Dynamic Response

Dynamic Response - ME375 Handouts Dynamic Response of...

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Unformatted text preview: ME375 Handouts Dynamic Response of Linear Systems • Linear System Response – – • Dynamic Response of 1st Order Systems – – • Characteristic Equation - Free Response Stable 1st Order System Response Dynamic Response of 2nd Order Systems – – • Superposition Principle Responses to Specific Inputs Characteristic Equation - Free Response Stable 2nd Order System Response Transient and Steady-State Response SteadySchool of Mechanical Engineering Purdue University ME375 Dynamic Response - 1 Linear System Response & & y ( n ) + a n −1 y ( n −1) + L + a2 && + a1 y + a 0 y = bm u ( m ) + L + b1u + b0 u y • Superposition Principle Input Output Linear System u1 (t) y1 (t) u2 (t) y2 (t) k1 u1 (t) + k2 u2 (t) The response of a linear system to a complicated input can be obtained by studying how the system responds to simple inputs, such as zero input, unit impulse , unit step, and sinusoidal 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) – Unit Step Response Response to unit step input (u (t) = 1). u(t) Time t – Sinusoidal Response Response to sinusoidal inputs at different frequencies. The steady state sinusoidal response is call the response is call the Frequency Frequency Response. Time t School of Mechanical Engineering Purdue University ME375 Dynamic Response - 3 Dynamic Response of 1st Order Systems • Characteristic Equation: & y + ay = b u s+a =0 ⇒ • Free Response [ yFree(t)]: (u = 0) 0) a>0 e.g. ⇒ y Free ( t ) = A e − a t = y ( 0 ) e − a t a=0 y Free ( t ) = A e − 4 ⋅ t y H (t ) e.g. a<0 y Free ( t ) = A e 0 ⋅ t y Free ( t ) = A e − ( − 4 ) ⋅ t y H (t ) y H (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 Ti K : Static (Steady State, DC) Gain Static – Unit Step Response (force response: u = 1 and zero ICs ) (force and Y (s) = K (τ s + 1) s y(t) y P (t ) = K K K τK − s τ s +1 K K = − 1 s s+ = Time t yH(t) = − K τ y (t ) = K − K e − e -t/τ 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) zero 1 & τy + y = Ku ⇒ y ( t ) = K (1 − e −tτ Normalized Response 0.9 ) Normalized (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 1τ 0 τ Time t ( 1− e − t/ τ 2τ 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 y(t ) −t = (1 − e τ ) K Slope at t = 0: 0: d yn ( t ) = dt d yn ( 0 ) = ⇒ dt ⇒ Normalized Response ⇒ y ( t ) = K (1 − e 1 −tτ 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Q: Can y ( 0 ) = 0 ? Can & 0 2 4 6 Time [sec] School of Mechanical Engineering Purdue University 8 10 ME375 Dynamic Response - 7 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): y(t ) −t ⇒ yn ( t ) = = (1 − e τ ) K Initial Slope 1 & yn ( 0 ) = τ K & y(0 ) = τ Normalized Response 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1τ School of Mechanical Engineering Purdue University 2τ 3τ Time [ t ] 4τ 5τ 6τ ME375 Dynamic Response - 8 4 ME375 Handouts 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)? u(t) t1 Time t (Hint: superposition principle ?!) School of Mechanical Engineering Purdue University ME375 Dynamic Response - 9 Dynamic Response of 2nd Order Systems & && + a1 y + a0 y = b1 u + b0 u & y • Characteristic Equation: s 2 + a1 s + a 0 = 0 ⇒ • Free Response [ yFree(t)]: (u = 0) 0) Determined by the roots of the characteristic equation: – Real and Distinct [ s1 & s2 ]: y F r e e ( t ) = A1 e s1 ⋅t + A 2 e s 2 ⋅t – Real and Identical [ s1 = s2 ]: – Complex [ s1,2 = α ± jβ ] y Free ( t ) = e α ⋅t y F r e e ( t ) = A 1 e s1 ⋅ t + A 2 t e s1 ⋅ t ( A1 cos( β t ) + A 2 sin( β t ) ) = A e α School of Mechanical Engineering Purdue University ⋅t cos( β t + φ ) ME375 Dynamic Response - 10 5 ME375 Handouts Dynamic Response of 2nd Order Systems Free Response (Two distinct real roots) y Free (t ) = A1 e s1⋅t + A2 e s2 ⋅t s1 < 0 & s2 < 0 y Free (t ) = A1 e s1⋅t + A2 e s2 ⋅t s1 < 0 & s2 = 0 Img. y Free (t ) = A1 e s1⋅t + A2 e s2 ⋅t s1 < 0 & s2 > 0 Img. Img. Real y H (t ) Real Real y H (t ) y H (t ) Time (t) Time (t) Time (t) School of Mechanical Engineering Purdue University ME375 Dynamic Response - 11 Dynamic Response of 2nd Order Systems Free Response (Two identical real roots ) y Free (t ) = A1 e s1⋅t + A2 te s1⋅t s1 = s2 < 0 y Free (t ) = A1 e s1⋅t + A2 te s1⋅t s1 = s2 = 0 y Free (t ) = A1 e s1⋅t + A2 te s1⋅t s1 = s2 > 0 Img. Img. Img. Real y H (t ) y H (t ) y H (t ) Time (t) Real Real 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 complex roots) y Free (t ) = A eα ⋅t cos( β t + φ ) s1, 2 = α ± j β & α < 0 y Free (t ) = A eα ⋅t cos( β t + φ ) s1, 2 = ± j β & α = 0 y Free (t ) = A eα ⋅t cos( β t + φ ) s1, 2 = α ± j β & α > 0 Img. Img. Img. Real Real y H (t ) Real y H (t ) y H (t ) Time (t) Time (t) Time (t) School of Mechanical Engineering Purdue University ME375 Dynamic Response - 13 Example – Automotive Suspension y m && & & my + by + ky = br + kr g k b r Response to Initial Conditions 0.02 b k && + y + y = 0 & y m m && + 28 y + 400 y = 0 & y 0 -0.02 Amplitude for free response: && & my + by + ky = 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 - 14 7 ME375 Handouts 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 - 15 Response of Stable 2nd Order System • Stable 2nd Order System && + a1 y + a 0 y = b u & y where ⇒ && + 2 ζ ω n y + ω n 2 y = K ω n 2 u & y ωn > 0 : Natural Frequency [rad/s] ζ > 0 : Damping Ratio K : Static (Steady State, DC) Gain Static 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 - 16 8 ME375 Handouts Response of Stable 2nd Order System Unit Step Response of Under-damped 2nd Order Systems ( u = 1 and zero ICs ) Under&& + 2 ζ ω n y + ω n 2 y = K ω n 2 u & y Characteristic equation: equation: s 2 + 2ζ ω n s + ω n 2 = 0 ⇒ s = −ζ ω n ± j ω n (1 − ζ 2 ) 1 4243 ωd y (t) = 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 ) UnderHow the solution was derived: Y (s) = 2 K ωn 2 s ( s 2 + 2ζω n s + ω n ) = Ks + 2 K ζω n K − 2 s s 2 + 2ζω n s + ω n = K ( s + ζωn ) + K ζωn K − s ( s + ζω ) 2 + (ω 1 − ζ 2 ) 2 n n y (t ) = K − Ke −ζωn cos ωn 1 − ζ 2 t − K ζ 1− ζ 2 School of Mechanical Engineering Purdue University e −ζωn sin ωn 1 − ζ 2 t 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: 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: % tS 1% 2% 5% 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 Pole location and Transient Response Img. ωn Real −ωn School of Mechanical Engineering Purdue University ME375 Dynamic Response - 22 11 ME375 Handouts 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 + B x + 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 - 23 School of Mechanical Engineering Purdue University ME375 Dynamic Response - 24 Examples 12 ME375 Handouts Transient and Steady State Response Ex: Let’s find the total response of a stable first order system: & y + 5 y = 10u to a ramp input: with IC: – Total Response u(t) = 5t y(0) = 2 Y ( s) = ⋅ U ( s) + ⋅ y(0), where 14 3 24 U (s) = L [5 t ] = Transfer Function G(s) ∴ Y ( s) = ⋅ + – PFE: Y (s) = A3 A1 A2 + + + 1444 24444 4 3 14 244 4 3 y (t ) = School of Mechanical Engineering Purdue University ME375 Dynamic Response - 25 Transient and Steady State Responses In general, the total response of a stable LTI system total stable (n) ( n −1) & & an y + an−1 y + L+ a1 y + a0 y = bm u ( m) + bm−1 u ( m−1) + L+ b1 u + b0 u bm s m + bm−1s m−1 + L + b1s + b0 N (s) bm (s − z1 )(s − z2 )L(s − zm ) = = an s n + an−1s n−1 + L + a1s + a0 D(s) an (s − p1 )(s − p2 )L(s − pn ) to an input u(t) can be decomposed into two parts: G( s ) ≡ 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 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. School of Mechanical Engineering Purdue University ME375 Dynamic Response - 26 13 ME375 Handouts Transient and Steady State Response Ex: Let’s find the total response of a stable second order system: && + 4 y + 3 y = 6 u & y 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 - 27 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), the all then then f(t) converges to a constant value f(∞). f(∞) can be obtained without converges to constant value ). can be obtained without knowing 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: && + 4 y + 1 2 y = 4 u + 3 u & & y (1) if a constant input u = 5 is applied at t = 0, determine whether the output y(t) will 0, converge to a constant value? converge to constant value? (2) If the output converges, what will be its steady state value? School of Mechanical Engineering Purdue University ME375 Dynamic Response - 28 14 ME375 Handouts Steady State Response Given a stable LTI system & & an y(n) + an−1 y(n−1) +L+ a1 y + a0 y = bmu(m) + bm−1u(m−1) +L+ b1u + b0 u The corresponding transfer function is corresponding transfer function is G (s) ≡ bm s m + bm −1 s m −1 + L + b1 s + b0 b ( s − z1 )( s − z 2 ) L ( s − z m ) =m a n s n + a n −1 s n −1 + L + a1 s + a 0 a n ( s − p1 )( s − p 2 ) L ( s − p n ) • Steady State Value of the Free Response Recall the free response of the system is: YFree (s) = F(s) = an sn + an−1sn−1 +L+ a1s + a0 Apply FVT: School of Mechanical Engineering Purdue University ME375 Dynamic Response - 29 Steady State Response • Steady State Value of the Unit Impulse Response Y (s) = G (s) ⋅U (s) = Apply FVT: 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 - 30 15 ...
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This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue.

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