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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?

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  • \$\begingroup\$ 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? \$\endgroup\$ – Shamtam Jan 27 '15 at 3:45
  • \$\begingroup\$ @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? \$\endgroup\$ – dfg Jan 27 '15 at 3:49
  • \$\begingroup\$ 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. \$\endgroup\$ – Shamtam Jan 27 '15 at 3:56
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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: enter image description here

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): enter image description here

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