MAE 142 Homework #2 (Design Project) SOLUTIONS. (a) The main body s kinematic relationship is: φ θ ψ. + C 3 (ψ) 0 + C 3 (ψ)c 1 (θ)

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1 MAE 42 Homework #2 (Design Project) SOLUTIONS. Top Dynamics (a) The main body s kinematic relationship is: ω b/a ω b/a + ω a /a + ω a /a ψâ 3 + θâ + â 3 ψˆb 3 + θâ + â 3 ψˆb 3 + C 3 (ψ) θâ ψ + C 3 (ψ)c (θ) â 3 θ + C 3 (ψ) + C 3 (ψ)c (θ) sin θ sin ψ cos ψ sin θ cos ψ sin ψ cos θ The momentum wheel s kinematic relationship is: θ ψ ω c/a ω c/b + ω b/a ω c/b + ω b/a + ω a /a + ω a /a γâ 3 + ψâ 3 + θâ + â 3 γĉ 3 + ψˆb 3 + θâ + â 3 γĉ 3 + C 3 (γ) ψˆb 3 + C 3 (γ)c 3 (ψ) γ + C 3 (γ) θâ + C 3 (γ)c 3 (ψ)c (θ) â 3 θ + C 3 (γ)c 3 (ψ) + C 3 (γ)c 3 (ψ)c (θ) ψ (cosγ sin ψ + sinγ cos ψ) sinθ + θ(cos γ cos ψ sin γ sin ψ) (cosγ cosψ sin γ sin ψ) sin θ + θ( sin γ cos ψ cos γ sin ψ) cosθ + ψ + γ (cosγ sin ψ + sinγ cos ψ) sinθ (cos γ cos ψ sin γ sin ψ) (cosγ cos ψ sin γ sin ψ) sin θ ( sin γ cos ψ cos γ sin ψ) cosθ (b) The Lagrangian is: θ ψ γ

2 L T U ( ) ω T 2 b/a J b ω b/a + ωc/aj T c ω c/a mgl cosθ [(J + J wt )( 2 2 sin θ 2 + θ 2 ) + J 3 ( ψ + cos θ) 2 + J w ( γ + ψ + ] cosθ) 2 mgl cos θ (c) Note that φ, ψ, and γ do not appear in the Lagrangian. Ergo, they are cyclic generalized coordinates. (d) Recall that Lagrange s method for each generalized coordinate is: d dt ( ) L q i L q i Q i There is only one non-cyclic generalized coordinate, the nutation angle: (J + J wt ) θ sinθ((j + J wt J 3 J w )) 2 cos θ (J 3 + J w ) ψ J w γ + mgl) (e) Recall that for ignorable coordinates (φ, ψ, and γ), the conjugate momenta is conserved: d dt ( ) L d ( ) T d q i dt q i dt (β i) β i const β φ (J 3 + J w J J wt ) cosθ 2 + (J + J wt ) + J w γ cos θ + (J 3 + J w ) ψ cosθ β ψ J w γ + (J 3 + J w ) cos θ + (J 3 + J w ) ψ β γ ( γ + ψ + cos θ)j w The generalized velocities are an affine function of the conjugate momenta. Therefore, the EOMs for the ignorable coordinates are: ψ γ ( J J wt+(j +J wt) cos θ 2 ) cos θ ( J J wt+(j +J wt) cos θ 2 ) (f) For steady precession, θ and const : cos θ ( J J wt+(j +J wt) cos θ 2 ) ( J J wt+(j +J wt J 3 ) cos θ 2 ) ( J J wt+(j +J wt) cos θ 2 )J 3 J 3 J 3 (J 3 +J w) J 3 J w (J + J wt J w J 3 ) 2 cos θ ((J 3 + J w ) ψ + J w γ) + mgl 2 β φ β ψ β γ

3 ((J 3 + J w ) ψ + J w γ) ± ((J 3 + J w ) ψ + J w γ) 2 4(J + J wt J w J 3 ) cos θmgl 2(J + J wt J w J 3 ) cos θ for real solutions to exist, the following must hold: ( θ π 2 ): ((J 3 + J w ) ψ + J w γ) 2 4(J + J wt J w J 3 ) cos θmgl ( π 2 < θ < π): J w γ ±2 (J + J wt J w J 3 ) cos θmgl (J 3 + J w ) ψ ((J 3 + J w ) ψ + J w γ) 2 4(J + J wt J w J 3 ) cos θmgl J w γ ±2 (J + J wt J w J 3 ) cos θmgl (J 3 + J w ) ψ Therefore, some design goals and operating conditions for TopBot can be deduced. For example, the relative spin rate ( ψ) of the main body should be zero to minimize energy dissipation through the foot bearings. For mass properties: J + J wt J 3 J w for the right-side up top (foot mounted on floor) and J + J wt J 3 J w for the upside down top (foot mounted on ceiling). In the case of a sleeping top (purely vertical, θ θ ), the momentum required from the momentum wheel will be the highest. Therefore, steady precession at larger nutation angles would be preferred. 3

