Time Response of Second Order Systems

Time Response of Second Order Systems d 2 y(t) dy(t) M = f (t) − B − Ky(t) dt2 dt M(s2Y(s) − sy(0) − y'(0)) = F(s) − B(sY(s) − y(0))− KY(s) Let y'(0) = 0 2 Ms Y(s) + BsY(s) + KY(s) = Msy0 + By0 + F(s) Or (Ms+ B)y F(s) Y(s) = 0 + Ms2 + Bs+ K Ms2 + Bs+ K Consider t he first term only: (s + B/ M)y (s + 2ςω )y Y(s) = 0 = n 0 2 2 2 s + (B/ M)s + K / M s + 2ςωns +ωn ς = dimensionl ess damping ratio ωn = the natural frequency Where K B ω = ; ς = n M 2 KM 2 2 The characteristic equation s + 2ςωns +ωn has two roots : 2 s1 = −ςωn + jωn 1−ς 2 s2 = −ςω n − jωn 1−ς If ς >1⇒ roots are real If ς <1⇒ roots are complex (under damped) If ς =1⇒ same roots and real (critically damped) For unit step response : 1 y(t) =1− e−ςωnt sin(βω t +θ ) β n β β = 1−ς 2 , θ = tan −1( ) = cos−1 ς ς 1 −ζω t Step response: y(t) =1− e n sin(ω βt +θ ) β n 2 −1 where β = 1− ζ , θ = cos ζ Showing the step response with different damping coefficients Standard Performance measures Performance measures are usually defined in terms of the step response of a system as below: Swiftness of the response is measured by rise time , and peak time Swiftness of the response is measured by rise time T r , and peak time Tp For underdamped system, the Rise time T r (0-100% rise time) is useful, For overdamped systems, the the Peak time is not defined, and the (10-90 % rise time) is normally used Tr1 Peak time: T p Steady-state error: ess Settling time: T s Percent of Overshoot: M − fv P.O. = pt × 100 % fv is the peak value Mpt fv is the final value of the response Percentage overshoot measures the closeness of the response to the desired response. The settling time T s is the time required for the system to settle within a certain percentage δ of the input amplitude. For second order system, we seek T s for which the response remains within 2% of the final value. This occurs approximately when: e−ζωnTs < 0.02 or : ζωnTs ≅ 4 Therefore: 4 Ts ≅ 4τ = ζωn Hence the settling time is defined as 4 time constants. Explicit relations for M and : Explicit relations for p t and Tp To find T p we can either differentiate y(t) directly or indirectly through Laplace Transform of y’(t): π π T = = p ω β 2 n ωn 1−ς The peak response is : 2 M =1+ e−ζπ / 1−ς pt Percentage overshoot : 2 P.O. =100e−ζπ / 1−ς Impulse response of the second order system: Laplace transform of the unit impulse is R(s)=1 2 ω n Y ( s ) = 2 2 Impulse response: ( s + 2ζω n s + ω n ) Transient response for the impulse function, which is simply is the derivative of the response to the unit step: ω y (t ) = n e − ζω n t sin( ω β t ) β n Responses and pole locations Time Responses and Pole Locations: 2 2 The characteristic equation s + 2ςωns +ωn has two roots : 2 s1 = −ςωn + jωn 1−ς 2 s2 = −ςωn − jωn 1−ς are the poles. s1, s2 4 Ts ≅ 4τ = ζωn Therefore the settling time is inversely proportional to the real part of the poles..

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