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Complex Variables
Complex numbers are really two numbers packaged into one entity (much like matrices). The two “numbers” are the real and imaginary portions of the complex number:
.jyxz
}.Im{
}.Re{
zy
zx
We may plot complex numbers in a complex plane: the horizontal axis corresponds to the real part and the vertical axis corresponds to the imaginary part.
Re{z}
Im{z}
x
yz = x + jy
Often, we wish to use polar coordinates to specify the complex number. Instead of horizontal x and vertical y, we have radius r and angle .
Re{z}
Im{z}
x
yz = x + jy
r
The best way to express a complex number in polar coordinates is to use Euler’s identity:
.sincos je j So,
,sincos jrrrez j and
.sin
.cos
ry
rx
We also have
.sincos 2222222 rrryx
A summary of the complex relationships is on the following slide.
Re{z}
Im{z}
.
.sincos
jyxz
jrrre j
x = r cos
y = r sin r
.tan
.22
x
y
yxr
The magnitude of a complex number is the square-root of the sum of the squares of the real and imaginary parts:
.22 yxjyxz
If we set the magnitude of a complex number equal to a constant, we have
,22 cyxjyxz
or,
.2222cyxz
This is the equation of a circle, centered at the origin, of radius c.
Re{z}
Im{z}
x
y
c2 = |z|2 = x2 + y2
c
z = x + jy
Suppose we wish to find the region corresponding to
.2222cyxz
This would be a disk, centered at the origin, of radius c.
Re{z}
Im{z}
x2 + y2 = |z|2 < c2
x
yc
Suppose we wish to find the region corresponding to
.22
0 czz
This would be a disk, centered at z0, of radius c.
.)()( 220
20
2
0 cyyxxzz
Re{z}
Im{z}(x-x0)2 + (y-y0)2 = |z-z0|2 < c2 = |z-z0|2
x0
y0
c
z0
Functions of Complex Variables
Since z is a complex number, w will be a complex number. Since z has real and imaginary parts, w will have real and imaginary parts.
Suppose we had a function of a complex variable, say
.)( 2zzfw
.2
2
)(
)(
22
22
2
2
xyjyx
yjxyx
jyx
zzfw
The standard notation for the real and imaginary parts of z are x and y respectively.
The standard notation for the real and imaginary parts of w are u and v respectively.
,
222
jvu
xyjyxw
where
Both u and v are functions of x and y.
.2
.22
xyv
yxu
So a complex function of one complex variable is really two real functions of two real variables.
).,(),()( yxjvyxuzfw
Exercise: Find u(x,y) and v(x,y) for each of the following complex functions:
.)(
.)(
.cos)(
.)(
.)(
.)( 3
z
jz
z
zzf
zzf
zzf
ezf
ezf
zzf
Continuity of Complex Functions
In order to perform operations such as differentiation and integration of complex functions, we must be able to verify of the complex function is continuous.
)(zfA complex function
is said to be continuous at a point z0 if as z approaches z0 (from any direction) then f(z) can be made arbitrarily close to f(z0).
)()( 0zfzf
for some such that
A more mathematical definition of continuity would be for any , we can make
.0 zz
Since we are dealing with complex numbers, the geometric interpretation of this statement is different from that of real numbers.
Re{z}
Im{z}
z0
The region |z-z0| < defines a disk in the complex plane of radius centered about z0.
Re{w}
Im{w}
f(z0)
So, if we wish |f(z)-f(z0)| < we must find a to make this so.
Example: Suppose
.)( 2zzf
Find such that
5.0)()( 0 zfzf
for
.210 jz
Solution:
.2)( 222 xyjyxjyxzf
22220 423)()( xyyxzfzf
.4321)( 20 jjzf
All we need to do is to find a value of such that if
,21 220 yxzz
then
.5.0423 2222 xyyx
We can do some calculations on a spreadsheet (continuity.xls).
A value of < 0.1 seems to do it.
x y
1.100 2.100 0.1414 0.65151.050 2.050 0.0707 0.32101.050 1.950 0.0707 0.31470.950 2.050 0.0707 0.31780.950 1.950 0.0707 0.31151.010 2.010 0.0141 0.0634
22 21 yx 2222 423 xyyx
A MATLAB plot (by continuity.m) of the previous example is shown on the following slide.
-4 -3 -2 -1 0 1 2-1
0
1
2
3
4
5
Re{z}
Im{z
} |z - z0| < = 0.1
|z2 - z02| < = 0.5
Differentiation of Complex Functions
How do we take derivatives of complex functions with respect to complex variables?
),(zfwIf
what is
?)(
dz
zdf
dz
dw
The differential dz can vary in one of two ways: along the real axis (dx) or along the imaginary axis (dy).
Re{z}
Im{z}
x
yy+dy
x+dx
dx
dy
As z varies in either direction, the derivative must be the same.
