{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

e483h5s_07

# e483h5s_07 - 1 ENSC 483 HW#5 Solutions by M Saif Simon...

This preview shows pages 1–4. Sign up to view the full content.

ENSC 483 HW#5 Solutions by M. Saif 1 Simon Fraser University School of Engineering Science ENSC 483 - Modern Control Systems 1. Electrical circuit– Obtain the state space description of the network shown in Figure 1. Use the voltages across the capacitors and the current through the inductor as the state variables. Note that v 1 is the input and v 2 is the output. Figure 1: An electrical network Solution: dv C 1 dt = 1 C 1 i C 1 dv C 2 dt = 1 C 2 i C 2 di L dt = 1 L v L R 1 i R 1 + v C 1 = v 1 R 2 i R 2 + R 3 i L + v L = v C 1 R 3 i L + v L = v C 2 v 2 = v C 2 i R 1 = i C 1 + i R 2 i R 2 = i L + i i = i C 2 Now if we take v C 1 ; v C 2 , and i L as our state variables, we get: ˙ v C 1 ˙ v C 2 ˙ i L = - 1 C 1 1 R 1 + 1 R 2 · 1 C 1 R 2 0 1 C 2 R 2 - 1 C 2 R 2 - 1 C 2 0 1 L - R 3 L v C 1 v C 2 i L + 1 R 1 C 1 0 0 v 1 v 2 = [ 0 1 0 ] v C 1 v C 2 i L 2. Mass-Spring Systems – Obtain the differential equations for the system shown in Figure 2. Draw a block diagram for this system and find the transfer function H ( s ) = X 2 ( s ) F ( s ) . Write the state variable formulation of this system using the state variables: z 1 = x 1 , z 2 = ˙ x 1 , z 3 = x 2 , z 4 = ˙ x 2 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
ENSC 483 HW#5 Solutions by M. Saif 2 Figure 2: A mass-spring system. Figure 3: Electrical analogue of the mass-spring system
ENSC 483 HW#5 Solutions by M. Saif 3 Solution: Draw the electrical analog of the mass-spring system as in Figure 3. From it we get, M 1 s 2 X 1 ( s ) + ( Ds + K 1 )( X 1 - X 2 ) = U ( s ) M 2 s 2 X 2 ( S ) + ( D 1 s + K 1 )( X 2 - X 1 ) + K 2 X 2 = 0 The block diagram representation of the system in terms of the input U ( s ) , and output X 2 ( s ) is shown in Figure 4. The transfer function of it is H ( s ) = Ds + K 1 M 1 M 2 s 4 + ( M 1 + M 2 ) Ds 3 + ( M 1 K 1 + M 2 K 1 + M 1 K 2 ) s 2 + K 2 Ds + K 1 K 2 Finally, the state space representation of the system is ˙ z ( t ) = d dt x 1 ˙ x 1 x 2 ˙ x 2 = 0 1 0 0 - K 1 M 1 - D M 1 K 1 M 1 D M 1 0 0 0 1 K 1 M 2 D M 2 - K 1 + K 2 M 2 - D M 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern