EXPEDITION

ELECTRONICS

sonic adventures!

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^33As^23As + 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)sA - 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

References and further reading

  • Electric Druid has more discussion of the Xpander / Matrix 12 modes, as well as schematics of a modern take on pole mixing and links to even more reading.