I have a this circuit:
I am asked to find |H(\$\omega\$)| in its simplest form.
I approached the problem by first recognizing that the first op am was a standard inverter. With that I basically rewrote the circuit so that the second op amp had a common input of -Vi.
Using a voltage divider for C1 and R6 I found the transfer function to be:
\$(2R_x/(R_x+Z_c)) = -(V_o -Vi)/V_i \$
assuming that I did that right which is probably not a good assumption, the questions asks choose R6 and C1 so \$ \angle H(2\pi*10^3) = -90^\circ \$
I'm not sure how to get the transfer function from a magnitude to an angle so I can sub 1K (a practical value for an op amp) in for R6 and solve for C1?
Any help with this homework assignment is appreciated.
Best Answer
You're on the right track but you need to get your transfer function into standard form.
Gathering terms and multiplying through, we get:
\$\dfrac{V_o}{V_i} = H(j\omega) = \dfrac{1 - j\omega R_6 C_1}{1 + j\omega R_6 C_1}\$
This is an all-pass filter with a magnitude response of 1. To verify, multiply by the complex conjugate to get \$|H(j \omega)|^2\$:
\$|H(j\omega)|^2 = H H^* = \dfrac{1 - j\omega R_6 C_1}{1 + j\omega R_6 C_1} \dfrac{1 + j\omega R_6 C_1}{1 - j\omega R_6 C_1} = 1\$
For the phase, multiply top and bottom by \$(1 - j\omega R_6 C_1)\$:
\$H(j \omega) = \dfrac{[1 - (\omega R_6 C_1)^2] - j2\omega R_6 C_1}{(1 + \omega R_6 C_1)^2}\$
In this form, the real and imaginary parts are manifest and, for \$\angle H(j \omega) = -90^\circ\$, the real part must be zero.
\$\angle H(j \omega) = \tan^{-1}\frac{-2\omega R_6 C_1}{1 - (\omega R_6 C_1)^2} \$
However, for this particular transfer function (in the first form I wrote), it's easy to see that the numerator and denominator have equal and opposite phase. Thus, the phase of this transfer function is simply twice the phase of the numerator. Then, by inspection, we see that \$\angle H(j \omega) = -90^\circ\$ when \$\omega = \frac{1}{R_6 C_1}\$.
(Thanks to Szymon Bęczkowski for the nice plots!)