# How is phase shift of an op amp calculated using polar coordinates?

I have a textbook (Op-Amp: Characteristics & Applications by Robert G. Irvine, 1981) and its frequency response chapter started by discussing the internal LP filters between the Input Differential Amplifier stage, the Voltage Level Shifting Stage, and the Push-Pull Output Stage.

It explains that the capacitor will create a -45 degree phase shift at its corner frequency (where attenuation is -3 dB) and corner frequency is $$\1/(2\pi RC)\$$.

It says that if the frequency on the input of the Op-Amp is much less than the pole frequency then the phase shift is negligible but uses the equation:

$$R_C=159,000 \Omega \times 0.1\text{ uF}.=15.9\text{ ms}$$

pole F: $$\1/(2\pi \times 0.0159)-1/0.1=10\text{ Hz}\$$

1. When F=0.1-Hz Vout/Vin=10/(10+j(0.1))= 1 @ -0.57-degrees

2. When F=10-Hz Vout/Vin=10/(10+j(10))= 10/(14.141 @ +45 Degrees) = 0.707 @ -45-degrees

3. When F=1,000-Hz Vout/Vin=10/(10+j(1000))= 10/(1000.05 @ +89.43 Degrees) = 0.01 @ -89.43-degrees

4. When F=10,000-Hz Vout/Vin=10/(10+j(10,000))= 0.001 @ -89.94-degrees

I can use Euler's formula to transpose polar coordinates to rectangular, but the textbook doesn't give any explanation as to how, say, $$\10+j10\$$ becomes 14.141 @ -45-degrees.

or how does $$\10+j1000\$$ become 1000.01 @ +89.43 degrees?

Basically, how do you calculate phase shift due to a a capacitor as a function of its frequency in polar and rectangular coordinates?

Your question suggests that you have unfortunately not learned complex arithmetic sufficiently well during your precalculus course. Consequently, you may struggle at the level at which you are now studying.

Nevertheless, this might help: the Argand diagram.

If you wish to know how the last figure you mention is calculated, then sqrt(102+10002) = 1000.05 and arctan(1000/10) = +89.43 degrees. To understand, or partly understand, the reason for the calculation, review the linked figure and recall your high-school trigonometry.

• @ Danny Use the right-triangle, where the long side, squared, is the sum of the two right-angle sides, each squared. Thus a right triangle with 3 and 4 as the right-angle sides, will have 5 as the long side. And the ratio of 3 and 4, or 3/4, gives you the tangent of the angle. In your case, 10 & 10, the ratio is 10/10 and that angle is 45 degrees. Welcome to stackX. Oh, yes, this is called "Pythagorian theorem" or "triangle". Nov 22, 2018 at 23:38
• @analogsystemsrf: rebuke accepted.
– thb
Nov 23, 2018 at 0:09
• I'm very comfortable with calculus and Trig: It's the only two classes at UCSB i got A+s in and I actually have made YouTube videos explaining them in relations to wavelengths I just didn't make the connection with Phase shift. So the Magnitude would be like adding vectors (The Value of the Hypotenuse of a right triangle whose Adjacent and Opposite sides are the Pole Frequency ("10") and the current Angular Frequency and the phase angle is found by plugging the slop of the hypotenuse into the ARCTAN function? Nov 24, 2018 at 13:52
• That will work. If you wish to be severely correct, a phasor is not actually a vector, though one handles it rather like a real-valued geometrical vector for most practical purposes. The distinction hardly matters until graduate-level study of continuum mechanics (if the major is mechanical engineering) or electromagnetic waves (electrical engineering), when complex-valued geometrical vectors arrive and one must remember that a phasor (that is, a complex number) is actually a scalar. But these are minor points. Your approach will work and is probably how many engineers think of it.
– thb
Nov 25, 2018 at 13:38