F07%20314%20HW%2002%20all-s

F07%20314%20HW%2002%20all-s - EECS 314 Fall 2007 HW 02...

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Homework 2 Problem 1 Part1 : 1) The problem statement gives us the following information: ! K R R P X ! " 2 b a Vout ! " In addition, we know because of voltage division that # # $ % & & ( " P X S R R V Vout . If you substitute the given information for X R , you get: # $ % & ( ! # # $ % & & ( " K R R V Vout P P S 2 1 # # $ % & & ( ! " P S R K V Vout 2 1 P S S R KV V Vout ! " 2 1 By matching coefficients, we can see that: V V a S 6 2 1 " " deg 12 . 0 deg 25 3 V V R KV b P S " " " EECS 314 Fall 2007 HW 02 Solutions Page 1 of 12
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2) -80 -60 -40 -20 0 20 40 60 80 -4 -2 0 2 4 6 8 10 12 14 16 alpha (degrees) Vout (V) Plot of Vout vs. Alpha 0V 12V x=linspace(-75,75,5000); y=6+ 0.12*x; plot(x,y); xlabel( 'alpha (degrees)' ); ylabel( 'Vout (V)' ); grid; For the plot, note that if you just plot the equation ! 25 3 6 ! " Vout , the output voltages exceed 0V and 12V at -50 and +50 degrees respectively. Since this isn’t possible, it implies that with angles lower than -50 degrees, the output voltage is 0V and with angles greater than 50 degrees, the output voltage is 12V. 3) If we rearrange 25 3 6 ! " Vout , we get: ) * 6 3 25 + " Vout . ) * , + - + " + " , 67 . 16 3 50 6 4 3 25 4 ) * , - " + " , 33 . 8 3 25 6 7 3 25 7 4) Vout . " . 3 25 ) * , " " . 083333 . 010 . 3 25 EECS 314 Fall 2007 HW 02 Solutions Page 2 of 12
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5) We can rewrite the error above as: b Vout . " . ! . We also know that P S R KV b " . When you substitute, you get: S P P S V K R Vout R KV Vout / / . " . " . . By inspection, you can see that . will decrease when P R is decreased, when K is increased, or when S V is increased. Part2: No , you cannot use the device above. The range of the output voltage exceeds the desired range. When 0 " , S V Vout 2 1 " . Since we know that we want V Vout 5 . 2 " when 0 " , we have: 5 . 2 2 1 " " S V Vout V V S 5 " This means that you must decrease S V Next, when 50 " , we would like to have . 5 V Vout " P S S R KV V Vout ! " 2 1 ) * ) * 50 5 5 2 1 5 P R K ! " 100 1 " P R K Therefore, as long as you keep the above ratio, there is more than one solution. For example, you could change K to 3 deg / 0 k and P R to 300 0 k . However, if K is increased, so must P R and if K is decreased, so must P R . EECS 314 Fall 2007 HW 02 Solutions Page 3 of 12
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Potentiometer R P is what effectively limits the range of detection for the circuit. The adjustable tap on the potentiometer gives fractional resistance R x = R P / 2 + K · ! . Because the range is limited as R x ! (0, R P ), this puts ultimate limits on the angle variable ! , given as: 0 R x R p 0 R p 2 K α R p R p 2 K α R p 2 K 40 α 40 The actuator is turned on when V out = V s . The comparator will make the output voltage V out = V s whenever the potential V 2
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F07%20314%20HW%2002%20all-s - EECS 314 Fall 2007 HW 02...

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