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Chapter 7 · 3 hours

Sensitivity

IOE past exam questions

Past questions and answers

16 questions set from this chapter, 5 of them more than once. Most asked first.

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

Answer

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

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%.

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Define sensitivity. What is the importance of sensitivity analysis in filter design? Perform the sensitivity analysis of Tow-Thomas low pass biquad filter (with respect to all the resistors and capacitors present in the circuit).

Answer

Definition

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 of sensitivity analysis

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 analysis of the Tow-Thomas low-pass biquad

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 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.

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.

Sensitivity of DC gain H=R1R3H = \dfrac{R_1}{R_3} (from T(0)T(0)): SR1H=1S^H_{R_1} = 1, SR3H=−1S^H_{R_3} = -1, all others 0.

ElementSωoS^{\omega_o}SQS^{Q}
R1R_1−1/2−1/2
R2R_201
R3R_300
R4R_4−1/2−1/2
C1C_1−1/21/2
C2C_2−1/2−1/2
Inverter rr00

All sensitivities are at most 1 in magnitude and do not depend on QQ, so the Tow-Thomas biquad is a low-sensitivity circuit; R2R_2 tunes QQ alone, and R3R_3 sets the gain alone.

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What is (define) sensitivity? What is the importance of sensitivity analysis in filter design? Perform sensitivity analysis of ωo in Sallen-Key low pass filter (with respect to all the resistors and capacitors).

Answer

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}.

For example, SRωo=−0.5S^{\omega_o}_R = -0.5 means a 1% increase in RR gives about a 0.5% decrease in ωo\omega_o.

Importance of sensitivity analysis

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 Sallen-Key low-pass filter

Sallen-Key low-pass biquad (gain K=1+RB/RAK = 1 + R_B/R_A):

              C1
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         RA
                         |
                        GND

Its transfer function is

T(s)=K/(R1R2C1C2)s2+s(1R1C1+1R2C1+1−KR2C2)+1R1R2C1C2T(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}

so

ωo=1R1R2C1C2=R1−1/2R2−1/2C1−1/2C2−1/2\omega_o = \frac{1}{\sqrt{R_1R_2C_1C_2}} = R_1^{-1/2}R_2^{-1/2}C_1^{-1/2}C_2^{-1/2}

Sensitivity with respect to R1R_1:

∂ωo∂R1=−12R1−3/2(R2C1C2)−1/2=−ωo2R1SR1ωo=R1ωo⋅(−ωo2R1)=−12\begin{aligned} \frac{\partial\omega_o}{\partial R_1} &= -\frac12 R_1^{-3/2}(R_2C_1C_2)^{-1/2} = -\frac{\omega_o}{2R_1} \\ S^{\omega_o}_{R_1} &= \frac{R_1}{\omega_o}\cdot\left(-\frac{\omega_o}{2R_1}\right) = -\frac12 \end{aligned}

In the same way:

SR1ωo=SR2ωo=SC1ωo=SC2ωo=−12S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12

ωo\omega_o does not contain KK, RAR_A or RBR_B, so

SKωo=SRAωo=SRBωo=0S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0
ElementSωoS^{\omega_o}
R1R_1, R2R_2−1/2 each
C1C_1, C2C_2−1/2 each
RAR_A, RBR_B (gain)0

So a 1% increase in any one R or C lowers ωo\omega_o by about 0.5%; if all four increase by 1%, ωo\omega_o falls by about 2%. (The Q of the Sallen-Key circuit is more sensitive: for the equal-element design SKQ=3Q−1S^Q_K = 3Q - 1, which grows with Q.)

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What is the importance of sensitivity analysis in filter design? Perform sensitivity analysis for ωo of Sallen-Key low pass filter with respect to all the resistors and capacitors present in the circuit.

Answer

Importance of sensitivity analysis

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.

Useful properties: Sxkx=1S^{kx}_x = 1, Sxxn=nS^{x^n}_x = n, Sx1/y=−SxyS^{1/y}_x = -S^y_x, Sxy1y2=Sxy1+Sxy2S^{y_1y_2}_x = S^{y_1}_x + S^{y_2}_x, Sxky=SxyS^{ky}_x = S^y_x (kk constant).

