c. if a) >0, a2 >0 (region (2) in Fig. 5.16), then <0, p< 0 and in 
this case R(r) alternate signs by r as in Fig. 5.17(B), and 
d. if aj >0, a) <0 (region @) in Fig. 5.16), then 44),42 have opposite 
signs. 
Therefore, for a and c, the autocovariance function shows the patterns (A) and (B) 
in Fig. 5.17, respectively. For b and d, the shape will be (A) or (B), depending on the 
size Of 1,2. 
Ry) of AR(2);a,>% 
(A): a, (a1 < 0,a2 > 0) (B) : ¢, (a, >0,a2 >0) 
Fig. 5.17. A(n) of AR(2); a2 <a?2/4. 
2. When a> > a7/4, (subzone [II] in Fig. 5.16), ;,2 are complex and conjugate to 
each other, and from Eqs. 5.67 and 5.75 
My M2 =a, fy =Vane® = py = Vane"? (5.97) 
and—a@, =; +42, which satisfies 
-a,= 2 Vax coso. (5.98) 
Thus coso = —a,/2 Jaz (5.98’) 
M1 —-b2=2 Jazisinoa. (5.99) 
and 
Using the relations of Eqs. 5.97 to 5.99 in Eq. 5.94 gives 
sin(r + 1) d—azsin(r—1)o 
an ea Dre eee 
(1 + a2) sino 
Here again setting 
l+a 
tang = | ——~ ] tana, (5.100) 
gives 
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