Replotting the Nyquist Plot: A New Visualization Proposal Predrag - PowerPoint PPT Presentation

Replotting the Nyquist Plot: A New Visualization Proposal Predrag Pejović

Introduction ◮ Nyquist stability criterion . . . ◮ started as a fix to Barkhausen “criterion” ◮ original derivation complicated . . . ◮ nowadays taught using Cauchy’s argument principle ◮ fundamental, fairly esoteric, deep math . . . but elegant! ◮ highly abstract topological criterion, reduces to: 1. CW encirclement of − 1 + j 0 adds one unstable pole 2. CCW encirclement of − 1 + j 0 removes one unstable pole while closing the loop; topological and relative ◮ hard to teach, requires focused (and competent) students ◮ frequently hard to visualize due to imaginary axis poles, “closed” curve is not closed, but it “encloses” . . . ◮ and this is the point where our story begins . . .

Nyquist Criterion Revisited: the tracking system + x e y Σ W ( s ) −

Nyquist Criterion Revisited: assumptions the transfer function W ( s ) = N ( s ) D ( s ) let N ( s ) and D ( s ) be polynomials such that deg ( N ( s )) ≤ deg ( D ( s )) which is satisfied for systems without algebraic degeneration the problem is whether W ( s ) H ( s ) = 1 + W ( s ) is stable or not?

A Word on Barkhausen . . . poles at 1 + W ( s ) = 0 i.e. − W ( s ) = 1 + j 0 WRONG generalization and a NONSENSE : stable if ❤❤❤❤❤❤❤❤ ✭ ✭✭✭✭✭✭✭✭ − W ( s ) ≤ 1 + j 0 ❤ or | W ( s ) | ≥ 1 BTW, which s ? For oscillators s = j ω 0 , where ℑ ( W ( s )) = 0 . . .

Nyquist Criterion Revisited: the first disaster, W 0 ( s ) = 1 s , stable for sure W 0 ( s ) = 1 s W 0 ( s ) 1 1 + W 0 ( s ) = 1 + s pole at s = − 1 + j 0 , definitely stable

Nyquist Criterion Revisited: the first disaster, W 0 ( s ) = 1 s , straightforward Nyquist Diagram 1e+06 500000 Imaginary Axis 0 -500000 -1e+06 -2 -1 0 1 2 Real Axis

Nyquist Criterion Revisited: the first disaster, W 0 ( s ) = 1 s , escape contour j ω max 4 2 j ω min ℑ ( s ) 0 − j ω min − 2 − 4 − j ω max − 4 − 2 0 2 4 ℜ ( s )

Nyquist Criterion Revisited: the first disaster, W 0 ( s ) = 1 s , here is what we need 2 . 0 1 . 5 1 . 0 0 . 5 ℑ ( W ( s )) 0 . 0 − 0 . 5 − 1 . 0 − 1 . 5 − 2 . 0 − 2 − 1 0 1 2 ℜ ( W ( s ))

What Do We Really Plot? W ( s ) = W r ( s ) + j W i ( s ) where W r ( s ) = ℜ ( W ( s )) , and W i ( s ) = ℑ ( W ( s )) in polar form � ( W r ( s )) 2 + ( W i ( s )) 2 r ( s ) = ϕ ( s ) = atan2 ( W i ( s ) , W r ( s )) and r ( s ) is the problem! idea: compress r ( s ) somehow?

Compression Function Requirements 1. ρ ( r ) in monotonic, to preserve topological properties of the critical point encirclements, 2. ρ (0) = 0 , to keep the same base point where the phase is irrelevant, 3. ρ (1) = 1 , to keep the critical point and visualization of the phase margin, 4. lim r →∞ ρ ( r ) is finite, to confine the diagram in a finite space.

