EE/ME/AE34: Dynamical Sytem Chapter 8: Tranfer Function Analyi
The Sytem Tranfer Function Conider the ytem decribed by the nth-order I/O eqn.: ( n) ( n 1) ( m) y + a y + + a y = b u + + bu n 1 0 m 0 Taking the Laplace tranform of the ytem eqn. with IC = 0 : ( n n 1 ) ( m + ) n 1 + + 0 = m + + 0 a a Y b b U The Tranfer Function (TF) i defined a: H () Factoring the TF yield: where p i m Y() b m + + b0 = n n 1 U() + a + + a IC= 0 n 1 0 ( z1)( z) ( z ) H K m = ( p1 )( p ) ( pn ) and are the ytem pole and zero, repectively z i
The Sytem Tranfer Function If the PFE of the TF ha the form: H A A An ( p ) ( p ) ( p ) 1 = + + + 1 where A are the reidue aociated with the ytem pole, i the zero-input ytem repone will have the form: pt 1 pt y (t) = K e + K e + + K e zi p t 1 i where the e term are called the ytem mode n pt n n The tability of the ytem repone i baed on the p i: Stable if R p < 0 for all { } i R{ p } Untable if R > 0 for any p i Marginally table (ocillatory) if R { p} = 0 for ditinct p i i i p i
Second Order Repone Aume a nd order TF of the form Y() G DC n DC n () = = U() + ζω n+ ωn H where 0 ω G ω < ζ < 1 i the damping ratio (unitle), i the DC gain and ω i the natural frequency The characteritic po n ly. can be factored a G DC [ rad/ec] ( ζω jω 1 ζ )( ζω j ω 1 ζ ) n n n n = + + + 1 1 = + + jωd + jωd τ τ
Second Order Repone ( ) Thi implie that the root (pole) of are: 1 1 1, = ζωn ± j ωn 1 ζ = ± j ω d τ 1 where τ i the time contant [ec] and ζω n ω ω ζ d n 1 i the damped frequency [rad/ec] It alo implie = + 1 = ζ ω ω ζ ω 1, n n n i the ditance from the complex pole to the origin of the -plane, auming ζ < 1
Second Order Step Repone We now viualize econd order ytem repone to unit tep input for G = ω = 1 a ζ varie GDCωn Note, Y() H() U() + ζω + DC = = ( ) n ωn n 1 ( t ) τ yt () = G 1 in DC e ω, dt φ 1 ζ where φ = 1 1 ζ tan for 0 < ζ < 1 ζ
Second Order Step Repone In the plot that follow for ζ 1: %Overhoot ζ = 100 exp ζπ 1 ζ %OS( 0) = 100%, %OS(.5) = 16.3%, %OS(.707) = 4.3%, %OS( 1) = 0% Second_Order_Repone.m on the cla web ite will be ued to generate pole zero plot and tep repone
ζ = 0, marginally table (undamped) y. θ = ζ = ζ 1 co 90, valid only for 1 ω ω ζ ω [ ] d n 1 = n = 1 rad θ T [ ] 6.3 ec ω = 1 π T T = 63 6.3 [ rad/ec]
ζ = 0.5, under-damped (table) y. θ = ζ = 1 co 60 θ
ζ = 0.707, under-damped damped (table) y. θ = = 1 co ζ 45 θ
ζ = θ 1, critically damped (table) 1 = co ζ = 0
ζ = 1.5, over-damped (table) y.
