Linear Control Systems Solution to Assignment #1

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1 Linear Control Systems Solution to Assignment # Instructor: H. Karimi Issued: Mehr 0, 389 Due: Mehr 8, 389 Solution to Exercise. a) Using the superposition property of linear systems we can compute the transfer functions from each input the the output individually and then obtain the overall transfer function by adding the individual contributions. Let s start with the transfer function from r to c. Disregard the other two inputs, i.e. set Ds) = 0 and Ns) = 0. rearranging: Cs) = Ks)Gs)Es) = Ks)Gs)Rs) Bs)) = Ks)Gs)Rs) Ks)Gs)Hs)Cs) Cs) = Ks)Gs) + Ks)Gs)Hs) Rs) In a similar fashion, considering one input at the time, for disturbance we get: and for noise: Cs) = Cs) = Gs) + Ks)Gs)Hs) Ds) Ks)Gs)Hs) + Ks)Gs)Hs) Ns) As a final result, the overall transfer function is: Cs) = [Gs)Ds) + Ks)Gs)Rs) Ks)Gs)Hs)Ns)] + Ks)Gs)Hs) b) If there is no noise and no disturbance, and Hs) =, the error signal is given by: Es) = Rs) Bs) = Rs) Cs) = Rs) KGs) + KGs) Rs) = + KGs) Rs) Linear Control Systems Solution to Assignment # Page of 0

2 c) If our input is the unit step, and if the conditions of the theorem are met, then the steady-state error is the following: e ss = lim t et) = lim s 0 ses) = lim s 0 s + KGs) s = lim s 0 + KGs) Final Value Theorem doesn t apply to E s), since it has poles on the imaginary axis. Actually, e ss doesn t exist, and e t) oscillates sinusoidally about 0. e 2ss = lim + K s 0 ss+2) = lim s 0 ss + 2) s 2 + 2s + K = 0 Solution to Exercise 2. a) Input: switch-on; output: clean clothes. Open loop. The washing machine does not feed back the cleanliness of clothes. b) Input: electrical signal from readout device e.g. laser in a CD player); output: microphone vibration. Open loop. The output audio level and quality is not used to determine the input to the speaker. c) Input: desired temperature; output: temperature. Closed loop. The vast majority of air conditioners have temperature sensors often, a simple bimetal switch) so they turn themselves on when the temperature is outside a prescribed range, off when it is inside the prescribed range. d) Manual gear train in an automobile Input: lever location set by driver; output: wheel RPM. Open loop. The wheel speed is not used in determining the gear setting though a good driver would probably apply a sense of the vehicle s present speed to decide the proper gear to apply so as not to stall or rev up the engine.) Automatic gear train in an automobile Unlike the manual gearbox, in the automatic gearbox as the name implies!) there is a feedback mechanism: the wheel RPM heats up a special fluid causing it to expand or contract. The engine gear ratio is changed due to the fluid volume change. Solution to Exercise 3. a) From force balance, you can derive the equation of motion. For simplicity, the system variable θt) is chosen with polar coordinates. Then you don t need to care about tension on the rod and centrifugal force. Free-body diagram: Assuming that the length of the rod is l, we obtain ml 2 θt) = mgl sin θt) + ft)l cos θt) Note that inertia of the mass with respect to the rotation axis is ml 2. It is a nonlinear differential equation because it has sin θt) term. If θt) is small enough to be assumed sin θt) θt) and cos θt), then it is linear equation as ml θt) = mgθt) + ft). Linear Control Systems Solution to Assignment # Page 2 of 0

3 Figure : Problem 3a) b) You have one input ft) and two outputs xt) and θt), which are related to each other. Unlike rolling with no-slipping case, you have to consider slipping because viscous friction happens when the relative velocity at the interface of two surface is nonzero. Free-body diagram of the mass: Figure 2: Problem 3b), Free-body diagram of the mass Free-body diagram of the inertia: Figure 3: Problem 3b), Free-body diagram of the inertia The viscous friction between the mass and inertia is proportional to the relative velocity ẋ r θ). From free-body diagrams, the equations of motion are Mẍt) + 2f v ẋt) + Kxt) f v r θt) = ft) J θt) + r 2 f v θt) rfv ẋt) = 0 Linear Control Systems Solution to Assignment # Page 3 of 0

