Laplace
Transformation
Time domain
unknown f(t), d/dt, Diff Eqs
Frequency domain
unknown F(s), Alg Eqs
Solve
Differential
Equations
Solve
Algebraic
Equations
Time domain
known f(t)
Frequency domain
known F(s)
Inverse
Laplace
Transform
slide1
EE 102 spring 2001-2002
Handout #10
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 = 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 − 1 has three roots: 1, (−1 + j 3)/2, (−1 − j 3)/2
3
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:
b(s) =
a(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
s3 + s2 − s − 1
a(s)
=
= s2 − 1 also has a zero at s = −1
•
s+1
s+1
a(s)
s3 + s2 − 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
n
n
λi
a0 = an (−λi), an−1 = −an
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:
λ1
λ3
λ4
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
r
m
(s − λi )
(s − λi)(s − λi )
a(s) = an
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:
a(s) = an
r
(s − λi)
i=1
m
(s2 + αis + βi)
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
,
s=
2
s = 1,
√
−1 − j 3
s=
2
• complex factored form
√
√
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 + b1s + · · · + bmsm
F (s) =
=
,
n
a(s)
a0 + a1s + · · · + ans
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
2
s + 3,
s
s3 + 3s + 3
1
+
= 3
2
s + 1 2s + 3 2s + 3s2 + 2s + 3
Rational functions and partial fraction expansion
5–10
rational function F (s) =
b(s)
a(s)
polynomials b and a are not uniquely determined, e.g.,
3
s2 + 3
1
=
=
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 + b1s + · · · + bmsm
=
F (s) =
,
n
a(s)
a0 + a1s + · · · + ans
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
has one zero at s = −2, two poles at s = −1
example: 2
s + 2s + 1
Rational functions and partial fraction expansion
5–12
factored or pole-zero form of F :
(s − z1) · · · (s − zm)
b0 + b1 s + · · · + bm s m
=k
F (s) =
a0 + a1s + · · · + ansn
(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
1
(× denotes a pole; ◦ denotes a zero)
this example is for
F (s) = k
(s + 1.5)(s + 1 + 2j)(s + 1 − 2j)
(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 + b1s + · · · + bmsm
=
F (s) =
a(s)
a0 + a1s + · · · + ansn
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:
−1
1
1
s2 − 2
=
+
+
s3 + 3s2 + 2s
s
s+1 s+2
let’s check:
1
1
−1(s + 1)(s + 2) + s(s + 2) + s(s + 1)
−1
+
+
=
s
s+1 s+2
s(s + 1)(s + 2)
in partial fraction form, inverse Laplace transform is easy:
r
r1
n
+ ··· +
L−1(F ) = L−1
s − λ1
s − λn
= r1eλ1t + · · · + rneλnt
(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
r1
b0 + b1 s + b2 s 2
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
r2(s − λ1 ) r3(s − λ1 )
(s − λ1)(b0 + b1s + b2s2)
+
= 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:
• residue r1:
r1 =
r1
r2
r3
s2 − 2
= +
+
s(s + 1)(s + 2)
s
s+1 s+2
r3s r2 s
+
r1 +
s+1 s+2 s=0
s −2
=
(s + 1)(s + 2) 2
s=0
= −1
• residue r2:
2
r3(s + 1) s − 2 r1(s + 1)
r2 =
+ r2 +
=
=1
s
s+2
s(s + 2) s=−1
s=−1
• residue r3:
2
s − 2 r1(s + 2) r2(s + 2)
r3 =
+
+ r3 =
=1
s
s+1
s(s + 1) s=−2
s=−2
so we have:
−1
1
1
s2 − 2
=
+
+
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:
rk =
b(λk )
a(λk )
to see this, note that
b(λk )
(s − λk )b(s)
b(s) + b(s)(s − λk )
= lim
=
rk = lim
s→λk
s→λk
a(s)
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,
s − 2 r1 = 2
3s + 4s + 2 Rational functions and partial fraction expansion
2
s=0
= −1
5–21
Example
let’s solve
v − v = 0,
v(0) = 1,
v (0) = v (0) = 0
1. take Laplace transform:
s3V (s) − s2 −V (s) = 0
L(v )
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
√ √
s + 1/2 − j 3/2
s − 1 = (s − 1) s + 1/2 + j 3/2
3
Rational functions and partial fraction expansion
5–22
4. now convert V to partial fraction form
V (s) =
r1
r2
r2
√ +
√
+
3
3
1
1
s−1 s+ +j
s
+
−
j
2
2
2
2
to find residues we’ll use
