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# Complex%20numbers

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### Complex%20numbers

1. 1. MATHEMATICS FOR ENGINEERING TUTORIAL 6 – COMPLEX NUMBERSThis tutorial is essential pre-requisite material for anyone studyingmechanical and electrical engineering. It follows on from tutorial 5 onvectors. This tutorial uses the principle of learning by example. Theapproach is practical rather than purely mathematical.On completion of this tutorial you should be able to do the following. • Explain and use complex numbers. • Explain and use the Argand diagram. • Explain and use vectors with polar co-ordinates. • Multiply and divide complex numbers. • Explain and use the conjugate number. • Explain and use phasors and phasor diagrams. • Explain the meaning of phase angle. • Explain and use complex impedance.It is presumed that students have already studied vectors, basic algebra andtrigonometry.© D.J.Dunn 1
2. 2. 1. ROOTS OF NEGATIVE NUMBERSOrdinary numbers can be added, subtracted and multiplied and are good enough for every day use.In engineering, we come across problems that can not be solved with ordinary numbers and one ofthese problems is how to handle the square root of a negative number. You should already knowthat 1 x 1 = 1 and that -1 x -1 = 1 so it follows that √1 is either 1 or -1. However there is no answerto the question ‘what is the root of -1?’ There is no number that can be multiplied by its self to give-1. To get around this problem in the first instance, we simply designate √-1 by the letter j and itfollows that j2 = -1 and j = √-1Consider the equation x2 = -9. Using conventional numbers, there is no solution but using this newidea, the solution becomes j3 since (j3)2 = j2 x 32 = -1 x 9 = -9.2. COMPLEX NUMBERConsider the number given as P = A + − B 2If we use the j operator this becomes P = A + − 1 x BPutting j = √-1we get P = A + jB and this is the form of a complex number. WORKED EXAMPLE No.1 Find the solution of P = 4 + − 9 and express the answer as a complex number. SOLUTION P = 4 + − 9 = 4 + j3 SELF ASSESSMENT EXERCISE No.11. Write down the solution to the following. x2 = -4 x= 2 x = -25 x= x2 = -10 x= (Answer j2, j5 and j√10)2. Express the following as complex numbers. P = 3 + − 16 P = 2 − − 81 P = −5 − − 12 Answers (3 + j4), (2 – j 9) and (-5 –j √12)© D.J.Dunn 2
3. 3. 3. FURTHER PROPERTIES OF THE OPERATOR jConsider a point A on a Cartesian plane situated at coordinates 4, 0 as shown. Now suppose thatmultiplying this point by j has the affect of rotating the line O–A 90o anticlockwise. This willproduce point B which should be designated as j4 to indicate it is on the vertical axis. Figure 1If we multiply point B by j we rotate again to get point C. This is located at -4 and was obtained bymultiplying A by j2.Since j2 = -1 then point C is at j24 = -4 which is correct.If we multiply by j again and we get point D and this is j34 = -j4 so point D is designated –j4.This work was produced by a French mathematician called Argand. We may simplify matters bylabelling the vertical axis ‘j’. Numbers on the horizontal axis are called REAL NUMBERS and onthe vertical axis are called IMAGINARY NUMBERS. Point A is +4, point B is j4, point C is –4and point C is –j4. This is fine for handling negative numbers but does not explain what a complexnumber is.4. ARGAND DIAGRAMA complex number A + jB could be considered to be twonumbers A and B that may be placed on the previousgraph with A on the real axis and B on the imaginary axis.Adding them together as though they were vectors wouldgive a point P as shown and this is how we represent acomplex number. The diagram is now called an ArgandDiagram. Figure 2If we draw a line from the origin to the point P it forms a vector and in some applications it is calleda phasor. The length of the line is called the MODULUS and the angle formed with the real axis iscalled the ARGUMENT. The complex number can hence be expressed in polar form as │OP│∠θConsider four such numbers on the Argand diagram, one in each quadrant as shown.© D.J.Dunn 3
