Lecture 5 Rational functions and partial fraction expansion

Transcription

1 S. Boyd EE102 Lecture 5 Rational functions and partial fraction expansion (review of) polynomials rational functions pole-zero plots partial fraction expansion repeated poles nonproper rational functions 5 1

2 Polynomials and roots polynomials a(s) = a 0 + a 1 s + + a n s n a is a polynomial in the variable s a i are the coefficients of a (usually real, but occasionally complex) n is the degree of a (assuming a n 0) roots (or zeros) of a polynomial a: λ C that satisfy a(λ) = 0 examples a(s) = 3 has no roots a(s) = s 3 1 has three roots: 1, ( 1 + j 3)/2, ( 1 j 3)/2 Rational functions and partial fraction expansion 5 2

3 factoring out roots of a polynomial if a has a root at s = λ we can factor out s λ: dividing a by s λ yields a polynomial: b(s) = a(s) s λ is a polynomial (of degree one less than the degree of a) we can express a as for some polynomial b a(s) = (s λ)b(s) example: s 3 1 has a root at s = 1 s 3 1 = (s 1)(s 2 + s + 1) Rational functions and partial fraction expansion 5 3

4 the multiplicity of a root λ is the number of factors s λ we can factor out, i.e., the largest k such that is a polynomial a(s) (s λ) k example: a(s) = s 3 + s 2 s 1 a has a zero at s = 1 a(s) s + 1 = s3 + s 2 s 1 s + 1 = s 2 1 also has a zero at s = 1 a(s) (s + 1) 2 = s3 + s 2 s 1 (s + 1) 2 = s 1 does not have a zero at s = 1 so the multiplicity of the zero at s = 1 is 2 Rational functions and partial fraction expansion 5 4

5 Fundamental theorem of algebra a polynomial of degree n has exactly n roots, counting multiplicities this means we can write a in factored form a(s) = a n s n + + a 0 = a n (s λ 1 ) (s λ n ) where λ 1,..., λ n are the n roots of a example: s 3 + s 2 s 1 = (s + 1) 2 (s 1) the relation between the coefficients a i and the λ i is complicated in general, but n n a 0 = a n ( λ i ), a n 1 = a n λ i i=1 are two identities that are worth remembering i=1 Rational functions and partial fraction expansion 5 5

6 Conjugate symmetry if the coefficients a 0,..., a n are real, and λ C is a root, i.e., a(λ) = a 0 + a 1 λ + + a n λ n = 0 then we have a(λ) = a 0 + a 1 λ + + a n λ n = (a 0 + a 1 λ + + a n λ n ) = a(λ) = 0 in other words: λ is also a root if λ is real this isn t interesting if λ is complex, it gives us another root for free complex roots come in complex conjugate pairs Rational functions and partial fraction expansion 5 6

7 example: PSfrag replacements I λ 1 λ 3 λ 4 λ 6 R λ 5 λ 2 λ 3 and λ 6 are real; λ 1, λ 2 are a complex conjugate pair; λ 4, λ 5 are a complex conjugate pair if a has real coefficients, we can factor it as ( r ) ( m a(s) = a n λ i ) i=1(s i=r+1 (s λ i )(s λ i ) where λ 1,..., λ r are the real roots; λ r+1, λ r+1,..., λ m, λ m are the complex roots ) Rational functions and partial fraction expansion 5 7

8 Real factored form (s λ)(s λ) = s 2 2(Rλ) s + λ 2 is a quadratic with real coefficients real factored form of a polynomial a: ( r ) ( m ) a(s) = a n λ i ) (s i=1(s 2 + α i s + β i ) i=r+1 λ 1,..., λ r are the real roots α i, β i are real and satisfy αi 2 < 4β i any polynomial with real coefficients can be factored into a product of degree one polynomials with real coefficients quadratic polynomials with real coefficients Rational functions and partial fraction expansion 5 8

9 example: s 3 1 has roots s = 1, s = 1 + j 3, s = 1 j complex factored form s 3 1 = (s 1) ( s + (1 + j ) ( 3)/2 s + (1 j ) 3)/2 real factored form s 3 1 = (s 1)(s 2 + s + 1) Rational functions and partial fraction expansion 5 9

