376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD. D(s) = we get the compensated system with :

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1 376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Therefore by applying the lead compenator with ome gain adjutment : D() = we get the compenated ytem with : PM =65, ω c = 22 rad/ec, o that ω BW & 25 rad/ec. The Bode plot with deigned compenator i : B ode D iagram 6 G m = db (at rad/ec),pm = deg.(at rad/ec) 4 2 Phae (deg);m agnitude (db ) Frequency (rad/ec) 45. For the ytem hown in Fig. 6.14, uppoe that G() = 5 ( +1)(/5+1). Deign a lead compenation D() with unity DC gain o that PM 4 uing Bode plot ketche, then verify and reþne your deign uing MATLAB. What i the approximate bandwidth of the ytem? Start with a lead compenator deign with : D() = T+1 αt+1 which ha unity DC gain with α < 1.

2 377 Figure 6.14: Control ytem for Problem 45 TheBodeplotofthegivenytemi: B ode D iagram 1 G m = db (at rad/ec),pm = deg.(at rad/ec) 5 Phae (deg);m agnitude (db ) Frequency (rad/ec) Since PM =3.9, let add phae lead 6.FromFig.6.53, 1 ' 2 = chooe α =.5 α To apply maximum phae lead at ω = 1 rad/ec, ω = 1 αt =1= 1 T =2.2, 1 αt =45 Therefore by applying the lead compenator : D() =

3 378 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD we get the compenated ytem with : PM =4, ω c =2.5 The Bode plot with deigned compenator i : B ode D iagram 1 G m =24.13 db (at rad/ec),pm =4.244 deg.(at rad/ec) 5 Phae (deg);m agnitude (db ) Frequency (rad/ec) From Fig. 6.5, we ee that ω BW ' 2 ω c ' 5 rad/ec. 46. Derive the tranfer function from T d to θ for the ytem in Fig Then apply the Final Value Theorem (auming T d = contant) to determine whether θ( ) i nonzero for the following two cae: (a) When D() ha no integral term: lim D() =contant; (b) When D() ha an integral term: D() = D (), where lim D () =contant. The tranfer function from T d to θ : where T d () = T d /..9 Θ() T d () = D()

4 379 (a) Uing the Þnal value theorem : θ( ) = lim θ(t) = limθ(t) = lim t (+2)+1.8D() 2 (+2) T d T d = lim D() = T d contant 6= Therefore, there will be a teady tate error in θ for a contant T d input if there i no integral term in D(). (b) θ( ) = lim θ(t) = limθ(t) =lim t = 1.8lim D () = (+2)+1.8D () 3 (+2) T d So when D() contain an integral term, a contant T d input will reult in a zero teady tate error in θ. 47. The inverted pendulum ha a tranfer function given by Eq. (2.35), which i imiliar to G() = (a) Deign a lead compenator to achieve a PM of 3 uing Bode plot ketche, then verify and reþne your deign uing MATLAB. (b) Sketch a root locu and correlate it with the Bode plot of the ytem. (c) Could you obtain the frequency repone of thi ytem experimentally? (a) Deign the lead compenator : D() =K T+1 αt+1 uch that the compenated ytem ha PM ' 3 & ω c ' 1 rad/ec. (Actually, the bandwidth or peed of repone wa not peciþed, o any croover frequency would atify the problem tatement.) α = 1 in(3 ) 1+in(3 ) =.32 To apply maximum phae lead at ω =1rad/ec, ω = 1 αt =1= 1 T =.57, 1 αt =1.77

5 38 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Therefore by applying the lead compenator : D() =K By adjuting the gain K o that the croover frequency i around 1 rad/ec, K =1.13 reult in : PM =3.8 TheBodeplotofcompenatedytemi: B ode D iagram 5 G m = db (at rad/ec),pm =3.848 deg.(at rad/ec) -5-1 Phae (deg);m agnitude (db ) Frequency (rad/ec) (b) Root Locu of the compenated ytem i :

6 381 and conþrm that the ytem yield all table root with reaonable damping. However, it would be a better deign if the gain wa raied ome in order to increae the peed of repone of the low real root. A mall decreae in the damping of the complex root will reult. (c) No, becaue the inuoid input will caue the ytem to blow up becaue the open loop ytem i untable. In fact, the ytem will blow up even without the inuoid applied. Or, a better decription would be that the pendulum will fall over until it hit the table. 48. The open-loop tranfer function of a unity feedback ytem i G() = K (/5+1)(/5 + 1). (a) Deign a lag compenator for G() uing Bode plot ketche o that the cloed-loop ytem atiþe the following peciþcation: i. The teady-tate error to a unit ramp reference input i le than.1. ii. PM =4 (b) Verify and reþne your deign uing MATLAB. Let deign the lag compenator : D() = T+1 αt+1, α > 1 From the Þrt peciþcation, Steady-tate error to unit ramp = lim D()G() 1 1+D()G() <.1 = 1 K <.1 = Chooe K = 15 Uncompenated, the croover frequency with K = 15 i too high for a good PM. With ome trial and error, we Þnd that the lag compenator, D() = will lower the croover frequency to ω c ' 4.46 rad/ec where the PM =

7 382 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD 4.7. B ode D iagram 15 G m = db (at rad/ec),pm =4.71 deg.(at rad/ec) 1 5 Phae (deg);m agnitude (db ) Frequency (rad/ec) 49. The open-loop tranfer function of a unity feedback ytem i K G() = (/5+1)(/2 + 1). (a) Deign a lead compenator for G() uing Bode plot ketche o that the cloed-loop ytem atiþe the following peciþcation: i. The teady-tate error to a unit ramp reference input i le than.1. ii. For the dominant cloed-loop pole the damping ratio ζ.4. (b) Verify and reþne your deign uing MATLAB including a direct computation of the damping of the dominant cloed-loop pole. Let deign the lead compenator : D() = T+1 αt+1, α < 1 From the Þrt peciþcation, Steady-tate error to unit ramp = lim D()G() 1+D()G() = 1 K <.1 = Chooe K = <.1

8 383 From the approximation ζ ' PM, the econd peciþcation implie PM 1 4. After trial and error, we Þnd that the compenator, D() = reult in a PM =42.5 and a croover frequency ω c ' 51.2 rad/ec a hown by the margin output: B ode D iagram 8 G m = db (at rad/ec),pm = deg.(at rad/ec) Phae (deg);m agnitude (db ) Frequency (rad/ec) and the ue of damp veriþe the damping to be ζ =.42 for the complex cloed-loop root which exceed the requirement. 5. A DC motor with negligible armature inductance i to be ued in a poition control ytem. It open-loop tranfer function i given by G() = 5 (/5+1). (a) Deign a compenator for the motor uing Bode plot ketche o that the cloed-loop ytem atiþe the following peciþcation: i. The teady-tate error to a unit ramp input i le than 1/2. ii. The unit tep repone ha an overhoot of le than 2%. iii. The bandwidth of the compenated ytem i no le than that of the uncompenated ytem.

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