1 Routh Array: 15 points
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1 EE C28 / ME34 Problem Set 3 Solution Fall 2 Routh Array: 5 point Conider the ytem below, with D() k(+), w(t), G() +2, and H y() (+). Find the cloed loop tranfer function Y () R(), and range of k for all cloed-loop pole in the left half plane. We can ue the tandard feedback formula to find the cloed loop tranfer function: Y () R() D()G() + D()G()H y () + k(+) k(+) (+) 2k( + )( + ) 2( + 2)( + ) + k( + )( ) 2k( ) ( ) + (k 3 + k 2 + 2k) + (k 2 + k + 2k) 2k( ) (2 + k) 3 + (24 + k) 2 + (4 + 3k) + 2k Now develop the Routh array. Grey cell come from the polynomial, white cell are calculated. cell (a): cell (b): k 4 + 3k k 2k (a) 96+2k+3k2 24+k (b) 2k (2 + k) (4 + 3k) (24 + k) 2k 24 + k ( (4 + 3k)(24 + k) 2k(2 + k) 24 + k (24 + k) ) 2k 2k 96+2k+3k 2 24+k 96+2k+3k 2 24+k k + 3k k (the white cell with zero in them are eay). Recall that all pole in LHP i equivalent to no ign change in firt column of Routh array. So we want either everything in the firt column poitive or everything in the firt column negative. Break the problem into two cae. trying to make everything poitive: 2 + k > k > 2 (*) 24 + k > k > 24 (*) The fractional term i eaier becaue we already know that the denominator mut be poitive! k + 3k k > and denominator poitive k + 3k 2 >
2 Ue the quadratic equation to find the root at k 2± { 2.245,.3984}. Tet point on either ide of the root to find that the quadratic expreion i poitive if x < or x > k + 3k 2 > k < or k >.3984 (*) 2k > k > (*) All four of the tarred condition have to be true; the lat one dominate all of them. k > i the final condition. trying to make everything negative: 2 + k < k < 2 (**) 24 + k < k < 24 (**) Since the denominator i negative, the numerator of the fractional term mut be poitive! k + 3k k < and denominator negative k + 3k 2 > k + 3k 2 > k < or k >.3984 (**) 2k < k < (**) All of the (**) tarred condition mut be true. Thi time it one of the quadratic condition that dominate the other. k < i the reulting condition. Either of the two final condition will work to make the ytem have all left-hand-plane pole. The final range i: 2 Linearization: 2 point k (, 2.245) (, ) For the ytem: ẋ ẋ 2 ẋ 3 ẋ 4 f f 2 f 3 f 4 x 2 β(x 3 + kx ) u α x 2 4 βx 2 () Linearize the ytem about x x 2 x 3, x 4, u, and expre in tate pace form: δx Aδx + Bδu. (β, k and α are contant.) The A matrix i compoed of the partial derivative of the ytem dynamic for each tate variable, evaluated 2
3 at the equilibrium A f x f x 2 f x 3 f x 4 x x 2 x 3 x 4 x x 2 x 3 x 4 x x 2 x 3 x 4 βk β 2αu x 3 4 2βx βk β 2αu (evaluated at equilibrium) Similarly the B matrix i from the partial w.r.t. the input B f u u u u α α x 2 4 (evaluated at equilibrium) Therefore the linearized ytem i: δx βk β 3 Steady State Error: 2 point δx + α δu For the ytem below, let D(), G() (+) (+2), H y() + 4, and w(t). a) What i the ytem type and appropriate tatic error contant? Thi problem i intereting becaue there i a enor H y which ditort the output. The problem hould have pecified which error to conider, either the error a meaured between the reference and the enor e(t) r(t) y (t) (referred to a e a (t) in Nie), or the error between the reference and the actual ytem output e y (t) r(t) y(t). (For thi problem, either anwer will be accepted.) Uually, a control ytem deigner would be concerned with e y (t). For example, if the enor i an airplane altimeter, the autopilot hould care about the true height above ground, not the enor output. Here we will calculate both e(t) and e y (t) and how why they are different type ytem. Note that the ytem i table, o that the final value theorem can be applied. 3
4 3. Error between reference and enor r(t) y (t) Since w(t) we know that we re intereted in the teady tate error (expreed at e(t)) a it relate to the reference input (r(t)). So we look for the tranfer function E() R(). Finding thi i very much like deriving the uual feedback formula. E() R() Y () Y () H y ()Y () Y () D()G()E() E() R() H y ()D()G()E() E() R() + H y ()D()G() + ( + 4) (+) (+2) ( + 2) ( + 2) + ( + 4)( + ) Now we can ue the Final Value Theorem to evaluate the repone of E() to a unit tep input: e tep ( ) lim R() E() R() lim ( + 2) ( + 2) + ( + 4)( + ) ( + 2) ( + 2) + ( + 4)( + ) 4 So we know the ytem i not Type. Next, we ue FVT to find the repone of E() to a unit ramp input: e ramp ( ) lim R() E() R() lim ( + 2) 2 ( + 2) + ( + 4)( + ) ( + 2) ( + 2) + ( + 4)( + ) Contant error from a ramp input mean that the ytem i Type. The tatic error contant for a Type ytem i K v (velocity error contant), and the relationhip between the teady tate error and the error contant i (ee table 7.2 on page 353 on Nie): e ramp ( ) K v o K v 2. b) What input waveform r(t) would yield a contant error? (e.g. tep, ramp, parabola, or?) c) What i the teady tate error for a unit input of r(t) from b)? Both of thee anwer fall out of our previou work. A ramp input give a contant error; and a unit input ramp will lead to teady tate error of Error reference and output r(t) y(t) Here we re intereted in the teady tate ouput error expreed a e y (t) r(t) y(t)) So we look for the tranfer function Ey() R(). Finding thi i very much like deriving the uual feedback formula. 4
