Phasors and Complex Numbers

1. Phasors and Complex Numbers Alan Murray

2. Complex NumbersRevision • j = √(-1), so j2 = -1 • 2,3, -4.75 are real numbers • 2j, 3j, -4.75j are imaginary numbers • (3.5 - 7.2j) is a complex number • With real and imaginary parts • +3.5 and -7.2j respectively Alan Murray – University of Edinburgh

3. Complex Arithmetic • C1 = a + jb • C2 = c + jd • Adding two complex numbers … • C1 + C2 = (a + c) + j(b + d) • add Re() and Im() parts separately • Multiplying two complex numbers … • C1 x C2 = (a + jb) x (c + jd) = (ac – bd) + j(bc + ad) • But … • C1 x C1 = (a + jb) x (a + jb) = (a2 – b2) + j x 2ab • So |C1| is NOT√(C12) as it is for real numbers (a + jb) x (c + jd) ac + jbc + j2bd + jda ac + jbc - bd + jda Alan Murray – University of Edinburgh

4. Imaginary Part C1 C2 b2 RealPart a2 Complex Numbers as Pictures C1 = a1 + jb1 C2 = a2 + jb2 This is an“Argand Diagram” b1 a1 C1 has real and imaginary components a1 and b1 respectively This is very similar to the x- and y- components of a 2D vector (remember them?). Alan Murray – University of Edinburgh

5. b3 Imaginary Part C3 b1 C1 C2 C2 b2 RealPart a3 a1 a2 Complex Numbers as Pictures Adding Complex Numbers …just like vectors, C3=C1+C2 C1 = a1 + jb1 C2 = a2 + jb2 C3 = (a1 + a2) + j(b1 + b2) Alan Murray – University of Edinburgh

6. Imaginary Part b C RealPart a Complex Numbers as Pictures Magnitude of Complex Numbers …almost like vectors If C were simply a vector C Pythagoras →|C| = √ (a2 + b2) |C| = √[C.C] = √[C2] Here, |C| also = √ (a2 + b2) But |C| ≠ √ [C2] |C| ≠ √[(a+jb)(a+jb)] |C| = √[(a+jb)(a–jb)] |C| = √(C x C*) C* = Complex Conjugate of C Alan Murray – University of Edinburgh

7. Imaginary Part RealPart I V`R VS VC I Complex Numbers as Pictures Argand Diagram Phasor Diagram C3 = C1 + C2 VS = VR + VC I, VR C1 C2 VC Alan Murray – University of Edinburgh

8. Trigonometry is conceptually simple Phasor diagrams are nice to analyse sin(A+B)+cos(C+D) etc is a pain! Phasor diagrams for complex circuits look like this Back to lots of nasty trigonometry AC Circuit analysis Alan Murray – University of Edinburgh

9. Rationale • Argand Diagrams = Phasor Diagrams? • Can we use complex number mathematics (L) to make geometrical things simpler to do (J)? • No pain, no gain! Alan Murray – University of Edinburgh

10. I, VR VS = VR + VC VC Remember how phasors work? Now spin it! Let’s look closer I VR VS VC I Alan Murray – University of Edinburgh

11. Ø Translation into complex numbers and Argand diagrams? Imaginary Part RealPart “Project on to x-axis” = “Take the real part” “Spin” to create cos(ωt) = allow Ø to run 0→360˚ Alan Murray – University of Edinburgh

12. b = C0sin(Ø) C0 = |C| = √(a2+b2) Ø a = C0cos(Ø) So we need to represent C = a+jb in terms of Ø and |C| = C0 Imaginary Part C = a+jb RealPart C = a+jb = C0cos(Ø) + jC0sin(Ø) = C0[cos(Ø)+jsin(Ø)] Alan Murray – University of Edinburgh

13. Imaginary Part C = a+jb RealPart Ø There is a neater, if superficially nastier, notation • C = a+jb = C0[cos(Ø)+jsin(Ø)] • It turns out, can be written:- • cos(Ø)+jsin(Ø) = ejØ • So C = C0ejØ (or C0ÐØ) • Means “C0 at an angle of Ø°” • ejØ does all the right things, when we take the real part to get a real number (voltage or current, in our case) out of it. • Our numbers will be real voltages and currents Alan Murray – University of Edinburgh

14. cos(Ø)+jsin(Ø) = ejØ … why? Here are some old friends(!) … Now substitute “jØ” for “x” Alan Murray – University of Edinburgh

15. cos(Ø)+jsin(Ø) = ejØ … proved. Now substitute “jØ” for “x” Alan Murray – University of Edinburgh

16. Some examples for you • C1 = 3 + 4j • C2 = 5 – 2j • C1+C2 = • C1*C2= • C1*C1 = • |C1| = 8 + 2j (15+8) + (20–6)j =23 + 14j (3+4j)(3+4j) = -7 + 24j √[(3+4j)(3-4j)] = 5 Alan Murray – University of Edinburgh

17. j 2 1 1 2 Some examples for you Alan Murray – University of Edinburgh

18. j f 1 2 -1 -2 Some examples for you Clicker time Alan Murray – University of Edinburgh

19. Take a complex number ejØ ejØ = cos(Ø)+jsin(Ø) Multiply by j j×ejØ = jcos(Ø)+j2sin(Ø) = -sin(Ø)+jcos(Ø) Multiply by j again j×j×ejØ = j2cos(Ø)-jsin(Ø) = -cos(Ø)-jsin(Ø) Multiply by j again j×j×j×ejØ = j3cos(Ø)-j2sin(Ø) = sin(Ø) -jcos(Ø) And finally j×j×j×j×ejØ = ejØ =cos(Ø)+jsin(Ø) jsin(Ø) ejØ jejØ Ø cos(Ø) j2ejØ j3ejØ And ×j does something interesting Imaginary Part j4ejØ RealPart Alan Murray – University of Edinburgh

20. ejØ = cos(Ø)+jsin(Ø) Multiply by j j×ejØ = jcos(Ø)+j2sin(Ø) = -sin(Ø)+jcos(Ø) Multiply by j j×j×ejØ = j2cos(Ø)-jsin(Ø) = -cos(Ø)-jsin(Ø) Multiply by j j×j×j×ejØ = j3cos(Ø)-j2sin(Ø) = sin(Ø)-jcos(Ø) Multiply by j j×j×j×j×ejØ = ejØ =cos(Ø)+jsin(Ø) And ×j does something interesting Re(ejØ) Real part in blue xj = 90° phase advance xj = 90° phase advance xj = 90° phase advance xj = 90° phase advance Alan Murray – University of Edinburgh

21. e jØ = cos(Ø) + jsin(Ø) xj rotates a complex number by 90° i.e. xj advances Re(e jØ) by 90° e jπ/2 = cos(π/2) + jx sin(π/2) = 0 + jx 1 So e jπ/2 = +j Summary …complex numbers …and some little cunning stunts Alan Murray – University of Edinburgh

22. ωt What does this have to do withV = V0cos(ωt) etc? Set Ф = ωt • VS = VS0ejωt • VS = VS0[cos(ωt) + jsin(ωt)] • For the real voltage VS • take the real part of VS • VS = VS0cos(ωt) • That is all there is to it – better understood via some examples • We will do the RC example on the left first, 3 times, to compare … • the ejωt method • a phasor diagram solution • dear old trigonometry VS = VR + VC I VR VS VC I Alan Murray – University of Edinburgh