Thread: Fourier Transform help

If S(t) is any non-periodic continuous function of t, show that:
(i) S(t) is real iff F(-omega) = F*(omega)
(ii) S(t) is pure imaginary iff F(-omega) = -F*(omega)
(iii) S(t) is even iff F(-omega) = F(omega)
(iv) S(t) is odd iff F(-omega) = -F(omega)

Originally Posted by superdael

If S(t) is any non-periodic continuous function of t, show that:
(i) S(t) is real iff F(-omega) = F*(omega)
(ii) S(t) is pure imaginary iff F(-omega) = -F*(omega)

I am note sure what these stars means.

(iii) S(t) is even iff F(-omega) = F(omega)
(iv) S(t) is odd iff F(-omega) = -F(omega)

(iii) and (iv) are similar. I will do (iii) and leave (iv) for you to do. For some reason I got directly the opposite of what you said. Are you sure you are right?

(Note the converse is basically the same idea).

EDIT I just realized why I made a mistake. When I made a variable substitution s=-t I forgot to attach the minus sign in front of the integral. I am just lazy to retype the entire thing. So you are right.

Originally Posted by superdael

If S(t) is any non-periodic continuous function of t, show that:
(i) S(t) is real iff F(-omega) = F*(omega)
(ii) S(t) is pure imaginary iff F(-omega) = -F*(omega)
(iii) S(t) is even iff F(-omega) = F(omega)
(iv) S(t) is odd iff F(-omega) = -F(omega)

The basic method of doing all of these is the same so I will do only one.

F(omega) = integral_{x=-infty to +infty} f(x)exp(-i omega x) dx

Now suppose f(x) is real, then:

F*(omega) = conj[integral_{x=-infty to +infty} f(x)exp(-i omega x) dx]

............... = integral_{x=-infty to +infty} conj[f(x)exp(-i omega x)] dx

............... = integral_{x=-infty to +infty} f(x)exp(i omega x) dx

............... = F(-omega)

So f real implies F*(omega) = F(-omega).

Now assume F*(omega) = F(-omega), then:

integral_{x=-infty to +infty} conj[f(x)exp(-i omega x)] dx

............... = integral_{x=-infty to +infty} f(x)exp(i omega x) dx

or:

integral_{x=-infty to +infty} f*(x)exp(i omega x)] dx

............... = integral_{x=-infty to +infty} f(x)exp(i omega x) dx

Expand these into a+i b form and equate real and imaginary parts of these
integrals gives im(f(x) = 0.

So:

So F*(omega) = F(-omega) implies f real, which together with the earlier
paer proves the result.

RonL