Via stub resonance intuition of insertion loss drop by formula
Some keywords or things to look up would be &quot;standing wave resonance.&quot;
https://www.allaboutcircuits.com/textbook/alternating-current/chpt-14/standing-waves-and-resonance/
When you have a stub and you have a signal propagating down the transmission line, you'll have part of the signal go down the stub and part of the signal going through the via (in the slide you show). If the signal is at a frequency (or wavelength) where it hits the end of the stub, reflects back, and cancels out at that point with the value of the incoming waveform, you'll have a resonance that dips down.
This is a relatively narrow-band effect, and you'll generally see several suckouts at regular spacing in the frequency domain. These are the harmonics of the wavelength that create those standing wave nodes.
Note that the quarter wavelength stub creates a standing wave/node because of the reflected signal from the (open) end of the stub. If you imagine a sine wave, the first &quot;standing wave&quot; length (or first place the value returns to zero) is 180 degrees out of phase. When you have a quarter wavelength, the void is caused by the signal propagating down to the end (1/4 wavelength = 90 degrees out of phase) and then bouncing back and reflecting. The distance traveled is 2X the 1/4 wavelength, which gives you half a wavelength.
To be explicit: The quarter wavelength stub is effectively 1/2 wavelength when you consider the REFLECTED waveform.
The suckout you're seeing isn't (strictly speaking) due to the impedance change, or characteristic impedance itself, it's more of a factor of standing wave type behavior. The magnitude of the dip will have to do (some) with the impedance (mis)matching.