.x
vj
x
u
x
w
dz
dw
x direction
.y
v
y
uj
y
wj
jy
w
dz
dw
y direction
So, we must have
.y
v
x
u
.x
v
y
u
These last two conditions
.y
v
x
u
.x
v
y
u
are called the Cauchy-Riemann equations. These equations are the criteria for a complex function to be differentiable (with respect to z = x + jy).
Example: Show that the function
.)( 2zzfw
is differentiable
.2
.22
xyv
yxu
Solution: We have shown that
.2
.2
xy
v
xx
u
.2
.2
yx
v
yy
u
.y
v
x
u
.x
v
y
u
Now that we have determined that this function is differentiable, the derivative can be found using
.x
vj
x
u
dz
dw
.y
uj
y
v
dz
dw
or
If we apply these formulas to
.2
.22
xyv
yxu
).,(),()( 2 yxjvyxuzzfw
where
.22
222
yjxx
xyj
x
yx
dz
dw
).(2)2(2
2 22
jyxyjx
y
yxj
y
xy
dz
dw
or
we have
.2)(2 zjyxdz
dw
The answer is what we would expect to get if z were treated as a real variable.
We see that the derivative in both cases is
As it turns out, for most well-behaved complex functions, the derivative can be found by treating z as if it were a real variable.
Example: Show that the function
}.Re{)( zzfw
is not differentiable
.0
.
v
xu
Solution: We have shown that
.0
.1
y
v
x
u
.0
.0
x
v
y
u
.y
v
x
u
.x
v
y
u
Exercise: Is
}Im{)( zzfw
differentiable?
Definition: A function
).(zfw
is said to be analytic if it is differentiable throughout a region in the complex plane.
Integration of Complex Functions
What happens when we try to take the integral of a complex function along some path in the complex plane?
.)( dzzfC
A complex integral is like a line integral in two dimensions.
.),(),(
),(),(
),(),()(
dyyxudxyxvj
dyyxvdxyxu
jdydxyxjvyxudzzf
CC
CC
CC
The real and the imaginary parts of the integral are nearly identical to classic line integrals.
Example: Integrate
2)( zzfw
over the real interval z = 0 + j0 to z = 2 + j0.
.2
.22
xyv
yxu
Solution: We have shown that
.2
2
),(),(
),(),()(
22
22
dyyxdxxyj
dyxydxyx
dyyxudxyxvj
dyyxvdxyxudzzf
CC
CC
CC
CCC
Since we are integrating along the real (x) axis, all integrals with respect to dy are zero. In addition y=0. So,
.3 3
8
2
0
3
2
0
2
22
x
dxx
dxxdzzCC
The result is exactly what we would expect to get if we simply integrated a real variable from 0 to 2.
Example: Integrate
2)( zzfw
over the imaginary interval z = 0 + j0 to z = 0 + j2.
Solution: The integral becomes
.)(2
0 3822 jdyyjdyyjdzzf
CC
The result is exactly what we would expect to get if we simply integrated
.)( 38
2
0
22jdyyjdjyjy
C
dzzC
2
where C = jy and where y=[0,2] :
Example: Integrate
2)( zzfw
over the complex path z = 0 + j2 to z = 2 + j2.
Re{z}2
2
Im{z}
.2
2)(
2
0
2
0
22
22
dxxyjdxyx
dxxyjdxyxdzzfCCC
Solution:
The value of y is that of the path: y=2.
.)2(2)2()(2
0
2
0
22 dxxjdxxdzzfC
.8243
)2(2)2()(
316
2
0
2
2
0
3
2
0
2
0
22
jxjxx
dxxjdxxdzzfC
Example: Integrate
2)( zzfw
over the complex path z = 2 + j0 to z = 2 + j2.
Re{z}2
2
Im{z}
Solution:
.83
42
)2()2(2
2)(
316
2
0
32
0
2
2
0
222
0
22
jy
yjy
dyyjdyy
dyyxjdyxydzzfCCC
Example: Integrate
2)( zzfw
over the complex path z = 0 + j0 to z = 2 + j2.
Solution: The path of integration is a line z = x + jy where x = y = t = [0,2].
Re{z}2
2
Im{z}
Solution: The integral is more complicated.
.2
2
2
2)(
2
0
222
0
2
0
2
0
22
22
22
dtttdtttj
dtttdttt
dyyxdxxyj
dyxydxyxdzzf
CC
CCC
.12
3
2
3
2
22)(
38
2
0
32
0
3
2
0
22
0
2
j
tj
t
dttjdttdzzfC
This result is the same as the sum of the integral from 0+j0 to 0+j2 with the integral from 0+j2 to 2+j2.
Re{z}2
2
Im{z}
This result also is the same as the sum of the integral from 0+j0 to 2+j0 with the integral from 2+j0 to 2+j2.
Re{z}2
2
Im{z}
So it seems that it does not matter what path is taken as long as the endpoints are the same.
