08_ComplexPairInverseW12

# 08_ComplexPairInverseW12 - Page 1 of 2 Complex Conjugate...

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Unformatted text preview: Page 1 of 2 Complex Conjugate Pair of Roots CHE 361 It sometimes happens in process dynamics problems that we obtain part of Y(s) that looks like: ⎡A A⎤ Y ( s) = … + ⎢ + + − ⎥ + … ⎣ s − r+ s − r− ⎦ After evaluation of the constants in this partial fraction expansion, this can be written as: ⎡ B + Cj B − Cj ⎤ Y ( s) = … + ⎢ + ⎥ +… ⎣ s − r − jω s − r + jω ⎦ Here the root (s = r+ ) has a real part (r), i.e. r+ = r + jω, with (ω > 0) and corresponds to the numerator A+ = B + Cj. Thus, we first find the 2 roots with root parameters (r,ω), then the numerator parameters (B, C). Then the Laplace inverse of this part of Y(s) corresponding to these two complex conjugate roots can be written as: L−1 [ 2 complex conjugate pair parts ] = 2e rt [ B cos(ωt ) − C sin(ωt )] Several equivalent forms can be written which utilize a single (sin) or (cos) function. ⎡B⎤ = 2 B 2 + C 2 e rt sin(ωt + φ1 ) , where φ1 = tan −1 ⎢ ⎣ −C ⎥ ⎦ ⎡ −B ⎤ = 2 B 2 + C 2 e rt sin(ωt − φ2 ) , where φ2 = tan −1 ⎢ ⎣ −C ⎥ ⎦ ⎡C ⎤ = 2 B 2 + C 2 e rt cos(ωt + φ3 ) , where φ3 = tan −1 ⎢ ⎥ ⎣B⎦ ⎡ −C ⎤ = 2 B 2 + C 2 e rt cos(ωt − φ4 ) , where φ4 = tan −1 ⎢ ⎣B⎥ ⎦ The following MATLAB commands show how to calculate the roots and coefficients to find y(t). Simulink can be used to plot y(t) as the step response in this case of a transfer function block model. First the m-file pfrac.m is shown, then the Command Window output from pfrac. Then those results are used in the second form above. A Simulink block diagram and the plot made using that block diagram simulation are shown as a check of these results. %pfrac.m num=[10]; den = [1 2 10 0]; [r,p,k]=residue(num,den) Y (s) = 10 N (s) 1 =K = 10 3 2 s ( s + 2 s + 10) D( s) (1) s + (2) s + (10) s + (0)1 2 Page 2 of 2 The output variables [r,p,k] are p = poles { the roots of D(s)=0 }, r = residues { the coefficients in the partial fraction expansion } and k = a constant in the partial fraction expansion, corresponding to k times the Dirac delta(t) in y(t). Here k = 0, so there is no term in the partial fraction expansion that is a simple constant. Here is the diary file for the above .m file. MATLAB results use i for j = the square root of -1. pfrac r= -0.5000 + 0.1667i -0.5000 - 0.1667i 1.0000 p= -1.0000 + 3.0000i -1.0000 - 3.0000i 0 k= p(1) = -1+3i = r+ = r + jω because of the +3i. The corresponding r(1) = A+ = B + Cj and thus B = -0.5000 and C = 0.1667 From p(1) you find r = -1.000 and ω = 3.000. Using the second form listed on the first page, you need to calculate φ2 . From –B = Δy = +0.5 and -C = Δx = -0.1667 in the complex plane, you know that the angle φ2 is between +π / 2 and +π radians (Make a sketch). Using your calculator correctly for an angle in this quadrant, you should find that φ2 = 1.8925 radians. Thus the result is: y (t ) = 1 + 1.0541 e − t sin(3t − 1.8925) Evaluating this at t =1, you get y(1) = 1 + (1.0541)(0.3679) sin(1.1075) = approximately 1.35 Next look at a Simulink block model for this problem which makes a plot the response. Note the value of this curve at t = 1 is about 1.35 as expected. Complex Conjugate Pair of Roots in Y(s) Example 2 10 Unit_Step s 2 +2s+10 Transfer_Function PC_graph Y' , deviation output variable 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 Time (min) 4 5 ...
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