r1 =
√
b(−1/2 − j 3/2)
1
√
r2 =
=
a(−1/2 − j 3/2) 3
1
b(1)
=
,
a (1) 3
so partial fraction form is
V (s) =
1
3
s−1
+
1
3
1
2
s+ +
√
j 23
+
1
3
1
2
s+ −j
√
3
2
(check this by just multiplying out . . . )
Rational functions and partial fraction expansion
5–23
5. take inverse Laplace transform to get v:
v(t) =
=
1 t 1 (− 1 −j
e + e 2
3
3
√
3
2 )t
√
1
1
+ e(− 2 +j
3
√
3
2 )t
1 t 2 −t
3
2
e + e cos
t
3
3
2
6. check that v − v = 0, v(0) = 1, v (0) = v (0) = 0
Rational functions and partial fraction expansion
5–24
Repeated poles
now suppose
F (s) =
b(s)
(s − λ1)k1 · · · (s − λl )kl
• the poles λi are distinct (λi = λj for i = j) and have multiplicity ki
• degree of b less than degree of a
partial fraction expansion has the form
F (s) =
r1,k1−1
r1,1
r1,k1
+
+ ··· +
(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
has expansion
example: F (s) = 2
s (s + 1)
F (s) =
r1 r2
r3
+
+
s2
s
s+1
inverse Laplace transform of partial fraction form is easy since
L
−1
r
(s − λ)k
r
=
tk−1eλt
(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:
r1 r2
r3
1
=
+ +
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
F (s) =
Rational functions and partial fraction expansion
1
1
1
+
−
s2 s s + 1
5–27
extension of method 2: to get ri,ki ,
• multiply on both sides by (s − λi)ki
• evaluate at s = λi
gives
F (s)(s − λi)
ki s=λi
= ri,ki
to get other r’s, we have extension
j
1 d
j! dsj
ki F (s)(s − λi) s=λi
= ri,ki−j
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.):
F (s) =
r1 r2
r3
+
+
s2
s
s+1
• multiply by s2:
1
r 3 s2
s F (s) =
= r1 + r2 s +
s+1
s+1
2
• evaluate at s = 0 to get r1 = 1
• differentiate with respect to s:
d
1
= r2 +
−
(s + 1)2
ds
2
r3 s
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 + b1s + · · · + bmsm
=
F (s) =
,
n
a(s)
a0 + a1s + · · · + ans
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,
then
d = d 0 + · · · + d k sk ,
k<n
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
F (s) =
so
2
5(s + 1) − 5 + 3
=5−
,
s+1
s+1
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:
F (s) =
s(s3 + 2s2 + s) − s(2s2 + s) + s3 − 2s2 + 1
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
r2
r3
−s2 + s + 1 r1
+
+
=
s(s + 1)2
s
s + 1 (s + 1)2
• determine r1:
−s + s + 1 r1 =
(s + 1)2 2
• determine r3:
−s + s + 1 r3 =
s
2
s=−1
• determine r2:
r2 =
s=0
=1
d
ds
−s + s + 1 s
2
=1
−s − 1 =
s2 2
s=−1
s=−1
= −2
(alternatively, just plug in some value of s other than s = 0 or s = −1:
¬
−s + s + 1 ¬¬
¬
s(s + 1)2 ¬
2
s=1
=
r3
r2
1
1
r2
= r1 +
+
=1+
+ =⇒ r2 = −2)
4
2
4
2
4
Rational functions and partial fraction expansion
5–33
3. inverse Laplace transform
L−1(F (s)) = L−1
2
1
1
+
s−1+ −
s s + 1 (s + 1)2
= δ (t) − δ(t) + 1 − 2e−t + te−t
Rational functions and partial fraction expansion
5–34
ph_dorf_app 4/24/01 4:54 AM Page 1
APPENDIX
C
Table C.1
Symbols, Units,
and Conversion Factors
Symbols and Units
Parameter or Variable Name
Symbol
SI
English
Acceleration, angular
a(t)
rad/s2
rad/s2
Acceleration, translational
a(t)
m/s2
ft/s2
Friction, rotational
b
Nm
rad>s
ft-lb
rad>s
Friction, translational
b
N
m>s
lb
ft>s
Inertia, rotational
J
Nm
rad>s2
rad>s2
Mass
M
kg
slugs
Position, rotational
u(t)
rad
rad
Position, translational
x(t)
m
ft
Speed, rotational
v(t)
rad/s
rad/s
Speed, translational
v(t)
m/s
ft/s
Torque
T(t)
Nm
ft-lb
ft-lb
1
ph_dorf_app 4/24/01 4:54 AM Page 2
2
Appendix C
Symbols, Units, and Conversion Factors
Table C.2
Conversion Factors
To Convert
Into
Multiply by
Btu
Btu
Btu/hr
Btu/hr
Btu/min
Btu/min
Btu/min
ft-lb
J
ft-lb/s
W
hp
kW
W
778.3
1054.8
0.2162
0.2931
0.02356
0.01757
17.57
cal
cm
cm
cm3
J
ft
in.
ft3
4.182
3.281 102
0.3937
3.531 105
deg (angle)
deg/s
dynes
dynes
dynes
rad
rpm
g
lb
N
0.01745
0.1667
1.020 103
2.248 106
105
ft/s
ft/s
ft-lb
ft-lb
ft-lb/min
ft-lb/s
ft-lb/s
ft-lb
rad>s
miles/hr
miles/min
g-cm
oz-in.
