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[SOLVED] Prove Wo = cutoff frequency when Q = 1/ square root 2

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HappyPenguin

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Heihei! S^2+Wo/Q S+(W0)^2 is the denominator of a transfer function..be it HPF or LPF
From the transfer function, we know that Wo is the cutoff frequency, but only if Q = 1/ square root 2...
So, how do i "Quantitatively" PROVE that when Q = 1/ sq2....wo is the cuttoff ?? Appretiate the help..haha xD
 

It's relatively simple.
Evaluate the magnitude of the denominator [sqrt(Re^2 + Im^2)] and equalize with sqrt(2).
Then, solve for Q. That's all.
 
okay...i did wat u said..and the Q = WWo / sqrt(2-(Wo^2-W^2)^2)
how does that prove that Wo is the cutoff frequency when Q = 1 / sqrt2...thx for helping haha
well, wat i tried is by assuming w=0, which is the max gain (for LPF) and then i divide the Av by sqrt2...and then dead end...
 

I didn't check your calculation. But I immediately see that it can't be correct because of the dimensions (units).
Please, recalculate and set w=wo. Then solve for Q.
That means: If w=wo , then Q=0.7071.
 
Thx!! so..wat i got is Wo^2/Q = sqrt2
But i dun understand y do we have to use this....<[sqrt(Re^2 + Im^2)] and equalize with sqrt(2)>
why equalize it with sqrt(2)??
 

HappyPenguin said:
Thx!! so..wat i got is Wo^2/Q = sqrt2
But i dun understand y do we have to use this....<[sqrt(Re^2 + Im^2)] and equalize with sqrt(2)>
why equalize it with sqrt(2)??
Click to expand...

Because |H(s)|=1/sqrt(2)=0.7071 (identical to -3 dB) is defined as the magnitude for the passband edge frequency.
Or - the other way round: If Q=1/sqrt(2) there is a 3 dB loss for w=wp (pole frequency). This defines the Butterworth response with wp=w(3 dB).
 
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