I. Define the Situation

Transcription

1 I. efine he Siuaion This exam explores he relaionship beween he applied volage o a permanen magne C moond he linear velociy (speed) of he elecric vehicle (EV) he moor drives. The moor oupu is fed ino a gear box for he purpose of increasing he oupu orque and decreasing he speed. The oupu of he gear box direcly drives a ire. The ire ranslaes roaional moion ino linear moion of he EV. The following is a concepual block diagram of he sysem under consideraion. i a v a - P C oor Gear Box Tire τ m ω m τ ω f EV (mass) f f γ dx/d x C oor Parameers K T : Nm A Torque consan : Vehicle Parameers 7.8 kg amping coefficien K V : V rad Volage consan : 17kg Vehicle mass : 330 µh Armaure inducance : 0.011Ω Armaure resisance Gear Box Parameer G R :.5 Gear raio : Tire Parameer 0.54m Radius of he ires Page 1 of 14

2 II. Goals A. eermine he ransfer funcion H(s) Γ(s)/V a (s) where Γ(s) is he Laplace ransform of γ() dx()/d and V a (s) is he Laplace ransform of v a (). B. eermine and graph he impulse response. Tha is, deermine γ δ () for v a () δ() and graph γ δ (). C. eermine and graph he sep response. Tha is, deermine γ u () for v a () u() and graph γ u ().. eermine he frequency response of he H(s), boh magniude and angle. III. Generae Ideas C oor Equaions i a Va - - e τ m ω m v a i a di a e d (1) e K v ω m () τ m K T i a (3) 4 Variables: e i a τ m ω m 1 inpu v a Page of 14

3 Gear Box Equaions Gear Box τ m ω m τ ω ω m G R ω T (4) τ T G R τ m (5) more variables: τ T ω T Tire and Vehicle Equaions Tire is he radius of he ire τ ω f EV (mass) f f f τ (6) γ dx/d x f dγ f F 0 f γ f f γ d dγ d (7) γ ω T (8) more variables: f γ 8 variables oal: e f γ i a τ m τ T ω m ω T 8 equaions 1 inpu: v a Page 3 of 14

4 IV. ake a Plan A. Take he Laplace ransform of Eq. 1 - Eq. 8. B. Form a GENI Table C. 1. Remove one variable a a ime by solving an equaion ha involves ha variable for ha variable. Subsiue he resuling expression ino he oher equaions.. Repea sep "C.1" unil Γ(s) is he only variable remaining. Form he ransfer funcion H(s) E. eermine he impulse response, γ δ () h() inverse Laplace ransform of H(s) F. Graph he impulse response, γ δ (). G. eermine he sep response, γ u () inverse Laplace ransform of H(s)/s H. Graph he sep response, γ u () I. Graph he magniude (in db) and angle (in degrees) of H(jω) versus log(ω) V. Take Acion A. Take he Laplace ransform of Eq. 1 - Eq. 8. (Assume no sored energy.) V a I a si a E ( L1) E K v Ω m ( L) Τ m K T I a ( L3) Ω m G R Ω T ( L4) Τ T G m ( L5) F Τ T ( L6) F Γ s Γ ( L7) Ω T Γ ( L8) Page 4 of 14

5 B. Form a GENI Table Goal Equaion Need Informaion Γ 1 Γ F ( L7) F is a parameer is a parameer F F 1 T T ( L6) T T is a parameer T T T T G ( L5) T G R is a parameer T T K T I a ( L3) I a K T is a parameer I a V a E L a I a ( L1) E V a is he inpu is a parameer is a parameer E E K v Ω m ( L) Ω m K v is a parameer Ω m Ω m G R Ω T ( L4) Ω T 1 Ω T Ω T Γ ( L8) C. 1. Remove one variable a a ime by solving an equaion ha involves ha variable for ha variable. Subsiue he resuling expression ino he oher equaions.. Repea sep "C.1" unil Γ(s) is he only variable remaining Page 5 of 14

