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Unformatted text preview: ME375 Handouts Dynamic Response of Linear Systems • Linear System Response – Superposition Principle – Responses to Specific Inputs • Dynamic Response of 1st Order Systems – Characteristic Equation - Free Response – Stable 1st Order System Response • Dynamic Response of 2nd Order Systems – Characteristic Equation - Free Response – Stable 2nd Order System Response • Transient and Steady-State Response School of Mechanical Engineering Purdue University ME375 Dynamic Response - 1 Linear System Response y ( n ) + a n −1 y ( n −1) + + a 2 y + a1 y + a 0 y = bm u ( m ) + + b1u + b0 u • Superposition Principle Input u1 (t) Output Linear System Linear System y1 (t) y2 (t) u2 (t) k1 u1 (t) + k2 u2 (t) The response of a linear system to a complicated input can be obtained by obtained studying how the system responds to simple inputs, such as zero input, unit input, impulse , unit step, and sinusoidal inputs. inputs. School of Mechanical Engineering Purdue University ME375 Dynamic Response - 2 1 ME375 Handouts Typical Responses • Free (Natural) Response – Response due to non-zero initial conditions (ICs) and zero input. non- • Forced Response – Response to non-zero input with zero ICs. non– Unit Impulse Response Response to unit impulse input. u(t) – Sinusoidal Response – Unit Step Response Response to unit step input (u (t) = 1). Response to sinusoidal inputs at different frequencies. The steady state sinusoidal response is call the Frequency Response. Response. u(t) Time t Time t School of Mechanical Engineering Purdue University ME375 Dynamic Response - 3 Dynamic Response of 1st Order Systems y + ay = b u • Characteristic Equation: s+a =0 ⇒ • Free Response [ yH(t)]: (u = 0) )]: a>0 e. g. ⇒ yH (t ) = A e − a t a=0 y H ( t ) = A e − 4 ⋅t yH (t) e. g. a<0 y H ( t ) = A e 0 ⋅t y H ( t ) = A e − ( − 4 ) ⋅t yH (t) yH (t) Time (t) e.g. Time (t) Time (t) Q: What determines whether the free response will converge to zero ? Q: How does the coefficient, a, affect the converging rate ? School of Mechanical Engineering Purdue University ME375 Dynamic Response - 4 2 ME375 Handouts Response of Stable 1st Order System • Stable 1st Order System y + ay = b u ⇒ τy + y = Ku where τ : Time Constant K : Static (Steady State, DC) Gain – Unit Step Response ( u = 1 and zero ICs ) y(t) y (t ) = y H (t ) + y P (t ) = Ae −t τ yP(t) = K K +K IC : y (0) = A + K Time t A= yH(t) = − K y (t ) = e -t/τ School of Mechanical Engineering Purdue University ME375 Dynamic Response - 5 Response of Stable 1st Order System Normalized Unit Step Response (u = 1 & zero ICs) 1 τy + y = Ku ⇒ y ( t ) = K (1 − e −tτ Normalized Response 0.9 ) Norm alized (such that as t → ∞ , y n → 1) : ⇒ yn ( t ) = y(t ) −t = (1 − e τ ) K 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 τ Time t ( 1− e −t/ τ 2τ 1τ 3τ 2τ 3τ Time [ t ] 4τ 4τ 5τ 6τ 5τ ) School of Mechanical Engineering Purdue University ME375 Dynamic Response - 6 3 ME375 Handouts Response of Stable 1st Order System Effect of Time Constant τ : τy + y = Ku Normalized: ⇒ yn ( t ) = 0.9 ) 0.8 Normalized Response ⇒ y ( t ) = K (1 − e 1 −tτ y(t ) −t = (1 − e τ ) K Slope at t = 0: d yn ( t ) = dt d yn ( 0 ) = ⇒ dt ⇒ 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 Q: What is your conclusion ? 