# Nyquist Stability Criterion

Discussion in 'Math' started by Distort10n, Dec 31, 2007.

1. ### Distort10n Thread Starter Active Member

Dec 25, 2006
429
1
I am reading up on the subject of Nyquist Stability Criterion. I did not learn this stability analysis technique in school - it was Bode Analysis. Body analysis is also used in all the op-amp literature I read as well.

I found a really good website that discusses Nyquist Stability Criterion, even bought a Schaum's Outline on Feedback Systems.

Something is bothering me about the examples below:

http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/Freq/Nyquist2.html

With the Java templates (about half way down the page) it starts out very basic. What I do not understand is why the first example with a pole shows a positive angle when rotating counter-clockwise yet the two pole example further down the page shows a negative angle. To me, the positive one would be correct.

What simple thing am I missing here?

2. ### studiot AAC Fanatic!

Nov 9, 2007
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I'm not convinced that the readouts are yielding the correct angles!

In the first frame

The text states G(s) = 1/(s+1), thus the argument (angle) of the complex number 1/(s+1) is the negative of the angle of the complex number (s+1). This is a fundamental property of complex numbers and is correct.
However the readout apportions a positive angle to position vector as it travels round the box. this should be negative.
A few quick checks on the values at the corners of the box agree the numerical values of the angles however. I just can't agree the sign.
The values correctly increase from zero to 360 and the vector does not pass through zero.

The second frame has a similar error where the signs are inverted. In this case the vector returns to zero at s=(-4,0) on the box, thereby reversing the sign.

I have prepared a simple Excel sheet to do the checks and note you have to do some jiggery-pokery to get the correct quadrant for the output angles. I expect the programmer had a similar problem but didn't solve it.

The third frame has the correct signage when the angles from both s+1 and s+4.5 are added.

3. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Complex analysis in the S-Plane is not the easiest introduction to Nyquist. The diagrams are meant to give experience of how the winding number works and are quite interesting for this purpose. Thank you for pointing them out. They are not, however, Nyquist Diagrams.

Nyquist diagrams are normally drawn for feedback situations and you should be warned that you will meet two versions of the stability criterion. This refers to either

the critical point (-1, 0) used in control engineering.

or

the critical point (+1, 0) used in feedback amplifier analysis.

This is because the series feedback signal is considered subtractive in control engineering and additive in feedback amp analysis.

4. ### Distort10n Thread Starter Active Member

Dec 25, 2006
429
1
"The negative of the angle of the complex number." I am a little confused. I am thinking about this as a rotating phasor. Counter-clockwise yields a positive angle, but that is angular velocity rather than phase correct?

5. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
This website approaches things mathematically by proving Nyquist from the complex transfer function.
Remember s is a complex number and G(s) is a complex function.

The plots show the S-Plane and the coordinates are the rectangular values of the complex number s in

s=x+iy

Any point on the S-Plane can be represented like this.

If we take two points on the S-Plane and draw a line between them the line is a vector and may be found from the difference between their coordinates

s(2) - s(1)

Usually s(1) is the origin but in this case we choose a different point.

The point about the red squares is that they are single closed loops; any closed loop will do but squares are easy.
The sequence of diagrams show what happens when the tip of a vector (phasor) from a fixed point is taken round a closed loop in the S-Plane.
The important point is that if the fixed point is inside the loop the path of the tip is a comple circle (360 degrees).
But if the fixed point lies outside the closed loop the tip path is just a line segment. It may even pass back on itself. Thus the counter does not add up to 360 degrees.

These statements are true for any fixed point. The clever bit is to choose useful points.
Now G(s) contains reciprocals of complex numbers - This is necessary to create a pole (resulting from division by zero). We choose the pole as our fixed point.
The negative angle arise directly from this because the in polar form of a complex number
the angle of the reciprocal is the negative of the angle of the original.

The angle of s = x+iy is inverse tan y/x
The angle of the reciprocal 1/s -(inversetan y/x)
This is much easier than calculating from 1/(x+iy)

I am out in the field for the rest of tthe day, but hope to have time to post more fully this evening, unless someone else wants to add?

6. ### Distort10n Thread Starter Active Member

Dec 25, 2006
429
1
Ahh...so 1/(s+1) is the reciprocal of s+1, if I get what you are saying. I never looked at it that way.

7. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes that's right.

However it is further complicated by the fact that for any given value of y/x, positive or negative.
inverse tan (y/x) has two solutions in the range 0-360.

Conventionally calculators and computer program functions yield and angle between -180 and +180, which is not what is wanted here.

This arises because we are actually solving two simultaneous equations

cos A = x/sqrt(x^2+y^2)

and

sin A = y/sqrt(x^2+y^2)

Solving these simultaneously, using the correct signs for x and y, will yield only one angle in the range 0-360

Alternatively we can use the invtan and proceed as follows

calculate A=invtan (y/x) regardless of signs of x and y.

Then

If
y+ve and x+ve output = A angle is first quadrant
If
y+ve and x-ve output = 180-A angle is second quadrant
If
y-ve and x-ve output = 180+A angle is third quadrant
if
y-ve and x+ve output = 360-A angle is fourth quadrant.

Remembering to change the sign of the output angle because of the reciprocal this is the final output.
When you have 2 poles the total angle is the arithmetic sum of the angles for the individual poles.

This is all to illustrate that traversing around a closed loop leads to less than 360 if the pivot lies outside the loop and 360 if the pivot lies inside. This is a geometric property of the S-Plane.

The diagrams illustrate this geometrically.

When the angles are regarded as phase changes (measured in angular measure) and the closed loop is the locus of the amplitude response plotted as the amplitude (length) of the vector, Nyquist uses this property to determine stability or otherwise from the relative position of the loop and the pivot (pole).

With some further complex arithmetic jiggery pokery He showed (1932) that the important pivots are at -1,0 or +1,0 depending upon whether you subtract or add the feedback voltage.