Signal Processing and Linear Systems-B.P.Lathi copy

T herefore e at d etermines t he b ehavior of e ajwt

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Unformatted text preview: t - + DO a nd 0 <0 0 >0 f ourth quadrants. I t will read tan-1(=~) as tan-1(~), which is clearly wrong. In computing inverse trigonometric functions, if t he angle appears in t he second or t hird q uadrant, t he answer of t he c alculator is off by 1800 • T he correct answer is o btained by adding or subtracting 180 0 t o t he value found with t he c alculator (either adding or subtracting yields t he correct answer). For this reason it is advisable to draw t he p oint in t he complex plane a nd d etermine t he q uadrant in which t he p oint lies. This issue will be clarified by t he following examples. • (B.14) r e j8 as a number a t a In future discussions you will find it very useful t o remember d istance r from t he origin a nd a t a n angle e w ith t he horizontal axis of t he complex plane. A Warning A bout Using Electronic Calculators in Computing Angles From t he C artesian form a + j b we c an readily compute t he p olar form r e j8 [see Eq. (B.6)]. Electronic calculators provide ready conversion of rectangular into polar a nd vice versa. However, if a calculator computes a n angle of a complex number using a n inverse trigonometric function e = t an - 1 (b / a ), proper attention must be paid to t he q uadrant in which t he n umber is located. For instance, e corresponding to the number - 2 - j3 is tan-1(=~). T his result is not the same as tan-1(~). T he former is - 123.7°, whereas t he l atter is 56.3 0 • An electronic calculator cannot make this distinction a nd can give a correct answer only for angles in t he first a nd E xample B .l Express the following numbers in polar form: ( a) 2 +j3 ( b) - 2+jl ( c) - 2-j3 ( d) I -j3 ( a) Izl = ) 22 + 32 = v'l3 Lz = t an- I (~) = 56.3° In this case the number is in the first quadrant, and a calculator will give the correct value of 56.3°. Therefore, (see Fig. B.4a) 2 + j 3 = v'l3 e j56.3° ( b) L z = tan -1 (!2) = 153.4° In this case the angle is in the second quadrant (see Fig. B.4b), and therefore the answer given by the calculator ( tan- I(!2) = - 26.6°) is off by 180°. The correct answer is ( -26.6 ± 1 80t = 153.4° or - 206.6°. Both values are correct because they represent the same angle. As a matter of convenience, we choose an angle whose numerical value is less than 180°, which in this case is 153.4°. Therefore, -2 + jl = V 5ejI53.4° Background 10 ( e) Izl == y'( _ 2)2 + ( -3)2 == B .l 11 Complex Numbers i m R e ..... _1l!. I n t his case t he a ngle a ppears in t he t hird q uadrant (see Fig. B.4c) , a nd t herefore t he a nswer obtained b y t he c alculator ( tan- l ( :::~) == 56.3°) is off by 180°. T he c orrect answer is (56.3 ± 180)° == 236.3° or - 123.7°. As a m atter o f convenience, we choose t he l atter a nd (see Fig. B.4c) - 2 - j3 == i 1m - 2{2 1m me 4 ....................... - 2{2 - j123.7° 4 e - i"3 R e ..... ( d) (b) (a) I n t his case t he a ngle a ppears in t he f ourth q uadrant (see Fig. B.4d) , a nd t herefore t he a nswer given b...
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