This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full Document
 Spring '11
 Francis
 Nyquist stability criterion, Nyquist contour

Click to edit the document details