Dynamic Response of Linear Systems

Transcription

1 Dyamic Respose of Liear Systems Liear System Respose Superpositio Priciple Resposes to Specific Iputs Dyamic Respose of st Order Systems Characteristic Equatio - Free Respose Stable st Order System Respose Dyamic Respose of 2d Order Systems Characteristic Equatio - Free Respose Stable 2d Order System Respose Trasiet ad Steady-State State Respose ME375 Dyamic Respose - Liear System Respose ( ) ( ) ( m) y + a y + + a y + a y + a y = b u + + bu + b u Superpositio Priciple Iput Liear System u u 2 k u + k 2 u 2 2 m Output y y 2 The respose of a liear system to a complicated iput ca be obtaied by studyig how the system respods to simple iputs, such as zero iput, uit impulse, uit step, ad siusoidal iputs. ME375 Dyamic Respose - 2

2 Typical Resposes Free (Natural) Respose Respose due to o-zero iitial coditios (ICs) ad zero iput. Forced Respose Respose to o-zero iput with zero ICs. Uit Impulse Respose Respose to uit impulse iput. u Uit Step Respose Respose espose to uit step iput (u( ) = ). u Siusoidal Respose Respose to siusoidal iputs at differet frequecies. The steady state siusoidal respose is call the Frequecy Respose. Time t Time t ME375 Dyamic Respose - 3 Dyamic Respose of st Order Systems Characteristic Equatio: y + ay = bu s + a = at Free Respose [ ]: (u u = ) yh ( t) = A e a > a = a < e.g. y ( t) = A e H 4 t e.g. y ( t) = A e H t e.g. y ( t) = A e H ( 4 ) t Q: What determies whether the free respose will coverge to zero? Q: How does the coefficiet, a, affect the covergig rate? ME375 Dyamic Respose - 4 2

3 Respose of Stable st Order System Stable st Order System y + ay = bu y + y = Ku where : Time Costat K : Static (Steady State, DC) Gai Uit Step Respose ( u = ad zero ICs ) y() t = y () t + y () t H t IC : y() = A + K A = y() t = P = Ae + K y K = K e -t/ y P = K Time t ME375 Dyamic Respose - 5 Respose of Stable st Order System Normalized Uit Step Respose (u = & zero ICs) y + y = Ku t () ( ) yt = K e Normalized (such that as t, ): y yt () y () t = = K t ( e ) Normalized Respose Time [ t ] Time t ( e t/ ) ME375 Dyamic Respose - 6 3

4 Respose of Stable st Order System Effect of Time Costat : y + y = K u t () ( ) yt = K e Normalized: yt () t y () t ( e = = ) K Slope at t = : d () dt y t = d ( ) dt y = Q: What is your coclusio? Normalized Respose Time [sec] ME375 Dyamic Respose - 7 Example Vehicle Acceleratio b m v F m v + v = F = v b b max 6 Stadig-Start Acceleratio; Dodge Viper SRT- 6 Stadig-Start Acceleratio; Licol Aviator SUV Speed (MPH) 8 6 Speed (MPH) Time (sec) Time (sec) ME375 Dyamic Respose - 8 4

5 Respose of Stable st Order System Normalized Uit Step Respose y + y = K u t () ( ) yt = K e Normalized (such that as t, ): y yt () y() t = = K Iitial Slope y ( ) = K y ( ) = t ( e ) Normalized Respose Time [ t ] ME375 Dyamic Respose - 9 Respose of Stable st Order System Q: How would you calculate the respose of a st order system to a uit pulse (ot uit impulse)? Q: How would you calculate the uit impulse respose of a st order system? u t Time t (Hit: superpositio priciple?!) Q: How would you calculate the siusoidal respose of a st order system? ME375 Dyamic Respose - 5

6 Dyamic Respose of 2d Order Systems y+ a y + ay = bu+ bu Characteristic Equatio: 2 s + a s + a = Free Respose [ ]: (u u = ) Determied by the roots of the characteristic equatio: ad Distict [ s & s 2 ]: y ( t) = A e + A e H s t s2 t 2 ad Idetical [ s = s 2 ]: s t s t ( t) Ae = + A2 te Complex [ s,2 = α ± jβ ] α t α t y ( t) = e ( A cos( β t) + A si( β t) ) = A e cos( β t + φ) H 2 ME375 Dyamic Respose - Dyamic Respose of 2d Order Systems Free Respose (Two distict real roots) s t s () 2 t s t yh t = Ae + A2e yh () t = Ae + A2e s < & s < s < & s = 2 2 s2 t s t s () 2 t H = + 2 s < & s2 > y t A e A e ME375 Dyamic Respose - 2 6

7 Dyamic Respose of 2d Order Systems Free Respose (Two idetical real roots ) s t s () t s t yh t = Ae + A2te yh () t = Ae + A2te s = s < s = s = 2 2 s t s t s t 2 = s2 > yh () t = A e + A te s ME375 Dyamic Respose - 3 Dyamic Respose of 2d Order Systems Free Respose (Two complex roots) t yh () t Ae α t = cos( β t + φ ) yh () t A e α = cos( β t + φ ) s = α ± jβ & α < s =± jβ & α =,2,2 t yh () t A e α = cos( β t + φ ) s = α ± jβ & α >,2 ME375 Dyamic Respose - 4 7

8 Example Automotive Suspesio m y g k b r my + by + ky = br + kr.2 Respose to Iitial Coditios for free respose: my + by + ky = b k y+ y + y = m m y+ 28y + 4y = Amplitude Time (sec) ME375 Dyamic Respose - 5 Dyamic Respose of 2d Order Systems Q: What part of the characteristic roots determies whether the free respose is bouded, covergig to zero or blowig up? Q: For a secod order system, what coditios will guaratee the system to be stable? (Hit: Check the characteristic roots ) Q: If the free respose of the system coverges to zero, what determies the covergece rate? ME375 Dyamic Respose - 6 8

