# Application of Frequency Response Theorem in Linear Circuits

One of the most important results of control theory for linear systems is the so called "Frequency Response Theorem" (here the references):

Briefly, it says that under the hypotesis of stability and linearity, if the input signal is sinusoidal, the output signal will be the original sine with phase and amplitude variations respectively equal to phase and amplitude of the transfer function of that system.

Now, let's consider for instance a RC low pass filter:

The output signal will be like this:

So, it is not sinusoidal, although both R and C are linear components. Where is the mistake?

• There’s no such thing as a “frequency response theorem”. Things have a frequency response period. – Andy aka Dec 27 '19 at 16:41
• Your conclusion (Briefly, it says...) is completely wrong. In nature, not even a resistor will make the output have the same phase as the input, since parasitic elements are everywhere. It's (one of) the reason(s) the word efficiency exists. – a concerned citizen Dec 27 '19 at 18:58
• That's NOT what the output of that circuit will be. – Scott Seidman Jan 17 at 13:56

The problem is that you statement: "The output signal will be like this:" is rather wrong.

What you have shown there is the output of a rectifier circuit. I don't know where that came from but it is NOT the output of the given circuit.

You get that output waveform if you replace the R with a diode, but in that case you no longer have a Linear Circuit.

The transfer function of the shown circuit is equal to:

$$\mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{out}\left(\text{s}\right)}{\text{V}_\text{in}\left(\text{s}\right)}=\frac{\frac{1}{\text{sC}}}{\frac{1}{\text{sC}}+\text{R}}=\frac{1}{1+\text{sCR}}\tag1$$

Now, when:

$$\text{v}_\text{in}\left(t\right)=\hat{\text{u}}\sin\left(\omega t+\varphi\right)\tag2$$

We get:

$$\text{V}_\text{in}\left(\text{s}\right)=\mathcal{L}_t\left[\hat{\text{u}}\sin\left(\omega t+\varphi\right)\right]_{\left(\text{s}\right)}=\hat{\text{u}}\cdot\frac{\text{s}\sin\left(\varphi\right)+\omega\cos\left(\varphi\right)}{\text{s}^2+\omega^2}\tag3$$

So, the output voltage is:

$$\text{V}_\text{out}\left(\text{s}\right)=\frac{1}{1+\text{sCR}}\cdot\text{V}_\text{in}\left(\text{s}\right)=\frac{1}{1+\text{sCR}}\cdot\hat{\text{u}}\cdot\frac{\text{s}\sin\left(\varphi\right)+\omega\cos\left(\varphi\right)}{\text{s}^2+\omega^2}\tag4$$

Assume that $$\\text{C}=\text{R}=\omega=\hat{\text{u}}=1\$$ and $$\\varphi=0\$$. In the time-domain we get:

$$\text{v}_\text{out}\left(t\right)=\frac{\exp\left(-t\right)}{2}+\frac{\sin\left(t\right)-\cos\left(t\right)}{2}\tag5$$