4. 4. Consider the vector labelled No.1. The horizontal component is 5 and the vertical component is 8 sothe vector may be written as P = 5 + j8.The angle of the vector is tan-1 (8/5) = tan-1 (1.6) = 58o Figure 3 SELF ASSESSMENT EXERCISE No. 21. Write down the other three vectors in the form A + jB and calculate their angles. 2 _______________________________________________ 3 _______________________________________________ 4 _______________________________________________© D.J.Dunn 4
5. 5. 5. MULTIPLYING COMPLEX NUMBERSUSING POLAR CO-ORDINATESA complex number may be expressed in polar co-ordinates as follows. Let the Modulus be R andthe argument θ. Consider the two shown. We have R1 ∠θ1 and R2 ∠θ2 Figure 4We should not confuse the multiplication of vectors (see dot and cross products in the vectortutorials) with the multiplication of complex numbers.The real and imaginary co-ordinates are A1 = R1 cosθ1 B1 = R1 sinθ1 A2 = R2 cosθ2 B2 = R2 sinθ2The complex number for each vector is: P1 = R 1cosθ1 + jR 1sinθ1 = R 1{cosθ1 + jsinθ1} P2 = R 2 cosθ 2 + jR 2 sinθ 2 = R 2 {cosθ 2 + jsinθ 2 }Multiplying them together and treating j as √-1 we get the following.P1 x P2 = R 1{cos θ1 + jsin θ1} x R 2 {cos θ 2 + jsin θ 2 }P1 x P2 = R 1R 2 [cos θ1cos θ 2 + cos θ1 jsin θ 2 + jsin θ1cos θ 2 + jsin θ1 jsin θ 2 ] [P1 x P2 = R 1R 2 cos θ1cos θ 2 + j(cos θ1sin θ 2 + sin θ1cos θ 2 ) + j2 sin θ1sin θ 2 ]P1 x P2 = R 1R 2 [cos θ1cos θ 2 − sin θ1sin θ 2 + j(cos θ1sin θ 2 + sin θ1cos θ 2 )]P1 x P2 = R 1R 2 [cos(θ1 + θ 2 ) + jsin(θ 2 + θ 2 )]This is a vector with a length R1R2 and angle θ1+θ2. The rule for multiplying is :The Modulus is the product of the other Modulii and the argument is the sum of the angles. Thisrule applies for any number of vectors multiplied together.© D.J.Dunn 5
6. 6. WORKED EXAMPLE No. 2 Find the result of (3 ∠45o) x (2 ∠30o) SOLUTION Modulus = 3 x 3 = 6 Argument is 45 + 30 = 75o The result is hence 6∠75 o SELF ASSESSMENT EXERCISE No.3 Find the vector that result for each below.1. 5 ∠50o x 3 ∠70o2. 7 ∠80o x 2 ∠30oUSING COMPLEX FORMConsider the following problem. Multiply 3 ∠45o x 2 ∠30o. The result is 6∠75 o. Figure 5To do this as complex numbers is more difficult as we shall now see. In the form A + j B wehavethe following.P1 has coordinates A1 = 3 cos 45 = 2.121 and B1 = 3 sin 45 = 2.121P2 has coordinates A2 = 2 cos 30 = 1.732 and B2 = 2 sin 30 = 1.0P1 = 2.121 + j 2.121 and P2 = 1.732 + j1© D.J.Dunn 6
7. 7. The result is a complex number P = P1 x P2 = (2.121 + j 2.121) x (1.732 + j1)P = P1 x P2 = (2.121 x 1.732) + (2.121 x j1) + (j 2.121 x 1.732) + (j2.121 x j)P = P1 x P2 = 3.673 + j2.121 + j3.673 + 2.121j2P = P1 x V2 = 3.673 + j5.794 – 2.121P = P1 x P2 = 1.5221 + j5.794This is shown on the diagram from which we deduce the following. Figure 6R = √(1.52212 + 5.7942) = 6θ = tan-1(5.794/1.5221) = 75oV = 6∠75oHence we have arrived at the same solution but in a more difficult way. WORKED EXAMPLE No.3 Find the result of multiplying the following complex numbers. P = (4 + j2) x (2 + j3) SOLUTION P = (4 + j2) x (2 + j3) = 8 + j12 +j4 + 6j2 = 8 + j16 – 6 = 2 + j16 SELF ASSESSMENT EXERCISE No.4 Find the result of multiplying the following complex numbers. 1. (3 + j3) x (5 – j2) Answer (21 + 9j ) 2. (12 + j2) x ( 2 – j3) Answer (30 - 32j )© D.J.Dunn 7
8. 8. 6. CONJUGATE NUMBERSThe conjugate of a complex number has the opposite sign for the j part.The conjugate of A + jB is A – jB.If a complex number is multiplied by its conjugate the result is a real number. WORKED EXAMPLE No.4 Find the result of multiplying (2 + j3) by its conjugate. SOLUTION The conjugate is (2 - j3) (2 + j3) x (2- j3) = 4 - 6j +6j -9j2 = 4 –(-9) = 13 SELF ASSESSMENT EXERCISE No.5 Find the result of multiplying the following by their conjugate. 1. (5 – j2) (Answer 29) 2. (-4 – j4) (Answer 32) 3. (7 + j6) (Answer 85)© D.J.Dunn 8
9. 9. 7. DIVISION OF COMPLEX NUMBERSSuppose V = 8 + j8 and I = 4 - j8 and we wish to find V/I. (This is Ohms Law for compleximpedance). This is done by multiplying the top and bottom by the conjugate of the bottom numberas follows. WORKED EXAMPLE No. 5 If A = 8 + j8 and B = 4 – j8 find the result of dividing A by B. SOLUTION V 8 + j8 = I 4 − j8 V 8 + j8 4 + j8 = x I 4 − j8 4 + j8 V 32 + j64 + j32 + j2 64 = I 16 + j32 - j32 - j2 64 V 32 + j96 - 64 - 32 + j96 ⎛ 32 ⎞ ⎛ 96 ⎞ = = = - ⎜ ⎟ + j⎜ ⎟ I 16 + 64 90 ⎝ 90 ⎠ ⎝ 90 ⎠ SELF ASSESSMENT EXERCISE No.6 Find the following results. V 1 + j2 1. = (Answer 0.7 – 0.1j) I 1 + j3 V 5 − j2 2. = (Answer 0.1 + 1.2j) I 2 − j4 x 1 − j9 3. = (Answer -1.3 - 0.6j) y 2 + j6© D.J.Dunn 9