Example: Integrate
.)( 2zzfw
over two paths:
(1) a semicircle z = ej, where = [0,].
(2) a semicircle z = e-j, where = [0,].
Show that the two integrals are the same.
Re{z}
Im{z}
C1
C2
= 0=
Solution: This integration is best handled using polar coordinates:
.)1(.)( 22222 reerrezzf jjj
.)1(.)( rdjedjrereddz jjj
.3
11
3
1
3
)(
3203
0
3
0
3
0
2
1
jj
jj
jj
C
ee
ej
jdej
djeedzzf
The integral around curve C1 is
The integral around curve C2 is
.3
11
3
1
3
)(
320
00
0
2
2
jj
jj
jj
C
ee
ej
jdej
djeedzzf
.03
11
3
1
3
)(
06
2
0
32
0
3
2
0
2
jj
jj
jj
C
ee
ej
jdej
djeedzzf
If we were to integrate around the whole circle C = ej for = [0, 2], we would get
The curve C can be thought of as C1 + (-C2 ).
Cauchy’s Integral Theorem: If a function f(z) is analytic over a region R enclosed by a (closed) path C, then
.0)( C dzzf
Re{z}
Im{z} C
R
Simple Proof:
.)( CCC
vdxudyjvdyudxdzzf
Both integrals are line integrals around a closed curve C. We can apply Green’s theorem (a special case of Stoke’s theorem) to these line integrals
.)(
RRC
dxdyy
v
x
ujydxd
y
u
x
vdzzf
If f(z) is analytic, then the Cauchy-Riemann equations apply:
RRC
dxdyy
v
x
ujydxd
y
u
x
vdzzf )(
.y
v
x
u
.x
v
y
u
If these are true, then both integrands of
are zero and the theorem is proved.
If
,0)( C dzzf
and C = C1 + C2, then
.0)()()(21
CCCdzzfdzzfdzzf
Re{z}
Im{z}C1
C2
We also have
.)(
)()(
2
21
C
CC
dzzf
dzzfdzzf
Re{z}
Im{z}C1
-C2
So it does not matter what path that you take so long as the endpoints are the same provided f(z) is analytic between any of the two paths.
If f(z) is not analytic at some point between two paths, then the path does matter.
Example: Integrate
zzfw
1)(
over a unit circle z = ej, where = [0,2].
Re{z}
Im{z}
Solution: As with the previous example, this integration is best handled using polar coordinates:
.)1(.111
)( reerrez
zf jjj
.)1(.)( rdjedjrereddz jjj
.2)(2
0
2
0
jdjdjeedzzf jj
C
This integral is not zero because there is a discontinuity (actually a pole) at z = 0.
Example: Integrate
1
1)(
z
zfw
over a unit circle z = 1 + ej, where = [0,2].
Re{z}
Im{z}
Solution: Let us first write the integral.
.1
)(
CC z
dzdzzf
To carry-out this integration, we first perform a substitution of variables. Let = z-1.
.)(' CC
ddzzf
The path C’ is equal to C minus one:
Re{z}
Im{z}
CC’
.2)('
jd
dzzfCC
We see that
This integral is not zero because there is a discontinuity (a pole) at z = 1 or = 0.
Exercise: Show that no matter what z0 is, if C is a circular path (of any radius) around z0 we will have
.20
C
jzz
dz
Cauchy’s Integral Formula: Let f(z) be analytic over a region R enclosed by a closed path C. If z0 is a point within R, then
).(2)(
00
zfjdzzz
zfC
Note that while f(z) is analytic throughout R, f(z)/(z-z0 ) is not analytic (z0 is a pole).
Re{z}
Im{z} C
z0R
Proof: We add and subtract f(z0) to the numerator of the integrand so as to split-up the integral into two terms:
.)()()(
)()()()(
00
0
0
0
00
0
C C
CC
zz
dzzfdz
zz
zfzf
dzzz
zfzfzfdz
zz
zf
If f(z) is analytic within the region R, then it is also continuous. So, for any , we can find a such that for
.)()( 0 zfzf
,0 zz
we have
Let us choose a such that z R, i.e., the disk |z-z0| < is totally within R. Let denote the path |z-z0| = by the symbol C’ .
Re{z}
Im{z} C
z0R
C’
So if
,0 zz
,)()( 0 zfzf
and
).C'path on the()()(
00
0
zzzz
zfzf
So for appropriate values of and , the integrand in
'0
0 )()(C
dzzz
zfzf
'0
0
0
0 .)()()()(
C Cdz
zz
zfzfdz
zz
zfzf
can be made arbitrarily small. Now since the integrand is analytic except at z = z0, we have
The integral
,2)(0)(
00
jzfdzzz
zfC
C zz
dz
0
is equal to 2j. So,
and the theorem is proved.
Example: Evaluate the integral
,1
1
C
dzz
z
Solution:
.1)(.10 zzfz
where C is a closed curve around z = 1.