Sensitivity of ωo\omega_o of the Sallen-Key low-pass filter

Sallen-Key low-pass biquad (gain K=1+RB/RAK = 1 + R_B/R_A):

              C1
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         RA
                         |
                        GND

Its transfer function is

T(s)=K/(R1R2C1C2)s2+s(1R1C1+1R2C1+1−KR2C2)+1R1R2C1C2T(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}

so

ωo=1R1R2C1C2=R1−1/2R2−1/2C1−1/2C2−1/2\omega_o = \frac{1}{\sqrt{R_1R_2C_1C_2}} = R_1^{-1/2}R_2^{-1/2}C_1^{-1/2}C_2^{-1/2}

Sensitivity with respect to R1R_1:

∂ωo∂R1=−12R1−3/2(R2C1C2)−1/2=−ωo2R1SR1ωo=R1ωo⋅(−ωo2R1)=−12\begin{aligned} \frac{\partial\omega_o}{\partial R_1} &= -\frac12 R_1^{-3/2}(R_2C_1C_2)^{-1/2} = -\frac{\omega_o}{2R_1} \\ S^{\omega_o}_{R_1} &= \frac{R_1}{\omega_o}\cdot\left(-\frac{\omega_o}{2R_1}\right) = -\frac12 \end{aligned}

In the same way:

SR1ωo=SR2ωo=SC1ωo=SC2ωo=−12S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12

ωo\omega_o does not contain KK, RAR_A or RBR_B, so

SKωo=SRAωo=SRBωo=0S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0
ElementSωoS^{\omega_o}
R1R_1, R2R_2−1/2 each
C1C_1, C2C_2−1/2 each
RAR_A, RBR_B (gain)0

So a 1% increase in any one R or C lowers ωo\omega_o by about 0.5%; if all four increase by 1%, ωo\omega_o falls by about 2%. (The Q of the Sallen-Key circuit is more sensitive: for the equal-element design SKQ=3Q−1S^Q_K = 3Q - 1, which grows with Q.)

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What is the sensitivity of a filter? Explain the single parameter and multi-parameter sensitivity. Perform the sensitivity analysis of ωo of Sallen-Key lowpass biquad filter.

Answer

Sensitivity of a filter

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}. It measures how much a filter parameter (ωo\omega_o, QQ, gain, or ∣T(jω)∣|T(j\omega)|) shifts when components deviate from their nominal values because of tolerance, temperature or ageing.

Single-parameter and multi-parameter sensitivity

Single-parameter sensitivity measures the effect of a change in one element while all others stay fixed:

Sxy=xy∂y∂x,Δyy≈SxyΔxxS^y_x = \frac{x}{y}\frac{\partial y}{\partial x}, \qquad \frac{\Delta y}{y} \approx S^y_x\frac{\Delta x}{x}

Examples are SR1ωoS^{\omega_o}_{R_1}, SKQS^Q_K, or the pole sensitivity SxpS^{p}_x. Properties: Sxxn=nS^{x^n}_x = n, Sx1/y=−SxyS^{1/y}_x = -S^y_x, Sxy1y2=Sxy1+Sxy2S^{y_1y_2}_x = S^{y_1}_x + S^{y_2}_x, Sxky=SxyS^{ky}_x = S^y_x.

Multi-parameter sensitivity gives the change in yy when all elements x1,x2,…,xnx_1, x_2, \dots, x_n change together, as they do in practice. From the total differential:

Δyy≈∑i=1nSxiyΔxixi\frac{\Delta y}{y} \approx \sum_{i=1}^{n} S^{y}_{x_i}\frac{\Delta x_i}{x_i}
  • Worst case: if every element can be off by up to εi\varepsilon_i, the largest error is ∣Δyy∣max=∑i∣Sxiy∣ εi\left|\dfrac{\Delta y}{y}\right|_{max} = \sum_i |S^y_{x_i}|\,\varepsilon_i.
  • Statistical (random, independent tolerances): the standard deviation is σy/y=∑i(Sxiy)2σxi2\sigma_{y}/y = \sqrt{\sum_i (S^y_{x_i})^2\sigma_{x_i}^2}, which is usually much smaller than the worst case.
  • The Schoeffler sensitivity ∑i(Sxiy)2\sum_i (S^y_{x_i})^2 is a single figure of merit for comparing circuits.

