Second order ransfer funcions nd Order ransfer funcion - Summary of resuls The canonical nd order ransfer funcion is expressed as H(s) s + ζ s + is he naural frequency; ζ is he damping coefficien. The Poles, s, s are s,s ( ζ ± ζ ) 3 On he righ is he roo locus for fixed and varying ζ. jω ζsinθ ζ0 s ζ< jωn -ωn θ jωn ζ> ζ> σ ζ ζ< -jωn ζ0 The ransfer funcion in he frequency domain is ω H(ω) n w + ζ w ( ω ) + 4 ζ ω ejφ The Bode plo is : For ζ we have he sharpes corner wihou a maximum. wih anφ ζ ω ω For ζ < here is a peak in he plo of magniude H max ζ ζ 5 H, db 0 0-0 -0-30 ζ ζ ζ 0. This is above he level a ω 0. For he H(s) given a 0 hen H The roll-off rae a high frequencies is 40 db per decade ( in frequency). The overall phase change is -80. For low ζ he swing in phase is all close o he frequency a he maximum H. The peak iself is a ω p ζ Beware; laer we inroduce ω d ω ζ -40 Phase, Φ degrees 0-45 -90-35 -0. 0 Frequency, rad s - ζ ζ 0. ζ -80 0.0 0. 0 50 Frequency, rad s - ( log scale ) w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions Impulse Responses ζ 0 Undamped y() /ωn π/ωn y() sin( ) -/ωn ζ > 0 Damped y() ζ ( exp(s ) exp(s ) ) ζ Criically damped 0.3 0. 0. -0. y() exp( s ) 3 4 5 y() exp( - ) -0. -0.3 exp( s ) Curve is very similar o damped response ζ < 0 Underdamped y() y() ω d exp(-ζ ) sin(ω d ) The logarihm of he raio of successive peaks is he logarihmic decremen or logdec for shor δ πζ ζ πζ if ζ << 0.8 0.6 0.4 0. -0. -0.4-0.6 5 0 5 0 5 w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 3 Sep responses ζ 0 Undamped y() ( cos( )) y() ζ > 0 Damped y() + ζ [ s exp(s ) s exp(s ) ] 0.8 0.6 0.4 0. y() 4 6 8 0 0 5 ζ Criically damped y() ( exp( ) [ + ] ) y() 4 6 8 0 ζ < 0 Underdamped y() exp( ζ ) cos( ω d ) + ζ ω sin(ω d d ) Firs peak *, Μ P + exp Proporional overshoo: Ο P y exp π ζ ζ π ζ ζ Time o be wihin δ, T s ln(δ) ζ For δ 0,0, i.e., % : T s 4 ζ + 4ζ Roo Toal square error 4 ζ * Some auhors call M P he overshoo. M P y() T p Period, T π ω d Overshoo, O P Frequency of oscillaion: ω d ζ Period of oscillaion: T π ω d δ 0 T s 0 π -ζ For leas roo square oal deviaion ζ /. Then O P 6% w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 4 Ramp response Now se illusraed. F s f() as f Perhaps unkindly I leave derivaions o he reader. The general response is: Y H(s) s s (s + ζ s + ) ζ 0 Undamped Y y s s + sin( ) Plo is for ; y sin() 0 8 6 4 y() Response y 4 6 8 0 Noe: s s ; s + s ζ ζ > 0 Damped : s, s ( ζ ) Y s + s +s s -ζ ± + s s s s-s s s s 6 4 y() Response y ζ + Plo is wih, ζ s s ( s e s s e s ) - -4 4 6 8 0 - ζ ωn ζ Criically Damped 8 Y() Y s s + ω s+ω + n b (s+ ) y ω n ( + ( + ) e ) The plo is for. 6 4 - Response 4-6 8 0 w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 5 ζ ω Y ω n s ζ ζ(s+ζ ) n s + ω ( ζ )ω d d (s+ζω n ) s+ ( ζ ) ζ < 0 Underdamped: ω d y ω n ζ+e ωnζ [ζcos(ω d ) ω ( ζ ) sin( ω d d )] 8 6 4 y() Response - ζ 4 6 8 0 The plo is wih and ζ 0 Second order ransfer funcion analysis As an example consider he mass, damper and spring aced on by a force, f(). We had for he displacemen, y : m y " + c y ' + k y f Take he Laplace ransform ( wih no iniial values ) : m s Y + c s Y + k Y F or s Y + c m s Y + k m Y F/m which wih k m and ζ c km becomes If we redefine F F/ m F/k: s Y + ζ s Y + F : or (s + ζ s + ) Y F and so H(s) F Y s + ζ s + This equaion is a sandard form for any second order ransfer funcion. Whaever he physical variables i helps o urn he expression o his forma early in he analysis. is he naural frequency; ζ is he damping coefficien. Poles Poles, s, s are he values of s for which he denominaor is zero. s,s ζ ± 4 ζ 4 ( ζ ± ) ζ 3 w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 6 Cases; () ζ 0: poles a s ± j, () ζ < : wo complex poles. These lie on a circle of radius (3) ζ : wo coinciden poles a s (4) ζ > : wo poles on he negaive real axis. These are illusraed on he roo locus diagram on he righ for