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Unformatted text preview: Problem Set 8 Solutions March 7, 2010 Problem 1 (a) We have: P ( s ) C ( s ) = 50 s ( s + 3)( s + 6) . The Nyquist contour is shown in Figure 1. D A B C Figure 1: Nyquist contour for 1(a) • AB : Set s = jω , where ω : 0 → ∞ . We have PC ( jω ) = 50 jω ( jω + 3)( jω + 6) = − 450 ω ω (9 + ω 2 )(36 + ω 2 ) + j 50 ω 2 − 900 ω (9 + ω 2 )(36 + ω 2 ) . Thus, we have ℜ ( PC ( jω )) = − 450 (9 + ω 2 )(36 + ω 2 ) ℑ ( PC ( jω )) = 50 ω 2 − 900 ω (9 + ω 2 )(36 + ω 2 ) . The plots of these functions for positive values of ω are shown in Figures 23. • BC : Set s = R e jθ where R → ∞ and θ : π 2 → − π 2 . Since PC is strictly proper, the semicircle at ∞ will collapse to zero. • CD : Use rules about complex conjugate and the plot for AB . 1 1 2 3 4 5 6 7 8108642 2 4 Figure 2: ℜ ( PC ) v.s. ω . 1 2 3 4 5 6 7 8108642 2 4 Figure 3: ℑ ( PC ) v.s. ω . • DA : Set s = ǫ e jθ where ǫ → 0 and θ : − π 2 → π 2 . This gives PC ( s )  s = ǫ e jθ ∼ 50 18 ǫ e jθ Now we can try a test point s = ǫ and we get PC ( ǫ ) ∼ 50 18 ǫ . This is a positive real number which tells us which way the infinite radius semicircle turns after departing from point D. Putting all this together, the Nyquist plot is shown in Figure 4. We extract the following data from the plot: p = 0 , n = 0 , z = p − n = 0 . Therefore the closed loop system is stable. (b) We have: P ( s ) C ( s ) = ( s + 4) s ( s + 1) . 2 D A B C Figure 4: Nyquist plot for 1(a) The Nyquist contour is shown in Figure 5. D A B C Figure 5: Nyquist contour for 1(b) • AB : Set s = jω , where ω : 0 → ∞ . We have PC ( jω ) = jω + 4 jω ( jω + 1) = − 3 ω ω (1 + ω 2 ) − j ω 2 + 4 ω (1 + ω 2 ) . Thus, we have ℜ ( PC ( jω )) = − 3 (1 + ω 2 ) ℑ ( PC ( jω )) = − 4 − ω 2 ω (1 + ω 2 ) . The plots of these functions for positive values of ω are shown in Figures 67. • BC : Set s = R e jθ where R → ∞ and θ : π 2 → − π 2 . Since PC is strictly proper, the semicircle at ∞ will collapse to zero. 3 1 2 3 4 5 6 7 8108642 2 4 Figure 6: ℜ ( PC ) v.s. ω . 1 2 3 4 5 6 7 8108642 2 4 Figure 7: ℑ ( PC ) v.s. ω . • CD : Use rules about complex conjugate and the plot for AB . • DA : Set s = ǫ e jθ where ǫ → 0 and θ : − π 2 → π 2 . This gives PC ( s )  s = ǫ e jθ ∼ 4 ǫ e jθ Now we can try a test point s = ǫ and we get PC ( ǫ ) ∼ 4 ǫ . This is a positive real number which tells us which way the infinite radius semicircle turns after departing from point D. Putting all this together, the Nyquist plot is shown in Figure 8. We extract the following data from the plot: p = 0 , n = 0 , z = p − n = 0 ....
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This note was uploaded on 01/30/2012 for the course ECE ECE311 taught by Professor Francis during the Spring '11 term at University of Toronto.
 Spring '11
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