2 4 6 Time [sec] School of Mechanical Engineering Purdue University 8 10 ME375 Dynamic Response - 7 Example – Vehicle Acceleration v F m b Standing-Start Acceleration; Lincoln Aviator SUV Standing-Start Acceleration; Dodge Viper SRT-10 160 140 140 120 120 100 100 Speed (MPH) 160 Speed (MPH) m 1 v + v = F = vmax b b 80 80 60 60 40 40 20 20 0 0 5 10 15 20 Time (sec) 25 30 35 40 0 0 5 School of Mechanical Engineering Purdue University 10 15 20 Time (sec) 25 30 35 40 ME375 Dynamic Response - 8 4 ME375 Handouts Response of Stable 1st Order System Normalized Unit Step Response τy + y = Ku ⇒ y ( t ) = K (1 − e −tτ 1 0.9 ) Normalized (such that as t → ∞ , yn → 1): ⇒ yn ( t ) = y(t ) −t = (1 − e τ ) K Normalized Response 0.8 Initial Slope 0.7 0.6 0.5 0.4 0.3 0.2 1 yn ( 0 ) = τ K y(0 ) = τ 0.1 0 0 1τ School of Mechanical Engineering Purdue University 2τ 3τ Time [ t ] 4τ 5τ 6τ ME375 Dynamic Response - 9 Response of Stable 1st Order System Q: How would you calculate the response of a 1st order system to a unit pulse (not unit impulse)? impulse)? Q: How would you calculate the unit impulse response of a 1st order system? system? u(t) t1 Time t (Hint: superposition principle ?!) Q: How would you calculate the sinusoidal response of a 1st order system? system? School of Mechanical Engineering Purdue University ME375 Dynamic Response - 10 5 ME375 Handouts Dynamic Response of 2nd Order Systems y + a1 y + a0 y = b1 u + b0 u • Characteristic Equation: s 2 + a1 s + a 0 = 0 ⇒ • Free Response [ yH(t)]: (u = 0) Determined by the roots of the characteristic equation: – Real and Distinct [ s1 & s2 ]: y H ( t ) = A1 e s1 ⋅t + A 2 e s 2 ⋅t – Real and Identical [ s1 = s2 ]: – Complex [ s1,2 = α ± jβ ] y H ( t ) = A 1 e s1 ⋅ t + A 2 t e s1 ⋅ t y H ( t ) = e α ⋅t ( A1 c o s ( β t ) + A 2 s in ( β t ) ) = A e α ⋅t c o s( β t + φ ) School of Mechanical Engineering Purdue University ME375 Dynamic Response - 11 Dynamic Response of 2nd Order Systems Free Response (Two distinct real roots) roots) y H (t ) = A1 e s1⋅t + A2 e s2 ⋅t s1 < 0 & s2 < 0 y H (t ) = A1 e s1⋅t + A2 e s2 ⋅t s1 < 0 & s2 = 0 Img. y H (t ) = A1 e s1⋅t + A2 e s2 ⋅t s1 < 0 & s2 > 0 Img. Img. Real yH (t) yH (t) Time (t) Real Real yH (t) Time (t) School of Mechanical Engineering Purdue University Time (t) ME375 Dynamic Response - 12 6 ME375 Handouts Dynamic Response of 2nd Order Systems Free Response (Two identical real roots ) y H (t ) = A1 e s1⋅t + A2 te s1⋅t s1 = s2 < 0 y H (t ) = A1 e s1⋅t + A2 te s1⋅t s1 = s2 = 0 y H (t ) = A1 e s1⋅t + A2 te s1⋅t s1 = s2 > 0 Img. Img. Img. Real yH (t) yH (t) yH (t) Real Real Time (t) Time (t) Time (t) School of Mechanical Engineering Purdue University ME375 Dynamic Response - 13 Dynamic Response of 2nd Order Systems Free Response (Two complex roots) roots) y H (t ) = A eα ⋅t cos( β t + φ ) s1,2 = α ± j β & α < 0 y H (t ) = A eα ⋅t cos( β t + φ ) s1,2 = ± j β & α = 0 y H (t ) = A eα ⋅t cos( β t + φ ) s1,2 = α ± j β & α > 0 Img. Img. Img. Real Real yH (t) yH (t) yH (t) Time (t) Real Time (t) School of Mechanical Engineering Purdue University Time (t) ME375 Dynamic Response - 14 7 ME375 Handouts Example – Automotive Suspension y m my + by + ky = br + kr g k b r Response to Initial Conditions 0.02 0 -0.02 Amplitude for free response: my + by + ky = 0 b k y+ y+ y =0 m m y + 28 y + 400 y = 0 -0.04 -0.06 -0.08 -0.1 0 0.05 0.1 School of Mechanical