Lecture 4 Transfer Function and Block Diagram Approach to Modeling Dynamic Systems System Analysis Spring 2015 1 The Concept of Transfer Function Consider the linear time-invariant system defined by the following differential equation : ()nn ( 1) ay01 ay ann 1 y ay ()mm ( 1) bu01 bu bmm 1 u b u() n m Where y is the output of the system, and x is the input. And the Laplace transform of the equation is, nn1 ()()aS01 aS ann 1 S a Y s mm1 ()()bS01 bS bmm 1 S b U s System Analysis Spring 2015 2 The Concept of Transfer Function The ratio of the Laplace transform of the output (response function) to the Laplace Transform of the input (driving function) under the assumption that all initial conditions are Zero. mm1 Ys() bS01 bS bmm 1 S b Transfer Function G() s nn1 Us() aS01 aS ann 1 S a Ps() ()nm Qs() System Analysis Spring 2015 3 Comments on Transfer Function 1. A mathematical model. 2. Property of system itself. - Independent of the input function and initial condition - Denominator of the transfer function is the characteristic polynomial, - TF tells us something about the intrinsic behavior of the model. 3. ODE equivalence - TF is equivalent to the ODE. We can reconstruct ODE from TF. 4. One TF for one input-output pair. : Single Input Single Output system. - If multiple inputs affect Obtain TF for each input x 6207()3()xx f t g t -If multiple outputs xxy 32 yy xut 3() System Analysis Spring 2015 4 Comments on Transfer Function 5. Analytic method and Experimental method u(t) y(t) Y(s) system G(s) U(s) U(s) Y(s) L.T.I 6. Different systems may have identical T.F ex) uT J y Y 1 J b T U T Js b Y R 1 Electrical System U Ls R L (if b=1, J=L/R) s 1 R System Analysis Spring 2015 5 Comments on Transfer Function A Ys() Gsus () () Impulse Input t t The Strength of an impulsive input : At The Dirac Delta function () t : An Impulsive function with a strength aquals to unity 0 ()tdt 1 0 [()]1 t ut() () t Ys() Gs () Transfer Function = unit impulse response us() 1 System Analysis Spring 2015 6 Collision, Impulse Response and Transfer Function Impulse: large force in a very short time Newton’s law of motion dv d F ma m () mv F dt d () mv dt dt The impulse-momentum principle dv d F ma m () mv F dt d () mv dt dt tv22 Fdt d() mv mv mv 21 tv11 System Analysis Spring 2015 7 The law of conservation of momentum And if There is no input force, then we get d()0, mv mv const . The law of conservation of momentum And also, I const. The law of conservation of angular momentum System Analysis Spring 2015 8 Inelastic collision xt() m1 k m1 Becomes embedded in mass m 2 after collision m2 v1 No external force, conservation of momentum t 2 m vv(0 ) 1 Fdt()(0)(0)0, m12 m v mv 11 1 mm12 t1 The equation of motion for the combined mass m1 ()0,(0)(0)mmxkxv12 x v 1 mm12 1 k xt() v (0) sin( wtnn ), w wmm12 System Analysis Spring 2015 9 Elastic collision xt() Perfectly elastic collision m 1 k m1 f ()t k m m2 2 v1 v v v2 0 3 4 No external force : conservation of momentum mv11 m 2 v 2 mv13 mv 24 Perfectly elastic collision : No energy loss, kinetic energy is conserved 111 mv22 mv mv 2 22211 13 2 4 mm12 2 m 1 vvvv3141, mm12 mm 12 System Analysis Spring 2015 10 The law of conservation of momentum 0 mm m f ()tdt mv mv m v v m12 1 v Mass 1 : 13 11 1 3 1 1 1 0 mm12 2m2 mv11 mm12 0 2mm ftdt() mv mv 12v Mass m 2 : 24 2 2 1 0 mm12 0 At () A Linear impulse 0 2m x(0 ) 0,vx (0 ) (0 ) 1 v xt() mxt2() kx 0 1 We can solve mm12 2 Or LT: msXs2 ( ( ) sx (0) x (0)) kXs ( ) 0 2mm 12 xx(0 ) (0 ) 0 mxt21() kx A () t v () t mm12 2 2mm12 msXs21(())() kXs v mm12 System Analysis Spring 2015 11 Block Diagram Input Signal Output Signal Transfer Function Summing Point Pick Off Point System Analysis Spring 2015 12 Series(Cascade) Connection Parallel Connection System Analysis Spring 2015 13 Associative Rule Commutative Rule System Analysis Spring 2015 14 Closed Loop Transfer Function Ys() GsEs () () Gs ()[() Rs HsYs () ()] mx bx kx h [1GsHs () ()]() Ys GsRs () () Ys() Gs() Transfer Function 1 Rs() 1 G1 () sHs () System Analysis Spring 2015 15 Closed Loop Transfer Function ex) Fs() 1 1 1 Xs() m s s b m k m 1 kb Fs() Xs () sXs () sXs2 () mmm Fs() kXs () bsXs () msXs2 () (ms22 bsXs ) () Fs () kXs (), ( ms bskXs ) () Fs () Xs() 1 Transfer Function Fs() ms2 bsk System Analysis Spring 2015 16 System Schematic Block Diagram Transfer Functions C(s) G(s) U(s) Y(s) C(s)G(s)H (s) H1(s) Y(s)= 1 U(s) 1+C(s)G(s)H2(s) H2(s) System Analysis Spring 2015 17 Example 2 ex) d y dy du mb2 kyu() dt dt dt d2 y dy du or m b ky b ku dt2 dt dt Assume the initial condition is 0, (ms2 bs k ) Y () s ( bs k ) U () s Ys() bsk Transfer Function Us() ms2 bsk If, m=10kg, b=20N-s/m, k=100N/m Ys( ) 20 s 100 2 s 10 1 ,()Us Us( ) 10 s22 20 s 100 s 2 s 10 s 2101ss 210 Ys() ss232210 sss 2 10 s System Analysis Spring 2015 18 Impulse response >>num=[8]; >>den=[50 100 2500]; >>sys=tf(num, den) >> impulse(sys) >> grid >>title(‘ ‘) >>xlabel(‘ ‘) >>ylabel(‘ ‘) System Analysis Spring 2015 19 Transient Response Analysis with MATLAB Ys( ) 20 s 100 2 s 10 1 ,()Us Us( ) 10 s22 20 s 100 s 2 s 10 s 2101ss 210 Ys() ss232210 sss 2 10 s Unit Step Response 1.5 t=0:0.01:8; num=[2 10]; den=[1 2 10]; 1 sys=tf(num,den); step(sys,t) grid title('Unit Step Response','Fontsize',15') Output 0.5 xlabel('t(sec)','Fontsize',15') ylabel('Output','Fontsize',15') 0 0 1 2 3 4 5 6 7 8 t (sec) System Analysis Spring 2015 20 Impulse Input ex) M 50kg ,mk 0.01g ,bNsmkNmv 100 / , 2500 / , (0 ) 800 ms / 1 50.01 0.01 800 7.9984 Xs() 50.01ss22 100 2500 50.02 50.01 ss 100 2500 Impulse Response of System 0.02 num=[7.9984]; 0.015 den=[50.01 100 2500]; 0.01 sys=tf(num,den); impulse(sys) 0.005 grid 0 title('Impulse Response of System','Fontsize',15') x(t)Response -0.005 xlabel('t(sec)','Fontsize',15') ylabel('Response x(t)','Fontsize',15') -0.01 -0.015 0 1 2 3 4 5 6 t (sec) System Analysis Spring 2015 21 Ramp Response Ys() 2 s 10 M=10kg, b=20Ns/m, k=100N/m Us() s2 2 s 10 u(): t Unit ramp input , u t , 1 Unit-Ramp Response num=[2 10]; 4.5 den=[1 2 10]; 4 sys=tf(num,den); t=0:0.01:4; 3.5 u=t; 3 lsim(sys,u,t) 2.5 grid title('Unit-Ramp Response','Fontsize',15) 2 xlabel('t') 1.5 ylabel('Output y(t) and Input u(t)=t','Fontsize',15) 1 text(0.8,0.4,'y','Fontsize',12) u y Output y(t) and Input u(t)=t text(0.4,0.8,'u','Fontsize',12) 0.5 y legend('y') 0 0 0.5 1 1.5 2 2.5 3 3.5 4 t (sec) System Analysis Spring 2015 22 Partial Fraction Expansion with MATLAB Bs() num b (0) Smm b (2) S1 bm ( ) As() den a (0) Snn a (2) S1 an () num[ b (0) b (1) b ( m )], den [ a (0) a (1) a ( n )] [r, p, k] = residue (num, den) >>num=[1 8 16 9 6]; B()srrrn (1) (2) () ks() >>den=[1 6 11 6]; A()sspspspn (1) (2) () >>[r,p,k]=residue(num,den) r= Bs() s43 8 s 16 s 2 9 s 6 -6.0000 ex) As() s32 6 s 11 s 6 -4.0000 3.0000 p= -3.0000 643 -2.0000 s 2 -1.0000 sss 321 k= 1 2 System Analysis Spring 2015 23 End of lecture 4 System Analysis Spring 2015 24.