Nyquist (2) Hany Ferdinando Dept. of Electrical Engineering Petra Christian University

Nyquist (2) - Hany Ferdinando 2 General Overview What is stability? What is stability? Mapping contour Mapping contour The Nyquist criterion The Nyquist criterion Relative Stability Relative Stability

Nyquist (2) - Hany Ferdinando 3 What is Stability? A B If both balls experience a force, which ball can back to it original position?

Nyquist (2) - Hany Ferdinando 4 Contour A unit square contour in s-plane can be mapped into another plane via a certain function. For example, mapping a unit square contour in s- plane to F(s) plane by function F(s) = 2s j -j A unit square contour

Nyquist (2) - Hany Ferdinando 5 F 1 (s) = 2s+1 s-plane j -j 1 A BC D F(s)-plane A BC D 2j -2j 3 s =  + j and F(s) = u + jv, F(s) = 2(+j) + 1 u = 2 + 1 and v = 2

Nyquist (2) - Hany Ferdinando 6 F 1 (s) = 2s+1 Zero of F 1 (s) = 2s+1 is F 1 (s) has no poles. Zero of F 1 (s) = 2s+1 is F 1 (s) has no poles. A unit square contour in s-plane encircles zero -0.5 once in clockwise direction. A unit square contour in s-plane encircles zero -0.5 once in clockwise direction. A mapping contour in F(s)-plane encircles origin once in clockwise direction. A mapping contour in F(s)-plane encircles origin once in clockwise direction.

Nyquist (2) - Hany Ferdinando 7 F 2 (s) = s/(s+2) s-plane j 1 DA -j CB A B C D F(s)-plane

Nyquist (2) - Hany Ferdinando 8 F 2 (s) = s/(s+2) Zero of F 1 (s) = s/(s+2) is 0 and its pole is -2. Zero of F 1 (s) = s/(s+2) is 0 and its pole is -2. A unit square contour in s-plane encircles zero 0 once in clockwise direction and encircles no poles. A unit square contour in s-plane encircles zero 0 once in clockwise direction and encircles no poles. A mapping contour in F(s)-plane encircles origin once in clockwise direction. A mapping contour in F(s)-plane encircles origin once in clockwise direction.

Nyquist (2) - Hany Ferdinando 9 Chaucy’s theorem Principle of the argument: If a contour  s in the s-plane encircles Z zeros and P poles of F(s) and does not pass through any poles and zeros of F(s) and the traversal is in the clockwise direction along the contour, the corresponding contour  F in the F(s)- plane encircles the origin of the F(s)- plane N = Z – P times in the clockwise direction

Nyquist (2) - Hany Ferdinando 10 Chaucy’s Theorem  z1  z2  p1  p2 ss FF FF s

Nyquist (2) - Hany Ferdinando 11 Chaucy’s Theorem  The net angle for  p1,  p2 and  z2 as s traverses along  s is zero, but  z1 is 2. If there are Z zeros are enclosed by  s in clockwise direction, then the net angle for zeros is 2Z. If there are Z zeros are enclosed by  s in clockwise direction, then the net angle for zeros is 2Z. If there are P poles are enclosed by  s in clockwise direction, then the net angle for poles is 2P. If there are P poles are enclosed by  s in clockwise direction, then the net angle for poles is 2P.

Nyquist (2) - Hany Ferdinando 12 Chaucy’s Theorem The net resultant angle of F(s),  F, is 2Z-2P. The net resultant angle of F(s),  F, is 2Z-2P.  F = 2Z – 2P, or  F = 2Z – 2P, or 2N = 2Z – 2P, or 2N = 2Z – 2P, or N = Z – P N = Z – P If N > 0,  F is in clockwise direction, Otherwise, is in counter clockwise. If N > 0,  F is in clockwise direction, Otherwise, is in counter clockwise.

Nyquist (2) - Hany Ferdinando 13 Chaucy’s Theorem Consequences ss Z = 3, P = 1 N = Z – P = 2

Nyquist (2) - Hany Ferdinando 14 The Nyquist Criterion A system is stable if all zeros of F(s) are in the left-hand s-plane A system is stable if all zeros of F(s) are in the left-hand s-plane  s is the entire right-hand s-plane, it encloses the entire right-hand s- plane, then determine if there are zeros of F(s) lie within  s  s is the entire right-hand s-plane, it encloses the entire right-hand s- plane, then determine if there are zeros of F(s) lie within  s It starts from the characteristic equation

Nyquist (2) - Hany Ferdinando 15 The Nyquist Criterion Usually, L(s) is in factored from but 1+L(s) is not Usually, L(s) is in factored from but 1+L(s) is not Instead of mapping to F(s)-plane we can do it for L(s)-plane Instead of mapping to F(s)-plane we can do it for L(s)-plane Now, how to map F(s) = 1 + L(s)?

Nyquist (2) - Hany Ferdinando 16 F(s)- into L(s)-plane? In F(s)-plane N is the number of clockwise encirclements of the origin In F(s)-plane N is the number of clockwise encirclements of the origin In L(s)-plane N is the number of clockwise encirclements of the -1 point in F(s), because F’(s)=F(s)-1 In L(s)-plane N is the number of clockwise encirclements of the -1 point in F(s), because F’(s)=F(s)-1

Nyquist (2) - Hany Ferdinando 17 Stability Analysis N = 0 N = 0 The system is stable if there is no pole of L(s) in the right-half plane The system is stable if there is no pole of L(s) in the right-half plane Otherwise, the system is unstable Otherwise, the system is unstable N < 0 N < 0 The system is stable if the number of counter clockwise encirclement is equal to the number of poles of L(s) in the right-half plane The system is stable if the number of counter clockwise encirclement is equal to the number of poles of L(s) in the right-half plane Otherwise, the system in unstable Otherwise, the system in unstable N > 0 N > 0 In this case, the system is unstable In this case, the system is unstable

Nyquist (2) - Hany Ferdinando 18 Example (1) S-planeL(s)-plane

Nyquist (2) - Hany Ferdinando 19 Example (2) S-planeL(s)-plane

Nyquist (2) - Hany Ferdinando 20 Example (3) S-planeL(s)-plane