.42)11(2)(1
10 jjjzfdz
z
zC
So,
The formula
Cdz
zz
zf
jzf
00
)(
2
1)(
makes calculating derivatives with respect to z0 relatively easy. We do not have to worry about z: it is independent of z0.
.
)(
2
1)(' 2
0
0
Cdz
zz
zf
jzf
In general it can be shown that
.
)(
2
!)( 1
0
0)(
C n
n dzzz
zf
j
nzf
This formula is very useful in deriving the Taylor series.
Example: Evaluate the integral
,
1
12
C
dzz
z
Solution:
.1.1)(.10 nzzfz
where C is a closed curve around z = 1.
.22)1(2)(
1
10
)1(2 jjjzfdz
z
zC
So,
Example: Evaluate the integral
,
1
13
C
dzz
z
Solution:
.2.1)(.10 nzzfz
where C is a closed curve around z = 1.
.0)0(
!2
2)(
1
10
)2(3
jj
zfdzz
zC
So,
Example: Evaluate the integral
,
2
13
3
Cdz
z
z
Solution:.2.1)(.2 3
0 nzzfz
where C is a closed curve around z = 2.
.12)2)(6(
!2
2)(
2
10
)2(3
3
jjj
zfdzz
zC
.6)()2( zzf
Exercise: Evaluate the integral
,
1
cos2 Cdz
z
z
where C is a closed curve around z = 1.
Inverse Laplace Transforms
.)()()}({
dttxesXtx stL
The forward Laplace transform was found using
We used a formula to calculate the forward Laplace transform, but we did not use a formula to calculate the inverse Laplace transform. Such a formula exists!
.)(2
1)()}({
dssXe
jtxsX st
1-L
The inverse Laplace transform can be found using a complex inversion integral formula:
We can evaluate the inverse Laplace transform using Cauchy’s integral formula.
Cauchy’s integral formula is for an integration around a closed loop. The inverse Laplace transform formula is an integral along an infinite line. This infinite line integral can actually be thought of as a loop.
Let us construct a closed curve C consisting of a line along the imaginary axis and a semicircle in the left-half plane.
Re{s}
Im{s}
r
s +j
s -j
C
s -
As the radius r of the semicircle approaches infinity, the closed loop approaches an infinite line (from s = -j to +j). As r approaches infinity, both the semicicular curve and the infinite line pass through something called the point at infinity.
The point at infinity can be reached from either the positive or negative half of the real or the imaginary axis. The limits s +j, s -j and s - are all the same.
Example: Find the inverse Laplace transform of
.1
1)(
s
sX
Solution:
.1
1
2
1
)(2
1)()}({
dss
ej
dssXej
txsX
st
st
1-L
.)(.10stesfs
.
2
12
1
1
2
1)}({
10
0
t
s
ts
st
e
ej
j
dss
ej
sX
1-L
Example: Find the inverse Laplace transform of
.1
1)(
2 s
sX
Solution:
.1
1
2
1
)(2
1)()}({
2
dss
ej
dssXej
txsX
st
st
1-L
The curve C is an extension of the s= -j to s = +j line.
C
st
st
dss
ej
dss
ej
sX
1
1
2
1
1
1
2
1)}({
2
2
1-L
The integral can be evaluated in the complex plane about a curve C:
Re{s}
Im{s}
C
}.,{0 jjs
.1
1
2
1
1
1
2
1)}({
1 222
C C
stst dss
ej
dss
ej
sX
1-L
There are actually two points of discontinuity:
We can evaluate the complex inversion integral by evaluating the integral around these two points of discontinuity. Let us call the paths around these two points C1 and C2.
Re{s}
Im{s}
s0 = +j
s0 = -j
C1
C2C
.sin2
111
2
12
2
12
1
2
11
2
1)}({
1 2
teej
ejj
ejj
ejsj
jejsj
j
dsjsjs
ej
dsjsjs
ej
sX
jtjtjtjt
js
st
js
st
C C
stst
1-L
Exercise: Using the complex inversion integral, find the inverse Laplace transforms of the following functions:
.1)1(
1)(
.1
)(
2
2
ssX
s
ssX
(First find the points of discontinuity and then evaluate the integral in paths around these points.)
Example: Find the inverse Laplace transform of
.cos
)(
s
ssX
Solution:
.cos
2
1
)(2
1)()}({
dss
se
j
dssXej
txsX
st
st
1-L
.0 s
.cos22
1
cos
2
1)}({
1
t
s
st
C
st
esejj
dss
se
jsX
1-L
There is one point of discontinuity:
Sequences and Series
Consider the sequence of values
,,,,1 41
31
21
each term in this sequence can be represented by
.1nnz
What happens as n goes to infinity? For this example zn goes to zero.