The Frequency Repone We now conider the teady-tate tate reponeofa of general ytem to a inuoidal input of the form: ω ut = in( ωt) U = + ω The ytem output can be expreed uing it TF a: ω Y() = H() U() = H() + ω ω = H () ( + jω )( jω )
The Frequency Repone Performing PFE, we have: Y * C C = + jω + jω + term aociated with the table pole of H The reidue C can be evaluated a hown in p, g C = jω Y Chapter 7, e.g.: () = j = ω ω H () = jω + = jω H jω j
The Frequency Repone jθ ω Dfi Define H jω H ω H ω = M ω e, where M ω and θ ω are real function repreenting the magnitude and phae angle of H jω repectively; then C * H j M e ω ω = = j j j θ ω Since the pole of H are aumed table, the teady-tatetate repone of thee pole i zero implying Y SS () jθω e j M ω M ω e = j jω j + jω θω
The Frequency Repone jθ ( ω) jθ( ω M ω ) e e YSS () = j ( j ω ) ( + j ω ) y ( t) = M SS e e j j ωt+ θ ω j ωt+ θ ω ( ω) = M ω ω + θ ω in ( t ) H( jω) = M ( ω) θ( ω) i known a the ytem' frequency repone function (or frequency repone)
The Frequency Repone Uing a imilar analyi, it can be hownthat input of the form: ut = Ain( ωt+ φ) and Aco( ωt+ φ) produce teady-tatetate output of the form: y () t = AM ω in ωt+ φ+ θ ω and SS AM ( t + + ) ω co ω φ θ ω, repectively
The Frequency Repone Thi implie that what come out of a linear ytem i imply a caled and hifted verion of what i input; If ut = A + Aco( ω nt+ φ ), then 0 n 0 n= 1 = DC Component + Fundamental frequency 0 Higher order harmonic n ω and ( φ ) y () t = A M 0 + SS 0 A nm ω 0n co ω 0nt+ φ n + θ ω 0n n= 1 Thi i reaon why Fourer Serie are commonly ued to analyze ytem and ignal (not cover in thi cla)!
Find A Frequency Repone Example y SS () t when the table ytem defined by 4 + 8 H () = i driven by the input + + 5 π u () t = 5in(4 t + ) 3 Evaluating the TF at the input frequency yield ild M ω = H () = =1.315 = jω= j4 4 + 8 + + 5 = jω= j4 >> ab j + j + j + ((4* *4 8)/(( *4) * *4 5))
and θ ω A Frequency Repone Example = H () = 4 + 8 + + 5 () = j ω= j 4 = jω= j4 = 1.4056 1 [rad] >> angle j + j + j + ((4* *4 8)/(( *4) * *4 5)) π y () 5 in 4 SS t = M ω t+ + θ ω 3 = 6.576in 4 t 0.3585
A Frequency Repone Example We can predict how thi ytem will pa any input frequency by plotting M ω and θ ω uing the Matlab bode command a hown: >>NUM=[4 8];DEN=[1 5];SYS=tf(NUM,DEN);bode(SYS) Note, a Bode plot ha a log cale on the frequency axi and a magnitude expreed in db, i.e., X [ db] = 0log 10 X 1 0 db, 0.707 3 db, 1.414 + 3 db
The Magnitude of the Frequency q yreponep at ω = 4 [rad/] i.3 [db] 1.3 a hown from TF The Magnitude of the Frequency Repone ( 4.1/0) ω = =1.6 at 0 [rad/] i 4.1 [db] 10 Compare thi to the teady-tate value of the ytem' unit tep repone (ee next lide) The Phae Angle of the Frequency Repone at ω = 4 [rad/] i -80 or -1.406 [rad], a indicated by from TF
By FVT from Chapter 7: y = lim y ( t) = lim Y USS U U t 0 H () Since YU =, 4 + 8 yuss = lim H = lim = 1.6 0 0 + + 5 >>NUM=[4 8];DEN=[1 5];SYS=tf(NUM,DEN);tep(SYS)
Remember, a hown in Chapter 7: 1 dy () () { } U t ht = L H = dt i.e., a ytem' impule repone i equal to the time derivative of it unit tep repone! >>NUM=[4 8];DEN=[1 5]; NUM [4 8];DEN [1 5]; >>SYS=tf(NUM,DEN);impule(SYS) >> H=[diff(Y)./diff(T);0];plot(T,H,'r:')