4 If you want to obtain the relation of single input ft) and single output xt), then you can eliminate xt) and combine the above two equations. From the second equation, ẋt) = J rf v θt) + r θt) This implies ẍt) = J rf v θ 3) t) + r θt) and xt) = J rf v θt) + rθt). Zero initial condition is assumed. If not, there is only offset θ 0 and it doesn t make any difference physically.) Using those relations, the equation of motion can be simplified to JMθ 3) t) + f v r 2 M + 2J) θt) + f v 2 r 2 + KJ) θt) + Kr 2 f v θt) = f v rft) It is a linear differential equation because it has only θt), θt), θt) and θ 3) t) terms. Solution to Exercise 4. a) Linear If the differential equation is linear and two functions x t) and x 2 t) are satisfied with ) 2π 7ẍ + 0.5ẋ + 5 sin 0 t x = f t) ) 2π 7ẍ ẋ sin 0 t x 2 = f 2 t) then x + x 2 also should be the solution of the differential equation with the input f t) + f 2 t), which is true because ) 2π 7ẍ + ẍ 2 ) + 0.5ẋ + ẋ 2 ) + 5 sin 0 t x + x 2 ) = f t) + f 2 t) b) Nonlinear With the same method as a), the two functions x t) and x 2 t) are the solutions of 7ẍ + 0.5ẋ x )x = f t) 7ẍ ẋ x 2 )x 2 = f 2 t) However, x +x 2 is not the solution of the differential equation with the input f t)+ f 2 t) because c) Nonlinear 7ẍ + ẍ 2 ) + 0.5ẋ + ẋ 2 ) x )x x 2 )x 2 7ẍ + ẍ 2 ) + 0.5ẋ + ẋ 2 ) x + x 2 ))x + x 2 ) d dt 2 mẋ2 + ) 2 kx2 = 0 ẋmẍ + kx) = 0 ẋ = 0 and/or mẍ + kx = 0 Linear Control Systems Solution to Assignment # Page 4 of 0

5 Let A and B denote the solution space of the homogeneous linear differential equations ẋ = 0 and mẍ + kx = 0, respectively. Clearly, A B. Choose x t) A A B and x 2 t) B A B; then both x t) and x 2 t) are solutions of d dt 2 mẋ2 + kx2) = 2 0, but x s t) = x t) + x 2 t) is not, since it satisfies none of the differential equations k ẋ = 0 and mẍ + kx = 0. A simple example is x t) = and x 2 t) = sin ). t m Some people may believe that the correct answer is Linear [under an assumption]. They may justify their answer either way:. By taking derivative first and inspecting. d dt 2 mẋ2 + ) 2 kx2 = ẋmẍ + kx) = 0 If we assume ẋ 0, then the equation is linear. The reason why the assumption is valid, if ẋ = 0 then x = cte and trivial.) 2. With the same method, we have two functions x and x 2 satisfied with d dt 2 mẋ2 + ) 2 kx 2 = 0 d dt 2 mẋ2 2 + ) 2 kx 2 2 = 0 From the first equation, and again assuming ẋ 0, ẋ 2 0 for nontrivial solution, From the second equation mẍ + kx = 0 mẍ 2 + kx 2 = 0 Now consider [ d dt 2 mẋ + ẋ 2 ) 2 + ] 2 kx + x 2 ) 2 = d [ dt 2 mẋ2 + ẋ ẋ ẋ 2 ) + ] 2 kx 2 + x x x 2 ) = mẋ ẍ + ẋ 2 ẍ 2 + ẋ ẍ 2 + ẍ ẋ 2 ) + kx ẋ + x 2 ẋ 2 + x ẋ 2 + ẋ x 2 ) = ẋ mẍ + kx ) + ẋ 2 mẍ 2 + kx 2 ) + ẋ mẍ 2 + kx 2 ) + ẋ 2 mẍ + kx ) = 0 Since the input x + x 2 also satisfies the equation of motion, the system is linear. Solution to Exercise 5. Let s define φt) as the rotation angle of the motor. From the relation of the gear pair, we know φn = θn 2 and φn = θn 2. T is the torque generated by the motor and it is scaled by N 2 /N at the inertia by the gear pair. Also the inertia of the motor is scaled by N 2 2 /N 2 by the gear pair. For the detail, please refer to section 2.7 of Nise) In my opinion, their assumption is not valid. Linear Control Systems Solution to Assignment # Page 5 of 0

6 From KCL at the node attached to the resistor, i s t) v e R T K m = 0. From torque balance at the inertia, J + N ) 2 2 J N 2 m θ = T N 2 b N θ Kθ. Converted torque from the motor drives the effective inertia inertia + converted inertia of the motor shaft), viscous friction and compliance.) Using the Laplace transform and v e = K v φt) = N Kv 2 N θt), the above two equations can be written as I s s) K v R J + N ) 2 2 J N 2 m s 2 + bs + K N 2 Θs)s T s) = 0 N K m ) Θs) = T s) N 2 N Solving the first equation with respect to T and plugging it into the second equation, we get the transfer function defined by Θs) I s s) = Solution to Exercise 6. df s) a) Use L{tft)} = ds ) J + N 2 2 J N 2 m s 2 + twice: N 2 /N )K m b + N 2 2 N 2 ). K v K m s + K R F s) = s + )3. 6 b) Note that sin t cos t = 2 sin2t).2 Then, F s) = s df s) c) Use the identity sint 3) = sin t cos 3 cos t sin 3 and L{tft)} = ds : F s) = cos 3 2s s 2 + ) 2 + sin 3 s2 + ) sin 3 2s 2 s 2 + ) 2. df s) d) Here you should use L{tft)} = first and then T > 0, L{ft T )t T )} = ds e T s F s): 2 avval ya dovome dabirestan! F s) = e 3s s ) 2. Linear Control Systems Solution to Assignment # Page 6 of 0

