The Nyquist plot given below represents which circuit ?

Concept:

1. Nyquist stability criteria state that the number of unstable closed-loop poles is equal to the number of unstable open-loop poles plus the number of encirclements of the origin of the Nyquist plot of the complex function D(s).

2. It can be slightly simplified if instead of plotting the function D(s) = 1 + G(s)H(s), we plot only the function G(s)H(s) around the point and count encirclement of the Nyquist plot of around the point (-1, j0).

3. From the principal of argument theorem, the number of encirclements about (-1, j0) is

N = P - Z

Where

Where P = Number of open-loop poles on the right half of s plane

Z = Number of closed-loop poles on the right half of s plane

Calculation:

The general transfer function of the all-pass filter is:

$T(s)=\frac{1-sT}{1+sT}$

So it has a pole at s = -1/T and

zero at s = 1/T.

$|T(j\omega )|=\frac{\sqrt{1+{\omega }^2}}{\sqrt{1+{\omega }^2}}=1$

20 log(1) = 0 db.

∠T(jω) = -tan^-1(jω) - tan-1(jω) = -2tan-1(jω)   

The magnitude plot will be:

The phase plot will be:

• If a closed-loop pole of the system is on the jω axis, then the magnitude equals 1 and the phase is 180 degrees at that frequency.
• If that occurs, the Nyquist plot will pass through the -1 point at that frequency.
• It indicates that there is an equal number of poles and zeros.

Hence option (3) is the correct answer.

NOTE: All other filters do not have the Nyquist point having magnitude 1 at all frequencies and passing through -1 point at the frequency.