Example: for the Sallen-Key ωo\omega_o, all four sensitivities are −1/2. With ±1% elements the worst-case error is 4×0.5×1%=2%4 \times 0.5 \times 1\% = 2\%, while the statistical spread is 4×0.25×1%=1%\sqrt{4 \times 0.25}\times 1\% = 1\%.

Sensitivity analysis of ωo\omega_o of the Sallen-Key low-pass biquad

Sallen-Key low-pass biquad (gain K=1+RB/RAK = 1 + R_B/R_A):

              C1
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         RA
                         |
                        GND

Its transfer function is

T(s)=K/(R1R2C1C2)s2+s(1R1C1+1R2C1+1−KR2C2)+1R1R2C1C2T(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}

so

ωo=1R1R2C1C2=R1−1/2R2−1/2C1−1/2C2−1/2\omega_o = \frac{1}{\sqrt{R_1R_2C_1C_2}} = R_1^{-1/2}R_2^{-1/2}C_1^{-1/2}C_2^{-1/2}

Sensitivity with respect to R1R_1:

∂ωo∂R1=−12R1−3/2(R2C1C2)−1/2=−ωo2R1SR1ωo=R1ωo⋅(−ωo2R1)=−12\begin{aligned} \frac{\partial\omega_o}{\partial R_1} &= -\frac12 R_1^{-3/2}(R_2C_1C_2)^{-1/2} = -\frac{\omega_o}{2R_1} \\ S^{\omega_o}_{R_1} &= \frac{R_1}{\omega_o}\cdot\left(-\frac{\omega_o}{2R_1}\right) = -\frac12 \end{aligned}

In the same way:

SR1ωo=SR2ωo=SC1ωo=SC2ωo=−12S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12

ωo\omega_o does not contain KK, RAR_A or RBR_B, so

SKωo=SRAωo=SRBωo=0S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0
ElementSωoS^{\omega_o}
R1R_1, R2R_2−1/2 each
C1C_1, C2C_2−1/2 each
RAR_A, RBR_B (gain)0

So a 1% increase in any one R or C lowers ωo\omega_o by about 0.5%; if all four increase by 1%, ωo\omega_o falls by about 2%. (The Q of the Sallen-Key circuit is more sensitive: for the equal-element design SKQ=3Q−1S^Q_K = 3Q - 1, which grows with Q.)

  • 2070 Chaitra · 1+4 marks

What is signal parameter sensitivity? Perform sensitivity analysis for center frequency (ωo) of Sallen-Key biquad with respect to all resistors and capacitors present in the circuit.

Answer

Signal parameter sensitivity

Signal parameter sensitivity is the sensitivity of a filter's performance parameter, such as the centre (pole) frequency ωo\omega_o, quality factor QQ or gain HH, to a change in an element value xx:

Sxωo=xωo∂ωo∂x,SxQ=xQ∂Q∂xS^{\omega_o}_x = \frac{x}{\omega_o}\frac{\partial\omega_o}{\partial x}, \qquad S^{Q}_x = \frac{x}{Q}\frac{\partial Q}{\partial x}

It is the per-unit change of the parameter per unit change of the element, Δωoωo≈SxωoΔxx\dfrac{\Delta\omega_o}{\omega_o} \approx S^{\omega_o}_x\dfrac{\Delta x}{x}. These parameters fix the pole positions, so they show directly how the response shifts.

Sensitivity of ωo\omega_o of the Sallen-Key biquad

Sallen-Key low-pass biquad (gain K=1+RB/RAK = 1 + R_B/R_A):

              C1
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         RA
                         |
                        GND

Its transfer function is

T(s)=K/(R1R2C1C2)s2+s(1R1C1+1R2C1+1−KR2C2)+1R1R2C1C2T(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}

so

ωo=1R1R2C1C2=R1−1/2R2−1/2C1−1/2C2−1/2\omega_o = \frac{1}{\sqrt{R_1R_2C_1C_2}} = R_1^{-1/2}R_2^{-1/2}C_1^{-1/2}C_2^{-1/2}

Sensitivity with respect to R1R_1:

∂ωo∂R1=−12R1−3/2(R2C1C2)−1/2=−ωo2R1SR1ωo=R1ωo⋅(−ωo2R1)=−12\begin{aligned} \frac{\partial\omega_o}{\partial R_1} &= -\frac12 R_1^{-3/2}(R_2C_1C_2)^{-1/2} = -\frac{\omega_o}{2R_1} \\ S^{\omega_o}_{R_1} &= \frac{R_1}{\omega_o}\cdot\left(-\frac{\omega_o}{2R_1}\right) = -\frac12 \end{aligned}

In the same way:

SR1ωo=SR2ωo=SC1ωo=SC2ωo=−12S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12

ωo\omega_o does not contain KK, RAR_A or RBR_B, so

SKωo=SRAωo=SRBωo=0S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0
ElementSωoS^{\omega_o}
R1R_1, R2R_2−1/2 each
C1C_1, C2C_2−1/2 each
RAR_A, RBR_B (gain)0

So a 1% increase in any one R or C lowers ωo\omega_o by about 0.5%; if all four increase by 1%, ωo\omega_o falls by about 2%. (The Q of the Sallen-Key circuit is more sensitive: for the equal-element design SKQ=3Q−1S^Q_K = 3Q - 1, which grows with Q.)

  • 2079 Baisakh · 1+4 marks

What information do you get when sensitivity of y with respect to x is 10? Perform sensitivity analysis for center frequency (ωo) of Sallen Key lowpass filter with respect to all the resistors and capacitors present in the circuit.

Answer

Meaning of Sxy=10S^y_x = 10

Sxy=Δy/yΔx/x=10  ⇒  Δyy≈10ΔxxS^y_x = \frac{\Delta y/y}{\Delta x/x} = 10 \;\Rightarrow\; \frac{\Delta y}{y} \approx 10\frac{\Delta x}{x}

A 1% increase in xx causes about a 10% increase in yy (same direction, since the sign is positive). yy is very sensitive to xx: the element xx must have a tight tolerance, or the circuit should be changed to a lower-sensitivity design.

Sensitivity of ωo\omega_o of the Sallen-Key low-pass filter

Sallen-Key low-pass biquad (gain K=1+RB/RAK = 1 + R_B/R_A):

              C1
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         RA
                         |
                        GND

Its transfer function is

T(s)=K/(R1R2C1C2)s2+s(1R1C1+1R2C1+1−KR2C2)+1R1R2C1C2T(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}

so

ωo=1R1R2C1C2=R1−1/2R2−1/2C1−1/2C2−1/2\omega_o = \frac{1}{\sqrt{R_1R_2C_1C_2}} = R_1^{-1/2}R_2^{-1/2}C_1^{-1/2}C_2^{-1/2}

Sensitivity with respect to R1R_1:

∂ωo∂R1=−12R1−3/2(R2C1C2)−1/2=−ωo2R1SR1ωo=R1ωo⋅(−ωo2R1)=−12\begin{aligned} \frac{\partial\omega_o}{\partial R_1} &= -\frac12 R_1^{-3/2}(R_2C_1C_2)^{-1/2} = -\frac{\omega_o}{2R_1} \\ S^{\omega_o}_{R_1} &= \frac{R_1}{\omega_o}\cdot\left(-\frac{\omega_o}{2R_1}\right) = -\frac12 \end{aligned}

In the same way:

SR1ωo=SR2ωo=SC1ωo=SC2ωo=−12S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12

ωo\omega_o does not contain KK, RAR_A or RBR_B, so

SKωo=SRAωo=SRBωo=0S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0
ElementSωoS^{\omega_o}
R1R_1, R2R_2−1/2 each
C1C_1, C2C_2−1/2 each
RAR_A, RBR_B (gain)0

So a 1% increase in any one R or C lowers ωo\omega_o by about 0.5%; if all four increase by 1%, ωo\omega_o falls by about 2%. (The Q of the Sallen-Key circuit is more sensitive: for the equal-element design SKQ=3Q−1S^Q_K = 3Q - 1, which grows with Q.)

  • 2075 Asoj · 1+4 marks

What information do you get when sensitivity of x with respect to y is -3? Perform sensitivity analysis for ωo of Sallen Key low pass filter with respect to all the resistors and capacitors present in the circuit.