Btu/min
hp
kW
oz-in.
rpm
0.6818
0.01136
1.383 104
192
1.286 103
1.818 103
1.356 103
g
g
g-cm2
g-cm
g-cm
dynes
lb
oz-in2
oz-in.
ft-lb
980.7
2.205 103
5.468 103
1.389 102
1.235 105
hp
hp
hp
hp
Btu/min
ft-lb/min
ft-lb/s
W
42.44
33,000
550.0
745.7
in.
in.
meters
cm
2.540 102
2.540
J
J
J
J
Btu
ergs
ft-lb
W-hr
9.480 104
107
0.7376
2.778 104
20.11
gram (g), joule (J), watt (W), newton (N), watt-hour (Wh)
kg
kg
To Convert
lb
slugs
Into
2.205
6.852 102
Multiply by
kW
kW
kW
Btu/min
ft-lb/min
hp
56.92
4.462 104
1.341
miles (statute)
mph
mph
mph
mils
mils
min (angles)
min (angles)
ft
ft/min
ft/s
m/s
cm
in.
deg
rad
5280
88
1.467
0.44704
2.540 103
0.001
0.01667
2.909 104
Nm
Nm
Nms
ft-lb
dyne-cm
W
0.73756
107
1.0
oz
oz-in.
oz-in2
oz-in.
oz-in.
g
dyne-cm
g-cm2
ft-lb
g-cm
28.349527
70,615.7
1.829 102
5.208 103
72.01
lb(force)
lb/ft3
lb-ft-s2
N
g/cm3
oz-in2
4.4482
0.01602
7.419 104
rad
rad
rad
rad/s
rad/s
rad/s
rpm
rpm
deg
min
s
deg/s
rpm
rps
deg/s
rad/s
57.30
3438
2.063 105
57.30
9.549
0.1592
6.0
0.1047
s (angle)
s (angle)
slugs (mass)
slug-ft2
deg
rad
kg
km2
2.778 104
4.848 106
14.594
1.3558
W
W
W
W
W
Wh
Btu/hr
Btu/min
ft-lb/min
hp
Nm/s
Btu
3.413
0.05688
44.27
1.341 103
1.0
3.413
ph_dorf_app 4/24/01 4:54 AM Page 3
APPENDIX
D
Laplace Transform Pairs
Table D.1
F(s)
f(t), t 0
1. 1
2. 1/s
n!
3. n+1
s
1
4.
1s + a2
1
5.
1s + a2 n
a
6.
s1s + a2
1
7.
1s + a2 1s + b2
s+a
8.
1s + a2 1s + b2
ab
9.
s1s + a21s + b2
d(t0), unit impulse at t t0
1, unit step
10.
11.
12.
1
1s + a2 1s + b21s + c2
s+a
1s + a2 1s + b21s + c2
ab1s + a2
s1s + a21s + b2
v
13. 2
s + v2
s
14. 2
s + v2
tn
eat
1
tn 1eat
1n - 12!
1 eat
1
(eat ebt)
1b - a2
1
[(a a)eat (a b)ebt]
1b - a2
a
b
1
eat ebt
1b - a2
1b - a2
e-at
e-bt
e-ct
1b - a21c - a2
1c - a21a - b2
1a - c21b - c2
1a - a2e-at
1a - b2e-bt
1a - c2e-ct
1b - a21c - a2
1c - b21a - b2
1a - c21b - c2
b1a - a2 at a1a - b2 bt
a
e e
1b - a2
1b - a2
sin vt
cos vt0
Table D.1 continued
3
ph_dorf_app 4/24/01 4:54 AM Page 4
4
Appendix D
Table D.1
Laplace Transform Pairs
Continued
F(s)
f(t), t 0
s+a
s2 + v2
v
16.
1s + a2 2 + v2
2a2 + v2
sin (vt f), f tan1 v/a
v
15.
17.
18.
19.
20.
21.
22.
23.
1s + a2
1s + a2 2 + v2
s+a
1s + a2 2 + v2
v2n
2
s + 2Zvns + v2n
1
s1s + a2 2 + v2
v2n
s1s + 2Zvns + v2n 2
2
1s + a2
s1s + a2 2 + v2
1
1s + c21s + a2 2 + v2
eat sin vt
eat cos vt
1
[(a a)2 v2]1/2 eat sin (vt f),
v
v
f tan1
a-a
vn
n
ezv t sin vn 21 - z2 t, z 1
21 - z2
1
1
eat sin (vt f),
a2 + v2 v2a2 + v2
v
f tan1
-a
1
n
1
ezv t sin 1vn 21 - z2 t + f2 ,
21 - z2
f cos1 z, z 1
a
1 1a - a2 + v 1/2 at
c
d e sin (vt f),
a2 + v2 v
a2 + v2
v
v
f tan1
tan1
a-a
-a
e-at sin 1vt + f2
e-ct
v
, f tan1
2
2
2
2 1>2
c
a
1c - a2 + v
v1c - a2 + v 2
2
ph_dorf_app 4/24/01 4:54 AM Page 5
APPENDIX
An Introduction
to Matrix Algebra
E
E.1 DEFINITIONS
In many situations, we must deal with rectangular arrays of numbers or functions.
