HW_11_SOL_F18

1 ME 340 HW#11 Will not be graded. SOLUTIONS 1. Let 𝑓𝑓 ( 𝑥𝑥 ) be the function depicted below… … and consider the differential equation 𝑥𝑥⃛ ( 𝑡𝑡 ) + 𝑓𝑓 ( 𝑥𝑥 ) = 0. Find a linear differential equation whose solution approximates the deviation variable ∆𝑥𝑥 ( 𝑡𝑡 ) = 𝑥𝑥 ( 𝑡𝑡 ) − 𝑥𝑥 0 when 𝑥𝑥 ( 𝑡𝑡 ) is close to 𝑥𝑥 0 . Solution: From the plot of 𝑓𝑓 ( 𝑥𝑥 ) we can see clearly that 𝑓𝑓 ( 𝑥𝑥 0) = 0 and 𝑥𝑥 0 = 5 . Since we want to approximate the deviation variable when 𝑥𝑥 ( 𝑡𝑡 ) is close to 𝑥𝑥 0 , our operating point becomes: 𝑥𝑥̅ = 𝑥𝑥 0 = 5 Computing the deviation dynamics: Δ𝑥𝑥 ( 𝑡𝑡 ) = 𝑥𝑥 ( 𝑡𝑡 ) − 𝑥𝑥̅ 𝑥𝑥 ( 𝑡𝑡 ) = Δ𝑥𝑥 ( 𝑡𝑡 ) + 𝑥𝑥̅ ⇒ 𝑥𝑥̇ ( 𝑡𝑡 ) = Δ𝑥𝑥 ̇ ( 𝑡𝑡 ), 𝑥𝑥̈ ( 𝑡𝑡 ) = Δ𝑥𝑥 ̈ ( 𝑡𝑡 ), 𝑥𝑥⃛ ( 𝑡𝑡 ) = Δ𝑥𝑥 ⃛ ( 𝑡𝑡 ) Substituting the above into our original equation: Δ𝑥𝑥 ⃛ ( 𝑡𝑡 ) + 𝑓𝑓 ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 5) = 0 Now, we will linearize the term 𝑓𝑓 ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 5) about Δ𝑥𝑥 ( 𝑡𝑡 ) = 0 . From the Taylor series expansion we know that: 𝑓𝑓 ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 5) = 𝑓𝑓 (5) + �Δ𝑥𝑥 ( 𝑡𝑡 ) + 5 − (0 + 5) ��𝑓𝑓 ′ ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 5) �� Δ𝑥𝑥=0 ⇒ 𝑓𝑓 ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 5) = 𝑓𝑓 (5) + ( Δ𝑥𝑥 ) 𝑓𝑓 ′ (5) From the plot, we see that 𝑓𝑓 ′ (5) ≈ − 4 . ⇒ 𝑓𝑓 ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 5) = 𝑓𝑓 (5) + Δ𝑥𝑥 ( − 4) Substituting this in the ODE above we get: Δ𝑥𝑥 ⃛ ( 𝑡𝑡 ) − 4 Δ𝑥𝑥 = 0 Replacing Δ𝑥𝑥 with 𝑥𝑥 � 𝑥𝑥 � ⃛ − 4 𝑥𝑥 � = 0 On solving the above linear ODE, we can get the 𝑥𝑥 � which approximates the value of Δ𝑥𝑥 ( 𝑡𝑡 ) .

