Hello there,
Electrician:
How did you known you could use sqrt(1/2) in your equation to solve for the cut off?
I ask because you did not show that.
All:
Matrix methods are good to know. This problem is simple though because for one thing all the component values are known.
To do the simplest analysis we only need to work in 2 dimensions throughout. That's because the complex impedance of any element is always of only two values:
(real,imag)
or stated as:
realpart+imagpart*j
and because we are given all the numerical values for every component, we can always reduce every step of the problem into just two dimensions. And because we want the output and the output is on the right side of the circuit, it would be convenient to work from left to right. We can do this using Norton and Thevenin theorems, working from left to right transforming the sources from one to the other and back again while keeping track of the complex impedances.
To start, we transform every C and L into it's complex impedance, we will call them then zC1, zC2, zL1. However it should be noted that these are simple NUMERICAL values once you use the values given. For example, zC1 is equal to -j/(w*C1) but because we already know the value of C1 it becomes simply:
-1e8*j/w
which has 0 real part and so we would write this as:
0-1e8*j/w.
That's the result of transforming C1 into it's complex impedance, and we note that this is just in 2 dimensions and that's how we can keep it throughout the remainder of the calculations.
So knowing this value and noting that V1 is in series with that, we can convert voltage source to a current source in parallel with C1. The current is V1/zC1 so we end up with:
I1=V1/(-1e8*j/w).
So we remove V1 and C1 and replace it with a parallel circuit with I1 and C1. That puts C1 directly in parallel with L1. Since we then have two impedances in parallel, we can compute their impedance Zp. We then have a current I1 in parallel with Zp, from which we can calculate V2 from V2=I1*Zp. We now have a circuit that has a single voltage source V2 in series with Zp, C2, and R1, so it is now reduced to a voltage divider. It is then simple to calculate the voltage across R1.
In the end we end up with a numerical transfer equation and we must PLOT this equation in order to determine what to do next because we do not yet know what kind of 'filter' it is such as:
1. Low pass
2. High pass
3. Band pass
We next determine the best equation to solve to determine the cut off.
I can show more detail if you would like. Basically what we do here is use Norton and Thevenin to reduce the circuit to a simple voltage divider, then calculate the voltage across R1, then determine what value to use in the equality equation to find the cut off. Then we actually find the cut off using the response. But the key idea is that we never have to calculate anything more than a single complex voltage or current and a single complex impedance to get the result, so it makes short work of these problems.