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?
Best Answer
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):