Unformatted text preview: t  + DO a nd 0 <0
0 >0 f ourth quadrants. I t will read tan1(=~) as tan1(~), 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 tan1(=~). T his result is not the same as tan1(~).
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)  2j3 ( 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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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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