Engineering Purdue University 0.15 0.2 Time (sec) 0.25 0.3 0.35 0.4 ME375 Dynamic Response - 15 Dynamic Response of 2nd Order Systems Q: What part of the characteristic roots determines whether the free response is bounded, converging to zero or “blowing up” ? Q: For a second order system, what conditions will guarantee the system to be stable ? (Hint: Check the characteristic roots ) Q: If the free response of the system converges to zero, what determines the convergence rate? School of Mechanical Engineering Purdue University ME375 Dynamic Response - 16 8 ME375 Handouts Response of Stable 2nd Order System • Stable 2nd Order System y + a1 y + a 0 y = b u where ⇒ y + 2ζ ω n y + ω n 2 y = K ω n 2 u ωn > 0 : Natural Frequency [rad/s] ζ > 0 : Damping Ratio K : Static (Steady State, DC) Gain Characteristic roots s 2 + 2ζω n s + ω n2 = 0 ⇒ s = −ζω n ± ω n Img. ωn ( ζ 2 − 1) ⎧ζ > 1 : ⎪ ⎨ζ = 1 : ⎪ζ < 1 : ⎩ Real −ωn School of Mechanical Engineering Purdue University ME375 Dynamic Response - 17 Response of Stable 2nd Order System Unit Step Response of Under-damped 2nd Order Systems ( u = 1 and zero ICs ) Undery + 2ζ ω n y + ω n 2 y = K ω n 2 u Characteristic equation: s 2 + 2ζ ω n s + ω n 2 = 0 ⇒ s = −ζ ω n ± j ω n (1 − ζ 2 ) ωd y (t) = School of Mechanical Engineering Purdue University ME375 Dynamic Response - 18 9 ME375 Handouts Response of Stable 2nd Order System Unit Step Response of 2nd Order Systems 1.6K yMAX 1.4K OS Unit Step Response 1.2K K 0.8K Td 0.6K 0.4K 0.2K 0 0 1 tP 2 3 4 5 6 7 8 Time [sec] 9 10 11 tS School of Mechanical Engineering Purdue University 12 13 14 15 ME375 Dynamic Response - 19 Response of Stable 2nd Order System • Peak Time (tP) Time when output y(t) reaches its maximum value yMAX. FG H y MAX = y ( t P ) = K ⋅ 1 + e y (t ) = ⎛ ⎡ ⎤⎞ ζω n K ⋅ ⎜ 1 − e −ζω n ⋅t ⎢ cos(ω d t ) + sin(ω d t ) ⎥ ⎟ ωd ⎣ ⎦⎠ ⎝ d y (t ) = dt Find t P such that y ( t P ) = 0 ⇒ • Percent Overshoot (%OS) At peak time tP the maximum output πζ − 1− ζ 2 IJ K The overshoot (OS) is: (OS) O S = y M AX − y SS = The percent overshoot is: % OS = tP = FG Hy IJ K OS 100% SS − y ( 0 ) = School of Mechanical Engineering Purdue University ME375 Dynamic Response - 20 10 ME375 Handouts Response of Stable 2nd Order System • Settling Time (tS) Time required for the response to be within a specific percent of the final (steady-state) value. (steadySome typical specifications for settling time are: 5%, 2% and 1%. Look at the envelope of the response: % 1% 2% 5% tS Q: What parameters in a 2nd order system affect the peak time? Q: What parameters in a 2nd order system affect the % OS? Q: What parameters in a 2nd order system affect the settling time? Q: Can you obtain the formula for a 3% settling time? School of Mechanical Engineering Purdue University ME375 Dynamic Response - 21 In Class Exercise • Mass-Spring-Damper System Mass- SpringQ: What is the static (steady-state) gain of (steadythe system ? x K M f(t) B I/O Model: M x + Bx + K x = f (t ) Q: How would the physical parameters (M, B, K) affect the response of the system ? (This is equivalent to asking you for the relationship between the physical parameters and the damping ratio, natural frequency and the static gain.) School of Mechanical Engineering Purdue University ME375 Dynamic Response - 22 11 ME375 Handouts Transient and Steady State Response Ex: Let’s