consan. ζ0 ζ< -ωn ζ> ζ> ζ ζ< ζ0 jω s jωn jωn σ -jωn w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 7 Bode Plo From eq. ω H(s) n s + ζ s + and puing s jω ω H(ω) n ω + j ζ ω 4a wih H(ω) and H(ω) exp( jφ) 4b ( ω ) + 4 ζ ω 4c anφ ζ ω ω ω 0 ω ω H ζ ω 4d Φ 0 π π For ζ < here is a peak in he H curve. I is easier o find he peak of H which is a he same value of ω as he peak of H. And easier sill o find he minimum of H whic occurs a eh same value of ω as he peak in H 4 d dω H so d dω d dω (( ω ) + 4 ζ ω ) ( ω )(-ω) + 8ζ ω 4ω ( ω ( ζ )) ζ H 0 if ω 0; or if ω In he laer case H ( + ζ ) + 4 ζ ( ζ ) 4 ζ ( ζ ) ζ 5 ζ Noe ha if ζ 0 hen he peak of H. This is he undamped case. Noe also ha he peak is a a frequency close o bu slighly less han. There is no peak if ζ > since hen he frequency is imaginary. This is illusraed in he nex figures wih. Noe ha he frequency scale has been exended in he phase plo o show he approach o he limis. In he magniude plo noe (i) he roll of rae is 40 db per decade. w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 8 (ii) he peak is a abou 4 db for ζ 0. Eq. 5 gives a peak of a facor of 5: so 0 log 0 (5) 3.97. The heory pans ou correcly. H, db 0 0-0 -0 ζ ζ ζ 0. -30-40 Phase, Φ degrees 0-0. 0 Frequency, rad s - ζ 0. -45-90 ζ -35 ζ -80 0.0 0. 0 50 Frequency, rad s - ( log scale ) Impulse Responses Consider he Dirac impulse, δ(), as an inpu. ζ 0. Y s + s + f() δ() so y() sin( ) There is no damping and he sysem rings forever. y() ωn π/ωn -ωn w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 9 ζ > The roos are real. We had eq. 3: Thus he response is Y and y() s,s ( ζ ± ζ ) ; s > s 3 (s s )(s-s ) s s s-s s s ζ ( exp(s ) exp(s ) ) The response below is for and ζ. The coefficiens of he componen exponenials is omied in he figure. The formula is : y 0 89( e 0 7 e 3 7 ) 0.3 0. 0. -0. -0. -0.3 y() exp( s ) 3 4 5 exp( s ) ζ Y (s + ) (s + ) so y() exp( - ) The plo is very like he previous one. ζ < This is a bi more complicaed. We need o complee squares in he denominaor. Y H(s) s + ζ s + (s + ζ ) ζ + (s + ζ ) + ω d 6 wih ω d ζ 7 This frequency appears ofen and should be noed. w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 0 Thus Y ω ω d d (s + ζ ) + ω d so, from he ables y() ω d exp(-ζ ) sin(ω d ) 8a ζ exp(-ζ) sin(ω d ) 8b The response for and ζ 0 is below. I is an exponenially decaying sine wave. The formula is : y.005 e 0. sin( 0 995 ) y() 0.8 0.6 0.4 0. -0. -0.4-0.6 5 0 5 0 5 The period of he oscillaion, T π ω d. The raio of succeeding posiive peaks is hen: exp exp(-ζ ) ζ + π ω d πζω exp n ω d The logarihm of his raio is δ πζ ω d πζ ζ so δ πζ ζ πζ if ζ << 9 δ is called he logarihmic decremen or log-dec for shor. I is a useful way of measuring he damping in a highly resonan sysem. Noe we use he symbol δ wih anoher meaning laer. w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions Sep funcion responses The inpu is f u() so f u() F s The response is, using H(s) from eq. Y s + ζ s + s 0 Cases ζ 0 Y s( s + ) A s + Bs + C s + Thus A(s + ω n ) + s (Bs+C) Pu s 0: ω n A ω n : A Coeff. s : 0 A + B : B A Coeff, s : 0 C : C 0 Y s s s + ω n cos( ) and y() ( ) The plo for is on he righ. The sysem oscillaes abou he mean indefiniely. y() 4 6 8 0 ζ > The ransfer funcion is facorisable and he roos are real. As before eq. 3 gives he roos: s,s ( ζ ± ζ ) ; s > s 3 and he response is Y (s s )(s-s ) s We have Y ` s(s s )(s-s ) A s + B s-s + B s-s w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions whence A(s s )(s s ) + Bs(s s ) + Cs(s s ) Pu s s : ω ω n B s (s s ) B n s (s s ) s s s Pu s s : ω ω n C s ( s s ) C n s (s s ) - s s s Pu s 0 : ω ω n A s s A n s s So Y s + ζ and y() + s s s s s s ) ; since s s ; s s ζ ζ [ s exp(s ) s exp(s ) ]) noing s,s ( ζ ± ζ ) ; s > s Wih and ζ his becomes y() + 0 077 e 3 73 077 e 0 7 which is illusraed on he righ. 0.8 0.6 0.4 0. y() 0 5 ω ζ Y n (s + ) s s(s + ) A s + B C s+ω + n (s+ ), say