2.010 Fall 2000 Solution of Homework Assignment 1

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1 2. Fall 2 Solution of Homework Assignment. Compact Disk Player. This is essentially a reprise of Problems and 2 from the Fall Homework Assignment 7. t is included here to encourage you to review your 2.3 Notes. A model of the system is shown in Fig.. The motor current i is generated by a controller, and the torque τ on the motor rotor is proportional to the current (τ = K T i). The torque coefficient K T is obtained from an experiment in which a torque of 2.25e-3 N-m was produced by a current of. amps. Therefore K T =2.25e-2 N-m/amp. n this experiment the steady-state rotational speed ω ss was found to be 4 revolutions per second (25.3 rad/sec). Since τ = B r ω ss in the steady state, this yields B r = τ = 2.25e-3 =8.95 N-m-sec/rad () ω ss 25.3 r d i τ = K T i τ B r m D Figure : Compact Disk Player. The statement that the same steady-state speed was reached with or without the disk in place implies that the damping coefficicient in Eq.() applies to both cases which means that the air drag on the disk is negligible in comparison with the motor rotor friction. The moment of inertia of the motor rotor is given as 9e-6 kg-m 2. The moment of inertia of the disk (See Problem, 2.3 Fall 999 Homework Solution) is CD = 2 m(r2 + r 2 )= m 8 (D2 + d 2 )=.28 8 ( )=5.2e-5 kg-m 2 (2) so the total rotational inertia, when the disk is mounted is total = 9e e-5 = 6.2e-5 kg-m 2 (3)

2 To answer the first two questions about the open loop system it is necessary to have a mathematical model of the system. When the motor rotor and the disk rotate at speed ω the torques acting to accelerate the total rotational inertia are the driving torque τ = K T i and the retarding friction torque B r ω, so the equation of motion is K T i B r ω = total dω or dω + B r total ω = K T total i (4) (a) n the steady state d/ soω ss is directly proportional to the input current. t was given that ω ss was 4 rev/sec for i =. amp, so for i =.2 amp the steady-state disk speed will be 8 rev/sec or 5.26 rad/sec. From Eq.(4), it can be seen that the decay time-constant of the model is τ d = total B r = 6.2e e-5 =.673secs (5) Now a decaying exponential decays to % of its initial value in 4.6 decay timeconstants so the time T 99% for the CD to reach 99% of the steady-state speed is T 99% = (4.6)(.673) = 3.9 secs. (b) From Eq.(4) one sees that for fixed input current i the product of the friction parameter B r and the steady-state speed ω ss must be constant: (B r ω ss ) original =(B r ω ss ) new (6) f (B r ) new =.(B r ) original then it is necessary that (ω ss ) new =. (ω ss) original =.99(ω ss ) original (7) A ten percent increase in motor friction causes a 9.% decrease in steady-state speed. The open-loop system is converted to a closed-loop system by sensing the speed ω and controlling the input current to be proportional to the difference between the sensed speed and an input reference speed r i = G(r ω) (8) where G is the controller gain with the dimensions of current per radian/sec. When this is inserted in Eq.(4) the equation of motion for the closed-loop system is dω + B r + GK T ω = GK T r (9) total total The closed-loop equation in (9) is similar in form to the open-loop equation in (4), but in the closed-loop system a wide variety of dynamic behaviors can easily be obtained by varying the gain parameter G. The following examples show that the decay time-constant can be shortened and the sensitivity of the operating speed on the friction can be decreased with an appropriate choice of G. 2

3 (c) With G =. amps per rad/sec, the closed-loop decay time-constant is τ d = total B r + GK T = 6.2e e-5 + (.)(2.25e-2 =.94secs () which is 28.4 % of the open-loop decay time-constant of Eq.(5). The time to reach 99 % of the steady-state speed is T 99% = (4.6)(.94) =.88 sec. (d) According to Eq.(9) the closed-loop steady state speed is ω ss = GK T B r + GK T r = r () + Br GK T f B r is increased by % then the ratio of the new steady-state speed to the original steady-state speed is (ω ss ) new (ω ss ) original = + Br GK T +. Br GK T =.972=.28 (2) which implies that in the closed-loop system, an increase of friction of % causes a decrease of only 2.8 % in the steady-state speed. Compare with 9. % decrease in the open-loop system. 2. Mechanical Transmission Dynamics. This problem uses the system of Problem 4 from Fall Homework Assignment 7 to review the concepts of statespace formulations, transfer functions, poles and zeros, and Bode plots that were developed in 2.3 and 2.4 to encourage you to review your Notes for these courses. The most difficult step is the formulation of the mathematical model. This is the engineer s most challenging task. Once a good model is available, programs such as MATLAB can produce the detailed formulas and graphs which provide the basis for engineering design decisions. τ m ωm ω l θ m Coupler θ l Motor Load Figure 2: Mechanical Power Transmission. 3

