PSet2Solutions

# PSet2Solutions - Problem Set #8 16.30: Estimation and...

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Problem Set #8 16.30: Estimation and Control of Aerospace Systems Posted: Feb 21, Due: Feb 28 1 Problem 8.10 in VDV. Let us look at the requirements: 1. “The steady-state error for step inputs must be zero:” the system’s transfer function G ( s ) must have a pole at the origin. A transfer function satisfying this constraint would be G 1 ( s ) = 1 s . 2. “The steady-state error following unit ramps may not exceed 2%:” this means that lim s 0 s 1 1 + G ( s ) 1 s 2 < 0 . 02 . and hence lim s 0 sG ( s ) > 50. This places a lower bound on the standard gain; a transfer function satisfying constraints 1 and 2 would be G 1 , 2 ( s ) = 50 s . 3. “The error in response to since inputs up to 10 rad/s may not exceed 10%:” Let us choose W 1 ( s ) = 10 for | s | ≤ 10 rad / s, W 1 ( s ) = 0 otherwise. Then the condition k W 1 ( s ) S ( s ) k < 1 translates into | 1+ G ( ) | > W 1 ( ) for all ω R . Remember that when | G ( ) | is large (e.g., at low frequencies), | 1 + G ( ) | ≈ | G ( ) | . Note that G 1 , 2 does not satisfy this constraints; in order to meet speciﬁcations 1-3, we can, e.g., increase the gain such that | G ( j 10) | ≥ 10, as in G 1 - 3 = 100 s . 4. “... the output in response to sine inputs above 250 rad/s may not exceed 10% of the input.” Let us choose W 2 ( s ) = 10 for | s | > 250 rad/s, W 2 ( s ) = 0 otherwise. Then the condition k W 2 ( s ) T ( s ) k < 1 translates into ± ± ± ± G ( ) 1 + G ( ) ± ± ± ± < | W 2 ( ) | - 1 for all frequencies. Remember that when | G ( ) | is small, | G ( ) / (1 + G ( )) | ≈ | G ( ) | . The transfer function G 1 - 3 does not meet this requirement. Since we need to lower the gain at high frequencies without touching the gain at low frequencies, we can add a pole at s = 10 rad/s 1 . This would result in | G ( j 100) | ≈ 1 / 10, which would satisfy the requirement. Hence, G 1 - 4 = 100 s ( s/ 10 + 1) . 1 From now on we will do our initial design taking into account asymptotic Bode plots, we will then ﬁne-tune the result at the end.

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5. “... he crossover frequency must be about 50 rad/s.” Crossover with the current design will occur approximately halfway between 10 and 100 rad/s (on the logarithmic plot), i.e., at a frequency approximately equal to 1000 = 31 . 6 rad/s. If we want to increase the crossover frequency we could, e.g., make the pole faster. For example, we could choose to move the pole to s = 25 rad/s . Crossover would result at a frequency approximately equal to 25 p 100 / 25 = 50, i.e., exactly our requirement! Note that in this case the magnitude of the frequency response at 250 rad/s would be approximately 1% of the magnitude at 25 rad/s, which in turn is about 4; this means that | G ( ) | ≤ 0 . 04 < 0 . 1 for all frequencies above 250 rad/s, still satisfying constraint 4. The transfer function candidate is now G 1 - 5 = 100 s ( s/ 25 + 1) . 6.
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## This note was uploaded on 03/17/2008 for the course COURSE 16.30 taught by Professor Frazzolli during the Spring '07 term at MIT.

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PSet2Solutions - Problem Set #8 16.30: Estimation and...

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