Fourier Serie Example (Not Teted) Aume the prior ytem H () i excited by a quare wave of period T defined for all integer p a: 0 1, for ( 1 ) < ( + 1 ) p T t p T 4 4 US () t = 1, for ( p+ 1 ) T ( 3 ) 0 t < p+ T 4 4 0 The Fourier Serie (FS) approx. of U t S 0 0 ()igi given by: N 4 nπ U ( t) = lim in co nω t where ω π = SFS 0 0 N n = 1 nπ T Approximate the repone of the prior ytem to U ( t) uing the firt twenty term of it FS (ee following graph) S 0
Fourier Serie Example (Not Teted) The red line are frequency component aociated with the FS approx. of U () t S
The Impedance Concept Impedance i a frequency domain concept that can be expreed for variou type of ytem a the TF (or ratio) of flow to force For electrical lytem, impedance i defined: d E () Z () I( ) 1 ZR = R, ZL = L, ZC = C Impedance can be treated t a a generalized reitance, even though they are function of
Electrical Impedance Ex.8.: RC Circuit Replacing circuit element with their equivalent impedance permit application of technique ued for pure reitive circuit, e.g., voltage-divider rule Combining parallel impedance at E (): Z R R = C = R 1 1 RC + + C o
Electrical Impedance Ex.8.: RC Circuit Uing voltage divider at E (): Z () E = E o () i () L + R1+ Z() 1 LC = L + RR 1 C R1 + R + LCR + LCR o ( L + R R C ) ( R + R ) E () 1 1 e + e + e = e t i 1 () o o o i If =, = LCR LCR LC ( R ) e AU t e i oss R + R A 1
Electrical Impedance Ex.8.4: Op Amp Circuit E it o() Find the TF for the op-amp circuit, H () = E () i
Electrical Impedance Ex.8.4: Op Amp Circuit R 1 + R Letting K, we can repeatedly apply R 1 the voltage-divider id rule to obtain: R 1 E E E 1 A () = o () = o () R1+ R K Z4 () EB() = EC() Z () + Z () 4 Due to the op-amp' virtual hort, E = E Z () + Z () 4 EC () = Eo () KZ4() A B
Electrical Impedance Ex.8.4: Op Amp Circuit KCL at Node C yield: 1 1 Z () Z () [ E () E () ] + [ E () E () ] C i C o 1 3 1 + EC () = 0 Z() + Z4() Eliminating E and implifying yield H = C 1 1 E o () E () KZ () Z () 3 4 Z () Z () + 1 K Z Z + Z Z + Z + Z [ ] 4 3 1 4 i
Sallen Key Low pa Filter When Z () = R, Z () = R, Z () = C, Z () = C, 1 A B 3 A 4 B the circuit i referred to a a Sallen-Key Low-pa filter
Sallen Key Low pa Filter Rf + Ri If RA = RB = R, CA = CB = C, with K = R i K ( RC ) Vo () H = ; V i () 3 K 1 + + RC RC the natural lfrequnecy of thi low-pa filter can be tuned by electing RC and damping by adjuting K
Frequency Repone of S K Low pa Filter For K = RC = 1 1 H =, a critically damped + + 1 ytem (a hown previouly) with ζ = ω n = 1
Mechanical Impedance For mechanical ytem, impedance i defined: F () Z () v () K ZB () = B, ZM () = M, ZK () = The mechanical impedance i a function of the frequency of the applied force and can vary greatly over frequency, e.g., for a pendulum (ee pg. 116): ( 1 ) F () G ω DC n Z() = ML = v () g + ζω + ω ( B ) + + ML L n n
Mechanical Impedance At reonant (natural) frequencie, the mechanical impedance will be lower meaning le force/power i needed to caue a tructure to move at a given velocity The implet example of thi i when a child puhe another on a wing; for the greatet wing amplitude the frequency of the puhe mut be more-or-le at the reonant (or natural) frequency of the ytem Note, ω n = g for the imple pendulum ytem L
Impedance (freq. repone) of the pendulum ytem with gmbl,,, et to 10 ω n = 1 [rad/] Note, the phae repone i not relevant to thi application
Quetion?