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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 timeinvariant. Shown above is a simple sensing scheme to measure the angle y . A power supply is connected between terminals A and C . A resistive semicircular 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
 Tomizuka

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