# EXPEDITION # ELECTRONICS

Let's look at the Xpander filter a little more closely. The structure is similar to the generic 4 pole filter. If you haven't gone through that, it should be considered a requisite. If you set the A term from the base 4 pole filter to 0, the two are structurally equivalent, although here we're using A as the amplification factor starting on V1. Additionally, in the Xpander 3 pole modes the first filter pole has its pole moved upwards by a factor of 330. We've included that pole in the calculations. 1. ### Solving for V0 to V4

 V0 = Vin - fV4 V1 = V0(.0030303s+1) Here, we're adding a factor to account for the way the Xpander switches out it's first low pass stage, by simply moving the pole to a very high frequency. The ratio of the capacitor values is 33nF:.1nF or 330:1. V2 = V1(s+1) = V0(.0030303s + 1)(s+1) V3 = V2(s+1) = V0(.0030303s + 1)(s+1)^2 V4 = V3(s+1) = V0(.0030303s + 1)(s+1)^3
Therefore
 V4 = Vin - fV4(.0030303s + 1)(s+1)^3 Let's define Den3 = (.0030303s + 1)(s+1)^3 V4 = Vin - fV4Den3 V4Den3 = Vin - fV4 V4Den3 + fV4 = Vin V4(Den3 + f) = Vin V4Vin = 1(Den3 + f) So we have the transfer function for V4. Now, it's helpful to divide the top and bottom by Den3. =  1 Den3 1 Den3 (Den3 + f) =  1 Den3 1 + f Den3 =  1 Den3 (1 + f Den3 ) This is a convenient form as it lets us rewrite the V4 transfer function in a form that looks like the standard 4 pole LP transfer function with an added F term. The original form with small f will be used later when expanding the common denominator 1 Den3 F where F = 1 + fDen3

With the transfer function in this form, we can easily define V1, V2, and V3 by adding succesive (s+1) terms to the numerator and reducing. We don't have a need for V0:
V4 = 1Den3 F = 1(.0030303s + 1)(s+1)^3 FVin
V3 = (s+1)Den3 F = 1(.0030303s + 1)(s+1)^2 FVin
V2 = (s+1)^2Den3 F = 1(.0030303s + 1)(s+1) FVin
V1 = (s+1)^3Den3 F = 1(.0030303s + 1) FVin

Now we can sum the individual responses into our mixing network.

2. ### Mix equation for 4 poles with feedback

We can see that Vout is just AV1 - BV2 + CV3 - DV4
Substituting the transfer functions we found previously gives us:
VoutVin = A(.0030303s + 1) F - B(.0030303s + 1)(s+1) F + C(.0030303s + 1)(s+1)^2 F - D(.0030303s + 1)(s+1)^3 F

the common denominator is (.0030303s + 1)(s+1)^3 F, so we can multiply each by the common denominator, cancel the common terms and expand the (s+1) exponential terms. Note that all of the F terms will cancel.
3. ### Numerators

 A(s+1)^3 => As^3 3As^2 3As + A -B(s+1)^2 => -Bs^2 - 2Bs - B C(s+1) => Cs + C D => - D + As^3 + (3A - B)s^2 + (3A - 2B + C)s A - B + C - D

As it turns out, this is the same numerator polynomial you get with or without feedback, due to the cancellations of the F terms. The F term will only appear in the denominator. We also don't end up with the shifted pole term (.0030303s + 1) in the numerator since we're not using V0.
4. ### Combined numerator coefficients

replacing s with jw, we can group the real and imaginary parts

real terms: R = -(3A - B)w^2 + (A - B + C - D)
imaginary terms: I = j [ Aw^3 - (3A - 2B + C)w ]

The magnitude of the numerator can then be found by Magn = √( R^2 + I^2 )
5. ### Denominator

The common denominator is Den3F. We showed before that
Den3 F = Den3 + f    (note big F on left and small f on right)

 Den3 + f = (.0030303s + 1)(s+1)^3 + f = (.0030303s + 1)(s^3 + 3s^2 + 3s + 1) + f = (.0030303s^4 + 1.0090909s^3 + 3.0090909s^2 + 3.0030303s + 1 + f
real terms: R = .0030303w^4 - 3.0090909w^2 + 1 + f
imaginary terms: I = j [ 1.0090909w^3 - 3.0030303w ]

The magnitude of the denominator can likewise be found by Magd = √( R^2 + I^2 )
6. The total magnitude is then just Magn / Magd