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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.

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Presentation on theme: "Complex Variables. Complex numbers are really two numbers packaged into one entity (much like matrices). The two numbers are the real and imaginary portions."— Presentation transcript:

1 Complex Variables

2 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:

3 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 y z = x + jy

4 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 y z = x + jy r

5 The best way to express a complex number in polar coordinates is to use Eulers identity: So, and

6 We also have A summary of the complex relationships is on the following slide.

7 Re{z} Im{z} x = r cos y = r sin r

8 The magnitude of a complex number is the square- root of the sum of the squares of the real and imaginary parts: If we set the magnitude of a complex number equal to a constant, we have

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

10 Re{z} Im{z} x y c 2 = |z| 2 = x 2 + y 2 c z = x + jy

11 Suppose we wish to find the region corresponding to This would be a disk, centered at the origin, of radius c.

12 Re{z} Im{z} x 2 + y 2 = |z| 2 < c 2 x y c

13 Suppose we wish to find the region corresponding to This would be a disk, centered at z 0, of radius c.

14 Re{z} Im{z} (x-x 0 ) 2 + (y-y 0 ) 2 = |z-z 0 | 2 < c 2 = |z-z 0 | 2 x0x0 y0y0 c z0z0

15 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

16 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.

17 where Both u and v are functions of x and y.

18 So a complex function of one complex variable is really two real functions of two real variables.

19 Exercise: Find u(x,y) and v(x,y) for each of the following complex functions:

20 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. A complex function is said to be continuous at a point z 0 if as z approaches z 0 (from any direction) then f(z) can be made arbitrarily close to f(z 0 ).

21 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.

22 Re{z} Im{z} z0z0 The region |z-z 0 | < defines a disk in the complex plane of radius centered about z 0.

23 Re{w} Im{w} f(z 0 ) So, if we wish |f(z)-f(z 0 )| < we must find a to make this so.

24 Example: Suppose Find such that for

25 Solution:

26 All we need to do is to find a value of such that if then

27 We can do some calculations on a spreadsheet (continuity.xls). A value of < 0.1 seems to do it.

28 A MATLAB plot (by continuity.m) of the previous example is shown on the following slide.

29

30 Differentiation of Complex Functions How do we take derivatives of complex functions with respect to complex variables? If what is

31 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 y y+dy x+dx dx dy

32 As z varies in either direction, the derivative must be the same. x direction y direction So, we must have

33 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).

34 Example: Show that the function is differentiable Solution: We have shown that

35

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

37 If we apply these formulas to where

38 or we have

39 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.

40 Example: Show that the function is not differentiable Solution: We have shown that

41

42 Exercise: Is differentiable?

43 Definition: A function is said to be analytic if it is differentiable throughout a region in the complex plane.

44 Integration of Complex Functions What happens when we try to take the integral of a complex function along some path in the complex plane?

45 A complex integral is like a line integral in two dimensions. The real and the imaginary parts of the integral are nearly identical to classic line integrals.

46 Example: Integrate over the real interval z = 0 + j0 to z = 2 + j0. Solution: We have shown that

47

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

49 The result is exactly what we would expect to get if we simply integrated a real variable from 0 to 2.

50 Example: Integrate over the imaginary interval z = 0 + j0 to z = 0 + j2. Solution: The integral becomes

51 The result is exactly what we would expect to get if we simply integrated where C = jy and where y=[0,2] :

52 Example: Integrate over the complex path z = 0 + j2 to z = 2 + j2. Re{z} 2 2 Im{z}

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

54

55 Example: Integrate over the complex path z = 2 + j0 to z = 2 + j2. Re{z} 2 2 Im{z}

56 Solution:

57 Example: Integrate 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}

58 Solution: The integral is more complicated.

59

60 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}

61 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.

62 Example: Integrate over two paths: (1) a semicircle z = e j, where = [0, ]. (2) a semicircle z = e -j, where = [0, ]. Show that the two integrals are the same.

63 Re{z} Im{z} C1C1 C2C2 = 0 =

64 Solution: This integration is best handled using polar coordinates:

65 The integral around curve C 1 is

66 The integral around curve C 2 is

67 If we were to integrate around the whole circle C = e j for = [0, 2 ], we would get The curve C can be thought of as C 1 + (-C 2 ).

68 Cauchys Integral Theorem: If a function f(z) is analytic over a region R enclosed by a (closed) path C, then Re{z} Im{z} C R

69 Simple Proof: Both integrals are line integrals around a closed curve C. We can apply Greens theorem (a special case of Stokes theorem) to these line integrals

70 If f(z) is analytic, then the Cauchy-Riemann equations apply: If these are true, then both integrands of are zero and the theorem is proved.

71 If and C = C 1 + C 2, then

72 Re{z} Im{z} C1C1 C2C2

73 We also have

74 Re{z} Im{z} C1C1 -C 2

75 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.

