Section iii s re j r 90 90 re k r e j

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ω ) → 0 / − (n − m)90° GH ( j where, n = the order of the denominator of GH m = the order of the numerator of GH This represents the normal frequency response of the loop system GH(jω). GH 12/13/2008 HSK - EPM 381 - Stability 22 The Nyquist Criterion: Sketching the GH(s) Function GH • Section IV: This section is the mirror image of Section II. • Section III: ( s = Re jθ , R → ∞ , θ = +90º → –90º ) Re θ K ( R e j + z1 )( L ) K jθ → ( n − m ) j ( n − m )θ GH ( R e ) = k j kθ jθ R e ( R e + p1 )( L ) R e = 0 /− (n − m)θ This corresponds to (n-m)×180 º counter-clockwise rotations about the origin. 12/13/2008 HSK - EPM 381 - Stability 23 The Nyquist Criterion: Example 1 K GH ( s ) = 2 s (s + a) Nyquist Path jω j∞ II III σ × I IV -j∞ 12/13/2008 Sec. II: s = j 0+ , GH(jω) → ∞ /-180º Sec. GH s = j ∞, GH(jω) → 0 /-270º GH Sec. III: no effect. Sec. IV: mirror image of II. Sec. I: (k = 2) Then GH(j0) rotates (k GH 360º clockwise from GH(j0–) to clockwise GH to GH(j0+) with a magnitude of ∞ . GH with HSK - EPM 381 - Stability 24 The Nyquist Criterion: Example 1 K GH ( s ) = 2 s (s + a) j∞ Im{GH} R→ ∞ • Stability Analysis: GH(j0+) GH(j0–) -1 -j∞ 12/13/2008 Re{GH} There are 2 clockwise rotations of GH(s) There GH about the “–1” point, ∴ N = 2. No poles of GH(s) in the RHS, ∴P=0 . No GH No poles of GH(s) in the RHS, ∴ P=0 . No GH Then, Z = N+P = 2 which means there Then, N+P are two poles of the closed-loop system in the RHS, i.e. the system is unstable. HSK - EPM 381 - Stability 25 The Nyquist Criterion: Example 2 K GH ( s...
View Full Document

This note was uploaded on 03/03/2010 for the course AUTOMATIC 335 taught by Professor ? during the Winter '10 term at Ain Shams University.

Ask a homework question - tutors are online