Transfer function of amplifier

Question

I would like to find transfer function between input signal \$U_1\$ and output signal \$U_2\$. So, I know how to find the transfer function of each op-amp, for example,

1 transfer function: $$\frac{v_o}{v_i} = -\frac{R_3}{R_1}\frac{1}{1+R_3C_3 s}$$ 2 transfer function: $$\frac{v_o}{v_i} = -\frac{1}{C_4 s R_4}$$ 3 transfer function: $$\frac{v_o}{v_i} = \frac{R_2}{2R}$$

Is that correct way to find $$G(s)=\frac{U_2}{U_1}$$? How can I do it?

Answer 1

For my opinion, the simplest solution makes use of the classical feedback formula from H. Black:

$$\frac{V_2}{V_1}=\frac{H(s)}{1-LG}$$

with:

• \$H(s)=H_1(s)H_2(s)\$=Forward transfer function for an open loop (in our case: \$H_1=V_3/V_1\$ for \$R_2\$>>infinite and \$H_2=V_2/V_3\$.)
• Loop gain \$LG\$=Product of all three transfer functions within the loop (with \$V_1=0\$ or \$R_1\$>>infinite).

Note that \$H(s)\$ is positive and the loop gain \$LG\$ must be negative (three inverting stages in series). The transfer functions of the three blocks are basic (inverting lowpass, inverting integrator, inverting amplifier).

Answer 2

Anyway, all that you have to do is to calculate some of voltage and current equation with integration or derivate in the s domaine (Laplace).

For example: $$U_3 = R_4(I_4+I_3)$$ $$\frac{d_{U_2}}{d_t} = - \frac{1}{C4}(I3+I4)$$ so $$I_4+I_3 = - \frac{d_{U_2}}{d_t}C_4$$ $$U_3 = -R_4\frac{d_{U_2}}{dt}C_4$$ in S domain $$U_3(s) = -R_4\times C_4\times S\times U_2(s)$$ so $$G_2(s)=\frac{U_2(s)}{U_3(s)} = -\frac{1}{R_4\times C_4}$$ in the same way yo calculate \$G_1(s)=\frac{U_3}{U_1}\$ and you multipliate the two TF to have the G(s) because you have two Tf in cascade.

Answer 3

The problem here is that the feedback network also feeds your first opamp. Specificly, the current I2 flows trough your feedback network of R3 and C3 creating a higher (less negative) voltage at U3. So this said, your first equation is not complete, because you also need to take I2 into account.

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I would start with the feedback network including the opamp, that produces current I2.

$$I_2 = -\frac{R_2}{U_2}$$

$$U_3 = -\left(\frac{U_{1}}{R_1} - I_2 \right ) \left( \frac{R_3}{1 + j\omega C_3 R_3}\right)$$

$$U_2 = -\left(\frac{U_3}{R_4} \right ) \left( \frac{1}{j\omega C_4} \right ) = -\frac{U_3}{j\omega C_4 R_4}$$

Then when you combine these, you get (if I don't make any typos or calculation errors): $$ U_2 = \frac{R_3}{-\omega^2 C_3 R_3 C_4 R_4} \left( \frac{R_2}{U_2} + \frac{U_1}{R_1} \right)$$