76 Example: Integrate over a unit circle z = e j, where = [0,2 ]. Re{z} Im{z}

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

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

79 Example: Integrate over a unit circle z = 1 + e j, where = [0,2 ]. Re{z} Im{z}

80 Solution: Let us first write the integral. To carry-out this integration, we first perform a substitution of variables. Let = z-1.

81 The path C is equal to C minus one: Re{z} Im{z} CC We see that

82 This integral is not zero because there is a discontinuity (a pole) at z = 1 or = 0.

83 Exercise: Show that no matter what z 0 is, if C is a circular path (of any radius) around z 0 we will have

84 Cauchys Integral Formula: Let f(z) be analytic over a region R enclosed by a closed path C. If z 0 is a point within R, then Note that while f(z) is analytic throughout R, f(z)/(z-z 0 ) is not analytic (z 0 is a pole).

85 Re{z} Im{z} C z0z0 R

86 Proof: We add and subtract f(z 0 ) to the numerator of the integrand so as to split-up the integral into two terms:

87 If f(z) is analytic within the region R, then it is also continuous. So, for any, we can find a such that for we have Let us choose a such that z R, i.e., the disk |z-z 0 | < is totally within R. Let denote the path |z- z 0 | = by the symbol C.

88 Re{z} Im{z} C z 0 R C

89 So if and

90 So for appropriate values of and, the integrand in can be made arbitrarily small. Now since the integrand is analytic except at z = z 0, we have

91 The integral is equal to 2 j. So, and the theorem is proved.

92 Example: Evaluate the integral Solution: where C is a closed curve around z = 1. So,

93 The formula makes calculating derivatives with respect to z 0 relatively easy. We do not have to worry about z: it is independent of z 0.

94 In general it can be shown that This formula is very useful in deriving the Taylor series.

95 Example: Evaluate the integral Solution: where C is a closed curve around z = 1. So,

96 Example: Evaluate the integral Solution: where C is a closed curve around z = 1. So,

97 Example: Evaluate the integral Solution: where C is a closed curve around z = 2.

98 Exercise: Evaluate the integral where C is a closed curve around z = 1.

99 Inverse Laplace Transforms 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!

100 The inverse Laplace transform can be found using a complex inversion integral formula: We can evaluate the inverse Laplace transform using Cauchys integral formula.

101 Cauchys 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.

102 Re{s} Im{s} r s +j s -j C s -

103 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.

104 Example: Find the inverse Laplace transform of Solution:

105

106 Example: Find the inverse Laplace transform of Solution:

107 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:

108 Re{s} Im{s} C

109 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 C 1 and C 2.

110 Re{s} Im{s} s 0 = +j s 0 = -j C1C1 C2C2 C

111

112 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.)

113 Example: Find the inverse Laplace transform of Solution:

114 There is one point of discontinuity:

115 Sequences and Series Consider the sequence of values each term in this sequence can be represented by

116 What happens as n goes to infinity? For this example z n goes to zero. How about the sequence of values each term in this sequence can be represented by

117 This sequence is said to converge to two (2): as n goes to infinity, z n goes to two. How about the sequence of values This series does not converge to any value, but rather it is said to diverge.

118 This sequence {z n } is said to converge to a value c if z n 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 we must have

119 The expression is for a complex number z n and defines a disk in the complex plane. Re{z} Im{z} c

120 Example: Plot the sequence of values in in the complex plane Solution: The radius of z n is 1/n and the angle of z n is 2 n/20. The plot is performed using MATLAB (sequence.m) and is shown on the following slide page.

121

122 This sequence converges to zero. The relationship between the sequence index n and distance to the limit is rather easy in this case. For we must have where 1/ corresponds to the smallest integer greater than 1/.

123 Similarly, for we must have

124 A necessary (but not sufficient) condition for a sequence to converge is that it be bounded. A bounded sequence is {z n } is one that for all n we have for some finite value B. If a sequence is not bounded, it will diverge.

125 Just because a sequence is bounded does not mean it converges. Many sequences which are bounded do not converge. For example, For this sequence

126 12345 n znzn

127 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.

128 Note that while {z n } does not converge, there are subsequences that do converge.

129 Series Suppose we were to add the numbers in a sequence: The term s n is called the partial sum of the series {z n }. The summation is a series.

130 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.

131 A necessary (but not sufficient) condition for the series {s n } to converge is that the sequence {z n } 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.

132 Example: Consider the (convergent) sequence The corresponding series does not converge (as n goes to infinity):

133 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

134 (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.

135 (1) Comparison Test: compare series term-by-term against a known convergent series. If we have a known convergent series then any series

136 such that also converges.

137 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

138

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

140 In order for q n to go to zero, we must have If q > 1, then the series diverges.

141 Example: Suppose q = ½.

142 Example: Suppose q = 9/10.

143 Example: Suppose q = -1/2.

144 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

145 If then the series converges.

146 So, As a justification (hardly a proof) for this test, consider definining z k in terms of w k :

147 The term

148 If then the series behaves like a convergent geometric series:

149 Example: Determine if converges. Solution: Taking the ratio test We see that the series converges.

150 If then we take (4) Root Test: Take the k-th root of the k-th term. If the limit is less than one, the series converges.