Answer

Meaning of Syx=−3S^x_y = -3

Syx=Δx/xΔy/y=−3  ⇒  Δxx≈−3ΔyyS^x_y = \frac{\Delta x/x}{\Delta y/y} = -3 \;\Rightarrow\; \frac{\Delta x}{x} \approx -3\frac{\Delta y}{y}

A 1% increase in yy causes about a 3% decrease in xx; the minus sign means xx moves opposite to yy. Since ∣S∣>1|S| > 1, the change is amplified three times, so xx is fairly sensitive to yy.

Sensitivity of ωo\omega_o of the Sallen-Key low-pass filter

Sallen-Key low-pass biquad (gain K=1+RB/RAK = 1 + R_B/R_A):

              C1
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         RA
                         |
                        GND

Its transfer function is

T(s)=K/(R1R2C1C2)s2+s(1R1C1+1R2C1+1−KR2C2)+1R1R2C1C2T(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}

so

ωo=1R1R2C1C2=R1−1/2R2−1/2C1−1/2C2−1/2\omega_o = \frac{1}{\sqrt{R_1R_2C_1C_2}} = R_1^{-1/2}R_2^{-1/2}C_1^{-1/2}C_2^{-1/2}

Sensitivity with respect to R1R_1:

∂ωo∂R1=−12R1−3/2(R2C1C2)−1/2=−ωo2R1SR1ωo=R1ωo⋅(−ωo2R1)=−12\begin{aligned} \frac{\partial\omega_o}{\partial R_1} &= -\frac12 R_1^{-3/2}(R_2C_1C_2)^{-1/2} = -\frac{\omega_o}{2R_1} \\ S^{\omega_o}_{R_1} &= \frac{R_1}{\omega_o}\cdot\left(-\frac{\omega_o}{2R_1}\right) = -\frac12 \end{aligned}

In the same way:

SR1ωo=SR2ωo=SC1ωo=SC2ωo=−12S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12

ωo\omega_o does not contain KK, RAR_A or RBR_B, so

SKωo=SRAωo=SRBωo=0S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0
ElementSωoS^{\omega_o}
R1R_1, R2R_2−1/2 each
C1C_1, C2C_2−1/2 each
RAR_A, RBR_B (gain)0

So a 1% increase in any one R or C lowers ωo\omega_o by about 0.5%; if all four increase by 1%, ωo\omega_o falls by about 2%. (The Q of the Sallen-Key circuit is more sensitive: for the equal-element design SKQ=3Q−1S^Q_K = 3Q - 1, which grows with Q.)

  • 2072 Chaitra · 1+4 marks

What information do you get when the sensitivity of x with respect to y is -5? Perform sensitivity analysis for center frequency (ωo) of the Sallen Key low pass filter with respect to all the resistors and capacitors present in the circuit.

Answer

Meaning of Syx=−5S^x_y = -5

Syx=Δx/xΔy/y=−5  ⇒  Δxx≈−5ΔyyS^x_y = \frac{\Delta x/x}{\Delta y/y} = -5 \;\Rightarrow\; \frac{\Delta x}{x} \approx -5\frac{\Delta y}{y}

A 1% increase in yy causes about a 5% decrease in xx (opposite direction, five times larger). xx is highly sensitive to yy, so yy needs a tight tolerance.

Sensitivity of ωo\omega_o of the Sallen-Key low-pass filter

Sallen-Key low-pass biquad (gain K=1+RB/RAK = 1 + R_B/R_A):

              C1
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         RA
                         |
                        GND

Its transfer function is

T(s)=K/(R1R2C1C2)s2+s(1R1C1+1R2C1+1−KR2C2)+1R1R2C1C2T(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}

so

ωo=1R1R2C1C2=R1−1/2R2−1/2C1−1/2C2−1/2\omega_o = \frac{1}{\sqrt{R_1R_2C_1C_2}} = R_1^{-1/2}R_2^{-1/2}C_1^{-1/2}C_2^{-1/2}

Sensitivity with respect to R1R_1:

∂ωo∂R1=−12R1−3/2(R2C1C2)−1/2=−ωo2R1SR1ωo=R1ωo⋅(−ωo2R1)=−12\begin{aligned} \frac{\partial\omega_o}{\partial R_1} &= -\frac12 R_1^{-3/2}(R_2C_1C_2)^{-1/2} = -\frac{\omega_o}{2R_1} \\ S^{\omega_o}_{R_1} &= \frac{R_1}{\omega_o}\cdot\left(-\frac{\omega_o}{2R_1}\right) = -\frac12 \end{aligned}

In the same way:

SR1ωo=SR2ωo=SC1ωo=SC2ωo=−12S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12

ωo\omega_o does not contain KK, RAR_A or RBR_B, so

SKωo=SRAωo=SRBωo=0S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0
ElementSωoS^{\omega_o}
R1R_1, R2R_2−1/2 each
C1C_1, C2C_2−1/2 each
RAR_A, RBR_B (gain)0

So a 1% increase in any one R or C lowers ωo\omega_o by about 0.5%; if all four increase by 1%, ωo\omega_o falls by about 2%. (The Q of the Sallen-Key circuit is more sensitive: for the equal-element design SKQ=3Q−1S^Q_K = 3Q - 1, which grows with Q.)

  • 2079 Bhadra

What do you understand when sensitivity of y with respect to x is 2. Perform sensitivity analysis for center frequency of Sallen-key low pass filter with respect to all elements presents in the circuit.

Answer

Meaning of Sxy=2S^y_x = 2

Sxy=Δy/yΔx/x=2  ⇒  Δyy≈2ΔxxS^y_x = \frac{\Delta y/y}{\Delta x/x} = 2 \;\Rightarrow\; \frac{\Delta y}{y} \approx 2\frac{\Delta x}{x}

A 1% increase in xx causes about a 2% increase in yy (same direction, twice as large). For example, if y∝x2y \propto x^2 then Sxy=2S^y_x = 2.

Sensitivity of the centre frequency of the Sallen-Key low-pass filter

Sallen-Key low-pass biquad (gain K=1+RB/RAK = 1 + R_B/R_A):

              C1
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         RA
                         |
                        GND

Its transfer function is

T(s)=K/(R1R2C1C2)s2+s(1R1C1+1R2C1+1−KR2C2)+1R1R2C1C2T(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}

so

ωo=1R1R2C1C2=R1−1/2R2−1/2C1−1/2C2−1/2\omega_o = \frac{1}{\sqrt{R_1R_2C_1C_2}} = R_1^{-1/2}R_2^{-1/2}C_1^{-1/2}C_2^{-1/2}

Sensitivity with respect to R1R_1:

∂ωo∂R1=−12R1−3/2(R2C1C2)−1/2=−ωo2R1SR1ωo=R1ωo⋅(−ωo2R1)=−12\begin{aligned} \frac{\partial\omega_o}{\partial R_1} &= -\frac12 R_1^{-3/2}(R_2C_1C_2)^{-1/2} = -\frac{\omega_o}{2R_1} \\ S^{\omega_o}_{R_1} &= \frac{R_1}{\omega_o}\cdot\left(-\frac{\omega_o}{2R_1}\right) = -\frac12 \end{aligned}

In the same way:

SR1ωo=SR2ωo=SC1ωo=SC2ωo=−12S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12

ωo\omega_o does not contain KK, RAR_A or RBR_B, so

SKωo=SRAωo=SRBωo=0S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0
ElementSωoS^{\omega_o}
R1R_1, R2R_2−1/2 each
C1C_1, C2C_2−1/2 each
RAR_A, RBR_B (gain)0

So a 1% increase in any one R or C lowers ωo\omega_o by about 0.5%; if all four increase by 1%, ωo\omega_o falls by about 2%. (The Q of the Sallen-Key circuit is more sensitive: for the equal-element design SKQ=3Q−1S^Q_K = 3Q - 1, which grows with Q.)

  • 2079 Bhadra · 1+4 marks

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

Answer

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).

        +------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).

  • 2080 Bhadra · 1+3 marks

What information do you get when the sensitivity of y with respect to x is 0.1? Perform sensitivity analysis for center frequency ωo of Tow Thomas low pass filter with respect to all the resistors and capacitors present in the circuit.

Answer

Meaning of Sxy=0.1S^y_x = 0.1

Δyy≈0.1Δxx\frac{\Delta y}{y} \approx 0.1\frac{\Delta x}{x}

A 1% increase in xx causes only about a 0.1% increase in yy. yy is insensitive to xx, so a cheap, wide-tolerance component can be used for xx. This is a desirable property.