4 Study the motion. The angular positions of the motor and load are θ m and θ l, respectively, and the angular velocities are ω m = dθ m / and ω l = dθ l /. The twist angle of the coupler is θ = θ m θ l, and its rate of change is d θ/ = ω m ω l. Constitutive equations. The coupler torque is K c θ+b c d θ/. The friction torques on the motor and the load are B m ω m and B l ω l, respectively, and both the motor rotor and the load rotor must satisfy the angular momentum equationστ = dω/. Study the forces (torques). The primary difficulty here is getting the signs correct in the equations of motion. When θ and d θ/ are both positive the torque at the motor end of the coupler acting on the coupler is in the same sense as positive θ m. The reaction to this acting on the motor rotor opposes rotation in the direction of positive θ m. The motor friction also opposes rotation in that direction, so the angular momentum equation for the motor rotor is dω m τ m K c θ B c (ω m ω l ) B m ω m = m (3) Similar reasoning indicates that the torque applied to the load by the coupler acts in the positive sense of θ l, while the load friction opposes rotation in that direction. The angular momentum equation for the load is dω l K c θ + B c (ω m ω l ) B l ω l = l (4) The angular momentum equations (3) and 4) are two equations for the three unknowns ω m, ω l and θ. A third equation is provided by the kinematic compatibility relation d θ = ω m ω l (5) (a) Note that the given data for this problem describes a symmetric system. Consequentially, the notation can be simplified. n the following, we use for m = l, and B for B m = B l. The three preceding equations can be cast in the standard state-space form d x = Ax + Bu y = Cx + Du (6) where and x = ω m ω l θ A = B+Bc B c B = 4 B c B+Bc Kc K c (7) C = { } D = (8)

5 with input u and output y where u = {τ m (t)} and y = {ω l (t)} (9) For the given data, the numerical values of the coefficients in the matrix A are A = (b) The time-domain equations in (6) can be transformed to s-domain equations by substituting s for the operation d/ and replacing ω(t) by ω(s), etc. The transfer function from τ(s) toω l (s) can then be obtained by algebraic maipulation of the equations to eliminate ω m (s) and θ(s). Perhaps the simplest way to do the algebra is to have the following session with MATLAB. To get started, type one of these: helpwin, helpdesk, or demo. For product information, visit A = [ ; ; - ]; B = [ 2 ; ; ]; C = [ ]; sys = ss(a,b,c,); tfsys = tf(sys) Transfer function: 8 s e s^ s^ s (2) So the transfer function is ω l (s) τ(s) = 8s s s s (2) The poles and zeros of this transfer function are most simply determined by returning to MATLAB and typing [p,z]= pzmap(sys) which produces p = -2. 5

6 i i z = -6.22e+2 A plot of these poles and zeros is produced by entering simply pzmap(sys). The plot is shown in Fig Pole-zero map mag Axis Real Axis Figure 3: Pole-Zero Map of Mechanical Power Transmission. (c) The frequency response to an excitation of the form τ(t) =A τm exp{jωt} is ω l (t) =A ωl exp{jωt}, where the complex amplitude ratio is obtained by substituting s = jω in the above transfer function A ωl A τm = j8ω j5.6ω 4.8Ω 2 jω 3 (22) The Bode plot which displays the magnitude in db and the phase in degrees is obtained from MATLAB by typing the command bode(sys). The plot produced is shown in Fig.4. 6

7 Bode Diagrams 5 Phase (deg); Magnitude (db) Frequency (rad/sec) Figure 4: Bode Plot for Mechanical Power Transmission. 3. Closed-Loop Control with Transmission Dynamics. When all the B- parameters in the transmission of Problem 2 are set equal to zero the A matrix in (6) reduces to Kc A = K c (23) (a) When the equation of motion with this matrix A is transformed into the s- domain the result is a set of three simultaneous algebraic equations s Kc ω m (s) K s c ω l (s) = s θ(s) which, when solved for ω l (s), produces the desired transfer function τ(s) (24) ω l (s) τ m (s) = K c s(s 2 +2 Kc ) (25) The poles are s = and s 2,3 = ±j 2 Kc = ±j22.3 rad/sec. (b) When there is no excitation, the time-domain equation of motion dx/ = Ax 7

