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Unformatted text preview: ME 132 Solutions # 7 1 Block diagrams Label e 1 and e 2 to be the signals at the inputs to the blocks C and A respectively. Then, from the block diagram we can write the relationships e 1 = u- n- y, e 2 = Ce 1 + Dy + Dd, y = BAe 2 Eliminating e 1 and e 2 from these equations yields y = bracketleftBig BAC 1+ BAC- BAD BAD 1+ BAC- BAD- BAC 1+ BAC- BAD bracketrightBig u d n 2 Modeling B C A This is a straightforward dynamics problem. My solution (which we will use later) is ( 4 L ( m 1 + m 2 )- 3 m 1 L sin 2 ( y ) ) ¨ y = 3 m 1 L ˙ y 2 sin( y )cos( y )- 6( m 1 + m 2 ) g cos( y ) + 6sin( y ) u Using the numbers supplied, this becomes ¨ y = ˙ y 2 sin( y )cos( y )- 80cos( y ) + 8sin( y ) u 4- sin 2 ( y ) The major source of modeling error is that we have neglected frictional effects. Disturbances might include external wind currents. As discussed in class, this model is dynamic, nonlinear, and time-invariant. Shown above is a simple sensing scheme to measure the angle y . A power supply is connected between terminals A and C . A resistive semi-circular loop is attached to the block, and light brush wires establish electrical contact between the loop and the point B through the pendulum. We now have a voltage divider, and the voltage drop between terminalspendulum....
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This note was uploaded on 10/20/2010 for the course ME 132 taught by Professor Tomizuka during the Spring '08 term at Berkeley.
- Spring '08