Sensitivity of ωo\omega_o of 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 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, C1C_1, C2C_2−1/2 each
R2R_2, R3R_3, rr0
  • 2071 Chaitra · 5 marks

Perform sensitivity analysis for center frequency (ωo) and quality factor (Q) of the Tow Thomas low pass filter with respect to all the resistors and capacitors present in the circuit.

Answer

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 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.

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.

Summary

ElementSωoS^{\omega_o}SQS^{Q}
R1R_1−1/2−1/2
R2R_201
R3R_300
R4R_4−1/2−1/2
C1C_1−1/21/2
C2C_2−1/2−1/2
Inverter rr00

All sensitivities are at most 1 in magnitude and do not depend on QQ, so the Tow-Thomas biquad is a low-sensitivity circuit; R2R_2 tunes QQ alone, and R3R_3 sets the gain alone.

  • 2069 Chaitra · 1+4 marks

What do you understand when the sensitivity of y with respect to x is equal to -3? Perform sensitivity analysis for Quality factor Q of the Tow Thomas low pass filter with respect to all the resistors and capacitors present in the circuit.

Answer

Meaning of Sxy=−3S^y_x = -3

Δyy≈−3Δxx\frac{\Delta y}{y} \approx -3\frac{\Delta x}{x}

A 1% increase in xx gives about a 3% decrease in yy: the change is three times larger and in the opposite direction. yy is quite sensitive to xx.

Sensitivity of Q of 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, R4R_4, C2C_2−1/2 each
R2R_21
C1C_11/2
R3R_3, inverter rr0
  • 2083 Baisakh · 1+4 marks

What do you understand when someone tells you that sensitivity of x with respect to y is -7? Compute the sensitivity expression for cutoff frequency and quality factor of RLC lowpass filter.

Answer

Meaning of Syx=−7S^x_y = -7

Δxx≈−7Δyy\frac{\Delta x}{x} \approx -7\frac{\Delta y}{y}

A 1% increase in yy causes about a 7% decrease in xx (opposite direction, seven times larger). xx is highly sensitive to yy, so yy must be held to a very tight tolerance.

Sensitivity of RLC low-pass filter

Take the series RLC low-pass circuit with the output across the capacitor:

Vin o--R--L--+--o Vout
             |
            === C
             |
GND o--------+--o
T(s)=1/(sC)R+sL+1/(sC)=1LCs2+RLs+1LCT(s) = \frac{1/(sC)}{R + sL + 1/(sC)} = \frac{\dfrac{1}{LC}}{s^2 + \dfrac{R}{L}s + \dfrac{1}{LC}}

Comparing with ωo2s2+(ωo/Q)s+ωo2\dfrac{\omega_o^2}{s^2 + (\omega_o/Q)s + \omega_o^2}:

ωo=1LC=L−1/2C−1/2,Q=ωoLR=1RLC=R−1L1/2C−1/2\omega_o = \frac{1}{\sqrt{LC}} = L^{-1/2}C^{-1/2}, \qquad Q = \frac{\omega_o L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}} = R^{-1}L^{1/2}C^{-1/2}

Here ωo\omega_o is the cutoff (pole) frequency; for the Butterworth case Q=0.707Q = 0.707 it is also the half-power frequency.

Cutoff frequency:

SLωo=Lωo∂ωo∂L=Lωo(−ωo2L)=−12SCωo=−12,SRωo=0\begin{aligned} S^{\omega_o}_L &= \frac{L}{\omega_o}\frac{\partial\omega_o}{\partial L} = \frac{L}{\omega_o}\left(-\frac{\omega_o}{2L}\right) = -\frac12 \\ S^{\omega_o}_C &= -\frac12, \qquad S^{\omega_o}_R = 0 \end{aligned}

Quality factor:

SRQ=RQ∂Q∂R=RQ(−QR)=−1SLQ=+12,SCQ=−12\begin{aligned} S^Q_R &= \frac{R}{Q}\frac{\partial Q}{\partial R} = \frac{R}{Q}\left(-\frac{Q}{R}\right) = -1 \\ S^Q_L &= +\frac12, \qquad S^Q_C = -\frac12 \end{aligned}
ElementSωoS^{\omega_o}SQS^Q
R0−1
L−1/21/2
C−1/2−1/2

So a 1% rise in L or C lowers the cutoff by 0.5%, while R affects only Q: a 1% rise in R lowers Q by 1%.

