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
Polynomials and roots
polynomials
a(s) = a0 + a1s + · · · + ansn
• a is a polynomial in the variable s
• ai are the coefficients of a (usually real, but occasionally complex)
• n is the degree of a (assuming an 6= 0)
roots (or zeros) of a polynomial a: λ ∈ C that satisfy
a(λ) = 0
examples
• a(s) = 3 has no roots
3
√ √
• a(s) = s − 1 has three roots: 1, (−1 + j 3)/2, (−1 − j 3)/2
Rational functions and partial fraction expansion 5–2
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:
a(s)
b(s) =
s−λ
is a polynomial (of degree one less than the degree of a)
• we can express a as
a(s) = (s − λ)b(s)
for some polynomial b
example: s3 − 1 has a root at s = 1
s3 − 1 = (s − 1)(s2 + s + 1)
Rational functions and partial fraction expansion 5–3
the multiplicity of a root λ is the number of factors s − λ we can factor
out, i.e., the largest k such that
a(s)
(s − λ)k
is a polynomial
example:
a(s) = s3 + s2 − s − 1
• a has a zero at s = −1
a(s) s3 + s 2 − s − 1
• = = s2 − 1 also has a zero at s = −1
s+1 s+1
a(s) s3 + s 2 − s − 1
• = = s − 1 does not have a zero at s = −1
(s + 1)2 (s + 1)2
so the multiplicity of the zero at s = −1 is 2
Rational functions and partial fraction expansion 5–4
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) = ansn + · · · + a0 = an(s − λ1) · · · (s − λn)
where λ1, . . . , λn are the n roots of a
example: s3 + s2 − s − 1 = (s + 1)2(s − 1)
the relation between the coefficients ai and the λi is complicated in
general, but
Yn Xn
a0 = an (−λi), an−1 = −an λi
i=1 i=1
are two identities that are worth remembering
Rational functions and partial fraction expansion 5–5
Conjugate symmetry
if the coefficients a0, . . . , an are real, and λ ∈ C is a root, i.e.,
a(λ) = a0 + a1λ + · · · + anλn = 0
then we have
n
a(λ) = a0 + a1λ + · · · + anλ = (a0 + a1λ + · · · + anλn) = 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
example:
PSfrag replacements =
λ1
λ4
λ3 λ6
<
λ5
λ2
λ3 and λ6 are real; λ1, λ2 are a complex conjugate pair; λ4, λ5 are a
complex conjugate pair
if a has real coefficients, we can factor it as
à r !à m !
Y Y
a(s) = an (s − λi) (s − λi)(s − λi)
i=1 i=r+1
where λ1, . . . , λr are the real roots; λr+1, λr+1, . . . , λm, λm are the
complex roots
Rational functions and partial fraction expansion 5–7
Real factored form
(s − λ)(s − λ) = s2 − 2(<λ) s + |λ|2
is a quadratic with real coefficients
real factored form of a polynomial a:
à r
!Ã m
!
Y Y
a(s) = an (s − λi) (s2 + αis + βi)
i=1 i=r+1
• λ1, . . . , λr are the real roots
• αi, βi are real and satisfy αi2 < 4βi
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
example: s3 − 1 has roots
√ √
−1 + j 3 −1 − j 3
s = 1, s= , s=
2 2
• complex factored form
³ √ ´³ √ ´
s3 − 1 = (s − 1) s + (1 + j 3)/2 s + (1 − j 3)/2
• real factored form
s3 − 1 = (s − 1)(s2 + s + 1)
Rational functions and partial fraction expansion 5–9
Rational functions
a rational function has the form
b(s) b0 + b 1 s + · · · + b m s m
F (s) = = n
,
a(s) a0 + a 1 s + · · · + a n s
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 2 1 s s3 + 3s + 3
, s + 3, + = 3
s+1 s + 1 2s + 3 2s + 3s2 + 2s + 3
2