How about the sequence of values
,1,1,,1 87
43
23
each term in this sequence can be represented by
.]1[2 21 n
nz
This sequence is said to converge to two (2): as n goes to infinity, zn goes to two.
How about the sequence of values
,,2,,1 25
23
This series does not converge to any value, but rather it is said to diverge.
This sequence {zn } is said to converge to a value c if zn can be made arbitrarily close to c for a large enough value of n.
A more formal definition of convergence of series would be for any positive value , we can find an integer N such that for
Nn we must have
. czn
The expression
. czn
is for a complex number zn and defines a disk in the complex plane.
Re{z}
Im{z}
c
Example: Plot the sequence of values in
20
21 nj
n en
z
in the complex plane
Solution: The radius of zn is 1/n and the angle of zn is 2n/20. The plot is performed using MATLAB (sequence.m) and is shown on the following slide page.
-0.2 0 0.2 0.4 0.6 0.8-0.2
0
0.2
0.4
0.6
0.8
1
Re{z}
Im{z
}
This sequence converges to zero. The relationship between the sequence index n and distance to the limit is rather easy in this case.
For
,1n
we must have
. czn
where 1/ corresponds to the smallest integer greater than 1/.
Similarly, for
,Nn we must have
.1
Nczn
A necessary (but not sufficient) condition for a sequence to converge is that it be bounded. A bounded sequence is {zn } is one that for all n we have
Bzn
for some finite value B. If a sequence is not bounded, it will diverge.
Just because a sequence is bounded does not mean it converges. Many sequences which are bounded do not converge. For example,
.,1,1,1,1 nz
For this sequence
.,1,0for1 nz nn
1 2 3 4 5n
zn
While this sequence does not have a limit, it does have an upper bound (+1) and a lower bound (-1) as n . These “bounds” are called the supremum and the infimum respectively. The supremum is the smallest upper bound (+1) and the infimum is the largest lower bound (-1). These bounds are abbreviated sup and inf respectively.
.1inflim
.1suplim
nn
nn
z
z
Note that while {zn } does not converge, there are subsequences
.,1,1
.,1,1
2
1
n
n
z
z
that do converge.
Series
Suppose we were to add the numbers in a sequence:
.1
n
kkn zs
The term sn is called the partial sum of the series {zn }. The summation is a series.
If we were to take the limit as n , we would get an infinite series:
.lim1
k
knn
zss
If the sequence of partial sums converges, we say that the series converges.
A necessary (but not sufficient) condition for the series {sn } to converge is that the sequence {zn } converges. The series is generally “wilder” that the sequence. If the series converges, the sequence must necessarily converge. If the series diverges, the sequence may or may not converge.
Example: Consider the (convergent) sequence
.1
nzn
The corresponding series does not converge (as n goes to infinity):
.1
00
n
k
n
kkn kzs
Now how do we find sufficient conditions for a series to converge? There are five (5) standard tests for series convergence:
(1) Comparison Test: compare series term-by-term against a known convergent series.
(2) Geometric Series Test: a geometric series converges if each geometric term is less than one
(3) Ratio Test: Take the limit of the ratio of one term to the previous term. If the limit is less than one, the series converges
(4) Root Test: Take the k-th root of the k-th term. If the limit is less than one, the series converges.
(5) Integral Test: compare the sequence to an integrand of a known integral.
(1) Comparison Test: compare series term-by-term against a known convergent series.
,1
n
kkn zs
If we have a known convergent series
n
kkn wr
1
then any series
such that
kk zw
also converges.
.1
n
k
kn qs
A geometric series is of the form
We can use this form to find a closed-form expression for the geometric series
(2) Geometric Series Test: a geometric series converges if each geometric term is less than one
.1
.
.
1
1
10
1
10
1
0
n
n
k
kn
k
knn
n
k
kn
k
kn
n
k
kn
q
qqqss
qqqs
qs
So,
.1
1
.11
1
q
qs
qqssn
n
nnn
If the geometric term q is such that qn goes to zero as n goes to infinity, then
.1
1lim
qsn
n
In order for qn to go to zero, we must have
.1q
If q > 1, then the series diverges.
Example: Suppose q = ½.
.21
1
...1
.
21
041
21
21
21
n
n
kn
nn
s
z
Example: Suppose q = 9/10.
.101
1
...1
.
109
010081
109
109
109
n
n
kn
nn
s
z
Example: Suppose q = -1/2.
.1
1
...1
.
32
21
041
21
21
21
n
n
kn
nn
s
z
,1
n
kkn zs
If
then we take
(3) Ratio Test: Take the limit of the ratio of one term to the previous term. If the limit is less than one, the series converges
.1
k
kk z
zw
If
then the series converges.
.1lim n
nw
So,
As a justification (hardly a proof) for this test, consider definining zk in terms of wk:
.1
k
kk z
zw
.1 kkk zwz
.
.