  • 2080 Bhadra · 5 marks

What do you understand when the sensitivity of y with respect to x is equal to 0.5? Discuss single parameter and multiple parameter sensitivities in details.

Answer

Meaning of Sxy=0.5S^y_x = 0.5

Δyy≈0.5Δxx\frac{\Delta y}{y} \approx 0.5\frac{\Delta x}{x}

A 1% increase in xx causes about a 0.5% increase in yy; the error is halved, so yy has low sensitivity to xx. This happens, for example, when y∝xy \propto \sqrt{x} (Sxx=1/2S^{\sqrt{x}}_x = 1/2), as with ωo\omega_o of an LC circuit (ωo∝1/LC\omega_o \propto 1/\sqrt{LC}, magnitude 1/2).

Single-parameter and multi-parameter sensitivity

Single-parameter sensitivity measures the effect of a change in one element while all others stay fixed:

Sxy=xy∂y∂x,Δyy≈SxyΔxxS^y_x = \frac{x}{y}\frac{\partial y}{\partial x}, \qquad \frac{\Delta y}{y} \approx S^y_x\frac{\Delta x}{x}

Examples are SR1ωoS^{\omega_o}_{R_1}, SKQS^Q_K, or the pole sensitivity SxpS^{p}_x. Properties: Sxxn=nS^{x^n}_x = n, Sx1/y=−SxyS^{1/y}_x = -S^y_x, Sxy1y2=Sxy1+Sxy2S^{y_1y_2}_x = S^{y_1}_x + S^{y_2}_x, Sxky=SxyS^{ky}_x = S^y_x.

Multi-parameter sensitivity gives the change in yy when all elements x1,x2,…,xnx_1, x_2, \dots, x_n change together, as they do in practice. From the total differential:

Δyy≈∑i=1nSxiyΔxixi\frac{\Delta y}{y} \approx \sum_{i=1}^{n} S^{y}_{x_i}\frac{\Delta x_i}{x_i}
  • Worst case: if every element can be off by up to εi\varepsilon_i, the largest error is ∣Δyy∣max=∑i∣Sxiy∣ εi\left|\dfrac{\Delta y}{y}\right|_{max} = \sum_i |S^y_{x_i}|\,\varepsilon_i.
  • Statistical (random, independent tolerances): the standard deviation is σy/y=∑i(Sxiy)2σxi2\sigma_{y}/y = \sqrt{\sum_i (S^y_{x_i})^2\sigma_{x_i}^2}, which is usually much smaller than the worst case.
  • The Schoeffler sensitivity ∑i(Sxiy)2\sum_i (S^y_{x_i})^2 is a single figure of merit for comparing circuits.

Example: for the Sallen-Key ωo\omega_o, all four sensitivities are −1/2. With ±1% elements the worst-case error is 4×0.5×1%=2%4 \times 0.5 \times 1\% = 2\%, while the statistical spread is 4×0.25×1%=1%\sqrt{4 \times 0.25}\times 1\% = 1\%.

Other single-parameter measures:

  • Function sensitivity SxT(jω)S^{T(j\omega)}_x: change of the whole response with xx; its real part gives the magnitude change, Sx∣T∣=Re SxTS^{|T|}_x = \text{Re}\,S^{T}_x.
  • Root (pole/zero) sensitivity Sxp=x ∂p∂xS^{p}_x = x\,\dfrac{\partial p}{\partial x} (unnormalized form): movement of a pole in the s-plane.

Multi-parameter use in design: single-parameter values show which element matters most, while the multi-parameter sum shows the total expected error from all elements together, which decides the component tolerances needed to meet the specification.

Questions from Old Question Collection (BEI EX 606 and BEX EX 704) (Scanned IOE papers: BEI EX 606 2078–2083 and BEX EX 704 2069–2076), Old Question Collection (EX 704) (IOE BEX EX 704 papers from 2069 to 2081) and 2080 course paper (ENEX 301) (IOE ENEX 301 new-course paper, 2082 Chaitra). Answers are written for this site; check them against your class notes.

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