Thus Α (s + ) + Bs(s + ) + Cs Pu s 0 : ω n A ω n A. Pu s : ω n C C Coeff. s : 0 A + B B A Y s ω s+ω n n (s + ) Thus from he Tables: y() exp( ) [ + ] y() For y() e ( +) This is illusraed on he righ. 4 6 8 0 ζ < The more complicaed bu more ineresing case. From eq. we have Y s + ζ s + s w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 3 s(s + ζ s + ) A s + Bs + C s + ζ s +, say. Thus A (s + ζ s + ω n ) + s (B s +C) Pu s 0 : ω n A ω n A Coeff. s : 0 A + B B A Coeff. s : 0 A ζ + C C ζ A ζ and Y s s + ζ s + ζ s + s s + ζ (s + ζ ) + ω d : compleing he square as in eq. 6 ζω (s + ζω n ) + n s ω ω d d (s + ζ ) + ω d : making he numeraor o sui he Tables wih ω d ζ Thus y() exp( ζ ) cos( ω d ) + ζ ω sin(ω d d ) For and ζ 0 y() we have y() e 0 [ cos( 0 98 ) + 0 sin( 0 98 ) ] This is ploed on he righ. + e - Noe ha he gradien is zero a 0 - e - 5 0 5 0 w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 4 There are properies of ineres of he curve. y() M P Overshoo, O P δ Period, T π ω d T p 0 T s 0 Overshoo The derivaive dy d ω d exp( ζ ) sin(ω d ) is remarkably simple. I is equal o he impulse response. Indeed i is generally he case ha he derivaive of he sep response of any linear sysem is he impulse response. dy d 0 when ω d nπ or when nπ ω d nπ -ζ nt where he period of he oscillaions is T π ω d π -ζ The value of y a hese exrema is Y + exp nπ ζ ζ The firs of hese when n gives he peak of he overshoo ( ofen simply called he overshoo. Μ P + exp π ζ ζ Express proporional overshoo as he excess as a proporion of he final seady sae value. Ο P y exp π ζ ζ a Overshoo, O p 0.8 0.6 0.4 0. occurring a T P π -ζ π ωd T b 0 0 0. 0.4 0.6 0.8 Damping, ζ w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 5 We may be ineresed in he ime o reach and say wihin a cerain 'disance' from he final value. The deviaion of he nh exremum is δ exp n π ζ ζ which occurs a n nπ ζ or nπ ζ n so and δ exp n π ζ ζ exp( ζ n ) T s n ln(δ) ζ Sricly n should be an inegral number of half periods bu he discrepancy is usually ignored. For δ 0 0, i.e., % and for δ 0 05, i.e., 5% T s T s 4 ζ 3 ζ We may also consider an overall expression of deviaion from he final value. The roo mean square seems a suiable choice. y() - (y-) 0.8-5 0 5 0 0.6 0.4 0. 5 0 5 0 The deviaion and is square are illusraed above for he previously shown response. We ake hen as a measure of he overall deviaion as We require E rs he oal square deviaion ( inegraed from 0 o ) / E rs ( y ) d 0 +4 ζ 4 ζ..8.6.4. This raher simple expression is valid for all ζ, i.e., 0 < ζ <. I has.0 a minimum value of when ζ. 3 4 5 E rs ζ w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 6 This minimum square deviaion condiion is illusraed on he righ wih a 6% overshoo...6 y() 5 0 5 w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 7 There are oher measures of he error. To pu emphasis on he approach o he final value on migh weigh erros by mupiplying by ime. The inegral of his de-emphasises he iniial approach o he final value.. Thus we have The roo of he inegral of ime weighed oal square erroe is / E rs ( y ) d 0 + 8 ζ4 ζ This has a minimum of ζ 8 /4 0 595 /4 when 0.8 0.6 0.4 (y - ) 0. 5 0 5 0 The overshoo is exp ζ π ζ exp π 9 8% w norris nd order ransfer funcions Monday 3 March 008
Second order ransfer funcions 8 Keeping facors consan Poles are a s ( ζ ± ζ ): Consan imaginary par. y() jω ω d ζ consan. So Consan frequency in response s σ Consan real par: y() ζ is consan. jω Fixed envelope y exp( ζ ) s σ and so ime o ge wihin a given limi is deermined by ζ. E.g., T δ% 4 ζ Consan damping raio, ζ y() jω Thus also consan overshoo: θ s O P exp π ζ ζ Noe: ζ sinθ where θ is marked in he figure. σ There is scarcely any overshoo when θ > 60. Fracional overshoo, O P.8.6 jω θ s σ.4. 0 40 60 90 θ w norris nd order ransfer funcions Monday 3 March 008
w norris nd order ransfer funcions Monday 3 March 008 Second order ransfer funcions 9