0
$\begingroup$

On my book there is the following proof relating the continuous Fourier Transform X(f) of a real and odd signal x(t)=x(-t) (goal X(f)=X(-f)):

$$X(f)=\int_{-\infty}^{+\infty}{x(t) \exp(-j2\pi ft)dt}$$

$$X(-f)=\int_{-\infty}^{+\infty}{x(t) \exp(j2\pi ft)dt}$$

Now $\alpha=-t$, thus:

$$X(-f)=-\int_{+\infty}^{-\infty}{x(-\alpha) \exp(-j2\pi f\alpha)d\alpha}=\int_{-\infty}^{+\infty}{x(-\alpha) \exp(-j2\pi f\alpha)d\alpha}=\int_{-\infty}^{+\infty}{x(\alpha) \exp(-j2\pi f\alpha)d\alpha}=X(f)$$

In the last operation I use $x(\alpha)=x(-\alpha)$ because $x(\alpha)$ is an odd signal.

Why $X(f)=\int_{-\infty}^{+\infty}{x(t) \exp(-j2\pi ft)dt}=\int_{-\infty}^{+\infty}{x(\alpha) \exp(-j2\pi f\alpha)d\alpha}$? $\alpha\neq t$

Thank you very much.

$\endgroup$
1
$\begingroup$

Firs of all, $x$ is an even signal, not an odd signal.

As for your question, $t$ and $\alpha$ are mute variables. You could use any symbol and the integral would be the same, like in the following example: $$ \int_0^1x\,dx=\int_0^1y\,dy=\int_0^1t\,dt=\int_0^1\square\,d\square=\frac12. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.