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Non-inverting realization of 4(s+200)/(s+100)

Short answer

Using non-inverting topology with Ra, Rb||C: choose C=1 μF, then Rb=1/(100×1 μF)=10 kΩ, R1=2.5 kΩ, Ra=7.5 kΩ. Yields 4(s+200)/(s+100), HF gain4, DC gain8.

Design the following transfer function using non-inverting op-amp configuration T(s) = 4(s+200)/(s+100)

Given T(s)=4s+200s+100T(s) = 4\dfrac{s+200}{s+100}: HF gain 4, DC gain 4×2=84\times2 = 8, zero at −200, pole at −100.

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}

Because the HF gain is greater than 1, use Z1=R1Z_1 = R_1 and Z2=Ra+(Rb∥C)Z_2 = R_a + (R_b \parallel C):

T(s)=R1+RaR1⋅s+1RbC(1+RbR1+Ra)s+1RbCT(s) = \frac{R_1 + R_a}{R_1}\cdot\frac{s + \frac{1}{R_bC}\left(1 + \frac{R_b}{R_1+R_a}\right)}{s + \frac{1}{R_bC}}
         +--Ra--+--Rb--+
         |      |      |
         |      +--||--+
         |         C   |
   +-----+-|-\         |
   |       |  >--------+--- Vo
  R1 Vin---|+/
   |
  GND

Matching:

1+RaR1=4⇒Ra=3R11RbC=100100(1+Rb4R1)=200⇒Rb=4R1\begin{aligned} 1 + \frac{R_a}{R_1} &= 4 \Rightarrow R_a = 3R_1 \\ \frac{1}{R_bC} &= 100 \\ 100\left(1 + \frac{R_b}{4R_1}\right) &= 200 \Rightarrow R_b = 4R_1 \end{aligned}

Choose C=1 μC = 1\ \muF:

Rb=1100×10−6=10 kΩ,R1=2.5 kΩ,Ra=7.5 kΩR_b = \frac{1}{100 \times 10^{-6}} = 10\ \text{k}\Omega,\quad R_1 = 2.5\ \text{k}\Omega,\quad R_a = 7.5\ \text{k}\Omega

Check: DC gain =1+(7.5+10)/2.5=8= 1 + (7.5 + 10)/2.5 = 8. Correct.

Answer: R1=2.5 kΩR_1 = 2.5\ \text{k}\Omega, Ra=7.5 kΩR_a = 7.5\ \text{k}\Omega, Rb=10 kΩR_b = 10\ \text{k}\Omega, C=1 μC = 1\ \muF.

✓ Answer checked by Claude Sonnet 5.5

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