9 Respose of Stable 2d Order System Stable 2d Order System 2 2 y + a y + a y = bu y + 2ζω y + ω y = Kω u where ω > : Natural Frequecy [rad/s] ζ > : Dampig Ratio K : Static (Steady State, DC) Gai Characteristic roots 2 2 s + 2ζ ω s + ω = s = ζ ω ± ω ( ζ ) 2 ω ζ > : ζ= : ζ< : ω ME375 Dyamic Respose - 7 Respose of Stable 2d Order System Uit Step Respose of Uder-damped damped 2d Order Systems ( u = ad zero ICs ) 2 2 y + 2ζ ω y + ω y = Kω u Characteristic equatio: 2 2 s + 2ζ ω s + ω = s = ζ ω ± jω ( ζ ) ω d 2 y ) = ME375 Dyamic Respose - 8 9

10 Respose of Stable 2d Order System Uit Step Respose of 2d Order Systems Uit Step Respose.6K y MAX.4K.2K K.8K.6K OS T d.4k.2k t P Time [sec] t S ME375 Dyamic Respose - 9 Respose of Stable 2d Order System Peak Time (t P ) Time whe output y ) reaches its maximum value y MAX. yt () = ζω ζω K e t cos( ω ) + dt si( ωdt) ωd d yt () = dt Fid t such that y ( t ) = P t = P P Percet Overshoot (%OS) At peak time t P the maximum output y = y( t ) = K + e MAX The overshoot (OS( OS) ) is: OS = y y = MAX P SS F HG The percet overshoot is: % OS = = F HG y SS I KJ πζ ζ2 OS y( ) % I KJ ME375 Dyamic Respose - 2

11 Respose of Stable 2d Order System Settlig Time (t S ) Time required for the respose to be withi a specific percet of the fial (steady-state) state) value. Some typical specificatios for settlig time are: 5%, 2% ad %. Q: What parameters i a 2d order system affect the peak time? Q: What parameters i a 2d order system affect the % OS? Look at the evelope of the respose: % % 2% 5% t S Q: What parameters i a 2d order system affect the settlig time? Q: Ca you obtai the formula for a 3% settlig time? ME375 Dyamic Respose - 2 I Class Exercise Mass-Sprig Sprig-Damper System K B I/O Model: M x f M x + Bx + K x = f () t Q: What is the static (steady-state) state) gai of the system? Q: How would the physical parameters (M, B, K) affect the respose of the system? (This This is equivalet to askig you for the relatioship betwee the physical parameters ad the dampig ratio, atural frequecy ad the static gai.) ME375 Dyamic Respose - 22

12 Trasiet ad Steady State Respose Ex: Let s fid the total respose of a stable first order system: y + 5 y = u to a ramp iput: u ) = 5t5 with IC: y() = 2 Total Respose Trasfer Fuctio Gs () [ ] Ys () = Us () + y(), where Us () = L5t= Ys ( ) = + PFE: A A2 A3 Y() s = yt () = ME375 Dyamic Respose - 23 Trasiet ad Steady State Resposes I geeral, the total respose of a stable LTI system ( ) ( ) ( m) ( m ) ay + a y + + ay + ay = bu m + bm u + + bu + bu m m bms + bm s + + bs + b Ns () bm( s z)( s z2) ( s zm) Gs () = = as + a s + + as + a D() s a( s p)( s p2) ( s p) to a iput u ) ca be decomposed ito two parts: yt () = y() () T t + y SS t where Trasiet Steady State Respose Respose Trasiet Respose (y T ) Cotais the free respose y Free of the system plus a portio of the forced respose. Will decay to zero at a rate that is determied by the characteristic roots (poles) of the system. Steady State Respose (y SS ) will take the same form as the forcig iput. Specifically, for a siusoidal iput, the steady state respose will be a siusoidal sigal with the same frequecy as the iput but with differet magitude e ad phase. ME375 Dyamic Respose

13 Trasiet ad Steady State Respose Ex: Let s fid the total respose of a stable secod order system: y + 4 y + 3y = 6 u to a step iput: u ) = 5 with IC: y ( ) = ad y( ) = 2 Total Respose PFE: ME375 Dyamic Respose - 25 Steady State Respose Fial Value Theorem (FVT) Give a sigal s LT F(s), if the poles of sf(s) all lie i the LHP (stable regio), the f ) coverges to a costat value f( ). f( ) ) ca be obtaied without kowig f ) by usig the FVT: f ( ) = lim f ( t) = lim sf( s) t s Ex: : A model of a liear system is determied to be: y + 4 y + 2 y = 4u + 3u () if a costat iput u = 5 is applied at t =, determie whether the output y ) will coverge to a costat value? (2) If the output coverges, what will be its steady state value? ME375 Dyamic Respose

14 Steady State Respose Give a stable LTI system ( ) ( ) ( m) ( m ) ay + a y + + ay + ay = bu m + bm u + + bu + bu The correspodig trasfer fuctio is m m bms + bm s + + bs+ b bm( s z)( s z2) ( s zm) G( s) = as + a s + + as+ a a( s p)( s p2) ( s p) Steady State Value of the Free Respose Recall the free respose of the system is: Fs ( ) YFree () s = = as + a s + + as+ a Apply FVT: ME375 Dyamic Respose - 27 Steady State Respose Steady State Value of the Uit Impulse Respose Y() s = G() s U() s = Apply FVT: Steady State Value of the Uit Step Respose Ys () = Gs () Us () = Apply FVT: st Order Systems: b Gs () = as+ a G() = 2d Order Systems: bs+ b Gs () = as as a G() = ME375 Dyamic Respose