2 2. Consider the system 𝑥𝑥̈ + 𝛼𝛼𝑥𝑥̇ + 𝑥𝑥 3 − 𝑒𝑒 1−𝑥𝑥 = 𝑢𝑢 , where 𝛼𝛼 is a constant. (a) Find 𝑢𝑢 � such that � 𝑥𝑥̅ 𝑥𝑥̇ ̅ 𝑢𝑢 � � = � 1 0 𝑢𝑢 � � is an operating point of the above system. Solution: At the operating point, all the dynamic terms should go to zero, including 𝑥𝑥̈ : 0 + 𝛼𝛼 (0) + 1 3 − 𝑒𝑒 1−1 = 𝑢𝑢 � ⇒ 𝑢𝑢 � = 0 (b) Linearize the above system about this operating point. Solution: The operating point found in (a) is � 𝑥𝑥̅ 𝑥𝑥̇ ̅ 𝑢𝑢 � � = � 1 0 0 � . Computing the deviation dynamics: Δ𝑥𝑥 ( 𝑡𝑡 ) = 𝑥𝑥 ( 𝑡𝑡 ) − 𝑥𝑥̅ 𝑥𝑥 ( 𝑡𝑡 ) = Δ𝑥𝑥 ( 𝑡𝑡 ) + 𝑥𝑥̅ 𝑢𝑢 ( 𝑡𝑡 ) = Δ𝑢𝑢 ( 𝑡𝑡 ) + 𝑢𝑢 � ⇒ 𝑥𝑥̇ ( 𝑡𝑡 ) = Δ𝑥𝑥 ̇ ( 𝑡𝑡 ), 𝑥𝑥̈ ( 𝑡𝑡 ) = Δ𝑥𝑥 ̈ ( 𝑡𝑡 ), 𝑥𝑥⃛ ( 𝑡𝑡 ) = Δ𝑥𝑥 ⃛ ( 𝑡𝑡 ), Δ𝑢𝑢 = 𝑢𝑢 Substituting the above into our original equation: Δ𝑥𝑥 ̈ ( 𝑡𝑡 ) + 𝛼𝛼Δ𝑥𝑥 ̇ ( 𝑡𝑡 ) + ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 𝑥𝑥̅ ) 3 − 𝑒𝑒 1− ( Δ𝑥𝑥 ( 𝑡𝑡 ) +𝑥𝑥̅ ) = Δ𝑢𝑢 Δ𝑥𝑥 ̈ ( 𝑡𝑡 ) + 𝛼𝛼Δ𝑥𝑥 ̇ ( 𝑡𝑡 ) + ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 1) 3 − 𝑒𝑒 1− ( Δ𝑥𝑥 ( 𝑡𝑡 ) +1 ) = Δ𝑢𝑢 Δ𝑥𝑥 ̈ ( 𝑡𝑡 ) + 𝛼𝛼Δ𝑥𝑥 ̇ ( 𝑡𝑡 ) + ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 1) 3 − 𝑒𝑒 −Δ𝑥𝑥 ( 𝑡𝑡 ) = Δ𝑢𝑢 Identify the non-linear terms and linearize them about Δ𝑥𝑥 ( 𝑡𝑡 ) = 0 : ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 𝑥𝑥̅ ) 3 ≈ (0 + 1) 3 + ( Δ𝑥𝑥 − 0)(3( Δ𝑥𝑥 + 1) 2 )| Δ𝑥𝑥=0 ( Δ𝑥𝑥 ( 𝑡𝑡 ) + 𝑥𝑥̅ ) 3 ≈ 1 + 3 Δ𝑥𝑥 𝑒𝑒 −Δ𝑥𝑥 ( 𝑡𝑡 ) ≈ 𝑒𝑒 −0 + ( Δ𝑥𝑥 − 0)( − 1) 𝑒𝑒 −Δ𝑥𝑥 ( 𝑡𝑡 ) � Δ𝑥𝑥=0 𝑒𝑒 −Δ𝑥𝑥 ( 𝑡𝑡 ) ≈ 1 − Δ𝑥𝑥 Substituting the linearized terms back in the ODE and removing any second or higher order terms: Δ𝑥𝑥 ̈ + 𝛼𝛼Δ𝑥𝑥 ̇ + 1 + 3 Δ𝑥𝑥 − (1 − Δ𝑥𝑥 ) ≈ Δ𝑢𝑢 ⇒ Δ𝑥𝑥 ̈ + 𝛼𝛼Δ𝑥𝑥 ̇ + 4 Δ𝑥𝑥 ≈ Δ𝑢𝑢 Replacing Δ𝑥𝑥 with 𝑥𝑥 � , and Δ𝑢𝑢 with 𝑢𝑢 − 𝑢𝑢 � = 𝑢𝑢 we have our linearized ODE: 𝑥𝑥 � ̈ + 𝛼𝛼𝑥𝑥 � ̇ + 4 𝑥𝑥 � = 𝑢𝑢 Upon solving the above ODE we would get 𝑥𝑥 � ( 𝑡𝑡 ) , and 𝑥𝑥̅ + 𝑥𝑥 � ( 𝑡𝑡 ) → 𝑥𝑥 ( 𝑡𝑡 ) . (c) Show that the equilibrium point � 𝑥𝑥̅ 𝑥𝑥̇ ̅ � = � 1 0 � for the homogenous system 𝑥𝑥̈ + 𝛼𝛼𝑥𝑥̇ + 