11) is given a step load input of 5 lb. The output voltage is allached in A forc

Question

11) is given a step load input of 5 lb. The output voltage is allached in A force transducer (K wo plots below: these are identical except the second one focuses on a naow load region. The voltage is identical to the foree transducer load measurement since K-1 Vlb. Use the second tigure to determine the following: (12 points) a) Rise time b) Settle time 7. · d) Damped natural frequency (os and fa) e Damping ratio 0 Undamped natural frequency (o and fai c) Damped period of oscillation 0001 00008 00012 00016 0 004 0008 00032 0006 000 0004 0.0048 000 00056 0 0 Time [s 0000 00008 00 00016 000 00024 0002s 00032 00036 0004 00044 00048 00052 00056 Time [s

Explanation / Answer

a) Let Rise time be Tr

Rise time is a time required to reach 90% of signal value after signal is applied to system.

Transducer sensitivity is 1 V/lb

Hence when 5 V is applied output will be V = 1*5 = 5V

90% of signal value = 0.9*5 V = 4.5 V

From the given graph it is clear that time required to reach 90% of signal value i.e. Rise time Tr = 0.0009 seconds

b) Let Settling time be Ts

Settling time is a time required to reach 98% of signal value after signal is applied to system and signal remains within that band.

98% of signal value = 0.98*5 V = 4.9 V

and on higher side signal can go to 5.1 V.

Hence signal remains in +/- 2% error band.

From

From the given graph it is clear that time required to setlle in 98% of signal value i.e. settling time Ts = 0.0048 seconds

c) Let Td be Damped period of oscillation

From the given graph first peak is reached at T1 = 0.0016 and second peak is reached at T2 = 0.0048 second

Td = T2 - T1 = 0.0048 - 0.0016 = 0.0032 seconds

This is the required Damped period of oscillation

d)Damped natural frequency

fd = 1/ Td = 312.5 Hz

wd = 2*Pi*fd = 2*3.14*312.5 = 1962.5 rad/s

This is the required undamped natural frequency

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