Lecture 0. MATH REVIEW for ECE : LINEAR CIRCUIT ANALYSIS II

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1 Lecture 0 MATH REVIEW for ECE 000 : LINEAR CIRCUIT ANALYSIS II Aung Kyi Sn Grdute Lecturer School of Electricl nd Computer Engineering Purdue University Summer 014

2 Lecture 0 : Mth Review Lecture 0 is intended for self-study. In this lecture, we will review some mthemticl tools tht re useful in studying ECE 000: Liner Circuit Anlysis II. Trigonometry, Complex numbers, Liner Algebr nd some useful Rules/Formuls will be reviewed. 0.1 Trigonometry Rdin nd Degree Mesures The mesure of n ngle cn be described in rdin(rd) or degree( ). A complete revolution bout the vertex (center of the circle) is 60 or rd. The ngle is positive for counterclockwise revolution nd negtive for clockwise revolution. Use the following to convert from one mesure to nother. rd = 180 Reltionship between Rdin Note tht if no unit is specified, rdin mesure is implied. nd Degree Mesures 0.1. Trigonometric Functions nd Right Tringle Definitions Referring to the Fig. 0.1, the six trigonometric functions of cute ngle θ re defined s follows. sin θ = O H, cos θ = A H, tn θ = O A, cot θ = A O, sec θ = H A, csc θ = H O These definitions re restricted to n cute ngle θ, or n ngle in qudrnt I. For n ngle in qundrnt II, III or IV, with reference to positive x-xis, these definitions need to be extended for their signs. Figure 0. shows the trigonometric functions tht re positive in ech qudrnt. A, S, T nd C stnd for ll, sin, tn nd cos mening ll trigonometric functions re positive in qudrnt I, sin (nd its inverse csc) is positive in qudrnt II, tn (nd its inverse cot) is positive in qudrnt III nd cos (nd its inverse sec) is positive in qudrnt IV. Figure 0.1 Right Tringle to define Trigonometric Functions. 1 Figure 0. Trigonometric Functions in different qudrnts.

3 0.1. Trigonometric Vlues for specil ngles sin(0) = sin0 = 0/ = cos90 = cos ( ) sin ( ) 6 sin ( ) 4 sin ( ) sin ( ) = sin0 = 1/ = cos60 = cos ( ) = sin45 = / = cos45 = cos ( ) 4 = sin60 = / = cos0 = cos ( ) 6 = sin90 = 4/ = cos0 = cos(0) Notice the pttern n/ where n=0,1,,,4 to help you remember the vlues Some Useful Trigonometric Identities/Formuls tn θ = sin θ cos θ Quotient sin θ + cos θ = 1, tn θ + 1 = sec θ, cot θ + 1 = csc θ Pythgoren sin (u ± v) = sin u cos v ± cos u sin v, cos (u ± v) = cos u cos v sin u sin v Sum-Difference sin θ = sin θ cos θ, cos θ = cos θ sin θ = cos θ 1 = 1 sin θ Double Angle sin 1 cos θ θ =, cos 1 + cos θ θ = sin ( θ) = sin θ, cos ( θ) = cos θ Hlf Angle Even-Odd sin ( θ + ) ( = +cos θ, cos θ + ) = sin θ Shift by sin (θ + ) = sin θ, cos (θ + ) = cos θ Shift by sin (θ + ) = +sin θ, cos (θ + ) = +cos θ Shift by sin θdθ = cos θ +C, cos θd θ = +sin θ +C Integrtion d d (sin θ) = cos θ, dθ (cos θ) = sin θ dθ Differentition