4 2. Controls (a) (ml 2 + J) θ + (mrl cos θ) φ mgl sin θ + τ (mrl cos θ) θ + ((m + m w )r 2 + J w ) φ mrl θ 2 sin θ τ Small angle approximation sin θ θ, cos θ, θ 2 (ml 2 + J) θ + (mrl) φ mglθ + τ (mrl) θ + ((m + m w )r 2 + J w ) φ τ (b) (ml 2 + J)Θ(s)s 2 + (mrl)φ(s)s 2 mglθ(s)s + τ(s) () (mrl)θ(s)s 2 + ((m + m w )r 2 + J w )Φ(s)s 2 τ(s) (2) Solving each equation for Φ(s), putting () and (2) together, and rearranging yields: Θ(s) τ(s) (m + m w )r 2 + J w + mrl [((m + m w )r 2 + J w )(ml 2 + J) (mrl) 2 ] s 2 mgl((m + m w )r 2 + J w )) 4

5 (c) Putting in the parameters, the open loop transfer function is: G(s) Θ(s) τ(s) 2 s 2 5 There are three requirements for controlled system: stability, a rise time of.2 seconds, and less than 2% overshoot. The stability requirement is usually satisfied in industry with a gain margin greater than 6dB and a phase margin greater than 45 degrees. The rise time requirement puts a lower bound on the bandwidth required. The overshoot requirement necisitates a certain level of damping on the closed loop system, which also implies a certain amount of phase required. The root locus and frequency response for G(s) are shown below. 2 Root Locus Bode Diagram.5 2 Imaginary Axis.5 Magnitude (db) Phase (deg) (a) Real Axis 8 2 Frequency (rad/sec) (b) Figure : G(s) To stabilize G(s), we need to pull the locus into the open left half plane on the root locus plot. On the Bode plot, it is obvious that we will need to add significant phase. Therefore, our compensator design starts with adding a lead compensator of the form: C(s) K s + z s + p, z < p Assuming our system will behave approximately like a second order system (the first closed-loop poles will dominate the response), we can make some initial guesses for the parameters in our compensator. Recall that for a lead compensator, the maximum amount of phase gained can be approximated by the following relationship (also plotted below): 5

6 ( p φ max sin ) z p + z (3) φ max p/z Figure 2: Maximum Phase Gained from a Lead Compensator For stability, we require φ > 45 deg. From the chart above this would imply a p ratio of at least 6. However, anticipating future needs (damping for overshoot) we will choose a much more conservative 7 deg phase addition which z corresponds to a p of 3. Another useful relationship tells us the frequency at z which the maximum phase occurs in a lead filter: ω max zp (4) We wish to place this frequency at the gain crossover frequency (bandwidth) to maximize stability. The rise time requirement yields the bandwidth suggestion based on this approximation: ωbw desired.8.8 t r.2 9rad sec Again, we would like to start off conservative, so we use.5ωbw desired Putting it all together: (5) instead. 6

7 ω bw.5ωbw desired 35 p z 3 (ω bw ) 2 z p z p p z z Note that this is an extremely fast pole. However, because we are using passive analog components in our controller, this is not a problem. Finally, we have to find the proportional gain K. As we wish to place the gain crossover at the frequency ω bw, we will need to move the entire gain response up by gain K. The current gain of C(s)G(s) at ω bw is:.333(jω bw ) C(jω bw )G(jω bw ).6667(jω bw ) (jω bw ) 2 (jω bw ) e-5 Therefore, to make the open loop frequency response equal to (db) at ω bw, we multiply C(s)G(s) by the reciprocal of C(jω bw )G(jω bw ) which implies that K is equal to: K C(jω bw )G(jω bw ) 2.46e-5 Our lead compensator is now designed: C(s) K s + z s + p e4 s s e4 (6) Note that the locus has been pulled into the left half plane, the phase margin is still roughly 7 degrees, and the gain margin is theoretically infinite (as the phase never cross below -8). The time domain response to a step input meets both the rise time and overshoot requirements. However, note that the response is converging to something other than. The suggested pre-filter K pre can alleviate this problem. We choose K pre to be the reciprocal of the steady-state gain of the closed loop system. Recall that the steady-state gain of a transfer function G(s) is simply G(). Therefore, (7) 7

8 3 Root Locus 5 Bode Diagram 2 Magnitude (db) 5 Imaginary Axis 9 2 Phase (deg) Real Axis (a) Frequency (rad/sec) (b) Figure 3: KC(s)G(s).4 Step Response.2.8 Amplitude Time (sec) Figure 4: Time Response of KC(s)G(s) +KC(s)G(s) 8

9 K pre KC()G() + KC()G() 6.65e4() +.639e6.6667() () e4() +.632e6.639e6.632e6.43 K pre.9957 Using the pre-filter K pre, the step response is now completely within spec..4 Step Response.2.8 Amplitude Time (sec) Figure 5: Time Response of K pre KC(s)G(s) +KC(s)G(s) Now, a, T, and therefore R, R 2, C can be deduced from the compensator. 9

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