Compression Function ρ ( r ) = 4 π arctan ( r ) “amplitude angle” r → ∞ r r = 1 ρ = 1 ρ = 2 π 4 ρ 0 1 0 1 0 1

A Family of Compression Functions r max r k ρ ( r ) = r max − 1 + r k r max is the radius of the circle the plot is confined into, this degree of freedom might be of some value k > 1 is a parameter r max = 2 and k = 4 π approximates the compression function applied in this paper the best, almost the same function, within 1.5% of r max (0.03)

Bode Plot, W 0 ( s ) = 1 s 120 80 40 20 log ( r ) 0 − 40 − 80 − 120 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 180 90 ϕ 0 − 90 − 180 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 ω

Amplitude Compression, r , r in decibels, and ρ r r [ dB ] ρ 1 0 1 10 20 1.8731 100 40 1.9873 1,000 60 1.9987 10,000 80 1.9999 100,000 100 2.0000

Bode Plot Alternative, W 0 ( s ) = 1 s 2 1 ρ 0 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 180 90 0 ϕ − 90 − 180 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 ω

Alternative Nyquist Plot, W 0 ( s ) = 1 s , strange, you’ve seen this graph before 2 . 0 1 . 5 1 . 0 0 . 5 ℑ ( W ( s )) 0 . 0 − 0 . 5 − 1 . 0 − 1 . 5 − 2 . 0 − 2 − 1 0 1 2 ℜ ( W ( s ))

1 2 of the Contour . . . 10 9 10 6 10 3 | s | 10 0 10 − 3 10 − 6 0 90 590 1090 1590 90 arg ( s ) [ ◦ ] 60 30 0 0 90 590 1090 1590 k

Argument Increase, W 0 ( s ) = 1 s 2 ρ − 1 1 0 no change in the number of unstable poles 180 ϕ − 1 0 − 180 0 90 590 1090 1590 k

1 Example 1, W 1 ( s ) = ( s + 1) 2 Pole-Zero Map 1 0.5 Imaginary Axis 0 -0.5 -1 -1.1 -1.05 -1 -0.95 -0.9 Real Axis

1 Example 1, W 1 ( s ) = ( s + 1) 2 Nyquist Diagram 2 1 Imaginary Axis 0 -1 -2 -2 -1 0 1 2 Real Axis

1 Example 1, W 1 ( s ) = ( s + 1) 2 2 . 0 1 . 5 1 . 0 0 . 5 ℑ ( W ( s )) 0 . 0 − 0 . 5 − 1 . 0 − 1 . 5 − 2 . 0 − 2 − 1 0 1 2 ℜ ( W ( s ))

1 Example 1, W 1 ( s ) = ( s + 1) 2 2 1 ρ 0 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 90 0 ϕ [ ◦ ] − 90 − 180 − 270 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 ω

1 Example 1, W 1 ( s ) = ( s + 1) 2 2 ρ − 1 1 0 no change in the number of unstable poles 180 ϕ − 1 0 − 180 0 90 590 1090 1590 k

1 Example 2, W 2 ( s ) = s ( s + 1) 2 Pole-Zero Map 1 0.5 Imaginary Axis 0 -0.5 -1 -1 -0.8 -0.6 -0.4 -0.2 0 Real Axis

1 Example 2, W 2 ( s ) = s ( s + 1) 2 Nyquist Diagram 1e+06 500000 Imaginary Axis 0 -500000 -1e+06 -2 -1.5 -1 -0.5 0 Real Axis

1 Example 2, W 2 ( s ) = s ( s + 1) 2 2 . 0 1 . 5 1 . 0 0 . 5 ℑ ( W ( s )) 0 . 0 − 0 . 5 − 1 . 0 − 1 . 5 − 2 . 0 − 2 − 1 0 1 2 ℜ ( W ( s ))

1 Example 2, W 2 ( s ) = s ( s + 1) 2 2 1 ρ 0 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 0 − 90 ϕ [ ◦ ] − 180 − 270 − 360 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 ω

1 Example 2, W 2 ( s ) = s ( s + 1) 2 2 ρ − 1 1 0 no change in the number of unstable poles 180 ϕ − 1 0 − 180 0 90 590 1090 1590 k

1 Example 2, W 2 ( s ) = s ( s + 1) 2 closed loop, stable Pole-Zero Map 1 0.5 Imaginary Axis 0 -0.5 -1 -2 -1.5 -1 -0.5 0 Real Axis

3 Example 2a, W 2 a ( s ) = 3 W 2 ( s ) = s ( s + 1) 2 Nyquist Diagram 3e+06 2e+06 Imaginary Axis 1e+06 0 -1e+06 -2e+06 -3e+06 -6 -5 -4 -3 -2 -1 0 Real Axis

3 Example 2a, W 2 a ( s ) = 3 W 2 ( s ) = s ( s + 1) 2 2 . 0 1 . 5 1 . 0 0 . 5 ℑ ( W ( s )) 0 . 0 − 0 . 5 − 1 . 0 − 1 . 5 − 2 . 0 − 2 − 1 0 1 2 ℜ ( W ( s ))

