resistor in a RC circuit is actually a decaying exponential.

Isn't this a contradiction? The transfer function shows that the output should be 0, but the differential equation shows that it's a decaying exponential?

• When you say the transfer function is "0," what do you mean? Note that it's a complex function, with magnitude and phase. Hint: what conditions are implied with phasors? Second hint: what's the difference between a transient response and a forced response? – Shamtam Jan 27 '15 at 3:45
• @Shamtam By 0, I mean that for the limit of $\omega -> 0$, $T(j\omega) -> 0$. It's a complex function, but it's magnitude approaches zero in the limit right? I think your hinting at the fact that phasors only give the steady state response right? – dfg Jan 27 '15 at 3:49
• That's correct (both, that the magnitude approaches zero, and that the phasors account for steady-state). The transfer function (in terms of $j\omega$) only accounts for steady-state responses. If you were to deal with it in terms of the Laplace variable $s$, you could transform the result back into a time-domain response that has the steady-state and transient responses accounted for, pending boundary conditions. – Shamtam Jan 27 '15 at 3:56

The key point is that it is 0 in steady state. That is, as $t\rightarrow\infty$, $V_{out}\rightarrow0$. When you take the differential equation, you're looking at output voltage with respect to time.

Let's try taking a look at an example of this system. I'm going to simplify $K=1$ and $RC=1$. Then, we are left with: $$\frac{V_{out}}{V_{in}}\left(j\omega\right)=T\left(j\omega\right)=\frac{1}{1-j/\omega}$$ To simplify our calculations, let's change the form of it and replace $j\omega$ with $s$. Then, we get: $$T\left(s\right)=\frac{s}{s+1}$$ Taking the step response of this (that is, $V_{out}$ when the input $V_{in}=u\left(t\right)$ where $u\left(t\right)$ is the Heaviside step function) we get this:

Note that this is in time domain, and in steady state, our output voltage is zero. The bode plot comes out as you'd expect (first order high-pass filter):