2.2. Complex numbers 11

Transcription

1 2.2. Complex numbers Generlized Integrls We will need to integrte functions over infinite intervls. Therefore, we need to generlize the concept integrls further. So, we define nd b f(x)dx f(x)dx, provided tht the limits do exist. The definite integrl f(x)dx demnds tht f(x) is continuous for x b. Wht if f(x) is continuous for < x b or x < b insted? Then we define f(x)dx. ǫ +ǫ with ǫ > in the first cse, nd ǫ ǫ f(x)dx. with ǫ > in the second cse, provided tht the limits do exist. 2.2 Complex numbers Complex numbers re widely used in electricl engineering. Especilly, complex numbers re often used torepresent sinusoidl signls. We refer to Adms [, AppendixI] for more detiled tretments of complex numbers. If the only numbers tht we re wre of re the nturl numbers, N = {,, 2,...}, we cn solve equtions like 2x = 4, but not 2x + 4 =. If we extend the set of numbers by introducing negtive numbers, thus being ble to use ll integers, Z = {..., 2,,,, 2,...}, we cn give the solution x = 2 of the second eqution. By modifying the first eqution slightly, e.g. 2x = 5, we get into trouble gin. There is no integer stisfying tht eqution. However, by introducing the set of rtionl numbers, Q = {/b :,b Z}, set tht contins Z, we cn give the solution x = 5/2. Now, we hve set tht cn be used to give solutions to ll polynomil equtions of degree one, with coefficients in N. Actully, tht lso holds if the coefficients re in Q. Polynomil equtions of degree two will pose problems even for rtionl numbers. We cn solve x 2 = 4, but not x 2 = 3. There is no rtionl number whose squre is three. For tht reson, the set of rels, R, hve been defined, which contins Q nd mong other elements contins n element denoted 3, whose squre is three.

2 2 Chpter 2. Mthemticl Bsics Irreducible Polynomils nd the Complex Field We hve seen tht by extending the set of numbers step by step, we cn solve more nd more polynomil equtions. Now, we still hve polynomil equtions over Q of degree two tht cnnot be solved using the rels. For instnce, we cnnot solve the eqution x 2 + = using the rels. We sy tht the polynomil x 2 + is irreducible over R. Insted, we define the set of complex numbers, C, by first ssuming tht there is solution to the eqution, nd we let j denote tht solution, s the trdition is in electricl engineering. So, j 2 = holds. Mthemticins usully use i insted, but electricl engineers trditionlly let i nd I denote currents. The set of complex numbers cn then be described s C = { + jb :,b R}, nd tht set contins R. Let A, B nd C be complex numbers. Then the following properties re esily shown. + A = A nd A = A (unit elements). A + B C nd A B C (closure). A + B = B + A nd A B = B A (commuttivity). (A + B) + C = A + (B + C) nd (A B) C = A (B C) (ssocitivity). A (B + C) = A B + A C (distributive lw). A B = implies A = or B = or both. In other words, ll lgebric lws tht hold for R lso hold for C. A set tht obeys those lws is usully clled field, nd C is so clled extension of R. It cn be shown tht ll solutions to polynomil equtions over C re in C, which is intimitely relted to the fct tht there re no irreducible polynomils over R of degree greter thn two. Rectngulr nd Polr Forms A complex number, A = + jb, with,b R, is normlly seen s point in two dimensions s in Figure 2.4. Here is clled the rel prt of A, denoted Re {A}, nd b is clled the imginry prt of A, denoted Im {A}. Note tht j identifies Im {A}, but it is not included in Im {A}. The Eucliden length of the stright line from the origin to A is referred to s the bsolute vlue of A nd is denoted A. There is right tringle in Figure 2.4, nd therefore A is given by the Pythgoren reltion A 2 = 2 + b 2. The ngle from the positive rel xis

3 2.2. Complex numbers 3 Imginry xis b A A rg{a} Rel xis Figure 2.4: A complex number A = + jb interpreted s point in two-dimensionl spce, with its bsolute vlue nd rgument displyed s well. to the mentioned stright line - considered positive in tht direction - is referred to s the rgument of A nd is denoted rg{a}. Bsed on simple geometry, we hve rctn(b/), >, π/2, =,b >, rg{a} = π/2, =,b <, rctn(b/) ± π, <, with unit rdins. The expression bove does not give vlue of rg{}. Tht is intentionl, since the ngle does not hve mening in tht cse. Thus, rg{} is bsiclly undefined, nd could be left tht wy. However, it is customry to define rg{} =, even though ny other definition of rg{} would mke just s much sense. The representtion A = + jb is clled rectngulr form. This representtion is especilly well suited for ddition nd subtrction. Adding two complex numbers is simply dding the rel prts nd the imginry prts seprtely, s in nd subtrction is simply ( + jb) + (c + jd) = ( + c) + j(b + d), ( + jb) (c + jd) = ( c) + j(b d).

4 4 Chpter 2. Mthemticl Bsics Imginry xis sin(x) j e jx x cos(x) Rel xis j Figure 2.5: The complex exponentil e jx = cos x+j sinx is point on the unit circle with rgument x. Multipliction is bit more complicted, nd we need to use the observtion j 2 =, s in ( + jb) (c + jd) = c + jd + jbc + j 2 bd = (c bd) + j(d + bc). There is lso representtion bsed on A nd rg{a}. Tht representtion uses the complex exponentil function, defined s with x R, see Figure 2.5. It cn be shown tht e jx cos(x) + j sin(x), d dx ejx = je jx holds. It cn lso be shown tht A = A e j rg{a} holds, which is the so clled polr form. This representtionis especilly well suitedfor multipliction nd division. Forthis representtion the equlities A B = A B e j(rg{a}+rg{b}) nd A B = A B ej(rg{a} rg{b})

5 2.2. Complex numbers 5 hold. Expressed differently: AB = A B rg{ab} = rg{a} + rg{b} A/B = A / B rg{a/b} = rg{a} rg{b} e jα = rg { e jα} = α Let A be complex number given by A = + jb. The relted complex number A = jb is clled the (complex) conjucte of A, nd in polr form we hve A = A e j rg{a} nd A 2 = A A. If A is root of polynomil p(x) over R, then A is lso root of p(x). The Euler Formuls Bsed on the definition of the complex exponentil function, it is esy to show the equlities cos(x) = ejx + e jx 2 nd sin(x) = ejx e jx, tht re known s the Euler formuls. Those formuls cn be useful for instnce for rewriting powers of trigonometric expressions. As n exmple, consider sin 3 (x) = = 4 ( ) e jx e jx 3 = (ejx e jx ) 3 (3 ejx e jx () 3 ej3x e j3x ) = ej3x 3e jx + 3e jx e j3x j8 = (3 sin(x) sin(3x)). 4 Furthermore, the Euler formuls cn be useful when clculting some integrls. As n exmple, consider the integrl e x cos(x)dx = e xejx + e jx dx = ( e (+j)x + e ( j)x) dx 2 2 = [ ] e (+j)x 2 + j + e( j)x = ( )] e [e x jx j 2 + j + e jx j = [e x( ] j)ejx + ( + j)e jx = ( )] e [e x jx + e jx + ejx e jx 2 ( + j)( j) 2 2 = 2 [ex (cos(x) + sin(x))] e (cos() + sin()) =.378 2