Complex Variables. Complex numbers are really two numbers packaged into one entity (much like...

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:

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.

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 .

yz = x + jy

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

We also have

.sincos 2222222 rrryx

A summary of the complex relationships is on the following slide.

.sincos

jrrre j

x = r cos

y = r sin r

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

.2222cyxz

This is the equation of a circle, centered at the origin, of radius c.

c2 = |z|2 = x2 + y2

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.

x2 + y2 = |z|2 < c2

Suppose we wish to find the region corresponding to

This would be a disk, centered at z0, of radius c.

.)()( 220

0 cyyxxzz

Im{z}(x-x0)2 + (y-y0)2 = |z-z0|2 < c2 = |z-z0|2

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

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.

xyjyxw

Both u and v are functions of x and y.

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)(

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

Since we are dealing with complex numbers, the geometric interpretation of this statement is different from that of real numbers.

The region |z-z0| < defines a disk in the complex plane of radius centered about 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

.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

.5.0423 2222 xyyx

We can do some calculations on a spreadsheet (continuity.xls).

A value of < 0.1 seems to do it.

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

} |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

The differential dz can vary in one of two ways: along the real axis (dx) or along the imaginary axis (dy).

As z varies in either direction, the derivative must be the same.

x direction

y direction

So, we must have

These last two conditions

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

Solution: We have shown that

Now that we have determined that this function is differentiable, the derivative can be found using

If we apply these formulas to

).,(),()( 2 yxjvyxuzzfw

).(2)2(2

jyxyjx

we have

.2)(2 zjyxdz

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

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

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.

),(),(

),(),()(

dyyxdxxyj

dyxydxyx

dyyxudxyxvj

dyyxvdxyxudzzf

Since we are integrating along the real (x) axis, all integrals with respect to dy are zero. In addition y=0. So,

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

0 3822 jdyyjdyyjdzzf

The result is exactly what we would expect to get if we simply integrated

.)( 38

22jdyyjdjyjy

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

dxxyjdxyx

dxxyjdxyxdzzfCCC

Solution:

The value of y is that of the path: y=2.

.)2(2)2()(2

22 dxxjdxxdzzfC

)2(2)2()(

dxxjdxxdzzfC

Example: Integrate

2)( zzfw

Re{z}2

Solution:

)2()2(2

dyyjdyy

dyyxjdyxydzzfCCC

Example: Integrate

2)( zzfw

Solution: The path of integration is a line z = x + jy where x = y = t = [0,2].

Re{z}2

Solution: The integral is more complicated.

dtttdtttj

dtttdttt

dyyxdxxyj

dyxydxyxdzzf

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

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

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.

Solution: This integration is best handled using polar coordinates:

.)1(.)( 22222 reerrezzf jjj

.)1(.)( rdjedjrereddz jjj

djeedzzf

The integral around curve C1 is

The integral around curve C2 is

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

Im{z} C

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

ujydxd

If f(z) is analytic, then the Cauchy-Riemann equations apply:

ujydxd

vdzzf )(

If these are true, then both integrands of

are zero and the theorem is proved.

,0)( C dzzf

and C = C1 + C2, then

.0)()()(21

CCCdzzfdzzfdzzf

Im{z}C1

We also have

dzzfdzzf

Im{z}C1

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

over a unit circle z = ej, where = [0,2].

Solution: As with the previous example, this integration is best handled using polar coordinates:

.)1(.111

)( reerrez

zf jjj

.)1(.)( rdjedjrereddz jjj

jdjdjeedzzf jj

This integral is not zero because there is a discontinuity (actually a pole) at z = 0.

Example: Integrate

over a unit circle z = 1 + ej, where = [0,2].

Solution: Let us first write the integral.

dzdzzf

To carry-out this integration, we first perform a substitution of variables. Let = z-1.