3 Example 2a, W 2 a ( s ) = 3 W 2 ( s ) = s ( s + 1) 2 2 1 ρ 0 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 0 − 90 ϕ [ ◦ ] − 180 − 270 − 360 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 ω

3 Example 2a, W 2 a ( s ) = 3 W 2 ( s ) = s ( s + 1) 2 2 ρ − 1 1 0 +2 unstable poles 180 0 ϕ − 1 − 180 − 360 − 540 0 90 590 1090 1590 k

3 Example 2a, W 2 a ( s ) = 3 W 2 ( s ) = s ( s + 1) 2 closed loop, unstable Pole-Zero Map 1.5 1 Imaginary Axis 0.5 0 -0.5 -1 -1.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 Real Axis

s + 1 Example 3, W 3 ( s ) = s (0 . 1 s − 1) Pole-Zero Map 1 0.5 Imaginary Axis 0 -0.5 -1 -2 0 2 4 6 8 10 Real Axis

s + 1 Example 3, W 3 ( s ) = s (0 . 1 s − 1) Nyquist Diagram 1e+06 500000 Imaginary Axis 0 -500000 -1e+06 -1 -0.8 -0.6 -0.4 -0.2 0 Real Axis

s + 1 Example 3, W 3 ( s ) = s (0 . 1 s − 1) 2 . 0 1 . 5 1 . 0 0 . 5 ℑ ( W ( s )) 0 . 0 − 0 . 5 − 1 . 0 − 1 . 5 − 2 . 0 − 2 − 1 0 1 2 ℜ ( W ( s ))

s + 1 Example 3, W 3 ( s ) = s (0 . 1 s − 1) 2 1 ρ 0 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 360 270 ϕ [ ◦ ] 180 90 0 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 ω

s + 1 Example 3, W 3 ( s ) = s (0 . 1 s − 1) 2 ρ − 1 1 0 − 1 unstable pole 180 0 ϕ − 1 − 180 − 360 0 90 590 1090 1590 k

s + 1 Example 3, W 3 ( s ) = s (0 . 1 s − 1) bypass ambiguity ∆ ϕ = − π ∆ ϕ = + π − 1 + j 0

s + 1 Example 3, W 3 ( s ) = s (0 . 1 s − 1) closed loop, on the stability boundary Pole-Zero Map 4 2 Imaginary Axis 0 -2 -4 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 Real Axis

s + 1 Example 3a, W 3 a ( s ) = 2 W 3 ( s ) = 2 s (0 . 1 s − 1) Nyquist Diagram 2e+06 1e+06 Imaginary Axis 0 -1e+06 -2e+06 -2 -1.5 -1 -0.5 0 Real Axis

s + 1 Example 3a, W 3 a ( s ) = 2 W 3 ( s ) = 2 s (0 . 1 s − 1) 2 . 0 1 . 5 1 . 0 0 . 5 ℑ ( W ( s )) 0 . 0 − 0 . 5 − 1 . 0 − 1 . 5 − 2 . 0 − 2 − 1 0 1 2 ℜ ( W ( s ))

s + 1 Example 3a, W 3 a ( s ) = 2 W 3 ( s ) = 2 s (0 . 1 s − 1) 2 1 ρ 0 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 360 270 ϕ [ ◦ ] 180 90 0 10 − 6 10 − 4 10 − 2 10 0 10 2 10 4 10 6 ω

s + 1 Example 3a, W 3 a ( s ) = 2 W 3 ( s ) = 2 s (0 . 1 s − 1) 2 ρ − 1 1 0 − 1 unstable pole 180 0 ϕ − 1 − 180 − 360 0 90 590 1090 1590 k

s + 1 Example 3a, W 3 a ( s ) = 2 W 3 ( s ) = 2 s (0 . 1 s − 1) closed loop, stable Pole-Zero Map 1 0.5 Imaginary Axis 0 -0.5 -1 -8 -7 -6 -5 -4 -3 -2 -1 Real Axis

Example 3b, W 3 b ( s ) = 1 2 W 3 ( s ) = 1 s + 1 2 s (0 . 1 s − 1) Nyquist Diagram 400000 Imaginary Axis 200000 0 -200000 -400000 -1 -0.8 -0.6 -0.4 -0.2 0 Real Axis