10 Rational functions a rational function has the form F (s) = b(s) a(s) = b 0 + b 1 s + + b m s m a 0 + a 1 s + + a n s n, i.e., a ratio of two polynomials (where a is not the zero polynomial) b is called the numerator polynomial a is called the denominator polynomial examples of rational functions: 1 s + 1, s2 + 3, 1 s s 2s + 3 = s 3 + 3s + 3 2s 3 + 3s 2 + 2s + 3 Rational functions and partial fraction expansion 5 10

11 rational function F (s) = b(s) a(s) polynomials b and a are not uniquely determined, e.g., (except at s = ±j 3... ) 1 s + 1 = 3 3s + 3 = s (s + 1)(s 2 + 3) rational functions are closed under addition, subtraction, multiplication, division (except by the rational function 0) Rational functions and partial fraction expansion 5 11

12 Poles & zeros F (s) = b(s) a(s) = b 0 + b 1 s + + b m s m a 0 + a 1 s + + a n s n, assume b and a have no common factors (cancel them out if they do... ) the m roots of b are called the zeros of F ; λ is a zero of F if F (λ) = 0 the n roots of a are called the poles of F ; λ is a pole of F if lim s λ F (s) = the multiplicity of a zero (or pole) λ of F is the multiplicity of the root λ of b (or a) example: 6s + 12 s 2 + 2s + 1 has one zero at s = 2, two poles at s = 1 Rational functions and partial fraction expansion 5 12

13 factored or pole-zero form of F : where F (s) = b 0 + b 1 s + + b m s m a 0 + a 1 s + + a n s n = k(s z 1) (s z m ) (s p 1 ) (s p n ) k = b m /a n z 1,..., z m are the zeros of F (i.e., roots of b) p 1,..., p n are the poles of F (i.e., roots of a) (assuming the coefficients of a and b are real) complex poles or zeros come in complex conjugate pairs can also have real factored form... Rational functions and partial fraction expansion 5 13

14 Pole-zero plots poles & zeros of a rational functions are often shown in a pole-zero plot I Sfrag replacements j 1 R ( denotes a pole; denotes a zero) this example is for F (s) = (s + 1.5)(s j)(s + 1 2j) k (s + 2.5)(s 2)(s 1 j)(s 1 + j) = (s + 1.5)(s 2 + 2s + 5) k (s + 2.5)(s 2)(s 2 2s + 2) (the plot doesn t tell us k) Rational functions and partial fraction expansion 5 14

15 let s assume (for now) Partial fraction expansion F (s) = b(s) a(s) = b 0 + b 1 s + + b m s m a 0 + a 1 s + + a n s n no poles are repeated, i.e., all roots of a have multiplicity one m < n then we can write F in the form F (s) = r 1 s λ called partial fraction expansion of F λ 1,..., λ n are the poles of F r n s λ n the numbers r 1,..., r n are called the residues when λ k = λ l, r k = r l Rational functions and partial fraction expansion 5 15

16 example: s 2 2 s 3 + 3s 2 + 2s = 1 s + 1 s s + 2 let s check: 1 s + 1 s s + 2 = 1(s + 1)(s + 2) + s(s + 2) + s(s + 1) s(s + 1)(s + 2) in partial fraction form, inverse Laplace transform is easy: ( L 1 (F ) = L 1 r1 + + r ) n s λ 1 s λ n = r 1 e λ 1t + + r n e λ nt (this is real since whenever the poles are conjugates, the corresponding residues are also) Rational functions and partial fraction expansion 5 16

17 Finding the partial fraction expansion two steps: find poles λ 1,..., λ n (i.e., factor a(s)) find residues r 1,..., r n (several methods) method 1: solve linear equations we ll illustrate for m = 2, n = 3 b 0 + b 1 s + b 2 s 2 (s λ 1 )(s λ 2 )(s λ 3 ) = r 1 s λ 1 + r 2 s λ 2 + r 3 s λ 3 Rational functions and partial fraction expansion 5 17

18 clear denominators: b 0 +b 1 s+b 2 s 2 = r 1 (s λ 2 )(s λ 3 )+r 2 (s λ 1 )(s λ 3 )+r 3 (s λ 1 )(s λ 2 ) equate coefficients: coefficient of s 0 : b 0 = (λ 2 λ 3 )r 1 + (λ 1 λ 3 )r 2 + (λ 1 λ 2 )r 3 coefficient of s 1 : b 1 = ( λ 2 λ 3 )r 1 + ( λ 1 λ 3 )r 2 + ( λ 1 λ 2 )r 3 coefficient of s 2 : b 2 = r 1 + r 2 + r 3 now solve for r 1, r 2, r 3 (three equations in three variables) Rational functions and partial fraction expansion 5 18