The rectangular array of numbers (or functions)
a11 a12
a
a22
A = D 21
o
o
am1 am2
p
p
p
a1n
a2n
T
o
amn
(E.1)
is known as a matrix. The numbers aij are called elements of the matrix, with the subscript i denoting the row and the subscript j denoting the column.
A matrix with m rows and n columns is said to be a matrix of order (m, n) or alternatively called an m n (m-by-n) matrix.When the number of the columns equals
the number of rows (m n), the matrix is called a square matrix of order n. It is common to use boldfaced capital letters to denote an m n matrix.
A matrix comprising only one column, that is, an m 1 matrix, is known as a
column matrix or, more commonly, a column vector. We will represent a column vector with boldfaced lowercase letters as
y1
y
y = D 2T
o
ym
(E.2)
Analogously, a row vector is an ordered collection of numbers written in a row—
that is, a 1 n matrix. We will use boldfaced lowercase letters to represent vectors.
Therefore a row vector will be written as
z = 3z1
z2
p
zn 4,
(E.3)
with n elements.
A few matrices with distinctive characteristics are given special names. A square
matrix in which all the elements are zero except those on the principal diagonal, a11,
a22, . . . , ann, is called a diagonal matrix. Then, for example, a 3 3 diagonal matrix
would be
5
ph_dorf_app 4/24/01 4:54 AM Page 6
6
Appendix E
An Introduction to Matrix Algebra
b11
B= C 0
0
0
b22
0
0
0 S.
(E.4)
b33
If all the elements of a diagonal matrix have the value 1, then the matrix is known as
the identity matrix I, which is written as
1
0
I=D
o
0
0
1
o
0
p
p
p
p
0
0
T.
o
1
(E.5)
When all the elements of a matrix are equal to zero, the matrix is called the zero, or
null matrix. When the elements of a matrix have a special relationship so that aij aji, it is called a symmetrical matrix. Thus, for example, the matrix
3
H = C -2
1
1
4S
8
-2
6
4
(E.6)
is a symmetrical matrix of order (3, 3).
E.2 ADDITION AND SUBTRACTION OF MATRICES
The addition of two matrices is possible only for matrices of the same order. The sum
of two matrices is obtained by adding the corresponding elements.Thus if the elements
of A are aij and the elements of B are bij, and if
C A B,
(E.7)
then the elements of C that are cij are obtained as
cij aij bij.
(E.8)
For example, the matrix addition for two 3 3 matrices is as follows:
2
C = C1
0
1
-1
6
0
8
3S + C1
2
4
2
3
2
1
10 3
0S = C 2 2
1
4 8
1
3S.
3
(E.9)
From the operation used for performing the operation of addition, we note that the
process is commutative; that is,
A B B A.
(E.10)
Also we note that the addition operation is associative, so that
(A B) C A (B C).
(E.11)
To perform the operation of subtraction, we note that if a matrix A is multiplied
by a constant a, then every element of the matrix is multiplied by this constant.Therefore we can write
ph_dorf_app 4/24/01 4:54 AM Page 7
Section E.3
7
Multiplication of Matrices
aa11 aa12
aa
aa22
aA = D 12
o
o
aam1 aam2
p
p
p
aa1n
aa2n
T.
o
aamn
(E.12)
Then to carry out a subtraction operation, we use a 1, and A is obtained by multiplying each element of A by 1. For example,
2
4
C=B-A=B
1
6
R-B
2
3
1
-4
R=B
1
1
0
R.
1
(E.13)
E.3 MULTIPLICATION OF MATRICES
The multiplication of two matrices AB requires that the number of columns of A
be equal to the number of rows of B. Thus if A is of order m n and B is of order n
q, then the product is of order m q. The elements of a product
C AB
(E.14)
are found by multiplying the ith row of A and the jth column of B and summing these
products to give the element cij. That is,
q
cij = ai1b1j + ai2b2j + p + aiqbqj = a aikbkj.
(E.15)
k=1
Thus we obtain c11, the first element of C, by multiplying the first row of A by the first
column of B and summing the products of the elements. We should note that, in general, matrix multiplication is not commutative; that is
AB BA.
(E.16)
Also we note that the multiplication of a matrix of m n by a column vector (order
n 1) results in a column vector of order m 1.
