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

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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...
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## 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.

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