4 0.1.5 Inverse Trigonometric Functions The three most common inverse trigonmetric functions with their corresponding domins nd rnges re defined s follows. Inverse Trigonometric Function Domin Rnge y = rcsin(x) = sin 1 (x) if nd only if x = sin(y) 1 x 1 y y = rccos(x) = cos 1 (x) if nd only if x = cos(y) 1 x 1 0 y y = rctn(x) = tn 1 (x) if nd only if x = tn(y) x y Inverse trigonometric functions re defined by plcing restrictions on domins of the corresponding trigonometric functions. rctn or tn 1 is the most often used inverse trigonometric function in circuit nlysis. 0. Complex Numbers 0..1 Complex Numbers Representtions A complex number z cn be described in rectngulr form z = + j b or polr form z = re j θ = r θ where z = complex number = rel prt of z nd usully denoted by Re{z} b = imginry prt of z nd usully denoted by I m{z} j = imginry unit nd j = 1, sometimes denoted by the symbol i s well r = mgnitude or bsolute vlue of z nd usully denoted by z θ = ngle or phse of z nd usully denoted by (z) Figure 0. Complex number in different representtions. Use the followings to chnge from one form to nother. = r cos θ, b = r sin θ Polr to Rectngulr Conversion ( ) r = + b, θ = tn 1 b Rectngulr to Polr Conversion

5 4 Notice tht the ngle of complex ( ) number is not unique but is found modulo nd the expression θ = tn 1 b does not give unique vlue of θ. To get unique vlue of θ, we typiclly restrict the ngle to be 0 θ < nd look for the qudrnt the complex number is in. ( ) First, find the vlue of θ using θ = tn 1 b (ignoring the signs of nd b), regrdless of the qudrnt the complex number is in nd then djust the nswer s follows. Qudrnt I : no djustment needed Qudrnt II: 180 θ or θ Qudrnt III: θ or + θ Qudrnt IV: 60 θ or θ Remember ll the djusted vlues re lso modulo. Exmple 0.1 Figure 0.4 Finding the ngle of complex number. Convert the following complex numbers into nother form. (1) z 1 = + j () z = 1 - j () z = 4 e j (4) z 4 = e j (1) r 1 = + =, θ 1 = tn 1 ( ) = 45 = 4 + j is in Qudrnt I nd thus no djustment needed. Therefore, z 1 = e j 4 = e j 45 () r = 1 + ( ) = 10, θ = tn 1 ( 1 ) = 60 = 1 j is in Qudrnt IV nd thus = 00 or = 5 Therefore, z = 10e j 5 = 10e j 00 = j 60 10e () = 4cos ( Therefore, z = + j ) = 4( 0.5) =,b = 4sin ( (4) 4 = 1 cos ( ) = (1)(0) = 0,b4 = (1)sin ( ) = 1 Therefore, z = 0 + j 1 = j Exmple 0. (1) Find the mgnitude nd ngle of z 5 =. () Find the rel nd imginry prts of z 6 = e j. ) ( ) = 4 = The ngle of positive rel number is 0. The ngle of negtive rel number is or -. The ngle of positive imginry number is. The ngle of negtive imginry number is or. Negtion of complex number dds ±180 phse. The mgnitude is lwys positive. In Qudrnt I, tn 1 (0) = 0 tn 1 ( very smll vlue ) 0 tn 1 ( 1/ ) = 0 = /6 tn 1 (1) = 45 = /4 tn 1 ( ) = 60 = / tn 1 ( very lrge vlue ) 90 tn 1 ( ) = 90 = / Tble 1 Some good to remember fcts. (1) r 5 =, θ 5 = or. () 6 = (1)cos() = 1, b 6 = (1)sin() = 0