A specific example of multiplication of a column vector by a matrix is
a
x = Ay = B 11
a21
a12
a22
y1
a13
1a y + a12y2 + a13y3 2
R C y2 S = B 11 1
R.
a23
1a21y1 + a22y2 + a23y3 2
y3
(E.17)
Note that A is of order 2 3, and y is of order 3 1. Therefore the resulting matrix
x is of order 2 1, which is a column vector with two rows. There are two elements
of x, and
x1 (a11y1 a12y2 a13y3)
(E.18)
is the first element obtained by multiplying the first row of A by the first (and only)
column of y.
Another example, which the reader should verify, is
2
-1
C = AB = B
3
-1
RB
2 -1
2
7
R=B
-2
-5
6
R.
-6
(E.19)
ph_dorf_app 4/24/01 4:54 AM Page 8
8
Appendix E
An Introduction to Matrix Algebra
For example, the element c22 is obtained as c22 1(2) 2(2) 6.
Now we are able to use this definition of multiplication in representing a set of
simultaneous linear algebraic equations by a matrix equation. Consider the following set of algebraic equations:
3x1 2x2 x3 u1,
2x1 x2 6x3 u2,
4x1 x2 2x3 u3.
(E.20)
We can identify two column vectors as
x1
x = C x2 S
x3
u1
u = C u2 S.
u3
and
(E.21)
Then we can write the matrix equation
Ax u,
(E.22)
where
3
2
A = C2
1
4 -1
1
6 S.
2
We immediately note the utility of the matrix equation as a compact form of a set of
simultaneous equations.
The multiplication of a row vector and a column vector can be written as
xy = 3x1 x2
p
y1
y2
xn 4 D T = x1 y1 + x2 y2 + p + xn yn.
o
yn
(E.23)
Thus we note that the multiplication of a row vector and a column vector results in
a number that is a sum of a product of specific elements of each vector.
As a final item in this section, we note that the multiplication of any matrix by
the identity matrix results in the original matrix, that is, AI A.
E.4 OTHER USEFUL MATRIX OPERATIONS AND DEFINITIONS
The transpose of a matrix A is denoted in this text as AT. One will often find the notation A' for AT in the literature. The transpose of a matrix A is obtained by interchanging the rows and columns of A. For example, if
6
A= C 1
-2
then
0
4
3
2
1 S,
-1
ph_dorf_app 4/24/01 4:54 AM Page 9
Section E.4
9
Other Useful Matrix Operations and Definitions
6 1
AT = C 0 4
2 1
-2
3 S.
-1
(E.24)
Therefore we are able to denote a row vector as the transpose of a column vector and
write
xT = 3x1
x2
p
xn 4.
(E.25)
Because xT is a row vector, we obtain a matrix multiplication of xT by x as follows:
xTx = 3x1
x2
p
x1
x2
xn 4 D T = x21 + x22 + p + x2n.
o
xn
(E.26)
Thus the multiplication xTx results in the sum of the squares of each element of x.
The transpose of the product of two matrices is the product in reverse order of
their transposes, so that
(AB)T BTAT.
(E.27)
The sum of the main diagonal elements of a square matrix A is called the trace
of A, written as
tr A a11 a22 … ann.
(E.28)
The determinant of a square matrix is obtained by enclosing the elements of the
matrix A within vertical bars; for example,
det A = 2
a11 a12
2 = a11a22 - a12a21.
a21 a21
(E.29)
If the determinant of A is equal to zero, then the determinant is said to be singular.
The value of a determinant is determined by obtaining the minors and cofactors of
the determinants. The minor of an element aij of a determinant of order n is a determinant of order (n 1) obtained by removing the row i and the column j of the original determinant.The cofactor of a given element of a determinant is the minor of the
element with either a plus or minus sign attached; hence
cofactor of aij aij (1)ijMij,
where Mij is the minor of aij. For example, the cofactor of the element a23 of
a11
det A = 3 a21
a31
a12
a22
a32
a13
a23 3
a33
(E.30)
is
a23 = 1-12 5M23 = - 2
a11
a31
a12
2.
a32
The value of a determinant of second order (2 2) is
(E.31)
ph_dorf_app 4/24/01 4:54 AM Page 10
10
Appendix E
An Introduction to Matrix Algebra
a11
a21
2
a12
2 = 1a11a22 - a21a12 2.
a22
(E.32)
The general nth-order determinant has a value given by
n
det A = a aijaij
with i chosen for one row,
(E.33)
with j chosen for one column.
(E.33)
j=1
or
n
det A = a aijaij
i=1
That is, the elements aij are chosen for a specific row (or column), and that entire row
(or column) is expanded according to Eq. (E.33). For example, the value of a specific
3 3 determinant is
2
det A = det C 1
2
=2 2
0
1
3
0
1
5
1S
0
1
3
2 -1 2
0
1
5
3
2 +2 2
0
0
= 21-12 - 1-52 + 2132 = 9,
5
2
1
(E.34)
where we have expanded in the first column.
