Is the cut off frequency of a passive and active second order filter (high and low pass) occur at -6db when C1= C2 and R1=R2?
I found this on the internet that says it is for a LPF second order but I don't understand the graphCutoff frequency is -3db. The fact the filter is second order just dictates the rate of roll off for the filter. The values of R & C you use will just move the -3db frequency in a passive circuit. In an active cofiguation you could also effect the gain of the circuit but this will be dependant on your circuit.
It has one that is -3db and the other that has -6dbWhat is it about the graph that you do not understand?
That is not just a label. That is a response.It has one that is -3db and the other that has -6db
The image is a bit deceiving as the 6dB doesn't really have a contextIt has one that is -3db and the other that has -6db
So?The image is a bit deceiving as the 6dB doesn't really have a context
You need to understand the difference between "Roll off Frequency" and "Cut off Frequency"
Simplistically - Roll off is the steepness of the filter response. (shown as -20db/Decade and -40db/Decade) in your image
Cut off frequency - The -3db point is typically the point at which a filter starts becoming effective valid for both curves in the image.
I think what that image is trying to show is that with a higher order filter you can achieve a faster roll off
So what?, I'm trying to provide something that may help him understand the graph, by all means if you wish to elaborate or share your expertise feel free.