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UK EE 422G - LECTURE NOTES

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EE 422G Notes: Chapter 5 Instructor: Cheung Page 5-225-3 Inversion of Rational Functions All the Laplace Transform you will encounter has the following form: snnnnmmmmeasasasbsbsbsbτ−−−−−+++++++1111110...... Rational function X(s) Delay Why? Rational functions come out naturally from the Laplace transform of ordinary differential equations, exponential, cosine and sine functions. Our strategy: breakdown a general rational function into simpler fractions and polynomials whose inverse transforms have already been computed in Table 5-3. How? I. Convert non-proper rational function into proper rational function Non-proper: degree of numerator ≥ degree of denominator Approach: Long Division Example: 4454412995)(2342345+++++++++=ssssssssssX ssssssssssssssssssssssss44454485444544129954454123423423452345234++++++++++++++++++++++ 445441)(234++++++=sssssssX Polynomial Proper fractionEE 422G Notes: Chapter 5 Instructor: Cheung Page 5-23It is easy to find the Laplace transform of polynomials: )()()1()()1()()1(111)(ttLsLsLtsnnδδδ+=+=+↔−−− II. Factorize the denominator polynomial: Example: ))(()2()1()2(4454222234jsjssssssss −++=++=++++ • Factorize into REAL linear and irreducible quadratic factors. Further break down REAL irreducible quadratic factors into conjugate roots. • Notice some factors may repeat. • No analytical formula for polynomial with degree 5 or other. Need to rely on numerical methods. >> roots([1 4 5 4 4]) ans = -2.0000 -2.0000 0.0000 + 1.0000i 0.0000 - 1.0000i >> factor(sym('s^4+4*s^3+5*s^2+4*s+4')) ans = (s+2)^2*(s^2+1) III. Break down the proper rational function into simple fractions. This technique is called Partial Fraction Expansion. Example: )1()2(4)(22++=ssssX First, write down the full partial fraction expansion with enough number of unknowns to represent all possible numerators: jsdjsdsbsasX−++++++=2)2(2)(EE 422G Notes: Chapter 5 Instructor: Cheung Page 5-24Four simple rules to write down unknowns: 1. ...)...(...+−→− askas for real root a 2. ...)...)((...+−+−→−− askaskasas for complex conjugate roots a and a 3. ...)(...)(...)(...221+−++−+−→−nnnaskaskaskas 4. ...)(.)(......)()(...11+−+−++−+−→−−nnnnnnaskaskaskaskasas IV. Solve for the unknowns. Many methods exist. I will talk about two: 1. Compare coefficients (The Dumb Way) Example: )1()2(4)(22++=ssssX )1()2())Im(82())Im(8)Re(8())Im(2)Re(82())Re(2()2(2)(22232+++++++++++++=−++++++=ssdbasddasddbasdajsdjsdsbsasX Notice that when you expand it out, no complex coefficients remain and you can ALWAYS do that if you follow the four simple rules above. You can ask Matlab to do this too: >> x=sym('a/(s+2)+b/((s+2)^2)+(rd+i*id)/(s+i)+(rd-i*id)/(s-i)'); >> x1 = simplify(x) % Multiply out x1 = (8*id*s+2*id*s^2+a*s+a*s^3+2*a*s^2+2*a+b*s^2+b+8*rd*s^2+8*rd*s+2*rd*s^3+8*id)/(s+2)^2/(s^2+1) >> x2 = sort(collect(x1)) %collect like terms in descending power x2 = ((a+2*rd)*s^3+(2*a+b+8*rd+2*id)*s^2+2*a+(a+8*rd+8*id)*s+b+8*id)/(s+2)^2/(s^2+1)EE 422G Notes: Chapter 5 Instructor: Cheung Page 5-25Since we are given )1()2(4)(22++=ssssX , we know that the numerator polynomial is just s4. By comparing coefficients, we get the following set of equations: 0)Im(824)Im(8)Re(80)Im(2)Re(820)Re(2=++=++=+++=+dbaddaddbada To solve this system of equations numerically, rewrite it in matrix form: =0400)Im()Re(8012880128120201ddba Then use matlab: >> A = [1 0 2 0;2 1 8 2;1 0 8 8; 2 1 0 8]; >> b = [0;0;4;0]; >> A\b % Solution to Ax=b ans = -0.4800 -1.6000 0.2400 0.3200 or 32.0)Im(,24.0)Re(,6.1,48.0==−=−=ddbaEE 422G Notes: Chapter 5 Instructor: Cheung Page 5-262. Heaviside’s Theorem: Alternatively, you can use the Heaviside’s Theorem: Example/ jsdjsdsbsassssX−++++++=++=222)2(2)1()2(4)( Multiply both sides by )( js+, jsjsddsjsbsjsajsss−++++++++=−+)()2()(2)()()2(422 Note that every term on the right has a factor (s+j). Substitute s=- j (the root for (s+j)): djjjj=−−+−−)()2()(42 or d = 0.24 + j0.32 A similar approach can solve for b as well. Multiply both sides by (s+2)2: jssdjssdbsass−+++++++=+222)2()2()2()1(4 Substituting s=-2: 6.1−=b However, we cannot use this approach find a. If we multiple )2(+s to both sides, we have jssdjssdsbasss−+++++++=++)2()2()2()2)(1(42 As (s+2) is a repeated factor, we still have one left after the multiplication. We can’t substitute s=-2 here as it will turn both sides to infinity. Instead, we still multiply (s+2)2 but then we take the derivative with respect to s: 2222222222)()2()2(2)()2()2(2)1(8)1(4)2)(1()2(4jssdjssdjssdjssdasssssssdsd−+−−++++−+++=+−+=+++ Taking the derivative kills b and exposes a while all the remaining terms will still have a factor of (s+2). Substitute s=-2, we get a =-0.48.EE 422G Notes: Chapter 5 Instructor: Cheung Page 5-27The same approach can be used for factors with multiplicity > 2. Example: Find the partial fraction expansion of )2()1(14)(3+++=ssssX Again, we first write down the partial fraction expansion )2()1(14)(3+++=ssssX Both methods arrive at the same answer: jsjjsjsssX−−+++++−++−=32.024.032.024.0)2(6.1248.0)(2 The previous four steps: converting to proper polynomial, factorizing denominator polynomial, writing down partial fraction expansion, and solving for the unknown coefficients can all done by a simple MATLAB command: residue From the Matlab Help-page: RESIDUE Partial-fraction expansion (residues). [R,P,K] = RESIDUE(B,A) finds the residues, poles and direct term of a partial fraction expansion of the ratio of two polynomials B(s)/A(s). If there are no multiple roots, B(s) R(1) R(2)


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