Lecture 6 Modelling Second Order Systems and Examples Process Control Prof. Kannan M. Moudgalya IIT Bombay Wednesday, 31 July 2013 1/39 Process Control Second Order Models and Response
Outline 1. Poles, zeros, response for different pole locations 2. Solution to second order systems 3. Integrating/capacitive systems 2/39 Process Control Second Order Models and Response
1. Poles, zeros, response for different pole locations 3/39 Process Control Second Order Models and Response
Transfer function terminologies ◮ Let the transfer function be G(s) = N(s) / D(s) where N(s) and D(s) are polynomials in s ◮ The roots of N(s) = 0 are called zeros of G(s) ◮ The roots of D(s) = 0 are called poles of G(s) 4/39 Process Control Second Order Models and Response
Transfer Function The a i in the following transfer function are known as G(s) = (s − a 1 ) · · · (s − a m ) (s − b 1 ) · · · (s − b n ) 1. Poles 2. Zeros 3. Gain 4. Time constants Answer: 2. Zeros 5/39 Process Control Second Order Models and Response
Transfer Function The b j in the following transfer function are known as G(s) = (s − a 1 ) · · · (s − a m ) (s − b 1 ) · · · (s − b n ) 1. Poles 2. Zeros 3. Gain 4. Time constants Answer: 1. Poles 6/39 Process Control Second Order Models and Response
Poles and zeros of two transfer functions ◮ Let the transfer function be G(s) = N(s) / D(s) where N(s) and D(s) are polynomials in s ◮ The poles of 1 + KG(s) are 1. same as the poles of G(s) 2. same as the zeros of G(s) 3. No relation between the poles/zeros of G(s) and 1 + KG(s) Answer: 1. same as the poles of G(s) 7/39 Process Control Second Order Models and Response
Poles and zeros of two transfer functions ◮ Let the transfer function be G(s) = N(s) / D(s) where N(s) and D(s) are polynomials in s ◮ Let the closed loop transfer function be T(s): KG(s) T(s) = 1 + KG(s) ◮ The poles of T(s) are 1. Poles of G(s) 2. Zeros of G(s) 3. Zeros of D(s) + KN(s) 4. Poles of D(s) + KN(s) Answer: 3. Zeros of D(s) + KN(s) 8/39 Process Control Second Order Models and Response
Partial fraction of second order system K M A s + b + C B ◮ Y(s) = s = s + a + s 2 + 2 ζω n s + ω 2 s n Given that y(t) is the output of a real life plant, and a is complex, 1. There is no relationship between a and b 2. b is real 3. a = − b 4. a and b are complex conjugates Answer: 4, i.e. a and b are complex conjugates If a and b are not complex conjugates, y(t) will be imaginary, not realistic for a real plant! 9/39 Process Control Second Order Models and Response
Partial fraction of second order system K M A s + b + C B ◮ Y(s) = s = s + a + s 2 + 2 ζω n s + ω 2 s n Given that y(t) is the output of a real life plant, and A is complex, 1. There is no relationship between A and B 2. B is real 3. A = − B 4. A and B are complex conjugates Answer: 4, i.e. A and B are complex conjugates If A and B are not complex conjugates, y(t) will be imaginary, not realistic for a real plant! 10/39 Process Control Second Order Models and Response
Response for different pole locations d c a b c d Poles are indicated by crosses 11/39 Process Control Second Order Models and Response
Step response for pole at a a ◮ For the pole at a place indicated by a, the response is of the form e − α 2 t ◮ The exponential part will decay, reaching a constant value 12/39 Process Control Second Order Models and Response
Scilab code to plot step response of a negative pole, a.sce s = %s 1 2 G a = 1/( s +3) t = 1 : 0 . 1 : 5 ; 3 y a = csim ( ’ step ’ , t , G a ) ; 4 p l o t 2 d ( t , y a ) 5 13/39 Process Control Second Order Models and Response
Response for pole at b b ◮ For the pole at a place indicated by b, the response is of the form e β 2 t ◮ The exponential part will grow unbounded 14/39 Process Control Second Order Models and Response
Scilab code to plot step response of a positive pole, b.sce s = %s 1 2 G b = 1/( s − 2) t = 0 : 0 . 0 1 : 6 ; 3 y b = csim ( ’ step ’ , t , G b ) ; 4 p l o t 2 d ( t , y b ) 5 15/39 Process Control Second Order Models and Response
Response for pole at c c ◮ For the poles at places indicated by c, c the response is of the form e γ 2 t × (sinusoidal terms) ◮ There will be growing oscillations 16/39 Process Control Second Order Models and Response
