--- title: "Sensitivity of ωo in Tow-Thomas LPF" subject: "Filter Design" chapter: 7 key: "semester_5/filter_design:sensitivity-wo-tow-thomas" source: https://www.bctengineer.com/q/semester_5/filter_design/sensitivity-wo-tow-thomas site: BCT Engineering --- # Sensitivity of ωo in Tow-Thomas LPF Filter Design · Chapter 7: Sensitivity - Asked in: 2079 Bhadra, 1+4 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/sensitivity-wo-tow-thomas - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-7 ## Question Why is sensitivity analysis important in filter design? Perform the sensitivity analysis of ωo in Tow Thomas low pass filter. ## 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. ## Answer #### Why sensitivity analysis is important **Sensitivity** is the per-unit (fractional) change in a filter parameter $y$ (such as $\omega_o$, $Q$ or gain) caused by a per-unit change in an element value $x$: $$ S^y_x = \frac{\partial y / y}{\partial x / x} = \frac{x}{y}\frac{\partial y}{\partial x} = \frac{\partial \ln y}{\partial \ln x} $$ So $\dfrac{\Delta y}{y} \approx S^y_x \dfrac{\Delta x}{x}$. **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 $\omega_o$, $Q$ and gain. - It lets the designer compare circuits that realize the same $T(s)$ 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-$Q$ filters are especially sensitive, so sensitivity decides whether a design will work after manufacture. #### Sensitivity of $\omega_o$ in the Tow-Thomas low-pass filter **Tow-Thomas low-pass biquad:** OA1 is a lossy integrator ($R_2 \parallel C_1$ in feedback, inputs $V_{in}$ through $R_3$ and $V_3$ through $R_1$), OA2 is an integrator ($R_4$, $C_2$), and OA3 is a unity-gain inverter ($V_3 = -V_2$) built with two equal resistors $r$ (the gain stays $-1$ when both change together). ```text +------R2------+ +------||------+ | C1 | +--||--+ Vin-R3--+--|-\ | | C2 | | | >--------+--R4----+-|-\ | | +|/ V1 (OA1) | | >-+-- V2 (LP) | GND GND+|/ | | (OA2) | +--R1-- V3 --[inverter -1]-----+ (OA3) ``` Node equations (ideal op-amps): $$ \begin{aligned} V_1\left(\frac{1}{R_2} + sC_1\right) &= -\frac{V_{in}}{R_3} - \frac{V_3}{R_1} \\ V_2 &= -\frac{V_1}{sR_4C_2}, \qquad V_3 = -V_2 \end{aligned} $$ Solving: $$ T(s) = \frac{V_2}{V_{in}} = \frac{\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_2C_1} + \dfrac{1}{R_1R_4C_1C_2}} $$ So $$ \omega_o = \frac{1}{\sqrt{R_1R_4C_1C_2}} = R_1^{-1/2}R_4^{-1/2}C_1^{-1/2}C_2^{-1/2}, \qquad Q = \omega_o R_2C_1 = R_2\sqrt{\frac{C_1}{R_1R_4C_2}} $$ **Sensitivity of $\omega_o$:** $\omega_o = R_1^{-1/2}R_4^{-1/2}C_1^{-1/2}C_2^{-1/2}$, so using $S^{x^n}_x = n$: $$ S^{\omega_o}_{R_1} = S^{\omega_o}_{R_4} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12, \qquad S^{\omega_o}_{R_2} = S^{\omega_o}_{R_3} = S^{\omega_o}_{r} = 0 $$ Example of one step: $S^{\omega_o}_{R_1} = \dfrac{R_1}{\omega_o}\dfrac{\partial\omega_o}{\partial R_1} = \dfrac{R_1}{\omega_o}\left(-\dfrac{1}{2}\dfrac{\omega_o}{R_1}\right) = -\dfrac12$. Check: sum over resistors $= -1$, sum over capacitors $= -1$, as expected for a frequency. | Element | $S^{\omega_o}$ | |---|---| | $R_1$, $R_4$ | −1/2 each | | $C_1$, $C_2$ | −1/2 each | | $R_2$, $R_3$, inverter $r$ | 0 | A 1% increase in $R_1$, $R_4$, $C_1$ or $C_2$ lowers $\omega_o$ by about 0.5%; $\omega_o$ is unaffected by $R_2$ (which sets $Q$) and $R_3$ (which sets gain). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is sensitivity analysis important in filter design? A. It shows how component tolerance and drift shift wo, Q and gain B. It increases the filter order C. It removes the need for op-amps D. It sets the filter's input impedance Answer: A. Real components have tolerances and drift, and sensitivity measures the effect on filter parameters, helping choose low-sensitivity designs. (Key point: Why sensitivity matters) 2. Which expression defines sensitivity of y to x? A. dy/dx B. (dy/y)/(dx/x) C. y/x D. (dx/x)/(dy/y) Answer: B. Sensitivity is the fractional change in y per fractional change in x, i.e. d ln y / d ln x. (Key point: Definition) 3. What is wo of the Tow-Thomas low-pass biquad? A. 1/sqrt(R1 R4 C1 C2) B. 1/(R2 C1) C. 1/sqrt(R2 R3 C1 C2) D. R2 sqrt(C1/(R1 R4 C2)) Answer: A. The s^0 term of the denominator is 1/(R1R4C1C2). (Key point: Tow-Thomas wo) 4. Which elements do not affect wo in the Tow-Thomas circuit? A. R1 and R4 B. C1 and C2 C. R2 and R3 D. R1 and C2 Answer: C. R2 sets Q and R3 sets gain; neither appears in wo, so their sensitivities are 0. 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