{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EE_302_2010_HW4_soln

# EE_302_2010_HW4_soln - EE302 HW4 Solution Q1 Sketch root...

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

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

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

Unformatted text preview: EE302 HW4 Solution Q1. Sketch root loci (by obtaining all relevant features) as K : 0 → ∞ for the unity feedback closed-loop systems with open-loop transfer functions given below. (a) G ( s ) = K s + 3 s ( s 2 + 2 s + 2) (b) G ( s ) = K s + 2 s 4- 1 (c) G ( s ) = K ( s + 1) 2 s 3 + 8 Sol’n. (a) We can write the characteristic equation (of the closed loop) as 1 + K ( s + 3) s ( s + 1 + j )( s + 1- j ) = 0 which has three roots. Therefore the root locus diagram (RLD) will have three branches. Open loop has m = 1 zero and n = 3 poles. Three branches will start from open-loop poles and two of them will sail to infinity along the asymptotes and the remaining one will approach the zero at s =- 3 as K → ∞ . Asymptotes will have angles ϕ =- 180 ◦ (2 ℓ + 1) n- m = ± 90 ◦ . Asymptotes will meet at the centroid σ ◦ which we calculate as σ ◦ = ∑ n i =1 p i- ∑ m i =1 z i n- m = (0- 1- j- 1 + j )- (- 3) 3- 1 = 1 2 where p i denotes open-loop pole and z i open-loop zero. Angle condition reveals that RLD will reside on the real axis only at the interval s ∈ [- 3 , 0]. To find where (if) the RLD crosses jω axis we exploit the characteristic equation. 0 = s 3 + 2 s 2 + ( K + 2) s + 3 K s = jω =- jω 3- 2 ω 2 + j ( K + 2) ω + 3 K = (- 2 ω 2 + 3 K )- j ( ω 3- ( K + 2) ω ) which implies- 2 ω 2 + 3 K = 0 ω 3- ( K + 2) ω = 0 whence either ω = ± √ 6 with K = 4 or ω = 0 with K = 0. Finally, we compute the departure angles. From the open loop pole p =- 1 + j RLD will depart by ψ which we obtain from ± 180 ◦ (2 ℓ + 1) = ∠ ( p + 3)- ( ∠ ( p + 0) + ∠ ( p + 1 + j ) + ψ ) = arctan(1 / 2)- (135 ◦ + 90 ◦ + ψ ) .-3.5-3-2.5-2-1.5-1-0.5 0.5-15-10-5 5 10 15 Root Locus Real Axis Imaginary Axis Figure 1: Root locus diagram for Q1(a). We compute ψ ≈ - 18 ◦ . Due to symmetry with respect to real axis, the departure angle from the conjugate pole p =- 1- j will be- ψ . This is all we have to have figured out to plot RLD for this example. The range of K for stability is 0 < K < 4....
View Full Document

{[ snackBarMessage ]}

### Page1 / 8

EE_302_2010_HW4_soln - EE302 HW4 Solution Q1 Sketch root...

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

View Full Document
Ask a homework question - tutors are online