5 E y () R() Y () Y () D()G()E() E y () R() D()G()E() D()G() R()( + H y ()D()G() ) E y () R() + D()G()(H y() ) + H y ()D()G() ( + 2) + ( + )( + 3) ( + 2) + ( + 4)( + ) Now we can ue the Final Value Theorem to evaluate the repone of E y () to a unit tep input: e ytep ( ) lim R() E y() R() lim ( + 2) + ( + )( + 3) ( + 2) + ( + 4)( + ) ( + 2) + ( + )( + 3) ( + 2) + ( + 4)( + ) 3 4 Since e ytep ( ) 3 4, the ytem type for output error i type. The error contant for a type ytem i K p, and ince e ytep ( ) 3 4 +K p, then K p 3. b) a unit tep give a contant error, with teady tate error of 3/ Comparion of e( ) e a ( ) and e y ( ) The teady tate output y( ) can be found from Y () R() D()G() + H y ()D()G() ( + ) ( + 2) + ( + 4)( + ) Now we can ue the Final Value Theorem to find y( ) for a unit tep input: y tep ( ) lim R() Y () R() lim ( + ) ( + 2) + ( + 4)( + ) ( + ) ( + 2) + ( + 4)( + ) 4 Since H y () (+4), then y (t) ẏ(t)+4y(t). Since we are in teady tate, with a tep input, ẏ( ). Thu y ( ) 4y( ). So everything work out. The enor error i e( ) r( ) y ( ) 4y( )
6 The output error i e ytep ( ) r( ) y( ) y( ) Since G() ha a pole at, it allow y( ) to be non-zero even with e( ). The teady tate error in y(t) hould not be too urpriing, ince we are in effect comparing the reference to 4y(t), and hence the output i only 4 what it would be with unity feedback. 4 PID Control: 25 point For the ytem below, let D() k p + k I + k D, H y (), and G() 2 +2ζω n+ω. n 2 a) With r(t), determine the ytem type (,, 2,...) and error contant with repect to tep diturbance input W (). Finding the cloed-loop tranfer function E() W () E() W () G() + D()G() 2 +2ζω n+ω 2 n by eq 7.6: + kp+ k I +k D 2 +2ζω n+ωn ζω n + ωn 2 + k p + k I + k D ( 2 + 2ζω n + ωn) 2 + k p + k I + k D 2 Q() Auming k I, the teady tate error due to tep diturbance by FVT i: lim Q() The teady tate error due to ramp diturbance i: lim Q() 2 k I Therefore, becaue the ytem can repond to a tep diturbance with error, but ha contant error from a ramp diturbance, it i Type. From Table 7.2, the error contant are K p,k v k I,K a. b) With w(t), determine the ytem type (,, 2,...) and error contant with repect to tep and ramp reference input R(). Since H y (), thi i a tandard Steady-State-Error problem, with a ytem tranfer function of D()G() k p + k I + k D 2 + 2ζω n + ωn 2 k p + k I + k D 2 ( 2 + 2ζω n + ω 2 n) Auming all contant are nonzero, the type of the ytem i. From Table 7.2, the error contant are K p, K v lim D()G() k I ω, K n 2 a. 6
7 5 Error and Diturbance: 2 point For the ytem below, let D() k(+2) +3, G() k2 (+4), and H y(). Find the value of k and k 2 uch that: a) the teady-tate error component e(t) due to a unit tep diturbance w(t) i 3 ; We need the tranfer function E() W (). Notice that we aume R() while olving for thi TF. Solve the uual way: E() Y () Y () Y () G() ( W () + U() ) U() D()E() E() G() ( W () + D()E() ) E() + G()D()E() G()W () E() W () G() + G()D() k 2 (+4) + k2 (+4) k (+2) +3 k 2 ( + 3) ( + 4)( + 3) + k k 2 ( + 2) Now ue the Final Value Theorem to evaluate the repone to a unit tep diturbance: e tep,w ( ) lim k 2 ( + 3) ( + 4)( + 3) + k k 2 ( + 2) k 2 ( + 3) ( + 4)( + 3) + k k 2 ( + 2) Given our deired value, we can olve for k : k 3 2( 3 ) 5 3k 2 2k k 2 3 2k k 2 i unretricted. Note that the integrator in G() i in effect calculating a teady-tate offet to cancel w(t), which i why the k 2 gain i a free parameter for thi problem. b) the teady tate error e(t) due to a unit ramp input r(t) i 3. Find the tranfer function E() R(), auming W (). E() R() Y () Y () H y ()Y () Y () G()D()E() E() ( + H y ()G()D() ) R() E() R() H y ()G()D()E() E() R() + H y ()G()D() ( + 4)( + 3) ( + 4)( + 3) + k k 2 ( + 2) + k2 k (+2) (+4) +3 7
8 Now ue FVT to find the repone to a ramp input: Given our deired value we can olve for k k 2 : e ramp,r ( ) lim ( + 4)( + 3) 2 ( + 4)( + 3) + k k 2 ( + 2) ( + 4)( + 3) ( + 4)( + 3) + k k 2 ( + 2) 2 2k k 2 k k Any combination of k, k 2 that atifie thi condition will do. If we want to atify both (a) and (b) imultaneouly, we find: k k
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