112223
112
zwwzwz
zwz
.1 1
11
n
k
k
mm
n
kkn wzzs
.)max(1
1
km
km
k
mm ww
The term
,1max1
mkmw
If
then the series behaves like a convergent geometric series:
.)max(1 1
11
11
1
n
k
n
k
km
km
k
mm
n
kkn wzwzzs
Example: Determine if
n
k
k
n ks
0 !
2
converges.
Solution: Taking the ratio test
.01
2
2
!
!1
2 11
k
k
kz
zw
k
k
k
kk
We see that the series converges.
,1
n
kkn zs
If
then we take
.kkk zw
(4) Root Test: Take the k-th root of the k-th term. If the limit is less than one, the series converges.
If
then the series converges.
.1lim n
nw
As a justification (better than that of the ratio test but still not a proof) for this test, consider
.kk
k zw
.kkk zw
.11
n
k
kk
n
kkn wzs
If wk < 1, then the series is a convergent geometric series.
Example: Determine if
n
n
k
n k
es
02
converges.
Solution: Taking the root test
.0/ln2/2ln/22e
e
e
e
e
e
e
k
e
k
ezw
kkkkkk
k
kkk
We see that the series diverges. (ek dominates k2.)
.1
n
kkn zs
The term zk is really a function of k. We can represent that function as z(k) or z(x). The convergence of the integral
1)( dxxz
(5) Integral Test: compare the sequence to an integrand of a known integral.
can be used to determine whether or not the series converges.
.1
1
n
kn ks
This series is greater than the integral
.1
1ndxx
Example: Consider the series
Since the integral diverges [ln n], the series diverges.
1 2 3 4 5 6 7 8 9 10
0.5
1
1.5
2
x
1/x
Integral 1dx/x Versus Series n=1
1/n
.1
11
1 222
n
k
n
kn kks
This series from k=2 to k=n is less than the integral
.1
1 2n
dxx
Example: Consider the series
Since the integral converges [1/n], the series converges.
1 2 3 4 5 6 7 8 9 10
0.5
1
1.5
2
x
1/x
Integral 1dx/x
2 Versus Series n=2
1 / n
2
Taylor Series
A Taylor series is a power series representation for a function.
.)(1
k
kk azczf
A Taylor series is much like a Fourier series (which is a harmonic series).
.)(
2
1)(
Cdz
f
jzf
To find a Taylor series, all we need to do is find the coefficients (much like Fourier series).
To find these coefficients let us start with Cauchy’s integral formula:
We will attempt to express this formula in a power series about z = a.
z1
We start with the fraction in the integral:
zaa 1
The object is to express this in terms of powers of (z – a). So, we try to get this fraction in the form of an infinite geometric series:
0
.1
1
111
k
k
a
az
a
aazaaza
The last is true if is on a curve C at a distance r from a and z is within the close curve.
Re{z}
Im{z} is on C
a
r
z
.aaz
.)(
2
1
)(
2
1)(
10
0
C kk
k
Ck
k
da
f
jaz
da
az
a
f
jzf
C n da
f
j
1
)(
2
1
The coefficient of (z – a)k inside the summation
.
)(
2
!)( 1
)(
C kk d
a
f
j
kaf
is similar to the k-th derivative of f(a) from the corollary to Cauchy’s integral theorem:
.!
)(
)(
2
1)(
0
)(
10
k
kk
C kk
k
k
afaz
da
f
jazzf
So,
Hence, we have our Taylor series coefficients.
.sin)(
.cos)(
.)(
3
2
1
zzf
zzf
ezf z
Example: Find the Taylor series for
. oddn cos1
evenn sin1)(
. oddn sin1
evenn cos1)(
.)(
2
2
2
2
)(3
)(2
)(1
z
zzf
z
zzf
ezf
n
n
n
n
n
n
zn
We will evaluate the Taylor series about z = 0 :
. oddn 1
evenn 0)(
.oddn 0
evenn 1)0(
.1)0(
2
2
)(3
)(2
)(1
n
n
zf
f
f
n
n
n
.!3
)(
.!4!2
1)(
.!4!3!2
1)(
3
3
42
2
432
1
zzzf
zzzf
zzzzzf
So,
You can use these Taylor series to prove
.sincos zjze jz
.1
)(z
zf
Example: Find the Taylor series for
Here, we will evaluate the series about a = 1.
.!
1)(1
)(n
nn
z
nzf
.11111
!4
1!4
!3
1!3
!2
1!2
!1
1!11
1
432
432
zzzz
zzzz
z
.ln)( zzf
Example: Find the Taylor series for
Here, we will evaluate the series about a = 1.
.1!1
1)( 1)(
nz
nzf
n
nn
.
4
1
3
1
2
1
1
1
!4
1!3
!3
1!2
!2
1!1
!1
10ln
432
432
zzzz
zzzzz
If z is not close to one, this series is very slow to converge.
.ln)( zzf
Exercise: Find the Taylor series for
Evaluate the series about an arbitrary a.