19 method 2: to get r 1, multiply both sides by s λ 1 to get (s λ 1 )(b 0 + b 1 s + b 2 s 2 ) (s λ 1 )(s λ 2 )(s λ 3 ) = r 1 + r 2(s λ 1 ) + r 3(s λ 1 ) s λ 2 s λ 3 cancel s λ 1 term on left and set s = λ 1 : b 0 + b 1 λ 1 + b 2 λ 2 1 (λ 1 λ 2 )(λ 1 λ 3 ) = r 1 an explicit formula for r 1! (can get r 2, r 3 the same way) in the general case we have the formula r k = (s λ k )F (s) s=λk which means: multiply F by s λ k then cancel s λ k from numerator and denominator then evaluate at s = λ k to get r k Rational functions and partial fraction expansion 5 19

20 example: s 2 2 s(s + 1)(s + 2) = r 1 s + r 2 s r 3 s + 2 residue r 1 : r 1 = ( r 1 + r 2s s r 3s s + 2) s=0 = s 2 2 (s + 1)(s + 2) = 1 s=0 residue r 2 : ( r1 (s + 1) r 2 = s + r 2 + r ) 3(s + 1) s + 2 s= 1 = s2 2 s(s + 2) = 1 s= 1 residue r 3 : ( r1 (s + 2) r 3 = s + r 2(s + 2) s + 1 ) + r 3 = s= 2 s2 2 s(s + 1) = 1 s= 2 so we have: s 2 2 s(s + 1)(s + 2) = 1 s + 1 s s + 2 Rational functions and partial fraction expansion 5 20

21 method 3: another explicit and useful expression for r k is: r k = b(λ k) a (λ k ) to see this, note that r k = lim s λ k (s λ k )b(s) a(s) = lim s λ k b(s) + b (s)(s λ k ) a (s) = b(λ k) a (λ k ) where we used l Hôpital s rule in second line example (previous page): s 2 2 s(s + 1)(s + 2) = s 2 2 s 3 + 2s 2 + 2s hence, r 1 = s 2 2 3s 2 + 4s + 2 = 1 s=0 Rational functions and partial fraction expansion 5 21

22 Example let s solve v v = 0, v(0) = 1, v (0) = v (0) = 0 1. take Laplace transform: s 3 V (s) s } {{ } 2 V (s) = 0 L(v ) 2. solve for V to get V (s) = s2 s the poles of V are the cuberoots of 1, i.e., e j2πk/3, k = 0, 1, 2 s 3 1 = (s 1) ( s + 1/2 + j ) ( 3/2 s + 1/2 j ) 3/2 Rational functions and partial fraction expansion 5 22

23 4. now convert V to partial fraction form V (s) = r 1 s 1 + r 2 s j r 2 s j 3 2 to find residues we ll use r 1 = b(1) a (1) = 1 3, r 2 = b( 1/2 j 3/2) a ( 1/2 j 3/2) = 1 3 so partial fraction form is V (s) = s s j s j 3 2 (check this by just multiplying out... ) Rational functions and partial fraction expansion 5 23

24 5. take inverse Laplace transform to get v: v(t) = 1 3 et e( 1 2 j 3 2 )t e( 1 2 +j 3 2 )t = 1 3 et e t 2 cos 3 2 t 6. check that v v = 0, v(0) = 1, v (0) = v (0) = 0 Rational functions and partial fraction expansion 5 24

25 Repeated poles now suppose F (s) = b(s) (s λ 1 ) k 1 (s λl ) k l the poles λ i are distinct (λ i λ j for i j) and have multiplicity k i degree of b less than degree of a partial fraction expansion has the form F (s) = r 1,k1 (s λ 1 ) k 1 + r 1,k 1 1 (s λ 1 ) k r 1,1 s λ 1 + r 2,k 2 (s λ 2 ) k 2 + r 2,k 2 1 (s λ 2 ) k r 2,1 s λ r l,k l (s λ l ) k l + r l,k l 1 (s λ l ) k l r l,1 s λ l n residues, just as before; terms involve higher powers of 1/(s λ) Rational functions and partial fraction expansion 5 25

