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Transfer function of op-amp phase-shift circuit

Short answer

The transfer function is Eo(s)/Ei(s) = (sR2C - 1)/(sR2C + 1). It is an all-pass filter with unity gain and phase shift from 180° to 0°.

Find the transfer function of given circuit. [Figure: op-amp circuit; input Ei is applied through resistor R1 to the inverting (−) input and through capacitor C to the non-inverting (+) input; R2 connects the non-inverting input to ground; feedback resistor R1 from output to the inverting input; output Eo]

Assumption: ideal op-amp (infinite gain and input impedance), so V−=V+V_- = V_+ and no current enters the inputs.

                R1 (feedback)
            +----/\/\/----+
            |             |
Ei --/\/\/--+---(-)|\     |
       R1          | >----+---- Eo
Ei ---||----+---(+)|/
       C    |
           [R2]
            |
           GND

Non-inverting input

CC and R2R_2 form a voltage divider:

V+=EiR2R2+1sC=EisCR21+sCR2V_+ = E_i\frac{R_2}{R_2 + \dfrac{1}{sC}} = E_i\frac{sCR_2}{1 + sCR_2}

Inverting input (KCL)

The current through the input R1R_1 equals the current through the feedback R1R_1, with V−=V+V_- = V_+:

Ei−V+R1=V+−EoR1  ⇒  Eo=2V+−Ei\frac{E_i - V_+}{R_1} = \frac{V_+ - E_o}{R_1} \;\Rightarrow\; E_o = 2V_+ - E_i

Transfer function

Eo=Ei(2sCR21+sCR2−1)=Ei sCR2−1sCR2+1Eo(s)Ei(s)=sR2C−1sR2C+1=−1−sR2C1+sR2C\begin{aligned} E_o &= E_i\left(\frac{2sCR_2}{1+sCR_2} - 1\right) = E_i\,\frac{sCR_2 - 1}{sCR_2 + 1} \\ \frac{E_o(s)}{E_i(s)} &= \frac{sR_2C - 1}{sR_2C + 1} = -\frac{1 - sR_2C}{1 + sR_2C} \end{aligned}

This is an all-pass (phase-shift) network: ∣Eo/Ei∣=1|E_o/E_i| = 1 at all frequencies, and the phase is ϕ=180∘−2tan⁡−1(ωR2C)\phi = 180^\circ - 2\tan^{-1}(\omega R_2C), going from 180∘180^\circ at ω=0\omega = 0 to 0∘0^\circ at high frequency (90∘90^\circ at ω=1/R2C\omega = 1/R_2C). It has a zero at s=+1/R2Cs = +1/R_2C (non-minimum phase) and a pole at s=−1/R2Cs = -1/R_2C.

✓ Answer checked by Claude Opus 5.5 (high effort)

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