𝑥𝑥 3 − 𝑒𝑒 1−𝑥𝑥 = 0, is stable only when 𝛼𝛼 > 0 . Solution: We know that if the real part of the roots of the characteristic equations for our ODE are negative, the system is stable. The characteristic equation for the linearized ODE from part (b) is: 𝜆𝜆 2 + 𝛼𝛼𝜆𝜆 + 4 = 0 ⇒ 𝜆𝜆 = −𝛼𝛼 ± √𝛼𝛼 2 − 16 2 From this we can see that the real part of the roots are only negative for 𝛼𝛼 > 0 , and therefore the system is only stable when 𝛼𝛼 > 0 .

4 ⇒ | 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 )| = 1 √𝑗𝑗 4 + 𝑗𝑗 2 ⇒ ∠𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ) = 𝜋𝜋 + tan −1 1 2 (c) What is the magnitude and phase of the steady state response when the input is sin(2 𝑡𝑡 ) ? Solution: For 𝑗𝑗 = 2 , | 𝐺𝐺 ( 𝑗𝑗 2)| = 1 √ 2 4 + 2 2 = 1 √ 20 = 1 5 √ 2 ∠𝐺𝐺 ( 𝑗𝑗 2) = 𝜋𝜋 + tan −1 1 2 5. The Bode plot for the system with transfer function 𝑋𝑋 ( 𝑠𝑠 ) 𝐺𝐺 ( 𝑠𝑠 ) = 𝑠𝑠 2 + 0.002 𝑠𝑠 + 10000 ( 𝑠𝑠 + 0.01)( 𝑠𝑠 + 1000) is shown below: Estimate the steady-state output 𝑥𝑥 𝑠𝑠𝑠𝑠 ( 𝑡𝑡 ) corresponding to the input 𝑓𝑓 ( 𝑡𝑡 ) = 0.0001 sin 0.0001 𝑡𝑡 + sin 100 𝑡𝑡 + sin 10000 𝑡𝑡 . Solution: We have three periodic terms, so we consult the Bode plot at each frequency to determine the gain and the phase shift of each. • The frequency 𝑗𝑗 = 0.0001 = 10 −4 is not included in the Bode plot, but the plot is level at 10 −3 so we assume the same gain of approximately 60 dB . Likewise, the phase shift is not shown, but appears to be very close to zero. Since 60 dB = 20 log(| 𝐺𝐺 |) ⇒ | 𝐺𝐺 | = 10 3 , we have an output component of 0.0001 ∗ 10 3 sin(10 −4 𝑡𝑡 ) = 0.1 sin(10 −4 𝑡𝑡 ) . • At 𝑗𝑗 = 100 = 10 2 , the gain is < − 150 dB . This gain effectively attenuates this signal to zero. • At 𝑗𝑗 = 10 4 , the gain is approximately 0 dB , which is a gain of 1. The phase shift is also effectively 0, so this gives a component of sin(10 4 𝑡𝑡 ) . Therefore, our output is approximately: 𝑥𝑥 ( 𝑡𝑡 ) = 0.1 sin(10 −4 𝑡𝑡 ) + sin(10 4 𝑡𝑡 )

