{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW7-Solution(2)

# HW7-Solution(2) - Flight Control Systems Homework#7...

This preview shows pages 1–3. Sign up to view the full content.

Flight Control Systems Homework #7 Solutions Dr. Bishop - Fall 2007 Reading Assignment: Modern Control Systems: Chapter 7 1. Exercises: E7.27 2. Problems: P7.39 3. Problems: AP7.10 4. Design Problems: DP7.12 5. Computer Problems: CP7.6. E7.27 The characteristic equation is 1 + p s s 2 + 4 s + 40 = 0 . The root locus shown in Figure 1 is stable for all 0 < p < . -12 -10 -8 -6 -4 -2 0 2 -8 -6 -4 -2 0 2 4 6 8 Root Locus Real Axis Imaginary Axis Figure 1: Root locus for 1 + p s s 2 +4 s +40 = 0. P7.39 The loop transfer function is G c ( s ) G ( s ) = 22 K ( s + 1)( s 2 + 8 s + 22) .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
-14 -12 -10 -8 -6 -4 -2 0 2 -10 -8 -6 -4 -2 0 2 4 6 8 10 Root Locus Real Axis Imaginary Axis Step Response Time (sec) Amplitude 0 0.5 1 1.5 2 2.5 3 3.5 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Figure 2: (a) Root locus for 22 K ( s +1)( s 2 +8 s +22) = 0. (b) Unit step response. When K = 0 . 529, the closed-loop poles are s 1 , 2 = - 3 . 34 ± 1 . 83 j and s 3 = - 2 . 32 and have the maximum damping ζ = 0 . 877. The root locus is shown in Figure 2a. The step respose is shown in Figure 2b. AP7.10 The characteristic equation (with K as the parameter) is 1 + K s 2 + 7 s + 20 s ( s 2 + 7 s + 10) = 0 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

HW7-Solution(2) - Flight Control Systems Homework#7...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online