X(w) = H(@)Z@). 
Taking the Fourier transform gives 
5a) = | h(z)6(t —t)dt. 
—-o 
Here 
l4@) = | H(w)e!*do, 
is the impulse response function. 
When the waves are expressed by a Fourier integral, 
E(t) = | Z(w)e!”'dw, 
-o 
1 
27 
the linear response x(r) will be 
x(t) = 
-o 
| X(w)e'dw. 
(Ost) 
(9.16) 
(9.17) 
(9.18) 
(9.19) 
Strictly speaking, these integrals (Eqs. 9.18 and 9.19) should be Fourier—Stieltjes 
integrals in the form of 
1 
2m 
where 
275 
1 
| azo», ae | e!?'dX@), 
(9.20) 
