# Bode diagram, why is the phase at origin 90 degrees constant?

The phase for a pole or root at the origin is given as either + or - atan(w)

So why is this a constant 90 degrees? for w=0 this should be 0 degrees, for w -> inf this should go to 90 degrees, yet we plot it as a constant 90.

I am confused.

• Active schemas can have constant 90 deg phase plot. About which schematic we're talking? – Looongcat Sep 8 '15 at 9:39
• A pole at the origin is $1/s$. Taking this into the frequency domain gives $1/j\omega = 0-j/\omega$, which in turn gives $\phi=arctan(-(1/\omega)/0)=-90^o$ for all values of $\omega$ – Chu Sep 8 '15 at 15:31

## 2 Answers

Let us say transfer function is H(jw)=1/jw=-j/w. Its a single pole system with pole at origin.

The phase of any transfer function is given by atan(IMAG(H(jw)/REAL(H(jw))). So

Phase(H(jw))= atan(-w/0) =-90deg for all w.

Hence, phase starts with 90deg and remains constant thereafter as maximum phase lag a pole can give is 90deg.

The same may not be true for a pole at any other frequency. Lets say that pole is at w=a. so, transfer function is

H(jw)=1/(1+j(w/a)).

Phase for this transfer function will be -atan(w/a)).

For w< Phase=0deg

For w>>a ==> Phase=-90deg

For w=a ==> Phase=-45deg.

Hence, in this case Phase is not constant, it starts with 0 and falls to -90deg.

Hope this helps..!!

Who told you that it would be "constant 90 deg"? It is not.

The open-loop gain of an opamp (phase-compensated) can be approximazed by a first-order lowpass (within the active region) with a rather low cut-off frequency (wc=10...100 Hz). Hence, the phase for w=0 is zero deg. The phase at w=wc is app. -45 deg (approximately, because there are other poles where the gain is <0 dB). Only for frequencies far above the cut-off wc the phase is app. -90deg.