Rational functions and partial fraction expansion 5–10
b(s)
rational function F (s) =
a(s)
polynomials b and a are not uniquely determined, e.g.,
1 3 s2 + 3
= =
s + 1 3s + 3 (s + 1)(s2 + 3)
√
(except at s = ±j 3. . . )
rational functions are closed under addition, subtraction, multiplication,
division (except by the rational function 0)
Rational functions and partial fraction expansion 5–11
Poles & zeros
b(s) b0 + b 1 s + · · · + b m s m
F (s) = = n
,
a(s) a0 + a 1 s + · · · + a n s
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
lims→λ |F (s)| = ∞
the multiplicity of a zero (or pole) λ of F is the multiplicity of the root λ
of b (or a)
6s + 12
example: 2 has one zero at s = −2, two poles at s = −1
s + 2s + 1
Rational functions and partial fraction expansion 5–12
factored or pole-zero form of F :
b0 + b 1 s + · · · + b m s m (s − z1) · · · (s − zm)
F (s) = =k
a0 + a 1 s + · · · + a n sn (s − p1) · · · (s − pn)
where
• k = bm/an
• z1, . . . , zm are the zeros of F (i.e., roots of b)
• p1, . . . , pn 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
Pole-zero plots
poles & zeros of a rational functions are often shown in a pole-zero plot
=
j
PSfrag replacements 1
<
(× denotes a pole; ◦ denotes a zero)
this example is for
(s + 1.5)(s + 1 + 2j)(s + 1 − 2j)
F (s) = k
(s + 2.5)(s − 2)(s − 1 − j)(s − 1 + j)
(s + 1.5)(s2 + 2s + 5)
= k
(s + 2.5)(s − 2)(s2 − 2s + 2)
(the plot doesn’t tell us k)
Rational functions and partial fraction expansion 5–14
Partial fraction expansion
b(s) b0 + b 1 s + · · · + b m s m
F (s) = =
a(s) a0 + a 1 s + · · · + a n sn
let’s assume (for now)
• no poles are repeated, i.e., all roots of a have multiplicity one
• m<n
then we can write F in the form
r1 rn
F (s) = + ··· +
s − λ1 s − λn
called partial fraction expansion of F
• λ1, . . . , λn are the poles of F
• the numbers r1, . . . , rn are called the residues
• when λk = λl, rk = rl
Rational functions and partial fraction expansion 5–15
example:
s2 − 2 −1 1 1
= + +
s3 + 3s2 + 2s s s+1 s+2
let’s check:
−1 1 1 −1(s + 1)(s + 2) + s(s + 2) + s(s + 1)
+ + =
s s+1 s+2 s(s + 1)(s + 2)
in partial fraction form, inverse Laplace transform is easy:
µ ¶
r1 rn
L−1(F ) = L−1 + ··· +
s − λ1 s − λn
= r 1 eλ1 t + · · · + r n eλn t
(this is real since whenever the poles are conjugates, the corresponding
residues are also)
Rational functions and partial fraction expansion 5–16
Finding the partial fraction expansion
two steps:
• find poles λ1, . . . , λn (i.e., factor a(s))
• find residues r1, . . . , rn (several methods)
method 1: solve linear equations
we’ll illustrate for m = 2, n = 3
b0 + b1s + b2s2 r1 r2 r3
= + +
(s − λ1)(s − λ2)(s − λ3) s − λ1 s − λ2 s − λ3
Rational functions and partial fraction expansion 5–17
clear denominators:
b0 + b1s + b2s2 = r1(s − λ2)(s − λ3) + r2(s − λ1)(s − λ3) + r3(s − λ1)(s − λ2)
equate coefficients:
• coefficient of s0:
b0 = (λ2λ3)r1 + (λ1λ3)r2 + (λ1λ2)r3
• coefficient of s1:
b1 = (−λ2 − λ3)r1 + (−λ1 − λ3)r2 + (−λ1 − λ2)r3
• coefficient of s2:
b2 = r 1 + r 2 + r 3
now solve for r1, r2, r3 (three equations in three variables)
Rational functions and partial fraction expansion 5–18
method 2: to get r1, multiply both sides by s − λ1 to get
(s − λ1)(b0 + b1s + b2s2) r2(s − λ1) r3(s − λ1)
= r1 + +
(s − λ1)(s − λ2)(s − λ3) s − λ2 s − λ3
cancel s − λ1 term on left and set s = λ1:
b0 + b1λ1 + b2λ21
= r1
(λ1 − λ2)(λ1 − λ3)
an explicit formula for r1! (can get r2, r3 the same way)
in the general case we have the formula
rk = (s − λk )F (s)|s=λk
which means:
• multiply F by s − λk
• then cancel s − λk from numerator and denominator
• then evaluate at s = λk to get rk
Rational functions and partial fraction expansion 5–19
example:
s2 − 2 r1 r2 r3
= + +
s(s + 1)(s + 2) s s+1 s+2
• residue r1:
µ ¶¯ 2
¯
r2 s r3s ¯¯ s −2 ¯
r1 = r1 + + = ¯ = −1
s+1 s+2 ¯ s=0 (s + 1)(s + 2) ¯ s=0
• residue r2:
µ ¶¯ 2
¯
r1(s + 1) r3(s + 1) ¯
¯ s − 2 ¯¯
r2 = + r2 + = =1
s s+2 ¯
s=−1 s(s + 2) s=−1
¯
• residue r3:
µ ¶¯ 2
¯
r1(s + 2) r2(s + 2) ¯ s − 2 ¯¯
r3 = + + r3 ¯
¯ = =1
s s+1 s=−2 s(s + 1) s=−2
¯
so we have:
s2 − 2 −1 1 1
= + +
s(s + 1)(s + 2) s s+1 s+2
Rational functions and partial fraction expansion 5–20
method 3: another explicit and useful expression for rk is:
b(λk )
rk =
a0(λk )
to see this, note that
(s − λk )b(s) b(s) + b0(s)(s − λk ) b(λk )
rk = lim = lim 0
= 0
s→λk a(s) s→λ k a (s) a (λk )
where we used l’Hôpital’s rule in second line
example (previous page):
s2 − 2 s2 − 2
=
s(s + 1)(s + 2) s3 + 2s2 + 2s
hence, ¯
2
s − 2 ¯¯
r1 = 2 = −1
3s + 4s + 2 ¯ s=0
Rational functions and partial fraction expansion 5–21
Example
let’s solve
v 000 − v = 0, v(0) = 1, v 0(0) = v 00(0) = 0
1. take Laplace transform:
s3V (s) − s2 −V (s) = 0
| {z }
L(v 000 )
2. solve for V to get
s2
V (s) = 3
s −1
3. the poles of V are the cuberoots of 1, i.e., ej2πk/3, k = 0, 1, 2
3
³ √ ´³ √ ´
s − 1 = (s − 1) s + 1/2 + j 3/2 s + 1/2 − j 3/2
Rational functions and partial fraction expansion 5–22
4. now convert V to partial fraction form
r1 r2 r2
V (s) = + √ + √
s−1 s+ +j
1 3
s + 1
− j 3
2 2 2 2
to find residues we’ll use
√
b(1) 1 b(−1/2 − j 3/2) 1
r1 = 0
= , r2 = √ =
a (1) 3 a0(−1/2 − j 3/2) 3
so partial fraction form is
1 1 1
3 3 3
V (s) = + √ + √
s−1 1
s+ +j 3 1
s+ −j 3
2 2 2 2
(check this by just multiplying out . . . )
Rational functions and partial fraction expansion 5–23
5. take inverse Laplace transform to get v:
√ √
1 t 1 (− 1 −j 3 1 1 3
v(t) = e + e 2 2 )t + e(− 2 +j 2 )t
3 3 3
√
1 t 2 −t 3
= e + e cos
2 t
3 3 2
6. check that v 000 − v = 0, v(0) = 1, v 0(0) = v 00(0) = 0
Rational functions and partial fraction expansion 5–24
Repeated poles
now suppose
b(s)
F (s) =
(s − λ1)k1 · · · (s − λl)kl
• the poles λi are distinct (λi 6= λj for i 6= j) and have multiplicity ki
• degree of b less than degree of a
partial fraction expansion has the form
r1,k1 r1,k1−1 r1,1
F (s) = + + ··· +
(s − λ1)k1 (s − λ1)k1−1 s − λ1
r2,k2 r2,k2−1 r2,1
+ + + · · · +
(s − λ2)k2 (s − λ2)k2−1 s − λ2
rl,kl rl,kl−1 rl,1
+ ··· + + + · · · +
(s − λl)kl (s − λl)kl−1 s − λl
n residues, just as before; terms involve higher powers of 1/(s − λ)
Rational functions and partial fraction expansion 5–25
1
example: F (s) = 2 has expansion
s (s + 1)
r1 r2 r3
F (s) = + +
s2 s s+1
inverse Laplace transform of partial fraction form is easy since
µ ¶
r r
L −1
= tk−1eλt
(s − λ)k (k − 1)!
same types of tricks work to find the ri,j ’s
• solve linear equations (method 1)
• can find the residues for nonrepeated poles as before
Rational functions and partial fraction expansion 5–26
example:
1 r1 r2 r3
= + +
s2(s + 1) s2 s s+1
we get (as before)
r3 = (s + 1)F (s)|s=−1 = 1
now clear denominators to get
r1(s + 1) + r2s(s + 1) + s2 = 1
(1 + r2)s2 + (r1 + r2)s + 1 = 1
which yields r2 = −1, r1 = 1, so
1 1 1
F (s) = − +
s2 s s + 1
Rational functions and partial fraction expansion 5–27
extension of method 2: to get ri,ki ,
• multiply on both sides by (s − λi)ki
• evaluate at s = λi
gives ¯
ki ¯
F (s)(s − λi) s=λi
= ri,ki
to get other r’s, we have extension
j
¯
1 d ¡ ¢
ki ¯
¯
F (s)(s − λi) ¯ = ri,ki−j
j! dsj s=λi
usually the ki’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
example (ctd.):
r1 r2 r3
F (s) = + +
s2 s s+1
• multiply by s2:
2 1 r3 s 2
s F (s) = = r1 + r2 s +
s+1 s+1
• evaluate at s = 0 to get r1 = 1
• differentiate with respect to s:
µ 2
¶
1 d r3 s
− = r 2 +
(s + 1)2 ds s+1
• evaluate at s = 0 to get r2 = −1
(same as what we got above)
Rational functions and partial fraction expansion 5–29
Nonproper rational functions
b(s) b0 + b 1 s + · · · + b m s m
F (s) = = n
,
a(s) a0 + a 1 s + · · · + a n s
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:
F (s) = b(s)/a(s) = c(s) + d(s)/a(s)
where
c(s) = c0 + · · · + cm−nsm−n, d = d 0 + · · · + d k sk , k<n
then
L−1(F ) = c0δ + · · · + cm−nδ (m−n) + L−1(d/a)
d/a is strictly proper, hence has partial fraction form
Rational functions and partial fraction expansion 5–30
example
5s + 3
F (s) =
s+1
is proper, but not strictly proper
5(s + 1) − 5 + 3 2
F (s) = =5− ,
s+1 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
Example
s4 + s3 − 2s2 + 1
F (s) =
s3 + 2s2 + s
1. write as a sum of a polynomial and a strictly proper rational function:
s(s3 + 2s2 + s) − s(2s2 + s) + s3 − 2s2 + 1
F (s) =
s3 + 2s2 + s
−s3 − 3s2 + 1
= s+ 3
s + 2s2 + s
−(s3 + 2s2 + s) + (2s2 + s) − 3s2 + 1
= s+
s3 + 2s2 + s
−s2 + s + 1
= s−1+ 3
s + 2s2 + s
−s2 + s + 1
= s−1+
s(s + 1)2
Rational functions and partial fraction expansion 5–32
2. partial fraction expansion
−s2 + s + 1 r1 r2 r3
= + +
s(s + 1)2 s s + 1 (s + 1)2
• determine r1: ¯
2
−s + s + 1 ¯¯
r1 = =1
(s + 1)2 ¯ s=0
• determine r3: ¯
2
−s + s + 1 ¯¯
r3 = =1
s ¯
s=−1
• determine r2:
µ 2
¶¯ 2
¯
d −s + s + 1 ¯¯ −s − 1 ¯¯
r2 = = = −2
ds s ¯
s=−1 s2 ¯ s=−1
(alternatively, just plug in some value of s other than s = 0 or s = −1:
¯
2
−s + s + 1 ¯¯ 1 r2 r3 r2 1
¯ = = r 1 + + = 1 + + =⇒ r2 = −2)
s(s + 1)2 ¯ 4 2 4 2 4
s=1
Rational functions and partial fraction expansion 5–33
3. inverse Laplace transform
µ ¶
1 2 1
L−1(F (s)) = L−1 s−1+ − +
s s + 1 (s + 1)2
= δ 0(t) − δ(t) + 1 − 2e−t + te−t
Rational functions and partial fraction expansion 5–34
