Is the Nyquist Plot Problem Solvable with a Stable Open Loop Transfer Function?

[rudra]

1. Problem is given in the following link:
http://s11.postimage.org/mrns0fu37/Problem.jpg

Homework Equations

In my opinion open loop transfer function is given stable. So poles on RHS for open loop t.f. will be 0. But nyquist plot rotates (0,0) point by -360 . So thare will be one RHS open loop zero.

Correct me if I am wrong.

 

[NascentOxygen]

In my opinion open loop transfer function is given stable. So poles on RHS for open loop t.f. will be 0. But nyquist plot rotates (0,0) point by -360 . So thare will be one RHS open loop zero.

Correct me if I am wrong.

Hi rudra, http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

It does look stable, and has no poles—is that what you are saying?

I have forgotten a lot about Nyquist plots ☹[/size][/color] but if there were a zero in the open loop transfer function, then surely the plot would go though (0,0)?

 

[rudra]

In my opinion open loop t.f. should have a R.H.S. zero. But I think that's the wrong answer. Because when I searched the answer keys for the problem. Every website or book gives option (c) i.e. G(s) is the impedance of passive network. But none of the book or websites explain the answer neither can I.

If you know the explanation please reply.

 

[NascentOxygen]

Let's first see what features we are in agreement on ...

What is the minimum forward gain, and what is the maximum forward gain?

What would you estimate to be the DC gain? — and why do you give this value?

To answer these, see the intro to this article on the Nyquist plot - http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/Freq/Freq6.html

 

[uart]

In my opinion open loop t.f. should have a R.H.S. zero. But I think that's the wrong answer. Because when I searched the answer keys for the problem. Every website or book gives option (c) i.e. G(s) is the impedance of passive network. But none of the book or websites explain the answer neither can I.

If you know the explanation please reply.

No it can't be the impedance of a passive network, because it has a negative real component for some values of frequency. A passive network can't have a negative resistance, so I think that your answer (b) is correct. Even by process of elimination, I think the only answer that is possible is "b".

Process of elimination.
(a) Allpass filter has constant gain, therefore constant radius (from origin) of Nyquist plot. So cannot be this.
(c) Cannot be impedance of passive network because it has a negative resistance at some frequencies.
(d) Cannot be marginally stable, as a pole on the jw axis would result in asymptotes (to infinity) at some points on the Nyquist plot.

 

[NascentOxygen]

(a) Allpass filter has constant gain, therefore constant radius (from origin) of Nyquist plot. So cannot be this.

It depends on what you expect from an allpass network. I contend that it is an allpass network—across the entire spectrum it has a uniform gain of unity, with a ripple ±6 dB.

 

[rudra]

At last someone gave some answer. Thank you uart for the help.