5 6. A simple device for harnessing energy from the North Sea, graveyard of the damned, is depicted below: Waves force the plate periodically; this forcing may be approximated by a sinusoid with period 2 𝜋𝜋 seconds. The motion of the plate generates electricity to help light the city of Copenhagen. (a) Assuming the system to be underdamped, derive an expression for its resonant frequency in terms of 𝑚𝑚 , 𝑏𝑏 , and 𝑘𝑘 . Solution: The resonant frequency is: 𝑗𝑗 𝑟𝑟 = 𝑗𝑗 𝑛𝑛 � 1 − 2 𝜁𝜁 2 ⇒ 𝑗𝑗 𝑟𝑟 = � 𝑘𝑘 𝑚𝑚 � 1 − 2 � 𝑏𝑏 2 √𝑘𝑘𝑚𝑚 � 2 ⇒ 𝑗𝑗 𝑟𝑟 = √ 4 𝑘𝑘𝑚𝑚 − 2 𝑏𝑏 2 2 𝑚𝑚 (b) Let 𝑚𝑚 = 10000 kg . You, the engineer, have the freedom to choose both a value for 𝑏𝑏 between 2000 kg/sec and 5000 kg/sec and a value for 𝑘𝑘 between 5000 kg/sec 2 and 20000 kg/sec 2 . Find the pair of values which will maximize the amplitude of the forced oscillation of the plate. Does this choice correspond to tuning the system’s resonant frequency to the forcing frequency? Feel free to use MATLAB’s bode command to explore the frequency response of the system for different values of 𝑏𝑏 and 𝑘𝑘 ; print out and turn in any plots you generate to support your final answer. Solution: The transfer function of the system is: 𝐺𝐺 ( 𝑠𝑠 ) = 1 𝑚𝑚𝑠𝑠 2 + 𝑏𝑏𝑠𝑠 + 𝑘𝑘 ⇒ 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ) = 1 ( 𝑘𝑘 − 𝑚𝑚𝑗𝑗 2 ) + 𝑗𝑗𝑏𝑏𝑗𝑗 The frequency of the input (force) is 1 rad/s , i.e. 𝑗𝑗 = 1 . Thus, | 𝐺𝐺 ( 𝑗𝑗 1)| = 1 � ( 𝑘𝑘 − 𝑚𝑚 ) 2 + 𝑏𝑏 2 𝑚𝑚 = 10000 kg is fixed. Hence we choose 𝑏𝑏 and 𝑘𝑘 to make | 𝐺𝐺 ( 𝑗𝑗 1)| as large as possible. That is, 𝑏𝑏 = 2000 and 𝑘𝑘 = 𝑚𝑚 = 10000 . In this case, the resonant frequency is: 𝑗𝑗 𝑟𝑟 = √ 4 𝑘𝑘𝑚𝑚 − 2 𝑏𝑏 2 2 𝑚𝑚 ≈ 0.995 rad/s The forcing frequency is 1, so our choice corresponds to tuning the system’s resonant frequency to the forcing frequency.

7 8. Match the Bode plots…

8 … with the step responses. Justify your answers.