The adjoint matrix of a square matrix A is formed by replacing each element aij
by the cofactor aij and transposing. Therefore
a11
a
adjoint A = D 21
o
an1
a12
a22
o
an2
p
p
p
a1n T
a2n
T =
o
ann
a11 a21
a
a22
D 12
o
o
a1n a2n
p
p
p
an1
an2
T.
o
ann
(E.35)
E.5 MATRIX INVERSION
The inverse of a square matrix A is written as A1 and is defined as satisfying the relationship
A1A AA1 I.
(E.36)
adjoint of A
det A
(E.37)
The inverse of a matrix A is
A-1 =
ph_dorf_app 4/24/01 4:54 AM Page 11
Section E.6
11
Matrices and Characteristic Roots
when the det A is not equal to zero. For a 2 2 matrix we have the adjoint matrix
adjoint A = B
a22
-a21
-a12
R,
a11
(E.38)
and the det A a11a22 a12a21. Consider the matrix
1
A = C2
0
2
-1
-1
3
4 S.
1
(E.39)
The determinant has a value det A 7. The cofactor a11 is
a11 = 1-12 2 2
-1
-1
4
2 = 3.
1
(E.40)
In a similar manner we obtain
-1
A
3 -5
adjoint A
1
=
= a- b C -2
1
det A
7
-2
1
11
2 S.
-5
(E.41)
E.6 MATRICES AND CHARACTERISTIC ROOTS
A set of simultaneous linear algebraic equations can be represented by the matrix
equation
y Ax,
(E.42)
where the y vector can be considered as a transformation of the vector x. We might
ask whether it may happen that a vector y may be a scalar multiple of x. Trying
y lx, where l is a scalar, we have
lx Ax.
(E.43)
Alternatively Eq. (E.43) can be written as
lx Ax (lI A)x 0,
(E.44)
where I identity matrix. Thus the solution for x exists if and only if
det (lI A) 0.
(E.45)
This determinant is called the characteristic determinant of A. Expansion of the determinant of Eq. (E.45) results in the characteristic equation. The characteristic equation is an nth-order polynomial in l. The n roots of this characteristic equation are
called the characteristic roots. For every possible value li (i 1, 2, . . . , n) of the nthorder characteristic equation, we can write
(liI A)xi 0.
(E.46)
The vector xi is the characteristic vector for the ith root. Let us consider the matrix
ph_dorf_app 4/24/01 4:54 AM Page 12
12
Appendix E
An Introduction to Matrix Algebra
2
A= C 2
-1
1
3
-1
1
4 S.
-2
(E.47)
The characteristic equation is found as follows:
1l - 22
det C
-2
1
-1
1l - 32
1
-1
-4 S = 1-l3 + 3l2 + l - 32 = 0.
1l + 22
(E.48)
Ax1 l1x1,
(E.49)
The roots of the characteristic equation are l1 1, l2 1, l3 3. When l l1 1, we find the first characteristic vector from the equation
and we have xT1 k 31 -1
equal to 1. Similarly, we find
04 , where k is an arbitrary constant usually chosen
xT2 30
1
-14,
and
xT3 = 32
3
(E.50)
-14.
E.7 THE CALCULUS OF MATRICES
The derivative of a matrix A A(t) is defined as
d
A1t2 = C
dt
da11 1t2>dt da12 1t2>dt
o
o
dan1 1t2>dt dan2 1t2>dt
p
p
da1n 1t2>dt
o S.
dann 1t2>dt
(E.51)
That is, the derivative of a matrix is simply the derivative of each element aij(t) of the
matrix.
The matrix exponential function is defined as the power series
exp A = eA = I +
A A2 p Ak p
Ak
+
+
+
+
= a
,
1!
2!
k!
k=0 k!
(E.52)
where A2 AA, and, similarly, Ak implies A multiplied k times. This series can be
shown to be convergent for all square matrices. Also a matrix exponential that is a
function of time is defined as
eAt = a
k=0
Aktk
.
k!
(E.53)
If we differentiate with respect to time, then we have
d At
1e 2 = AeAt.
dt
(E.54)
ph_dorf_app 4/24/01 4:54 AM Page 13
Section E.7
13
The Calculus of Matrices
Therefore for a differential equation
dx
= Ax,
dt
(E.55)
we might postulate a solution x eAtc fc, where the matrix f is f eAt, and c is
an unknown column vector. Then we have
dx
= Ax,
dt
(E.56)
AeAt AeAt,
(E.57)
or
and we have in fact satisfied the relationship, Eq. (E.55). Then the value of c is simply x(0), the initial value of x, because when t 0, we have x(0) c. Therefore the
solution to Eq. (E.55) is
x(t) eAtx(0).
(E.58)
ph_dorf_app 4/24/01 4:54 AM Page 14
ph_dorf_app 4/24/01 4:54 AM Page 15
APPENDIX
F
Decibel Conversion
Table F.1
M
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.