6 5 0.. Opertions on Complex Numbers Addition/Subtrction Do it in rectngulr form. Add or subtrct rel prts nd imginry prts seprtely. Exmple 0. If z 7 = + j nd z 8 = 5 j 9, find z 7 + z 8 nd z 7 z 8. z 7 + z 8 = ( + 5) + j ( + ( 9)) = j 6 z 7 z 8 = ( 5) + j ( ( 9)) = 7 + j 1 Multipliction In rectngulr form: Tret ech complex number s binomil nd use j = 1. In polr form: Use lw of exponent. z 1 z = r 1 e j θ 1 r e j θ = r 1 r e j (θ 1+θ ). Note tht the mgnitudes multiplied nd the ngles dded in complex numbers multipliction. For complex multipliction, polr form is esier to del with. Exmple 0.4 (1) If z 1 = 1 + j b 1 nd z = + j b, then find z 1 z. () If z = e j nd z 4 = 5e j 4 then find z z 4. (1) z 1 z = ( 1 + j b 1 )( + j b ) = 1 + j 1 b + j b 1 + j b 1 b = ( 1 b 1 b ) + j ( 1 b + b 1 ). () z z 4 = ()(5)e j e j 4 = 10e j 4 Complex Conjugtion The complex conjugte of complex number z, usully denoted by either z or z is obtined by chnging the sign of imginry prt of the complex number. Exmple 0.5 (1) If z 1 = 5 + j nd z = 5 j, then find z 1 nd z. () If z = e j nd z 4 = 5e j 4 then find z nd z 4. (1) z 1 = 5 j nd z = 5 + j () z = e j nd z 4 = 5e j 4

7 6 Division In rectngulr form: Multiply the numertor nd denomintor by the complex conjugte of the denomintor. z 1 In polr form: Use lw of exponent. = r 1e j θ1 z r e j θ = r 1 r e j (θ 1 θ ). Note tht the mgnitudes divided nd the ngles subtrcted in complex numbers division. For complex division, polr form is esier to del with. Exmple 0.6 (1) If z 1 = + j nd z = 4 + j, then find z 1 z. () If z = e j nd z 4 = 5e j 4 then find z z 4. (1) z 1 ( + j ) ( + j ) (4 j ) = = z (4 + j ) (4 + j ) (4 j ) ( + j )(4 j ) = (4 + j )(4 j ) = 14 + j = j 5 () z = z 4 5 e j e j 4 = 0.4e j 4 0. Liner Algebr 0..1 Mtrix Formuls Mtrices [ b If M = [ b If M =, then det(m) = c, then M 1 = 1 d bc b d = d bc [ d b c [ b If M = [ e f nd N = g h [ e + bg f + bh, then MN = ce + dg c f + dh Mtrices If M = b c b c, then det(m) = = e h f i b d g f i + g e h = ei f h bdi+b f g +cdh ceg

8 7 0.. Systems of Simulteneous Linerly Independent Equtions nd Mtrix Representtion A system with two simulteneous linerly independent equtions with two independent vribles x nd y, two dependent vribles m nd n, nd,b,c,d coefficients of the independent vribles cn be described s below. x + by = m cx + d y = n In mtrix form, it cn be written s [ b [ x y [ m = n The independent vribles x nd y cn be solved using inverse mtrix. [ x y [ b = 1 [ m n Alterntively, method of elimintion cn lso be used to solve x nd y. A system with three simulteneous linerly independent equtions with three independent vribles x, y nd z, three dependent vribles r, s nd t, nd, b, c, d, e, f, g, h, i coefficients of the independent vribles cn be described s below. x + by + cz = r + e y + f z = s g x + hy + i z = t In mtrix form, it cn be written s b c x y z = r s t The independent vribles x, y nd z cn be solved using inverse mtrix but finding n inverse of mtrix is not simple. Alterntively, method of elimintion or Crmer s Rule cn be used to solve for x, y nd z.

9 8 Crmer s Rule Given then b c x y z = r b c r c b r s e f b s f b e s t h i c t i c h t x =, y =, z = b c b c b c r s t, 0.4 Some Useful Rules/Formuls Stright line eqution : line nd c is the y-intercept. y = m x + c, where m is the slope ( = rise ) of the stright run Qudrtic formul : For x + bx + c = 0, the roots re x = b ± b 4c Chin Rule : If z is function of y which is function of x, then z is lso function of x nd the chin rule my be written s dz = dz d y.d y Product Rule: If u nd v re functions of x nd differentible, then d dv (u.v) = u + v du Quotient Rule: If u nd v re functions of x, differentible nd v 0, then Euler s Formul : d ( u v ) v du = u dv v e j θ = cosθ + j sinθ