Hello Winterstone,
Yes thanks for bringing that up. We actually have to calculate the amplitude before we can start to solve for any frequencies or even plot the function itself!
So the procedure would go like this:
1. Find the transfer function T(s), solve for the amplitude function Ampl(w).
2. Plot this function. Plotting shows us what kind of transfer function we have, as to it's basic nature.
3. Solve Ampl(w) for the required pass band frequency, call it w1. This could be 0 or infinity or anywhere in between.
4. Using w1, solve for the pass band gain G. This gives an exact value for G.
5. Set Ampl(w1) equal to G/sqrt(2).
6. Solve for the set of wc, the cutoff frequencies.
Here's a more complete example using a much simpler circuit so we can see in detail how this works...
(see attachment)
First the transfer function:
Ts=R3/(s*C1*R1*R3+R3+s*C1*R1*R2+R2+R1)
and then replacing s with j*w:
Tjw=R3/(j*w*C1*R1*R3+R3+j*w*C1*R1*R2+R2+R1)
Now calculate the amplitude using complex number theory:
Ampl(w)=R3/sqrt((w*C1*R1*R3+w*C1*R1*R2)^2+(R3+R2+R1)^2)
Plot the amplitude Ampl(w) and look at the plot.
Looking at the plot we see it is a low pass filter so we want to calculate G from Ampl(w) by setting w=0:
G=Ampl(0)=R3/(R1+R2+R3)
Now to find the -3db point we divide that by sqrt(2) and set it equal to the amplitude:
G/sqrt(2)=Ampl(w)
which is equal to:
(R3/(R1+R2+R3))/sqrt(2)=R3/sqrt((w*C1*R1*R3+w*C1*R1*R2)^2+(R3+R2+R1)^2)
Now at this point it is easier to insert the values for all the components, so we get:
1/(3*sqrt(2))=1000/sqrt(4.0*w^2+9000000)
and solving this for w we get:
w=-1500, and
w=1500
from which we reject the negative solution. So the -3db point frequency is:
fc=1500/(2*pi)
and this is approximately equal to 238.7 Hz.
Please note that when we are given the values for all of the components we can start with the transfer function using all the component values, so that immediately reduces all the algebra to that of using just complex numbers. For this example i waited more toward the end to insert the actual values because it is much cleaner to show the algebra than the numerical data. But when doing this calculations by hand it is almost always better to start with the numerical values in the transfer function right away because it reduces much quicker.
I leave this as an exercise for the reader here
Also note that when we used Ampl(0) above that is because we plotted the amplitude function and noted that it was a low pass filter. However, there may be times when the filter is not so well behaved so we may actually have to pick the value of w for this calculation using a little guesswork. This wont happen for most functions, but it is a possibility. Of course there is also the possibility that we are using this for a limited frequency range and then we might again have to pick this number by hand.