.)(' CC

The path C’ is equal to C minus one:

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

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)(

zfjdzzz

Note that while f(z) is analytic throughout R, f(z)/(z-z0 ) is not analytic (z0 is a pole).

Im{z} C

Proof: We add and subtract f(z0) to the numerator of the integrand so as to split-up the integral into two terms:

.)()()(

)()()()(

dzzfdz

zfzfzfdz

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

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’ .

Im{z} C

,)()( 0 zfzf

).C'path on the()()(

So for appropriate values of and , the integrand in

0 )()(C

0 .)()()()(

zfzfdz

can be made arbitrarily small. Now since the integrand is analytic except at z = z0, we have

The integral

,2)(0)(

jzfdzzz

is equal to 2j. So,

and the theorem is proved.

Example: Evaluate the integral

Solution:

.1)(.10 zzfz

where C is a closed curve around z = 1.

.42)11(2)(1

10 jjjzfdz

The formula

makes calculating derivatives with respect to z0 relatively easy. We do not have to worry about z: it is independent of z0.

1)(' 2

In general it can be shown that

n dzzz

This formula is very useful in deriving the Taylor series.

Solution:

.1.1)(.10 nzzfz

.22)1(2)(

)1(2 jjjzfdz

Solution:

.2.1)(.10 nzzfz

Solution:.2.1)(.2 3

0 nzzfz

.12)2)(6(

.6)()2( zzf

Exercise: Evaluate the integral

cos2 Cdz

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!

1)()}({

jtxsX st

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.

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

Solution:

1)()}({

dssXej

.)(.10stesfs

Solution:

1)()}({

dssXej

The curve C is an extension of the s= -j to s = +j line.

The integral can be evaluated in the complex plane about a curve C:

}.,{0 jjs

stst dss

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.

s0 = +j

s0 = -j

dsjsjs

jtjtjtjt

Exercise: Using the complex inversion integral, find the inverse Laplace transforms of the following functions:

(First find the points of discontinuity and then evaluate the integral in paths around these points.)

Solution:

1)()}({

dssXej

.cos22

There is one point of discontinuity:

Sequences and Series

Consider the sequence of values

,,,,1 41

each term in this sequence can be represented by

What happens as n goes to infinity? For this example zn goes to zero.

How about the sequence of values

,1,1,,1 87

each term in this sequence can be represented by

.]1[2 21 n

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

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

The expression

is for a complex number zn and defines a disk in the complex plane.

Example: Plot the sequence of values in

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

This sequence converges to zero. The relationship between the sequence index n and distance to the limit is rather easy in this case.

we must have

where 1/ corresponds to the smallest integer greater than 1/.

Similarly, for

,Nn we must have

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

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

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

Note that while {zn } does not converge, there are subsequences

that do converge.

Series

Suppose we were to add the numbers in a sequence:

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:

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

The corresponding series does not converge (as n goes to infinity):

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.

kkn zs

If we have a known convergent series

kkn wr

then any series

such that

also converges.

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

If the geometric term q is such that qn goes to zero as n goes to infinity, then

In order for qn to go to zero, we must have

If q > 1, then the series diverges.

Example: Suppose q = ½.

Example: Suppose q = 9/10.

010081

Example: Suppose q = -1/2.

kkn zs

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

then the series converges.

.1lim n

As a justification (hardly a proof) for this test, consider definining zk in terms of wk:

.1 kkk zwz

112223

zwwzwz

kkn wzzs

.)max(1

The term

,1max1

then the series behaves like a convergent geometric series:

.)max(1 1

kkn wzwzzs

Example: Determine if

converges.

Solution: Taking the ratio test

We see that the series converges.

kkn zs

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.

then the series converges.

.1lim n

As a justification (better than that of the ratio test but still not a proof) for this test, consider

.kkk zw

kkn wzs

If wk < 1, then the series is a convergent geometric series.

Example: Determine if

converges.