151 If then the series converges.

152 As a justification (better than that of the ratio test but still not a proof) for this test, consider If w k < 1, then the series is a convergent geometric series.

153 Example: Determine if converges. Solution: Taking the root test We see that the series diverges. (e k dominates k 2.)

154 The term z k is really a function of k. We can represent that function as z(k) or z(x). The convergence of the integral (5) Integral Test: compare the sequence to an integrand of a known integral. can be used to determine whether or not the series converges.

155 This series is greater than the integral Example: Consider the series Since the integral diverges [ln n], the series diverges.

156

157 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.

158

159 Taylor Series A Taylor series is a power series representation for a function. A Taylor series is much like a Fourier series (which is a harmonic series).

160 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 Cauchys integral formula: We will attempt to express this formula in a power series about z = a.

161 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:

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

163 Re{z} Im{z} is on C a r z

164

165 The coefficient of (z – a) k inside the summation is similar to the k-th derivative of f(a) from the corollary to Cauchys integral theorem:

166 So, Hence, we have our Taylor series coefficients.

167 Example: Find the Taylor series for

168

169 We will evaluate the Taylor series about z = 0 :

170 So,

171 You can use these Taylor series to prove

172 Example: Find the Taylor series for Here, we will evaluate the series about a = 1.

173

174 Example: Find the Taylor series for Here, we will evaluate the series about a = 1.

175 If z is not close to one, this series is very slow to converge.

176 Exercise: Find the Taylor series for Evaluate the series about an arbitrary a.

177 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).

178 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.

179 x = Re{z} y = Im{z} x=1x=2x=3x=4 y=1 y=2 y=3 y=4

180 To what do these curves correspond to in the u-v plane? Let us start with a simple example

181 u = Re{w} v = Im{w} x=1x=2 y=1 y=2

182 Example: Find the conformal map for We expand e z from the real and imaginary parts of z.

183 This expansion is best handled using polar coordinates. where

184 The resultant curves will be a set of circles of radii e x. Constant values of y correspond to rays at angle y.

185 u = Re{w} v = Im{w} x=1 x=2 y=1 y=2

186 Negative values of x correspond to circles of radius e -|x|. Negative values of y correspond to rays at angle -|y|.

187 u = Re{w} v = Im{w} x = -1 x=2 y = -1y = -2 x = -2

188 Example: Find the conformal map for We represent z and w in terms of their real and imaginary components:

189 We then try to make the denominator real:

190 A plot of the constant x and the constant y curves (in the u-v plane) is shown on the following slide.

191

192 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.

193 The Argument Principle Let 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?

194 Let z follow a closed path in the z-plane. Re{z} Im{z} C

195 As z follows this closed path, what path will w=f(z) follow? As an example, let As z goes around the circle once, w goes around the same circle twice.

196 Re{z} Im{z} C Re{w} Im{w} C

197 As another example, let As z goes around the circle counter-clockwise, w goes around the same circle clockwise.

198 Re{z} Im{z} C Re{w} Im{w} C

199 Let us choose two points on the z-path: z 1 and z 2 Re{z} Im{z} z1z1 z2z2 The two points z 1 and z 2 are very close together; their radii are the same, but their angles are different. C

200

201 The corresponding points in the w-plane f(z 1 ) and f(z 2 ) are very close together. Like z, the radii of f(z 1 ) and f(z 2 ) are the same, but their angles are different.

202 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).

203 The difference between log w 1 and log w 2 is

204 Now log w can be written as an indefinite integral: The difference between log w 1 and log w 2 can be written as a definite integral:

205 The points 1 and 2 in the integral correspond to z 1 and z 2. So,

206 Combining this expression with we have

207 So, the integral from z 1 to z 2 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.

208 Since w=f(z) follows a closed path in the w-plane, the angular rotation, 2 – 1, must be an integral multiple of 2.

209 Suppose G(s) is a polynomial fraction: Let s take on an infinite circular path in the right- half complex plane.

210 Re{s} Im{s} r s +j s -j C s +

211 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.

212 We will try to evaluate this integral by removing poles and zeroes in the right-half plane. Suppose p 1 is a right-half plane pole. We can define a new function H(s) with this pole removed: So,

213 If we have

214 If we have and

215 The integral of is equal to

216 By the Cauchy Integral Formula, the integral So

217 Now suppose z 1 is a right-half plane zero. We can define a new function F(s) with this zero removed: So,

218 If we have

219 If we have and

220 The integral of is equal to

221 By the Cauchy Integral Formula, the integral So

222 If we continue eliminating poles and zeroes, we get a term of -2 j for every pole and a term of +2 j for every zero. So where Z is the number of zeroes, and P is the number of poles.

223 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.

224 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:

225 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.

226 Equivalently, for 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).

227 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.

228 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.

229

230 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).

231 If we did a plot of 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).

232 The number of clockwise rotations around the origin of is equal to the number of clockwise rotations of around s = -1.

233 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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