3.
4.
5.
6.
7.
8.
9.
0
m
20.00
13.98
10.46
7.96
6.02
4.44
3.10
1.94
0.92
0.00
0.83
1.58
2.28
2.92
3.52
4.08
4.61
5.11
5.58
6.02
9.54
12.04
13.98
15.56
16.90
18.06
19.08
0.
1
40.00
19.17
13.56
10.17
7.74
5.85
4.29
2.97
1.83
0.82
0.09
0.91
1.66
2.35
2.98
3.58
4.14
4.66
5.15
5.62
6.44
9.83
12.26
14.15
15.71
17.03
18.17
19.18
1.
2
33.98
18.42
13.15
9.90
7.54
5.68
4.15
2.85
1.72
0.72
0.17
0.98
1.73
2.41
3.05
3.64
4.19
4.71
5.20
5.67
6.85
10.10
12.46
14.32
15.85
17.15
18.28
19.28
2.
3
30.46
17.72
12.77
9.63
7.33
5.51
4.01
2.73
1.62
0.63
0.26
1.06
1.80
2.48
3.11
3.69
4.24
4.76
5.25
5.71
7.23
10.37
12.67
14.49
15.99
17.27
18.38
19.37
3.
4
27.96
17.08
12.40
9.37
7.13
5.35
3.88
2.62
1.51
0.54
0.34
1.14
1.87
2.54
3.17
3.75
4.30
4.81
5.30
5.76
7.60
10.63
12.87
14.65
16.12
17.38
18.49
19.46
4.
5
26.02
16.48
12.04
9.12
6.94
5.19
3.74
2.50
1.41
0.45
0.42
1.21
1.94
2.61
3.23
3.81
4.35
4.86
5.34
5.80
7.96
10.88
13.06
14.81
16.26
17.50
18.59
19.55
5.
6
24.44
15.92
11.70
8.87
6.74
5.04
3.61
2.38
1.31
0.35
0.51
1.29
2.01
2.67
3.29
3.86
4.40
4.91
5.39
5.85
8.30
11.13
13.26
14.96
16.39
17.62
18.69
19.65
6.
7
23.10
15.39
11.37
8.64
6.56
4.88
3.48
2.27
1.21
0.26
0.59
1.36
2.08
2.73
3.35
3.92
4.45
4.96
5.44
5.89
8.63
11.36
13.44
15.12
16.52
17.73
18.79
19.74
7.
8
21.94
14.89
11.06
8.40
6.38
4.73
3.35
2.16
1.11
0.18
0.67
1.44
2.14
2.80
3.41
3.97
4.51
5.01
5.48
5.93
8.94
11.60
13.62
15.27
16.65
17.84
18.89
19.82
8.
9
20.92
14.42
10.75
8.18
6.20
4.58
3.22
2.05
1.01
0.09
0.75
1.51
2.21
2.86
3.46
4.03
4.56
5.06
5.53
5.98
9.25
11.82
13.80
15.42
16.78
17.95
18.99
19.91
9.
Decibels 20 log10 M.
15
ph_dorf_app 4/24/01 4:54 AM Page 16
ph_dorf_app 4/24/01 4:54 AM Page 17
APPENDIX
G
Complex Numbers
G.1 A COMPLEX NUMBER
We all are familiar with the solution of the algebraic equation
x2 1 0,
(G.1)
which is x 1. However, we often encounter the equation
x2 1 0.
(G.2)
A number that satisfies Eq. (G.2) is not a real number. We note that Eq. (G.2) may
be written as
x2 1,
(G.3)
and we denote the solution of Eq. (G.3) by the use of an imaginary number j1, so
that
j 2 1,
(G.4)
j = 4-1.
(G.5)
and
An imaginary number is defined as the product of the imaginary unit j with a real
number.Thus we may, for example, write an imaginary number as jb.A complex number is the sum of a real number and an imaginary number, so that
c = a + jb
(G.6)
where a and b are real numbers. We designate a as the real part of the complex number and b as the imaginary part and use the notation
Re{c} a,
(G.7)
Im{c} b.
(G.8)
and
17
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18
Appendix G
Complex Numbers
G.2 RECTANGULAR, EXPONENTIAL, AND POLAR FORMS
The complex number a jb may be represented on a rectangular coordinate place
called a complex plane. The complex plane has a real axis and an imaginary axis, as
shown in Fig. G.1. The complex number c is the directed line identified as c with coordinates a, b.The rectangular form is expressed in Eq. (G.6) and pictured in Fig. G.1.
An alternative way to express the complex number c is to use the distance from
the origin and the angle u, as shown in Fig. G.2. The exponential form is written as
c re ju,
(G.9)
r (a2 b2)1/2,
(G.10)
u tan1(b/a).
(G.11)
where
and
Note that a r cos u and b r sin u.
