# How do coaxial cable notch filters work?

I am trying to understand how a coax-cable notch filter works. A schematic of the filter is here: It consists of a transmission (coax) line with a deviation (D) where some part of the wave energy partially goes. The tick dashed lines are the inner conductors. By adjusting the length of the deviation undesirable frequency in the transmitted signal are suppressed due to a destructive interference which occurs between the signal which passed through the deviation and transmitted signals (signal reflects from the top of the deviation).

Question Assume we have only one frequency. Depending on the phase shift ($\delta$) created by D how much of the undesirable wave can be transmitted through the filter and what is the best phase shift (length of the deviation)?

My thoughts: The initial wave splits at D. A wave with the amplitude $\sqrt R \cdot E_\mathrm{in}$ enters in D. $R=1/2$ if we assume that the energy always splits equally. The transmitted wave has amplitude $\sqrt R \cdot E_\mathrm{in}$. The wave reflected from the open-end passing towards the receiver has the amplitude $\sqrt R \sqrt {R} E_\mathrm{in} e^{i\delta}$ (two times $\sqrt R$ because it splits again at the connection point and the accumulated phase shift). Now to see how much signal goes towards the receiver we find $E^2$ as follows:

$$E^2 = E \cdot E^* = (\sqrt R + \sqrt R \sqrt R e^{i\delta}) \cdot (\sqrt R + \sqrt R \sqrt R e^{-i\delta}) = R + R^2 + 2R \sqrt R \cos(\delta)$$

Then the transmission is minimal at $\delta=\pi$

• You're basically asking for someone to compute the S parameters for this circuit, right? Aug 30, 2016 at 20:07
• yes, you are right Aug 30, 2016 at 20:51
• Ok well, you can look that up in a book. Do you have a conceptual question? Aug 30, 2016 at 21:58