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Unformatted text preview: ME375 Handouts Dynamic Response of Linear Systems Dynamic response: total response of a dynamic system. response total response of dynamic system. It can be decomposed by two methods: • Forced and Free Response – – – – • Forced response of 1st and 2nd order systems due to a unit step input (i.e. unit step response) Performance specifications of unit step response Free response of 1st and 2nd order systems Pole location and free response Transient and Steady-State Response – – Transient response decay to zero SteadySteady-state response takes the form of the forcing input School of Mechanical Engineering Purdue University ME375 Dynamic Response - 1 1 ME375 Handouts 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 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 2 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 Frequency Frequency Response. Time t School of Mechanical Engineering Purdue University ME375 Dynamic Response - 3 3 ME375 Handouts Forced Response of Stable 1st Order System • Stable 1st Order System y + ay = b u where ⇒ τy + y = Ku τ : Time Constant Time K : Static (Steady State, DC) Gain Static (force and – Unit Step Response (force response: u = 1 and zero ICs ) Y (s) = K (τ s + 1) s y(t) yP(t) = K K τK K − s τ s +1 K K = − 1 s s+ = τ y (t ) = K − K e − Time t yH(t) = − K e -t/τ t τ School of Mechanical Engineering Purdue University ME375 Dynamic Response - 4 4 ME375 Handouts Forced 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 t hat as t → ∞ , y n → 1) : that a s 1) y(t ) −t ⇒ yn ( 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 - 5 5 ME375 Handouts Forced Response of Stable 1st Order System Normalized Unit Step Response τy + y = Ku −tτ 0.9 ) 0.8 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 ⇒ y ( t ) = K (1 − e 1 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Time [sec] School of Mechanical Engineering Purdue University ME375 Dynamic Response - 6 6 ME375 Handouts Forced Response of Stable 1st Order System Q: How would you calculate the response of 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 - 7 7 ME375 Handouts Forced Response of Stable 2nd Order System • Stable 2nd Order System y + a1 y + a 0 y = b u where ⇒ y + 2ζω 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 Static Characteristic roots 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 - 8 8 ME375 Handouts Forced 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 ⎡ ⎛ ⎞⎤ ⎢1 − e−ζωnt ⎜ cos ωd t + ζ y (t ) = K sin ωd t ⎟ ⎥ 2 ⎜ ⎟⎥ ⎢ 1− ζ ⎝ ⎠⎦ ⎣ Response: Response: = K ⎡1 − e−ζωnt cos (ωd t + φ ) ⎤ ⎣ ⎦ Question: how does it compare to the response of first order system? τ y + y = Ku ⇒ y ( t ) = K (1 − e −t τ) School of Mechanical Engineering Purdue University ME375 Dynamic Response - 9 9 ME375 Handouts Forced Response of Stable 2nd Order System How 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 2 s s + 2ζω n s + ω n = K ( s + ζ ωn ) + K ζ ωn 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 e−ζωn sin ωn 1 − ζ 2 t 1− ζ 2 ⎡ ⎛ ⎞⎤ ζ −ζωnt ⎜ cos ωd t + = K ⎢1 − e sin ωd t ⎟ ⎥ 2 ⎜ ⎟⎥ ⎢ 1− ζ ⎝ ⎠⎦ ⎣ School of Mechanical Engineering Purdue University ME375 Dynamic Response - 10 10 ME375 Handouts Forced Response of Stable 2nd Order System Unit Step Response of stable under-damped 2nd Order Systems under1.6K 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 School of Mechanical Engineering Purdue University 10 11 tS 12 13 14 15 ME375 Dynamic Response - 11 11 ME375 Handouts Forced Response of Stable 2nd Order System • Peak Time (tP) Time when output y(t) reaches its maximum value yMAX. Percent Overshoot (%OS) At peak time tP the maximum output πζ FG H y MAX = y ( t P ) = K ⋅ 1 + e y (t ) = ⎛ ⎡ ⎤⎞ ζω n sin(ω d t ) ⎥ ⎟ K ⋅ ⎜ 1 − e −ζω n ⋅t ⎢ cos(ω d t ) + ωd ⎣ ⎦⎠ ⎝ d y (t ) = dt Find t P such that y ( t P ) = 0 ⇒ • − 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 - 12 12 ME375 Handouts Forced 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? settling time? School of Mechanical Engineering Purdue University ME375 Dynamic Response - 13 13 ME375 Handouts Pole location and Performance specifications Img. ωn Real −ωn School of Mechanical Engineering Purdue University ME375 Dynamic Response - 14 14 ME375 Handouts System parameters and performance specifications • Mass-SpringMass-Spring-Damper System Q: 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 (M, B, K) affect the response of the K) affect the response of the system 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 - 15 15 ME375 Handouts Example School of Mechanical Engineering Purdue University ME375 Dynamic Response - 16 16 ME375 Handouts Summary on unit step response School of Mechanical Engineering Purdue University ME375 Dynamic Response - 17 17 ME375 Handouts Summary on pole location and performance specifications performance Img. ωn Real −ωn School of Mechanical Engineering Purdue University ME375 Dynamic Response - 18 18 ME375 Handouts Free Response of 1st Order Systems y + ay = b u • Characteristic Equation: 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 yH (t) e.g. a<0 y Free ( t ) = A e 0 ⋅t y Free ( 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 - 19 19 ME375 Handouts Free 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 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 ]: y F r e e ( t ) = A 1 e s1 ⋅ t + A 2 t e s1 ⋅ t – Complex [ s1,2 = α ± jβ ] y Free ( t ) = e α ⋅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 - 20 20 ME375 Handouts Free 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 yH (t) yH (t) Time (t) Real Real yH (t) Time (t) School of Mechanical Engineering Purdue University Time (t) ME375 Dynamic Response - 21 21 ME375 Handouts Free 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 yH (t) yH (t) yH (t) Time (t) Real Real Time (t) School of Mechanical Engineering Purdue University Time (t) ME375 Dynamic Response - 22 22 ME375 Handouts Free 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 yH (t) yH (t) yH (t) Time (t) Real Time (t) School of Mechanical Engineering Purdue University Time (t) ME375 Dynamic Response - 23 23 ME375 Handouts Free 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” ? “bl Q: For a second order system, what conditions will guarantee the system to be stable ? 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 the free response of the system converges to zero what determines the convergence rate? School of Mechanical Engineering Purdue University ME375 Dynamic Response - 24 24 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 to a ramp input: with IC: – Total Response u(t) = 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 - 25 25 ME375 Handouts 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 + + a1 y + a0 y = bm u ( m) + bm−1 u ( m−1) + + b1 u + b0 u bm s m + bm−1s m−1 + + 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 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 26 ME375 Handouts Transient and Steady State Response Ex: Let’s find the total response of a stable second order system: 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 - 27 27 ME375 Handouts 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 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: 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 0, converge to a constant value? (2) If the output converges, what will be its steady state value? School of Mechanical Engineering Purdue University ME375 Dynamic Response - 28 28 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 b m s m + bm −1 s m − 1 + G (s) ≡ 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 Recall the free response of the system is: YFree (s) = F(s) = an s + an−1s + + a1s + a0 n n−1 For stable system, apply FVT: School of Mechanical Engineering Purdue University ME375 Dynamic Response - 29 29 ME375 Handouts Steady State Response • Steady State Value of the Unit Impulse Response Y ( s) = G ( s) ⋅U ( s) = For stable system, apply FVT: • Steady State Value of the Unit Step Response Y (s) = G(s) ⋅U (s) = For stable system, apply FVT: 1st Order Systems: G(s) = ⇒ G(0) = 2nd Order Systems: b0 a1s + a0 G( s ) = b1s + b0 a2 s + a1s + a0 2 ⇒ G(0) = School of Mechanical Engineering Purdue University ME375 Dynamic Response - 30 30 ME375 Handouts Steady State Response • Steady State Value of the Harmonic Response For forced response due to harmonic input, u(t) = sinωt ⎡ω⎤ Y(s) = G(s)U(s) = G(s) ⎢ 2 2 ⎥ ⎣ s +ω ⎦ For stable systems, apply FVT: FV For ⎡ sω ⎤ yss = limsY (s) = limG(s) ⎢ 2 2 ⎥ = ?? s→0 s→0 ⎣ s +ω ⎦ School of Mechanical Engineering Purdue University ME375 Dynamic Response - 31 31 ...
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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 University-West Lafayette.

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