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

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

Ask a homework question - tutors are online