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What would be the transfer function of the following circuit, assuming an ideal op-amp?

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    \$\begingroup\$ hint: split \$V_{in}\$ in two separate generators, one connected only to \$R_1\$ and one only to \$R_2\$, then use superimposition. \$\endgroup\$ May 26, 2014 at 13:23
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    \$\begingroup\$ Note that R2 and C form a low pass filter. Analyze separately at low and high frequencies. "Low" means well before the filter starts attenuating, and "high" means well after. \$\endgroup\$ May 26, 2014 at 14:02
  • \$\begingroup\$ Olin's way is faster than mine but you've got to be a little experience with circuits. \$\endgroup\$ May 26, 2014 at 14:04

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This looks like a homework problem, so I'm going to demonstrate the setup, and leave the algebra to you. As a note, you've got two resistors called \$R_1\$, so I'm going to denote them by \$R_{1L}\$ and \$R_{1R}\$ for left and right, respectively.

You have negative feedback, so (assuming an ideal op amp as stated) the input voltages to the op amp are equal. This is a natural consequence of negative feedback on an op amp, and part of what makes it so useful. If the positive input is higher than the negative input, the output goes up, but that pulls the negative input up too. The only stable point is when the inputs are equal.

$$V_a = V_b$$

\$R_2\$ and \$C\$ make a voltage divider. We use the complex impedance of the capacitor in the standard voltage divider equation.

$$ V_b = V_{in}\frac{\frac{1}{sC}}{R_2 + \frac{1}{sC}} $$

We know the voltage on both sides of the left \$R_{1L}\$, so we know the current through that \$R_{1L}\$.

$$ I_1 = \frac{V_a-V_{in}}{R_{1L}} $$

The op amp is assumed to have infinite input impedance, so all the current flowing through \$R_{1L}\$ must also flow through \$R_{1R}\$. It has nowhere else go to!

$$ I_2 = I_1 $$

We know the voltage on the left side of \$R_{1R}\$, and the current through \$R_{1R}\$, so we know the voltage on the other side of \$R_{1R}\$.

$$ V_{out} = V_a + I_1R_2 $$

(Note: the transfer function of the voltage divider at point b can be problematic for DC signals. You could take the limit of the expression I gave as \$s \to 0\$. However, you should know what a capacitor looks like in a DC circuit, and be able to write the equation directly from that.)

You can run the algebra yourself. But don't just take the answer and turn it in! Learn from the steps so you can do it yourself, next time or ten years from now. You want to be a good engineer, right?

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    \$\begingroup\$ Another recommendation: When you have found the transfer function, set R1=R2=R and evaluate the specific properties of the circuit. It is of common use for some applications. \$\endgroup\$
    – LvW
    May 26, 2014 at 14:10
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This is a Phase Shifter, with transfer function given by:

$$\frac{V_o(s)}{V_i(s)} = H(s) = \frac{-R_2Cs+1}{R_2Cs+1}$$

Gain: \$1 V/V\$ (passes all signals without altering their amplitude).

Phase lag: \$0^0\$ to \$-180^0\$, with \$-90^0\$ at \$\omega = \omega_0=\frac{1}{R_2C}\$.

PhaseShifter

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What you have drawn is an allpass filter (see http://www.analog.com/media/en/training-seminars/tutorials/MT-202.pdf) which actually mimics a 1st-order Padé approximate of a pure delay. You can derive the transfer function either by using superposition (you split \$V_{in}\$ in \$V_{in1}\$ and \$V_{in2}\$) or using the fast analytical circuits techniques or FACTs. This is a 1st-order circuit (only 1 energy-storing element) and the generalized transfer function linking \$V_{out}\$ to \$V_{in}\$ is given by:

\$H(s)=\frac{H_0+sH^1\tau_1}{1+s\tau_1}\$

\$H_0\$ is the gain linking \$V_{out}\$ to \$V_{in}\$ determined for \$s=0\$ in which the capacitor is open. The time constant \$\tau_1\$ is determined by reducing the excitation voltage \$V_{in}\$ to 0 V and "looking" at the resistance driving capacitor \$C_1\$. Then, place this capacitor in its hi-frequency state (a short circuit) and calculate the gain \$H^1\$. Once you have these elements, assemble them according to the generalized transfer function expression. All the steps are drawn in the below figure:

enter image description here

Once you have captured the component values and expressions into a Mathcad sheet you can see how combining a left-half-plane pole (LHPP) \$\omega_p\$ with a right-half-plane zero (RHPP) \$\omega_z\$ produces a constant 0-dB gain but with a phase lagging down to 180°. This is a good way to model a delay in a transfer function. A delay for instance incurred to an A/D conversion time or a comparator transition time.

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The dynamic response is here:

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The FACTs helped me derive the transfer function in a very short time without manipulating complex expressions. You can find an introduction to the technique via the below links:

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

and

http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf

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