10 9. Sketch the Bode plot for the transfer function 𝐺𝐺 ( 𝑠𝑠 ) = 𝑋𝑋 ( 𝑠𝑠 ) 𝐹𝐹 ( 𝑠𝑠 ) = 𝑠𝑠 3 ( 𝑠𝑠 + 10000) �𝑠𝑠 + 1 10 � 2 ( 𝑠𝑠 + 100) . Solution: We write all the terms in the first and second order systems that we have seen in the class. 𝐺𝐺 ( 𝑠𝑠 ) = 𝑠𝑠 3 × 10 4 (10 −4 𝑠𝑠 + 1) 10 −4 (10 𝑠𝑠 + 1) 2 × 100(10 −2 𝑠𝑠 + 1) = 10 6 𝑠𝑠 3 (10 −4 𝑠𝑠 + 1) (10 𝑠𝑠 + 1) 2 (10 −2 𝑠𝑠 + 1) = 10 6 × 𝑠𝑠 3 × 1 1 10 −4 𝑠𝑠 + 1 × 1 10 𝑠𝑠 + 1 × 1 10 𝑠𝑠 + 1 × 1 10 −2 𝑠𝑠 + 1 . Therefore Bode plot of 𝐺𝐺 ( 𝑠𝑠 ) is Bode plot of 10 6 + Bode ployt of 𝑠𝑠 3 − Bode plot of 1 10 −4 𝑠𝑠+1 + � 2 × Bode Plot of 1 10𝑠𝑠+1 � + Bode Plot of 1 10 −2 𝑠𝑠+1 .

11 10. Suppose 𝐺𝐺 ( 𝑠𝑠 ) represent a transfer function. (a) Determine two distinct transfer functions 𝐻𝐻 1 ( 𝑠𝑠 ) 𝑎𝑎𝑎𝑎𝑑𝑑 𝐻𝐻 2 ( 𝑠𝑠 ) such that their magnitude plots match are the same as that of 𝐺𝐺 ( 𝑠𝑠 ); that is | 𝐻𝐻 1 ( 𝑗𝑗𝑗𝑗 )|= | 𝐻𝐻 2 ( 𝑗𝑗𝑗𝑗 )| = | 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 )|. (b) Determine two distinct transfer functions 𝐻𝐻 3 ( 𝑠𝑠 ) 𝑎𝑎𝑎𝑎𝑑𝑑 𝐻𝐻 4 ( 𝑠𝑠 ) such that their phase plots match are the same as that of 𝐺𝐺 ( 𝑠𝑠 ); that is ∠𝐻𝐻 1 ( 𝑗𝑗𝑗𝑗 ) = ∠𝐻𝐻 2 ( 𝑗𝑗𝑗𝑗 ) = ∠ 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ). Solution: (a) Let 𝐻𝐻 1 ( 𝑠𝑠 ) = −𝐺𝐺 ( 𝑠𝑠 ) ⇒ | 𝐻𝐻 1 ( 𝑗𝑗𝑗𝑗 )| = | −𝐺𝐺 ( 𝑗𝑗𝑗𝑗 )| = | − 1|| 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 )| = 1 ⋅ | 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 )| = | 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 )|. Similarly let 𝐻𝐻 2 ( 𝑠𝑠 ) = 𝑒𝑒 −𝑠𝑠 𝐺𝐺 ( 𝑠𝑠 ) ⇒ | 𝐻𝐻 2 ( 𝑗𝑗𝑗𝑗 )| = �𝑒𝑒 −𝑗𝑗𝑗𝑗 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ) � = �𝑒𝑒 −𝑗𝑗𝑗𝑗 � | 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 )| = � cos 2 𝑗𝑗 + sin 2 𝑗𝑗 ⋅ | 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 )| = | 𝐺𝐺 ( 𝑗𝑗𝑗𝑗 )|. (b) Let 𝐻𝐻 3 ( 𝑠𝑠 ) = 10 𝐺𝐺 ( 𝑠𝑠 ) ⇒ ∠𝐻𝐻 3 ( 𝑗𝑗𝑗𝑗 ) = ∠10 + ∠𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ) = 0 + ∠𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ) = ∠𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ) . Let 𝐻𝐻 4 ( 𝑠𝑠 ) = 𝑒𝑒 𝑠𝑠 2 𝐺𝐺 ( 𝑠𝑠 ) ⇒ ∠𝐻𝐻 3 ( 𝑗𝑗𝑗𝑗 ) = ∠𝑒𝑒 −𝑗𝑗 2 + ∠𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ) = 0 + ∠𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ) = ∠𝐺𝐺 ( 𝑗𝑗𝑗𝑗 ) .