Conformal Mapping
How do we “graph” complex functions? The difficulty lies in the dimensionality: we have two independent variables (x,y) and two dependent variables (u,v).
).,(),()( yxjvyxuzfw
To “graph” this function, we start with a family of curves corresponding to constant values of x and constant values of y. These curves are represented by dashed green lines on the following slide.
x = Re{z}
y = Im{z}
x=1 x=2 x=3 x=4
y=1
y=2
y=3
y=4
To what do these curves correspond to in the u-v plane?
Let us start with a simple example
.2
.2
.22
.2)(
yv
xu
yjxjvuw
zzfw
u = Re{w}
v = Im{w}
x=1 x=2
y=1
y=2
.)( zezfw
Example: Find the conformal map for
We expand ez from the real and imaginary parts of z.
.)( jyxjyxz eeeezfw
This expansion is best handled using polar coordinates.
, jjyx eeew
where
.
,
y
ex
The resultant curves will be a set of circles of radii ex. Constant values of y correspond to rays at angle y.
u = Re{w}
v = Im{w}
x=1
x=2
y=1y=2
Negative values of x correspond to circles of radius e-|x|. Negative values of y correspond to rays at angle -|y|.
u = Re{w}
v = Im{w}
x = -1
x=2
y = -1y = -2
x = -2
.1
1)(
z
zzfw
Example: Find the conformal map for
We represent z and w in terms of their real and imaginary components:
.1)(
1)(
jyx
jyxjvuw
We then try to make the denominator real:
.)1(
21
)1(
])1[1()1)(1(
)1(
)1(
)1(
)1(
22
22
22
2
yx
yjyx
yx
xxjyyxx
jyx
jyx
jyx
jyxjvu
.)1(
2
.)1(
1
22
22
22
yx
yv
yx
yxu
A plot of the constant x and the constant y curves (in the u-v plane) is shown on the following slide.
-1 -0.5 0 0.5 1-1
-0.5
0
0.5
1
u
jv
w = (z-1)/(z+1)
x = 0
x = 1
x = 2
y = -2y = -1
y = 0
y = 1y = 2
The resultant graph is that of a Smith Chart. This chart is used in radio-frequency electronics. The graph is a conformal map of line impedance onto complex reflection coefficient.
The Argument Principle
Let
)(zfw
be a function of the complex variable z.
As z follows a path in the z-plane, what path does w follow in the w-plane?
Let z follow a closed path in the z-plane.
).20,1( rrez j
Re{z}
Im{z}
C
As z follows this closed path, what path will w=f(z) follow?
As an example, let
.)( 2zzf
.)( 2222 jj eerzzfw
As z goes around the circle once, w goes around the same circle twice.
Re{z}
Im{z}
C
Re{w}
Im{w}
C’
.2zw
As another example, let
.1
)(z
zf
.11
)( jj eerz
zfw
As z goes around the circle counter-clockwise, w goes around the same circle clockwise.
Re{z}
Im{z}
C
Re{w}
Im{w}
C’
.1
zw
Let us choose two points on the z-path: z1 and z2
Re{z}
Im{z}
z1
z2
The two points z1 and z2 are very close together; their radii are the same, but their angles are different.
C
The two points z1 and z2 are very close together; their radii are the same, but their angles are different.
.
,2
1
22
11
j
j
erz
erz
.2,1
,0,1
21
11
r
r
The corresponding points in the w-plane f(z1 ) and f(z2
) are very close together. Like z, the radii of f(z1 ) and f(z2 ) are the same, but their angles are different.
.
,2
1
22
11
j
j
ew
ew
.
.
21
21
As z follows the circular closed path, what path will f(z) follow?
To get an idea of the path f(z) will follow, let us look at log f(z).
.loglog)(log jezf j
.loglog)(loglog 111111 jezfw j
.loglog)(loglog 222222 jezfw j
The difference between log w1 and log w2 is
.
loglogloglog
12
121212
j
jww
Now log w can be written as an indefinite integral:
w
dwwlog
The difference between log w1 and log w2 can be written as a definite integral:
2
112 loglogw
dwww
The points 1 and 2 in the integral
2
1 w
dw
correspond to z1 and z2. So,
.)(
)(
)(
)(loglog
2
1
2
1
'
12 z
z
z
z zf
dzzf
zf
zdfww
Combining this expression with
.)(
)(loglog 12
'
12
2
1
jzf
dzzfww
z
z
,loglog 1212 jww
we have
So, the integral from z1 to z2 is the same as the integral around the closed curve in the z-plane.
.)(
)(12
'
jzf
dzzfC
To summarize, as z follows a closed path in the z-plane, w=f(z) follows a closed path in the w-plane. The angular rotation that w takes is equal to 2 – 1.
.2)(
)('Nj
zf
dzzfC
Since w=f(z) follows a closed path in the w-plane, the angular rotation, 2 – 1, must be an integral multiple of 2.