26 example: F (s) = 1 s 2 (s + 1) has expansion F (s) = r 1 s 2 + r 2 s + r 3 s + 1 inverse Laplace transform of partial fraction form is easy since ( ) L 1 r (s λ) k = r (k 1)! tk 1 e λt same types of tricks work to find the r i,j s solve linear equations (method 1) can find the residues for nonrepeated poles as before Rational functions and partial fraction expansion 5 26

27 example: we get (as before) 1 s 2 (s + 1) = r 1 s 2 + r 2 s + r 3 s + 1 r 3 = (s + 1)F (s) s= 1 = 1 now clear denominators to get which yields r 2 = 1, r 1 = 1, so r 1 (s + 1) + r 2 s(s + 1) + s 2 = 1 (1 + r 2 )s 2 + (r 1 + r 2 )s + 1 = 1 F (s) = 1 s 2 1 s + 1 s + 1 Rational functions and partial fraction expansion 5 27

28 extension of method 2: to get r i,ki, multiply on both sides by (s λ i ) k i evaluate at s = λ i gives to get other r s, we have extension 1 j! F (s)(s λ i ) k i s=λi = r i,ki d j ( F (s)(s λi ds j ) k ) i = r i,ki j s=λi usually the k i s are small (e.g., 1 or 2), so fortunately this doesn t come up too often Rational functions and partial fraction expansion 5 28

29 example (ctd.): F (s) = r 1 s 2 + r 2 s + r 3 s + 1 multiply by s 2 : evaluate at s = 0 to get r 1 = 1 differentiate with respect to s: s 2 F (s) = 1 s + 1 = r 1 + r 2 s + r 3s 2 s (s + 1) 2 = r 2 + d ds ( r3 s 2 s + 1 ) evaluate at s = 0 to get r 2 = 1 (same as what we got above) Rational functions and partial fraction expansion 5 29

30 Nonproper rational functions F (s) = b(s) a(s) = b 0 + b 1 s + + b m s m a 0 + a 1 s + + a n s n, is called proper if m n, strictly proper if m < n, nonproper if m > n partial fraction expansion requires strictly proper F ; to find L 1 (F ) for other cases, divide a into b: where F (s) = b(s)/a(s) = c(s) + d(s)/a(s) c(s) = c c m n s m n, d = d d k s k, k < n then L 1 (F ) = c 0 δ + + c m n δ (m n) + L 1 (d/a) d/a is strictly proper, hence has partial fraction form Rational functions and partial fraction expansion 5 30

31 example F (s) = 5s + 3 s + 1 is proper, but not strictly proper F (s) = 5(s + 1) s + 1 = 5 2 s + 1, so L 1 (F ) = 5δ(t) 2e t in general, F strictly proper f has no impulses at t = 0 F proper, not strictly proper f has an impulse at t = 0 F nonproper f has higher-order impulses at t = 0 m n determines order of impulse at t = 0 Rational functions and partial fraction expansion 5 31

32 Example F (s) = s4 + s 3 2s s 3 + 2s 2 + s 1. write as a sum of a polynomial and a strictly proper rational function: F (s) = s(s3 + 2s 2 + s) s(2s 2 + s) + s 3 2s s 3 + 2s 2 + s = s + s3 3s s 3 + 2s 2 + s = s + (s3 + 2s 2 + s) + (2s 2 + s) 3s s 3 + 2s 2 + s = s 1 + s2 + s + 1 s 3 + 2s 2 + s = s 1 + s2 + s + 1 s(s + 1) 2 Rational functions and partial fraction expansion 5 32

33 2. partial fraction expansion s 2 + s + 1 s(s + 1) 2 = r 1 s + r 2 s r 3 (s + 1) 2 determine r 1 : determine r 3 : determine r 2 : r 2 = d ds r 1 = s2 + s + 1 (s + 1) 2 r 3 = s2 + s + 1 s ( ) s 2 + s + 1 s s= 1 = 1 s=0 = 1 s= 1 = s2 1 s 2 = 2 s= 1 (alternatively, just plug in some value of s other than s = 0 or s = 1: s 2 + s + 1 = 1 s(s + 1) 2 4 = r 1 + r r 3 4 = 1 + r = r 2 = 2) s=1 Rational functions and partial fraction expansion 5 33

34 3. inverse Laplace transform ( L 1 (F (s)) = L 1 s s 2 ) s (s + 1) 2 = δ (t) δ(t) + 1 2e t + te t Rational functions and partial fraction expansion 5 34