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Sensitivity of ωo in Tow-Thomas LPF

Short answer

Sensitivity analysis is important because component tolerances and drift shift filter parameters; it identifies low-sensitivity designs and critical components. For Tow-Thomas ωo: R1, R4, C1, C2 each have sensitivity -1/2; R2, R3 and inverter resistors have zero sensitivity.

Why is sensitivity analysis important in filter design? Perform the sensitivity analysis of ωo in Tow Thomas low pass filter.

Why sensitivity analysis is important

Sensitivity is the per-unit (fractional) change in a filter parameter yy (such as ωo\omega_o, QQ or gain) caused by a per-unit change in an element value xx:

Sxy=∂y/y∂x/x=xy∂y∂x=∂ln⁡y∂ln⁡xS^y_x = \frac{\partial y / y}{\partial x / x} = \frac{x}{y}\frac{\partial y}{\partial x} = \frac{\partial \ln y}{\partial \ln x}

So Δyy≈SxyΔxx\dfrac{\Delta y}{y} \approx S^y_x \dfrac{\Delta x}{x}.

Importance in filter design:

  • Real components have tolerances (e.g. ±5% resistors, ±10% capacitors) and drift with temperature and ageing. Sensitivity tells how much these errors shift ωo\omega_o, QQ and gain.
  • It lets the designer compare circuits that realize the same T(s)T(s) and choose the one least affected by element changes (low-sensitivity design).
  • It shows which elements need tight tolerance or trimming, which controls cost.
  • High-QQ filters are especially sensitive, so sensitivity decides whether a design will work after manufacture.

Sensitivity of ωo\omega_o in the Tow-Thomas low-pass filter

Tow-Thomas low-pass biquad: OA1 is a lossy integrator (R2∥C1R_2 \parallel C_1 in feedback, inputs VinV_{in} through R3R_3 and V3V_3 through R1R_1), OA2 is an integrator (R4R_4, C2C_2), and OA3 is a unity-gain inverter (V3=−V2V_3 = -V_2) built with two equal resistors rr (the gain stays −1-1 when both change together).

        +------R2------+
        +------||------+
        |      C1      |        +--||--+
Vin-R3--+--|-\         |        |  C2  |
        |  |  >--------+--R4----+-|-\  |
        |  +|/  V1     (OA1)    | |  >-+-- V2 (LP)
        | GND                GND+|/    |
        |                     (OA2)    |
        +--R1-- V3 --[inverter -1]-----+
                         (OA3)

Node equations (ideal op-amps):

V1(1R2+sC1)=−VinR3−V3R1V2=−V1sR4C2,V3=−V2\begin{aligned} V_1\left(\frac{1}{R_2} + sC_1\right) &= -\frac{V_{in}}{R_3} - \frac{V_3}{R_1} \\ V_2 &= -\frac{V_1}{sR_4C_2}, \qquad V_3 = -V_2 \end{aligned}

Solving:

T(s)=V2Vin=1R3R4C1C2s2+sR2C1+1R1R4C1C2T(s) = \frac{V_2}{V_{in}} = \frac{\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_2C_1} + \dfrac{1}{R_1R_4C_1C_2}}

So

ωo=1R1R4C1C2=R1−1/2R4−1/2C1−1/2C2−1/2,Q=ωoR2C1=R2C1R1R4C2\omega_o = \frac{1}{\sqrt{R_1R_4C_1C_2}} = R_1^{-1/2}R_4^{-1/2}C_1^{-1/2}C_2^{-1/2}, \qquad Q = \omega_o R_2C_1 = R_2\sqrt{\frac{C_1}{R_1R_4C_2}}

Sensitivity of ωo\omega_o: ωo=R1−1/2R4−1/2C1−1/2C2−1/2\omega_o = R_1^{-1/2}R_4^{-1/2}C_1^{-1/2}C_2^{-1/2}, so using Sxxn=nS^{x^n}_x = n:

SR1ωo=SR4ωo=SC1ωo=SC2ωo=−12,SR2ωo=SR3ωo=Srωo=0S^{\omega_o}_{R_1} = S^{\omega_o}_{R_4} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12, \qquad S^{\omega_o}_{R_2} = S^{\omega_o}_{R_3} = S^{\omega_o}_{r} = 0

Example of one step: SR1ωo=R1ωo∂ωo∂R1=R1ωo(−12ωoR1)=−12S^{\omega_o}_{R_1} = \dfrac{R_1}{\omega_o}\dfrac{\partial\omega_o}{\partial R_1} = \dfrac{R_1}{\omega_o}\left(-\dfrac{1}{2}\dfrac{\omega_o}{R_1}\right) = -\dfrac12.

Check: sum over resistors =−1= -1, sum over capacitors =−1= -1, as expected for a frequency.

ElementSωoS^{\omega_o}
R1R_1, R4R_4−1/2 each
C1C_1, C2C_2−1/2 each
R2R_2, R3R_3, inverter rr0

A 1% increase in R1R_1, R4R_4, C1C_1 or C2C_2 lowers ωo\omega_o by about 0.5%; ωo\omega_o is unaffected by R2R_2 (which sets QQ) and R3R_3 (which sets gain).

✓ Answer checked by Claude Sonnet 5.5

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