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Sensitivity of Q in Tow-Thomas LPF

Short answer

Sensitivity S_x^y=(x/y)(∂y/∂x) measures how component variations shift a parameter, guiding tolerance allocation and circuit choice, especially in high-Q designs. In the Tow-Thomas LP biquad, Q=R2√(C1/(R1R4C2)), so S_{R2}^Q=1, S_{C1}^Q=1/2, S_{R1}^Q=S_{R4}^Q=S_{C2}^Q=−1/2; all are small and Q-independent.

What is sensitivity? Describe its importance in filter design. Perform sensitivity analysis of quality factor (Q) in Tow-Thomas low pass filter.

What is sensitivity

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 Q 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).

        +------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 QQ: Q=R2 C11/2 R1−1/2R4−1/2C2−1/2Q = R_2\,C_1^{1/2}\,R_1^{-1/2}R_4^{-1/2}C_2^{-1/2}, so

SR2Q=1,SC1Q=12,SR1Q=SR4Q=SC2Q=−12,SR3Q=SrQ=0S^{Q}_{R_2} = 1, \quad S^{Q}_{C_1} = \frac12, \quad S^{Q}_{R_1} = S^{Q}_{R_4} = S^{Q}_{C_2} = -\frac12, \quad S^Q_{R_3} = S^Q_r = 0

For example, SR2Q=R2Q⋅∂Q∂R2=R2Q⋅QR2=1S^Q_{R_2} = \dfrac{R_2}{Q}\cdot\dfrac{\partial Q}{\partial R_2} = \dfrac{R_2}{Q}\cdot\dfrac{Q}{R_2} = 1.

Check: sum over resistors =1−12−12=0= 1 - \frac12 - \frac12 = 0, sum over capacitors =12−12=0= \frac12 - \frac12 = 0, as expected for a dimensionless quantity.

ElementSQS^Q
R1R_1−1/2
R2R_21
R3R_30
R4R_4−1/2
C1C_11/2
C2C_2−1/2

All QQ-sensitivities are small (magnitude ≤ 1) and independent of QQ, so the Tow-Thomas biquad can realize high-QQ poles reliably. A 1% rise in R2R_2 raises QQ by 1%.

✓ Answer checked by Claude Sonnet 5.5

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