RHP-zero understanding

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sinatra39

Joined Oct 18, 2018
15
I have trouble understanding why a RHP zero gives me a phase lag (behaves like a pole)

We can write the transfer function (fundamental theorem of algebra) as:

H(jw) = K * (jw-z1)*(jw-z2)..../(jw-p1)*(jw-p2)....

where z1,z2...are the zeros and P1, P2....the poles respectively.

My understanding now is that each component of the transfer function, e.g. jw-z1 we can write as A*exp(j arg), where A is |jw-z1| and arg=arctan( (w-im(z1))/real(z1)), which is a positive angle in the left half plane.

Now we could write the above transfer function as:

H(jw) = K * A*exp(j arg1) * B*exp(j arg2)...../C*exp(j arg 3) * D*exp(j arg 4)

Until here everything is clear: if have a certain frequency exp(jw) and we multiply it with the transfer function H(jw) the zeros will give us a positive phase shift if arg is positive and the poles a negative one, since they are in the denominator.

But:

Now if we have a zero let's say in the right real half plane, our argument of the exponential term of e.g. jw-z1 would become > 90 degrees but smaller than 180 degrees and therefore our argument would be in the 2nd quadrant of the Cartesian coordinate system and therefore the exponential form would have an argument of 180 - arctan(w/real(z1)), which would result in a positive phase shift between 90 and 180 degrees.

In all textbooks I read that we get a phase LAG of -arctan(w/real(z1)), which describes a completely different behaviour.

The only difference are the 180 degrees. If we let them away the results are identical but I think they we can't so what is correct?

Thanks a lot for support and thoughts!
 
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