6 L6 > L7 Γ 1 1 T T 1 R T T T (L9) L5 > L9 Γ G R R T T (L10) L3 > L10 Γ K T G R R T I a (L11) L1 > L11 Γ K T G R R T ( V a E) (L1) L > L1 Γ K T G R R T ( V a K V Ω m ) (L13) L4 > L13 Γ K T G R R T ( V a K V G R Ω T ) (L14) L8 > L13 Γ K T G R R T V a K V G R Γ (L15) Page 6 of 14

7 Eq. L15 repeaed here for ease of reference K T G R K V G R V R T Γ a R T Γ (L15) r a The only variable in Eq. L15 is Γ.. Form he ransfer funcion H(s) K T G R K V G R Γ V R T Γ (L16) a K T G R K T G R K V G R Γ V R T Γ (L17) a R T Γ K V K T G R K T G R Γ V R a (L18) T s r a K V K T G R K T G R Γ V R a (L19) T Hs () Γ() s V a () s K T G R (L0) s r a K T K v G R s L a L a Page 7 of 14

8 Check unis s : 1 K T G R N : N m V 3 : 1 1 s : L a 1 s r a K T K V G R 0 : Hs () s N 1 s 0 Hs () : s m V 3 Page 8 of 14

9 E. eermine he impulse response, γ δ () h() inverse Laplace ransform of H(s) v a () δ()v V a () s : 1V Γ δ () s Hs ()V complee he square Γ δ () s : 1 N α : ω d : 0 α α ω d rad Γ δ () s ( ) γ δ ( ) : 1.616e sin( ) F. Graph he impulse response, γ δ () ( ) Φ() m Impulse Responce N ω d rad V m Inverse Laplace > division by. 0.8 Velociy (m/s) γ δ () Time (onds) Page 9 of 14

10 G. eermine he sep response, γ u () inverse Laplace ransform of H(s)/s Γ u () s Hs () s s 0 : 0 N s s s 1 0 c : s s s 94 c : 1 ( ) 1 c : 0 0 roos : polyroos( c) roos i i s 1 : roos 0 s : roos 1 s i s i Γ u () s k 1 s k s s 1 k s s k 1 : 94 k k : s 1 s 1 s ( ) k i 10 3 k arg( k ) deg γ u ( ) : e cos( deg) Φ() mm H. Graph he sep response, γ u () 40 Sep Response Velociy (m/) γ u () Time (onds) Page 10 of 14

11 I. Graph he magniude (in db) and angle (in degrees) of H(jω) versus log(ω) NPTS : 100 n : 0, 1.. NPTS k1 : 0 k : 3 j : 1 k1 n NPTS ( k k1 ) 10 ω vn : G dbn 0 log H jω vn V : angle : arg H jω m n vn 180 π 0 Frequency Response 40 Gain (decibels) G db ω v Frequency (rad/) 0 Frequency Response 0 40 Angle (degrees) 60 angle ω v Frequency (rad/) Page 11 of 14

12 VI. Review, Reflec, and Refine The sep response is no realisic. I is oo fas and underdamped. Elecric vehicles do no accelerae o full speed in less han 50 m. An underdamped response is no desireable. Try decreasing he gear raio and increase he damping,. This problem is enered ino ahlab o enable easy modificaion o parameers and viewing of differen variables. VII. Summary Transfer Funcion Hs () : s m V 3 Page 1 of 14

13 Impulse response, γ δ () γ δ ( ) : 1.616e sin( ) Φ() m Impulse Responce 0.8 Velociy (m/s) γ δ () Time (onds) Sep response, γ u () γ u ( ) : e cos( deg) Φ() mm 40 Sep Response Velociy (m/) γ u () Time (onds) Page 13 of 14

14 0 Frequency Response Gain (decibels) G db ω v Frequency (rad/) 0 Frequency Response Angle (degrees) angle ω v Frequency (rad/) Page 14 of 14