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First order filter: pole -100, zero -1000, gain 10

Short answer

With Z1=R1, Z2=R2||C2 in a non-inverting op-amp, set pole 1/(R2C2)=100 and DC gain 1+R2/R1=10. Choosing C2=1 μF gives R2=10 kΩ, R1=1.11 kΩ; HF gain is 1.

Design a first order filter having a pole at -100 and a zero at -1000 with DC gain of 10 using non-inverting op-amp configuration.

Required function: pole at s=−100s=-100, zero at s=−1000s=-1000:

T(s)=Ks+1000s+100,T(0)=10K=10⇒K=1T(s) = K\frac{s+1000}{s+100}, \qquad T(0) = 10K = 10 \Rightarrow K = 1

So T(s)=s+1000s+100T(s) = \dfrac{s + 1000}{s + 100}: DC gain 10, HF gain 1.

Non-inverting configuration: Z1Z_1 from the inverting input to ground, Z2Z_2 from output to inverting input:

T(s)=1+Z2Z1T(s) = 1 + \frac{Z_2}{Z_1}

Use Z1=R1Z_1 = R_1 and Z2=R2∥C2Z_2 = R_2 \parallel C_2:

T(s)=s+1R2C2(1+R2R1)s+1R2C2T(s) = \frac{s + \frac{1}{R_2C_2}\left(1 + \frac{R_2}{R_1}\right)}{s + \frac{1}{R_2C_2}} 
          +---R2---+
          |        |
          +---||---+
          |   C2   |
   +------+-|-\    |
   |        |  >---+---- Vo
  R1  Vin---|+/
   |
  GND

Matching:

1R2C2=1001+R2R1=1000100=10⇒R2=9R1\begin{aligned} \frac{1}{R_2C_2} &= 100 \\ 1 + \frac{R_2}{R_1} &= \frac{1000}{100} = 10 \Rightarrow R_2 = 9R_1 \end{aligned}

Choose C2=1 μC_2 = 1\ \muF:

R2=1100×10−6=10 kΩ,R1=109=1.111 kΩR_2 = \frac{1}{100\times10^{-6}} = 10\ \text{k}\Omega, \qquad R_1 = \frac{10}{9} = 1.111\ \text{k}\Omega

Answer: R1=1.11 kΩR_1 = 1.11\ \text{k}\Omega, R2=10 kΩR_2 = 10\ \text{k}\Omega, C2=1 μC_2 = 1\ \muF. DC gain =1+R2/R1=10= 1 + R_2/R_1 = 10, HF gain 1.

✓ Answer checked by Claude Sonnet 5.5

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