Electronic – Don’t understand high pass filters

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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?

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: 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