Suppose G(s) is a polynomial fraction:
.)(
21
21
psps
zszssG
Let s take on an infinite circular path in the right-half complex plane.
Re{s}
Im{s}
r
s +j
s -j
C
s +
What kind of path will G(s) take?
To answer this question, let us try to evaluate the integral
,)(
)('
C sG
dssG
where C is the curve described on the previous slide.
).()(
2
211 sH
ps
zszssGps
We will try to evaluate this integral by removing poles and zeroes in the right-half plane. Suppose p1 is a right-half plane pole. We can define a new function H(s) with this pole removed:
So,
.)(
)(1ps
sHsG
If
,)(
)(1ps
sHsG
we have
.
)()()(
1
'
21
'
ps
sH
ps
sHsG
If
,)(
)(1ps
sHsG
we have
.
)()()(
1
'
21
'
ps
sH
ps
sHsG
and
.
)(
)(1
)(
)( '
1
'
sH
sH
pssG
sG
The integral of
.
)(
)(1
)(
)( '
1
'
sH
sH
pssG
sG
is equal to
.
)(
)(
)(
)( '
1
'
CCC sH
dssH
ps
ds
sG
dssG
By the Cauchy Integral Formula, the integral
So
.)(
)(2
)(
)( ''
CC sH
dssHj
sG
dssG
.2
1
jps
dsC
).()(
21
2
1
sFpsps
zs
zs
sG
Now suppose z1 is a right-half plane zero. We can define a new function F(s) with this zero removed:
So,
).()( 1 sFpssG
If
we have
).()()( '1
' sFzssFsG
).()( 1 sFzssG
If
we have
and
.
)(
)(1
)(
)( '
1
'
sF
sF
zssG
sG
).()()( '1
' sFzssFsG
).()( 1 sFzssG
The integral of
.
)(
)(1
)(
)( '
1
'
sF
sF
zssG
sG
is equal to
.
)(
)(
)(
)( '
1
'
CCC sF
dssF
zs
ds
sG
dssG
By the Cauchy Integral Formula, the integral
So
.)(
)(2
)(
)( ''
CC sf
dssfj
sG
dssG
.2
1
jps
dsC
If we continue eliminating poles and zeroes, we get a term of -2j for every pole and a term of +2j for every zero.
So
),(2)(
)('PZj
sG
dssGC
where Z is the number of zeroes, and P is the number of poles.
Now from our previous result,
we see that
,PZN
where N is the number of rotations of G(s), Z is the number of right-half plane zeroes, and P is the number of right-half plane poles.
,2)(
)(2
)(
)( ''
NjsG
dssGNj
zf
dzzfCC
By looking at the number of clockwise rotations of G(s), we can find the number of right-half plane zeroes minus the number of right-half plane poles.
We have already done two examples:
.1
)(z
zf
.)( 2zzf
We have already done two examples:
.1
)(z
zf
.)( 2zzf
The first function did two counter-clockwise rotations for each counter-clockwise rotation of z.
The second function did one clockwise rotation for each counter-clockwise rotation of z.
Equivalently, for
.1
)(s
sG
.)( 2ssG
The first function will do two clockwise rotations for each clockwise rotation of s. These clockwise rotations correspond to two zeroes (at zero).
The second function will do one counter-clockwise rotation for each clockwise rotation of s. This counter-clockwise rotation corresponds to one pole (at zero).
Example: How many times does the following transfer function circle the origin as s goes from -j to +j ?
.1
2)(
s
ssG
Solution: There is one RHP pole and no RHP zeros (The term s+2 corresponds to a LHP zero: s = -2.) The plot of G(s) will circle the origin in the counter-clockwise direction once.
In MATLAB, we can plot G(s) using the control function nyquist().
>> EDU» s = tf('s'); >> H = (s+2)/(s-1); >> nyquist(H)
The Nyquist plot is on the following slide.
Real Axis
Imag
inar
y A
xis
Nyquist Diagrams
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2
-1.5
-1
-0.5
0
0.5
1
1.5
2From: U(1)
To:
Y(1
)
In control systems, we are often concerned about the poles of a closed-loop transfer function
.)(1
)()(
sG
sGsT
The poles of T(s) are the zeros of 1+G(s).
The right-half plane poles of T(s) are the right-half plane zeros of 1+G(s).
If we did a plot of
)(1 sG
and 1+G(s) had no right-half plane poles, then the number of clockwise rotations around the origin is equal to the number of right-half plane zeros of 1+G(s) or the number of right-half plane poles of T(s) .
The number of clockwise rotations around the origin of
)(1 sG
is equal to the number of clockwise rotations of
around s = -1.
)(sG
Therefore, if 1+G(s) has no right-half plane poles, the number of clockwise rotations around s = -1 of
)(sG
is equal to the number right-half plane poles of
.)(1
)()(
sG
sGsT