7 e) Since hence L{te at } = s + a) 2 L{t cos at} = s2 a 2 s 2 + a 2 ) 2 Solution to Exercise 7. a) Let F s) = L{te at + 2t cos t} = L{te at } + L{2t cos t} = s 2 + s + s + )s + 2)s + 3). Write s + a) + 2 s2 2 s 2 + ) 2 F s) = a s + + a 2 s a 3 s + 3 ; a = s2 + s + s + 2)s + 3) = s= 2, a 2 = s2 + s + s + )s + 3) = 3, s= 2 a 3 = s2 + s + s + )s + 2) = 7 s= 3 2. f t) = 2 e t 3e 2t + 7 ) 2 e 3t t). Since T > 0, L{xt T )t T )} = e T s Xs), hence ft) = f t )t ) = 2 e t ) 3e 2t ) + 7 ) 2 e 3t ) t ). b) { } { L } = L 4 ss + 2) 2 s 4 s s + 2) 2 = 4 { e 2t 2te 2t} t) c) It can be re-written as F s) = s 2 3s ) Hs). s Therefore its inverse Laplace transform is ft) = d2 ht) dt 3 dht) dt t + ht) + 2 hτ) dτ. 0 Linear Control Systems Solution to Assignment # Page 7 of 0

8 d) F s) = ω 2 n s s 2 + 2ζω n s + ω 2 n) = s F s) F s) = ω 2 n s 2 + 2ζω n s + ω 2 n For all 0 < ζ <, f t) = L {F s)} ω n ) = ζ 2 e ζω nt sin ω n ζ2 t ft) = ) ζ 2 e ζω nt sin ω n ζ2 t + cos ζ Hint: You should use integration by parts twice. e) F s) = e s s 2 + e s ) = s 2 F s) F s) = e s + e s = e s e s ) e s + = e s e s) e s + e 2s ) = e s 2e 2s + 2e 3s f t) = δt ) 2δt 2) + 2δt 3) = ft) = t )t ) 2t 2)t 2) + 2t 3)t 3) Note that expanding the term / + e s ) [directly] to a geometric series, leads to an invalid non-causal answer; we are using one-sided Laplace transform definition. Solution to Exercise 8. Linear Control Systems Solution to Assignment # Page 8 of 0

9 a) ) ) Y s) = s + s 2 + 2s + 5 [ ] [ ] = s + s + + 2j)s + 2j) = c s + + c 2 s + + 2j) + c 3 s + 2j) c = 2j. 2j = 4 c = 2j. 4j = 8 c = c 2 = [ 8 yt) = 4 e t 8 e 2j)t ] 8 e +2j)t t) = [ e t 4 2 e t e 2jt + e 2jt)] t) = 4 e t [ cos 2t] t) b) 3s + 4 Y s) = s 4 + 4s 3 + 8s 2 + 8s + 4 s = 3s + 4 s [s + ) 2 + ] 2 = k s + c s + + j + c 2 s + + j) 2 + d s + j + d 2 s + j) 2 k = c = 2 4 j d = c = j c 2 = 4 2 j d 2 = c 2 = j [ yt) = ) j e j)t ) j te j)t ) j e +j)t ) ] j te +j)t t) = [ e cos t t + 2 sin t + 2 )] t cos t + t sin t t) Solution to Exercise 9 Nise, 4th Edition, Chapter 2, Problem 53). The relationship between the nonlinear spring s displacement, x s t), and its force, f s t), is x s t) = e f st). Linear Control Systems Solution to Assignment # Page 9 of 0

10 Solving for the force, f s t) = ln x s t)) ) Writing the differential equation for the system by summing forces, d 2 xt) dt 2 + dxt) dt ln xt)) = ft). 2) Letting xt) = x 0 + δx and ft) = + δf, linearize ln xt)): d ln x) ln x) ln x 0 ) = dx δx. x=x0 Solving for ln x), ln x) = ln x 0 ) d ln x) dx δx = ln x 0 ) δx. 3) x=x0 x 0 When f =, δx = 0. Thus from ), = ln x 0 ). Solving for x 0, Substituting x 0 = into 3), x 0 = e ln x) = ln 0.632) x 0 = δx = eδx Placing this value into 2) along with xt) = x 0 + δx and ft) = + δf, yields the linearized differential equation or d 2 δx dt 2 + dδx dt d 2 δx dt 2 + dδx dt + + eδx = + δf + eδx = δf. Taking the Laplace transform and rearranging yield the transfer function Xs) F s) = s 2 + s + e Good Luck! Linear Control Systems Solution to Assignment # Page 0 of 0

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