Scilab code to plot step response of complex conjugate poles in right half plane, c.sce s = %s 1 2 G c = 1/( s − (1+2 ∗ %i) ) /( s − (1 − 2 ∗ %i) ) t = 0 : 0 . 0 1 : 1 0 . 1 5 ; 3 y c = csim ( ’ step ’ , t , G c ) ; 4 p l o t 2 d ( t , y c ) 5 17/39 Process Control Second Order Models and Response
Response for poles at d d ◮ For the poles at places indicated by d, the response is of the form e − δ 2 t × (sinusoidal terms) ◮ There will be d decaying oscillations 18/39 Process Control Second Order Models and Response
Scilab code to plot step response of complex conjugate poles in left half plane, d.sce s = %s 1 2 G d = 1/( s − ( − 1+3 ∗ %i) ) /( s − ( − 1 − 3 ∗ %i) ) t = 0 : 0 . 0 1 : 6 ; 3 y d = csim ( ’ step ’ , t , G d ) ; 4 p l o t 2 d ( t , y d ) 5 19/39 Process Control Second Order Models and Response
2. Second order system 20/39 Process Control Second Order Models and Response
Second order system K ◮ Y(s) = G(s)U(s), G(s) = τ 2 s 2 + 2 ζτ s + 1 ◮ τ is the time constant ◮ ζ is the damping coefficient ◮ K is the steady state gain K M ◮ Step response: Y(s) = τ 2 s 2 + 2 ζτ s + 1 s A s + b + C B ◮ = s + a + s ◮ Is there a relation between a and b, A and B? ◮ For ζ < 1, a = b ∗ , A = B ∗ 21/39 Process Control Second Order Models and Response
Step response of a second order system K M A s + b + C B ◮ Y(s) = s = s + a + τ 2 s 2 + 2 ζτ s + 1 s ◮ a and b are roots of τ 2 s 2 + 2 ζτ s + 1 = 0 4 ζ 2 τ 2 − 4 τ 2 � ◮ a , b = − 2 ζτ ± 2 τ 2 ◮ = − ζ τ ± 1 ζ 2 − 1 � τ ◮ ζ ≥ 1, a, b are real ◮ ζ < 1, complex, a = b ∗ , A = B ∗ 22/39 Process Control Second Order Models and Response
Step response for ζ > 1 K M Y(s) = τ 2 s 2 + 2 ζτ s + 1 s ◮ Known as an overdamped system ( ζ > 1) ◮ Its solution is given by � 1 − e ( − ζ t /τ ) ∗ y(t) = KM � �� ζ 2 − 1 � �� ζ 2 − 1 �� ζ cosh t + sinh t } ζ 2 − 1 � τ τ 23/39 Process Control Second Order Models and Response
Step response for ζ = 1 K M Y(s) = τ 2 s 2 + 2 ζτ s + 1 s ◮ Known as critically damped system ( ζ = 1) ◮ Its solution is given by 1 + t � � � � e − t /τ y(t) = KM 1 − τ 24/39 Process Control Second Order Models and Response
Step response for ζ < 1 K M Y(s) = τ 2 s 2 + 2 ζτ s + 1 s ◮ Known as underdamped system ( ζ < 1) ◮ Its solution is given by � 1 − e ( − ζ t /τ ) ∗ y(t) = KM � �� � �� �� 1 − ζ 2 1 − ζ 2 ζ cos t + 1 − ζ 2 sin t } � τ τ 25/39 Process Control Second Order Models and Response
Another form of II order system K G(s) = τ 2 s 2 + 2 ζτ s + 1 will also be written as ω 2 n G(s) = s 2 + 2 ζω n s + ω 2 n ◮ K is taken as 1 ◮ ω n : natural frequency, ζ : damping factor. ◮ For different ζ , get underdamped ( ζ < 1), critically damped ( ζ = 1) and overdamped ( ζ > 1) systems 26/39 Process Control Second Order Models and Response
You should be able to derive all the previous expressions for y(t) 27/39 Process Control Second Order Models and Response
Solution to underdamped system Underdamped second order system: ω 2 n G(s) = s 2 + 2 ζω n s + ω 2 n � 1 − ζ 2 For ζ < 1, rts of den. = − ζω n ± j ω n The step response is, y(t) = 1 � � � − e − ζω n t 1 − ζ 2 � 1 − ζ 2 t + tan − 1 1 − ζ 2 sin ω n � ζ 28/39 Process Control Second Order Models and Response
Step response of underdamped system M p y ( t ) t r t p t s 29/39 Process Control Second Order Models and Response
Scilab code step-II-order.sce exec ( ’ second . s c i ’ ) ; 1 s = %s 2 3 omegan = 1 ; t = 1 : 0 . 1 : 1 5 ; 4 y = second ( omegan , 0 . 1 , t ) ; 5 y = second ( omegan , 0 . 5 , t ) ; 6 y = second ( omegan , 1 , t ) ; 7 y = second ( omegan , 2 , t ) ; f u n c t i o n y = second ( omegan , zeta , t ) 1 2 G = omegan ˆ2/( s ˆ2+2 ∗ zeta ∗ omegan ∗ s+ omegan ˆ2) 3 y = csim ( ’ step ’ , t ,G) ; p l o t 2 d ( t , y ) 4 e n d f u n c t i o n 5 30/39 Process Control Second Order Models and Response
3. Integrating/capacitive systems 32/39 Process Control Second Order Models and Response
Integrating Process Q i ( t ) ◮ Adh dt = Q i − Q ◮ In deviational variables, h ( t ) Q ◮ Ad∆h = ∆Q i (t) − ∆Q(t) dt ◮ Taking Laplace transform, As∆h(s) = ∆Q i (s) − ∆Q(s) ◮ ∆h(s) = As∆Q i (s) − 1 1 As∆Q(s) 33/39 Process Control Second Order Models and Response
Integrating Process Q i ( t ) In case of a step distur- bance in feed flow, tank will h ( t ) 1. overflow Q 2. run dry 3. overflow or run dry 4. stabilise at some suitable value Answer: 3: overflow or run dry 34/39 Process Control Second Order Models and Response
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