The number r is also called the magnitude of c, denoted as c.The angle u can also
be denoted by the form lu. Thus we may represent the complex number in polar
form as
c = c2 lu = rlu.
(G.12)
EXAMPLE G.1 Exponential and polar forms
Express c 4 j3 in exponential and polar form.
Solution First sketch the complex plane diagram as shown in Fig. G.3.Then find
r as
r (42 32)1/2 5,
and u as
u tan1(3/4) 36.9°.
Imaginary axis
Im
Im
c a jb
b
c re j␪
b
j3
r
␪
0
a
Real axis
FIGURE G.1 Rectangular form of
a complex number.
0
a
Re
FIGURE G.2 Exponential form of
a complex number.
␪
0
4
Re
FIGURE G.3 Complex plane for
Example G.1.
ph_dorf_app 4/24/01 4:54 AM Page 19
Section G.3
19
Mathematical Operations
The exponential form is then
c 5e j36.9°.
The polar form is
c = 5l36.9°. G.3 MATHEMATICAL OPERATIONS
The conjugate of the complex number c a jb is called c* and is defined as
c* a jb.
(G.13)
c* = rl-u.
(G.14)
In polar form we have
To add or subtract two complex numbers, we add (or subtract) their real parts and
their imaginary parts. Therefore if c a jb and d f jg, then
c d (a jb) (f jg) (a f) j(b g).
(G.15)
The multiplication of two complex numbers is obtained as follows (note j 1):
2
cd = 1a + jb21f + jg2
= af + jag + jbf + j2bg
= 1af - bg2 + j1ag + bf2.
(G.16)
Alternatively we use the polar form to obtain
cd = 1r1lu1 21r2lu2 2 = r1r2lu1 + u2,
(G.17)
where
c = r1lu1,
and
d = r2lu2.
Division of one complex number by another complex number is easily obtained using
the polar form as follows:
c r1lu1 r1
=
= lu1 - u2.
d r2lu2 r2
(G.18)
It is easiest to add and subtract complex numbers in rectangular form and to
multiply and divide them in polar form.
A few useful relations for complex numbers are summarized in Table G.1.
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20
Appendix G
COMPLEX NUMBERS
Table G.1
Useful
Relationships for
Complex Numbers
(1)
1
j
j
(2) (j)( j) 1
(3) j2 1
(4) 1l
>2 j
(5) ck rklk
EXAMPLE G.2 Complex number operations
Find c d, c d, cd, and c/d when c 4 j 3 and d 1 j.
Solution First we will express c and d in polar form as
c = 5l36.9°,
and
d = 22l-45°.
Then, for addition, we have
c d (4 j3) (1 j) 5 j2.
For subtraction we have
c d (4 j3) (1 j) 3 j4.
For multiplication we use the polar form to obtain
cd = 15l36.9°21 22l-45°2 = 522l-8.1°.
For division we have
5l36.9°
c
5
l81.9°. =
=
d 22l-45° 22
ph_dorf_app 4/24/01 4:54 AM Page 21
APPENDIX
H
z-Transform Pairs
Table H.1
x(t)
1 t = 0,
1. d(t) e
0 t = kT, k 0
1 t = kT,
2. d(t kT) e
0 t kT
X(s)
X(z)
1
1
ekTs
zk
3. u(t), unit step
1/s
4. t
1/s2
5. t2
2/s3
6. eat
1
s+a
7. 1 eat
a
s1s + a2
8. teat
1
1s + a2 2
9. t2eat
2
1s + a2 3
10. bebt aeat
11. sin vt
12. cos vt
13. eat sin vt
1b - a2s
1s + a21s + b2
v
s + v2
s
s2 + v2
2
v
1s + a2 2 + v2
z
z-1
Tz
1z - 12 2
T2z1z + 12
1z - 12 3
z
z - e-aT
11 - e-aT 2z
1z - 121z - e-aT 2
Tze-aT
1z - e-aT 2 2
T2e-aTz1z + e-aT 2
1z - e-aT 2 3
zz1b - a2 - 1be-aT - ae-bT 2
1z - e-aT 21z - e-bT 2
z sin vT
z - 2z cos vT + 1
z1z - cos vT2
2
z2 - 2z cos vT + 1
1ze-aT sin vT2
z2 - 2ze-aT cos vT + e-2aT
21
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22
Appendix H
Z-TRANSFORM
PAIRS
Table H.1 (continued)
14. eat cos vt
15. 1 eat acos bt +
a
sin btb
b
s+a
1s + a2 2 + v2
a2 + b2
s1s + a2 2 + b2
z2 - ze-aT cos vT
z - 2ze-aT cos vT + e-2aT
z1Az + B2
2
1z - 12z2 - 2e-aT 1cos bT2z + e-2aT
a
A 1 eaT cos bT eaT sin bT
b
a
B e2aT eaT sin bT eaT cos bT
b