Second Order Systems independent energy storage elements => Resonance: inertance & capacitance trade energy, kinetic to potential Example: Automobile Suspension x z vertical motions suspension spring shock absorber wheel Xroad(z) Bond Graph
Extract State Equations: p = kδ + b v road (t) p δ = v road (t) p coupled st order differential equations Rearrange into standard form: x x + ζω x +ω = f (t) n n Differentiate nd, substitute st & nd into result: δ = d dt v (t) road p = d kδ + b v road (t) dt v (t) road δ = d dt v (t) kδ b road δ p δ + b δ + k δ = d dt v (t) road Compare equations
f (t) = d dt v road, ζ = b, = k, ζ = b = System parameters natural frequency damping ratio ζ b k = k Solution by superposition x = omogeneous + x particular Homogeneous differential equation ( set forcing term f(t)=0 ) x x + ζω x +ω = 0 n n Homogeneous solution: = 0, system at rest. Want nonzero. Guess: = Ae λt, A 0
Test (substitute): 0 = x h + ζ x h + satisfied if = d (Ae λt ) dt λ + ζ λ + + ζω d(ae λt ) + n (Ae λt ) = Ae λt (λ + ζ λ + ) dt = 0 know as characteristic equation Eigenvalues λ, λ special values of parameter λ that permit nonzero solutions : λ,λ = ζ ± (ζ ) 4 = ζ ± ζ From superposition and = Ae λt, homogeneous solution of form: e λ t + A e λ t Stability: If either λ, λ has positive real part, (λ i = Re λ i + Im λ i, Re λ i >0), Ae λ it will increase indefinitely. System unstable!
Eigenvalues: λ,λ = ζ ± ζ Sign of ζ determines character of solutions = Ae λt, via ζ. 3 cases:. ζ > 0, ζ real => both eigenvalues λ, λ real. Solution becomes: e λ t + A e λ t e (ζ ζ )t + A e (ζ + ζ )t Note: ζ ζ > 0, two terms of exponential decay, two time constants τ, = /λ, τ =/ λ. ζ = 0 => ζ = & λ = λ one solution: e λ t e t nd solution ( for nd order system) = A te λ t = A te t Note: x h + x h + = d (A te ωnt ) d(a + ω te ωnt ) dt n dt + (A te t ) = A te t [( )+ ( )+ ] = 0 = + e t + A te t
3. ζ < 0 => ζ imaginary complex eigenvalues: j = λ,λ = ζ ± j ζ e (ζ j ζ )t + A e (ζ + j ζ )t = e ζt (A e j ζ t = cosθ + jsinθ Euler s identity: e jθ Let constants be complex: A = (A R + ja I )/, A = (A R ja I )/ Apply: = e ζ t { A R + ja I + A R ja I + A e j ζ t ) [cos( ζ t)+ jsin( ζ t)]+ [cos( ζ t) jsin( ζ t)]} Rearrange: = e ζ t {A R cos( ζ t) A I sin( ζ t)} = e ζt{a R cos(ω d t) A I sin(ω d t)} Damped natural frequency: ω d = ζ
Particular Solution Unit step function input f(t) =f o u s (t) u s (t) 0 t Differential equation (t > 0): x + ζ x + x = f o u s (t) = f o Particular solution (guess): x p = C Test: 0 + ζ (0)+ C = f o x p = f o
Total solution: x = + x p. x(t) = + x p e λ t + A e λ t + f o. x(t) e t + A te t + f o 3. x(t) = e ζt{a R cos( ζ t) A I sin( ζ t)}+ f o Note: need initial conditions to determine arbitrary constants (A s) Apply: x(0) = 0, x (0) = 0 (initially at rest). 0 = x(0) + A + f o, 0 = x (0) λ + A λ 3. 0 = x(0) = A R + f o 0 = x (0) = ζ A R ζ A I Find constants, solution to step excitation: x(t) = f o f o + ζ ζ e ζt ζ e (ζ + ζ )t ζ + ζ cos( f o t [ (+ t)e ],ζ = ζ t)+ (ζ ζ )t,ζ > ζ e ωn ζ ζ sin(ω ζ t) n,0 ζ <
Step Response, Second Order System x f o / ζ = 0 0. 0.3 0.5 0.7 5 4 8 t System to steady state value x ss (t) = f o /. Curve depends on ζ value: small ζ or small damping => oscillation Application to Automobile Suspension: Hit a bump or hole @ t = 0: via u s (t) Displacement δ = x => rough ride, especially for small ζ Design choice for ζ?