Solution: Taking the root test

.0/ln2/2ln/22e

kkkkkk

We see that the series diverges. (ek dominates k2.)

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.

This series is greater than the integral

Example: Consider the series

Since the integral diverges [ln n], the series diverges.

1 2 3 4 5 6 7 8 9 10

Integral 1dx/x Versus Series n=1

kn kks

This series from k=2 to k=n is less than the integral

Example: Consider the series

Since the integral converges [1/n], the series converges.

1 2 3 4 5 6 7 8 9 10

Integral 1dx/x

2 Versus Series n=2

Taylor Series

A Taylor series is a power series representation for a function.

kk azczf

A Taylor series is much like a Fourier series (which is a harmonic series).

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.

We start with the fraction in the integral:

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:

aazaaza

The last is true if is on a curve C at a distance r from a and z is within the close curve.

Im{z} is on C

C n da

The coefficient of (z – a)k inside the summation

C kk d

is similar to the k-th derivative of f(a) from the corollary to Cauchy’s integral theorem:

Hence, we have our Taylor series coefficients.

.sin)(

.cos)(

Example: Find the Taylor series for

. oddn cos1

evenn sin1)(

. oddn sin1

evenn cos1)(

We will evaluate the Taylor series about z = 0 :

. oddn 1

evenn 0)(

.oddn 0

evenn 1)0(

.!4!3!2

zzzzzf

You can use these Taylor series to prove

.sincos zjze jz

Here, we will evaluate the series about a = 1.

.11111

.ln)( zzf

Here, we will evaluate the series about a = 1.

1)( 1)(

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

To what do these curves correspond to in the u-v plane?

Let us start with a simple example

yjxjvuw

u = Re{w}

v = Im{w}

x=1 x=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

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}

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

y = -1y = -2

x = -2

Example: Find the conformal map for

We represent z and w in terms of their real and imaginary components:

jyxjvuw

We then try to make the denominator real:

])1[1()1)(1(

xxjyyxx

jyxjvu

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

w = (z-1)/(z+1)

y = -2y = -1

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

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

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.

As another example, let

)( jj eerz

As z goes around the circle counter-clockwise, w goes around the same circle clockwise.

Let us choose two points on the z-path: z1 and z2

The two points z1 and z2 are very close together; their radii are the same, but their angles are different.

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.

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

121212

Now log w can be written as an indefinite integral:

dwwlog

The difference between log w1 and log w2 can be written as a definite integral:

112 loglogw

The points 1 and 2 in the integral

correspond to z1 and z2. So,

)(loglog

Combining this expression with

)(loglog 12

dzzfww

,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.

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.

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:

zszssG

Let s take on an infinite circular path in the right-half complex plane.

What kind of path will G(s) take?

To answer this question, let us try to evaluate the integral

where C is the curve described on the previous slide.

211 sH

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:

we have

)()()(

we have

)()()(

The integral of

is equal to

CCC sH

By the Cauchy Integral Formula, the integral

sFpsps

Now suppose z1 is a right-half plane zero. We can define a new function F(s) with this zero removed:

).()( 1 sFpssG

we have

).()()( '1

' sFzssFsG

).()( 1 sFzssG

we have

).()()( '1

' sFzssFsG

).()( 1 sFzssG

The integral of

is equal to

CCC sF

By the Cauchy Integral Formula, the integral

If we continue eliminating poles and zeroes, we get a term of -2j for every pole and a term of +2j for every zero.

),(2)(

)('PZj

where Z is the number of zeroes, and P is the number of poles.

Now from our previous result,

we see that

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.

dssGNj

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:

.)( 2zzf

We have already done two examples:

.)( 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

.)( 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 ?

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

Nyquist Diagrams

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

2From: U(1)

In control systems, we are often concerned about the poles of a closed-loop transfer function

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.

Therefore, if 1+G(s) has no right-half plane poles, the number of clockwise rotations around s = -1 of

is equal to the number right-half plane poles of

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