8 is ω m Kc ω m d ω l = K c ω l (26) θ θ The free vibrations of the system can be assumed to have the form ω m a x = ω l = a 2 exp{λt} = vexp{λt} (27) θ a 3 Substitution of (27) into (26) yields the eigenvalue problem Av = λv Kc a a K c a 2 = λ a 2 (28) a 3 a 3 which can be solved by MATLAB, or analytically as follows. Move the left side of (28) to the right to get K λ c a λ Kc a 2 = (29) λ The characteristic equation for the eigenvalue λ is obtained by setting the determinant of the matrix on the left equal to zero =λ 3 +2λ K c = (3) The roots of the characteristic equation are λ =, and λ 2,3 = ±jω o (3) where ω o = 2 Kc = 2.24e-2 = 22.3 rad/sec (32) 5e-5 The mode shapes are obtained by back-substituting these values of λ into (29) and solving for the pattern of a i s. Because the matrix is singular, only the relative sizes of the a i s can be found. f we take the arbitrary magnitude unity for ω m we can solve for ω l and θ for each value of λ and get λ =,v = λ 2 = jω o, v 2 = λ 3 = jω o, v 3 = 8 2 jω o a 3 2 jω o (33)

9 The first mode describes uniform rotation. The motor rotor and the load have the same constant angular velocity. The second and third mode describe oscillating twist. The two independent motions x 2 = exp{jω o t} exp{jω o t} x 3 = exp{ jω o t} exp{ jω o t} (34) 2 jω o exp{jω o t} 2 jω o exp{ jω o t} are difficult to interpret because the quantities are complex. An equivalent pair of independent motions, expressed in real quantities, is cos ω o t sin ω o t x 4 = 2 (x 2 + x 3 )= cos ω o t x 5 = 2j (x 2 x 3 )= sin ω o t 2 ω o sin ω o t 2 ω o cos ω o t (35) The motions x 4 and x 5 are clearly the same twist mode with a phase difference of 9 degrees. The pole at s = is associated with the uniform rotation mode, and the pair of poles at ±jω o are associated with the twist mode. (c) A sinusoidally varying motor torque can be represented as τ m (t) =A τm exp{jωt} and the steady-state response of the load speed can be represented as A ωl exp{jωt}. The frequency response function can be obtained by setting s = jω in the corresponding transfer function (25) ( ) A ωl (Ω) A τm (Ω) = ωl (s) τ m (s) s=jω = 2 jω 248 (36) 496 Ω 2 When sketching Bode plots by hand, it is convenient to break up the frequency response into a number of simple factors and sketch the Bode plots of each factor separately, and then add the plots graphically to obtain the fiinal result. n the present case there are two factors, as indicated in (36). The first factor (a) has a magnitude that slopes downward at 2 db per decade and has the value of 86 db re one rad/sec per N-m at Ω=rad/sec. The phase of the first factor is -9 degrees. The second factor (b) has a low-frequency asymptote with constant magnitude.5 (-6 db re unity), and a high frequency asymptote that slopes downward at 4 db per decade. The two asymptotes intersect at the break-point frequency Ω = 22.7 rad/sec. There is an undamped resonance at the break point. The phase of the second factor is zero for frequencies smaller than the break-point frequency, and -8 degrees for frequencies greater than the break-point frequency. The Bode plots for the two factors are sketched as dotted lines in Fig.5 and their sum is sketched as a solid line. 9

10 2 Phase db re. rad/sec per N-m db/decade (a) (b). Ω -6 db/decade -4 db/decade. (a) Ω (b) Figure 5: Bode Plot of Load Speed Response. (d) Closed-loop proportional control is now added to the undamped transmission model. When τ m = G(r ω l ) the equations of motion become ω m G Kc ω m d ω l = K c ω l + θ θ G r (37) With G =.e-5 N-m per rad/sec, and r(t) a step function of magnitude rad/sec, the matrices which MATLAB needs to perform the step command

11 are A fb = B fb = 2 C fb = { } (38) When these matrices have been entered into MATLAB, the first three seconds of the response of ω l (t) to a rad/sec step of the reference variable r(t) is obtained by entering the command sysfb = ss(afb, Bfb, Cfb, ), followed by step(sysfb,3). The result is the plot shown in Fig.6. Step Response Amplitude Time (sec.) Figure 6: Step Response of Load Speed in System with Feedback. The load speed increases from zero toward the desired speed of rad/sec. The ripple superposed on the smooth rise is due to the twist mode with a natural frequency ω o =22.7 rad/sec (3.6 cycles/sec). The fact that the ripple is due to the twist mode is emphasized when the step responses of both the load and the motor rotor are compared in Fig.7. The motor rotor starts first, but after a half cycle the load speed overtakes the motor rotor speed, and thereafter the two speeds alternate in leading and lagging, as the coupler twists back and forth while the average speed increases smoothly. t may be noted that the amplitude of the ripple appears to increase slowly with time. This indicates that closing the loop with proportional control has caused a stable open-loop system to become unstable.

12 Step Response Amplitude Time (sec.) Figure 7: Step Response of Motor Rotor Speed and Load Speed. 2

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