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Unformatted text preview: Problem Set #4 Solution 16.30: Estimation and Control of Aerospace Systems Posted: Mar 14, Due: Mar 21 1 Problem 8.30 in VDV. The Bode plots for the (openloop) plant with the three given values for the time delay are shown below. Notice that the magnitude plot is the same for all cases, i.e., it is not affected by the time delay. Phase margin = 50 ◦ ω = 2 . 3 ω = 8 . 1  G ( j ω )  = 0 . 12  G ( j ω )  = 0 . 4 ω ω  G ( j ω )  ∠ G ( j ω ) In order to answer part (b) of the problem, identify on the phase plot the frequency at which the phase first drops below 130 degrees. That is where you want to put the crossover point. In order to do that, you will have to shift the magnitude plot up by a factor that is equal to 1 /G ( jω ) at that particular frequency. We get that, • For T d = 0 . 5, ω c ≈ 2 . 3 rad/s, and the corresponding gain is k ≈ 1 / . 4 = 2 . 5. • For T d = 0 . 1, ω c ≈ 8 . 1 rad/s, and the corresponding gain is k ≈ 1 / . 12 ≈ 8 . 3. 2 Problem 8.31 in VDV. This problem can be solved in a similar way as 8.30. The Bode plots are as follows: Phase margin = 50 ◦ ω ω  G ( j ω )  ∠ G ( j ω ) ω = 2 . 8  G ( j ω )  = 0 . 29  G ( j ω )  = 0 . 1 ω = 6 . 1 and the desired results are: • For T d = 0 . 2, ω c ≈ 2 . 8 rad/s, and the corresponding gain is k ≈ 1 / . 29 ≈ 3 . 4. • For T d = 0, ω c ≈ 6 . 1 rad/s, and the corresponding gain is k ≈ 1 / . 1 = 10. 3 Problem 4.10 in DFT. The nominal closedloop system is stable for all k > 1, as you can easily check, e.g., by inspecting the poles of the complementary sensitivity function T ( s ) = C ( s ) P ( s ) 1 + C ( s ) P ( s ) = k s + k 1 . Since we are dealing with multiplicative uncertainty, the robust stability condition takes the form (see, e.g., the lecture notes): k Δ( s ) W 2 ( s ) T ( s ) k ∞ < 1 ....
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 Spring '07
 Frazzolli
 vdv, corresponding gain

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