10. 10. 8. PHASOR DIAGRAMSA phasor is used to represent harmonic quantities such as alternating electricity and oscillatingmechanical systems.The phasor is a rotating vector with a constant length and the speed of rotation is the same as theangular frequency of the quantity (always anticlockwise). Projecting a rotating vector onto thevertical scale of a graph with angle plotted horizontally will generate a sinusoidal waveform. If thevector represents voltage or current it is called a PHASOR. The rotation is anti-clockwise. Figure 7Suppose we wish to represent a sinusoidal voltage by a phasor. The maximum voltage is V and thevoltage at any moment in time is v. The phasor is drawn with a length V and angle θ as shown.The angle is given by θ = ωt where t is the time and ω is the angular frequency in radian/s.The voltage at any moment in time is the vertical projection such that v = V sin (ωt)Since there is no necessity to start plotting the graph at the moment θ = 0 a more general equation isv = V sin (ωt+ φ) where φ is the starting angle and is often referred to as the phase angle. A phasormay also be given in polar form as V∠(θ+φ)If the phasor is drawn on an Argand diagram, the vertical component is the imaginary part and thehorizontal component is the real part. It follows that a harmonic quantity can be represented as acomplex number. An Argand diagram may be used to show a phasor at a particular moment in time.It might show more than one phasor. For example when the voltage across an inductor is showntogether with the current through it, the current is ¼ cycle behind the current so at a given momentin time the relationship might be like this. Figure 8© D.J.Dunn 10
11. 11. WORKED EXAMPLE No.5 A sinusoidal voltage has a peak value of 200 V and a phase angle of 20o. Represent it as a polar vector and a complex number. Sketch the phasor when θ = 50o. SOLUTION The polar form is v = 200∠θ + 20o The vertical component is v = 200 sin(θ+20o) The horizontal component is 200 cos(θ+20o) The complex number is hence v = 200 cos(θ+20o) + {200 sin(θ+20o)}j Putting θ = 50o this becomes v = 68.4 + 187.9 j The angle is tan-1(187.9/68.4) = 70o as expected. Figure 9© D.J.Dunn 11
12. 12. 9. REPRESENTING IMPEDANCE AS A COMPLEX NUMBERWhen an electric circuit with alternating current contains resistance, inductance and capacitance, thecurrent and voltage will not vary in time together but one will lead the other. The above diagramshows one example of this. The impedance of an electric circuit is defined as Z = V/I and in order todivide V by I we must represent the phasor as a complex number. Suppose V = 8 + j8 and I = 4 -j8. V 8 + j8 = I 4 - j8Multiply the top and bottom 4 + j8. This is called the conjugate number and it turns the bottom Theline into a real number. This does not change the equality.V 8 + j8 4 + j8 32 + j64 + j32 + j2 64 32 + j96 - 64 - 32 + j96 32 96 = x = = = =- + jI 4 - j8 4 + j8 16 + j32 − j32 - j 64 2 16 + 64 80 80 80The Impedance is Z = –(32/80) + j (96/80)On an Argand diagram the voltage and current are like this. Figure 10Impedance is also a phasor and looks like this. Figure 11The real part is called the Resistance part (R) and the imaginary part is called the reactive part X. ⎛X⎞It follows that. Z = X 2 + R 2 and φ = tan −1 ⎜ ⎟ and φ is called the phase angle. ⎝R⎠It follows that complex impedance may be written in the form Z = R + jX and impedances may beadded or subtracted.© D.J.Dunn 12
13. 13. SELF ASSESSMENT EXERCISE No.71. An electric circuit has a complex impedance of Z = 300 + j40. What is the resistance of the circuit and what is the reactance? (300 Ω and 40Ω) What is the phase angle? (7.6o)2. Two electric circuits are connected in series. The complex impedance of the first is Z = 50 +j3 and the second is Z = -20 + j2. What is the combined impedance? (Add them). (30 +5j) What is the resistance and reactance of the combined circuit? (30Ω and 5Ω) What is the phase angle? (9.5o)3. The current in an electric circuit is represented by I = 5 + j2 and the voltage is V = 100 + j5. Determine the impedance as a complex number. (17.6 + 6j) Determine the resistance and reactance of the circuit. (17.6Ω and 6Ω) Determine the phase angle. (18.8o)© D.J.Dunn 13