Syyy(@ 1,@2) exp[i(@1T, + W2T2)] dw dw (11.16) 
Ryyy(1, 72) = SXXY(W 1,2) exp[i(W 1T1 + W2T2)] dw\dw2. (11.17) 
When the input process X(r) is a Gaussian process, then 
Ryyx(T1,T2) = E[X(t+7,) X(t+T2) X(t] = 0. (11.18) 
Therefore 
Sxxx(@1,@2) = 0, (11.19) 
The bispectrum does not exist for a linear Gaussian process. 
For a quadratic (second order nonlinear) process Y(t), Ryyy(t1,T2) = 0, 
Rxxy(t1,T2) # O and the bispectrum exists as shown by Eq. 11.14 and Eq. 11.15. 
Now, assuming Y(f) is a stationary process and its Fourier—Stieltjes integral is 
foe) 
{Y(t) — my} = | dZ(w) exp(iw ft), (11.20) 
then {Y(t +1) — my} = | dZ(w 1) exp(iw jt + iw 4,7), 
{Y(t +72) — my} = | daZ(w2) exp(iw rt + iw2T2). 
Therefore 
Ryyy(t1, 72) = E[{(Y(t + 71) — my]{¥(¢ + t2) — my}{¥@) — my}] 
fo) 
= | | | expt, +W2+3)t] - exp[i(@1T1 + @2T2) 
E[dZ(w1)dZ(w2)dz(w3)] (11.21) 
[w is expressed as w3]. 
Because of its stationarity, Ryyy(T),T2) is a function only of T;,72 ; it is not a func- 
tion of ¢ but is independent of r. Accordingly, the integral has a value only when 
@,+@2+@3 =0 or 03 =-@)-@2 (11.22) 
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