find the total response of a stable first order system: Let’ y + 5 y = 10u to a ramp input: with IC: – Total Response u(t) = 5t 5t y(0) = 2 Y (s) = ⋅U ( s) + ⋅ y(0), where U (s) = L [5 t ] = Transfer Function G(s) ∴ Y ( s) = ⋅ + – PFE: Y (s) = + A1 + A2 + A3 y (t ) = School of Mechanical Engineering Purdue University ME375 Dynamic Response - 23 Transient and Steady State Responses In general, the total response of a stable LTI system an y ( n) + an−1 y ( n−1) + + a1 y + a0 y = bm u ( m) + bm−1 u ( m−1) + + b1 u + b0 u m−1 bm s + bm−1s + + b1s + b0 N (s) bm (s − z1 )(s − z2 ) (s − zm ) = = an s n + an−1s n−1 + + a1s + a0 D(s) an (s − p1 )(s − p2 ) (s − pn ) to an input u(t) can be decomposed into two parts: G( s ) ≡ m y (t ) = yT (t ) + ySS (t ) Transient where Response • Transient Response (yT(t)) Steady State Response – Contains the free response yFree(t) of the system plus a portion of the forced response. – Will decay to zero at a rate that is determined by the characteristic roots (poles) of the characteristic system. • Steady State Response (ySS(t)) – will take the same form as the forcing input. – Specifically, for a sinusoidal input, the steady state response will be a sinusoidal signal with the same frequency as the input but with different magnitude and phase. magnitude School of Mechanical Engineering Purdue University ME375 Dynamic Response - 24 12 ME375 Handouts Transient and Steady State Response Ex: Let’s find the total response of a stable second order system: Let’ y + 4y + 3y = 6u to a step input: with IC: – Total Response u(t) = 5 y ( 0 ) = 0 and y( 0) = 2 – PFE: School of Mechanical Engineering Purdue University ME375 Dynamic Response - 25 Steady State Response • Final Value Theorem (FVT) Given a signal’s LT F(s), if the poles of sF(s) all lie in the LHP (stable region), signal’ sF( region), then f(t) converges to a constant value f(∞). f(∞) can be obtained without knowing f(t) by using the FVT: f ( ∞ ) = lim f (t ) = lim sF ( s ) t →∞ s→0 Ex: A model of a linear system is determined to be: Ex: y + 4 y + 1 2 y = 4u + 3u (1) if a constant input u = 5 is applied at t = 0, determine whether the output y(t) will converge to a constant value? (2) If the output converges, what will be its steady state value? value? School of Mechanical Engineering Purdue University ME375 Dynamic Response - 26 13 ME375 Handouts Steady State Response Given a stable LTI system an y(n) + an−1 y(n−1) + + a1 y + a0 y = bmu(m) + bm−1u(m−1) + + b1u + b0 u The corresponding transfer function is G (s) ≡ b m s m + bm − 1 s m − 1 + a n s n + a n −1 s n −1 + + b1 s + b0 b ( s − z1 )( s − z 2 ) = m + a1 s + a 0 a n ( s − p1 )( s − p 2 ) ( s − zm ) ( s − pn ) • Steady State Value of the Free Response Recall the free response of the system is: YFree (s) = F(s) = an sn + an−1sn−1 + + a1s + a0 Apply FVT: School of Mechanical Engineering Purdue University ME375 Dynamic Response - 27 Steady State Response • Steady State Value of the Unit Impulse Response Y (s) = G (s) ⋅U (s) = Apply FVT: • Steady State Value of the Unit Step Response Y (s) = G(s) ⋅U (s) = Apply FVT: 1st Order Systems: G(s) = ⇒ G(0) = 2nd Order Systems: b0 a1s + a0 G( s ) = b1s + b0 a2 s 2 + a1s + a0 ⇒ G(0) = School of Mechanical Engineering Purdue University ME375 Dynamic Response - 28 14 ...
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