Question

Z(2) sin(22/2) 12/2 Given that Z(92) is real, its phase is either zero when Z(92) > 0 and when Z(92) < 0 (using these values so that the phase is an odd function of S2). Since z(t) = xy(t +0.5), then Z(2) Xi (S210.50 so that X1(92) = -0.52Z(2)

i can't understand the reference about phase. Why the phase is zero when z(w)>=0 and why it is +-phi when z(w)<0

Answer 1

When Z(w) is zero or positive then the corresposnding phase is zero.

If Z(w)=-x (Z(w) is negative i.e. Z(w)<0), where x is real and positive. The magnitude of Z(w) is x and the phase of Z(w) is Z(w)=arctan(0/-x)= 180 or -180=+pi or -pi.

sin (-2) Z (2) [zca) in yeal] given it ス(n) シ =) Let ニ0 Z (e) =0° 12(_e)| Z (-2) (ス(レ) 0) = X+jo Cx is real =) +ve) |Z(R_) = 0°[tan'() :o°[tan"() real & + ve ) z(2) = - x+jo( x (z(L) <0) |Z(4)|ニエ [z4) t180° oL -1820 phase of zC2)=[ZL2) = + TT So 50 Z(-2) Z(re) くo -> Iza) = ±T When z (x) = -X+jo, zn) = tan' () = ± 180° ズ ecl k tve