2
$\begingroup$

How would one go about obtaining the Fourier Transform of a general signal $x(-2t+4)$? I know one can use a table of Fourier Transform properties to easily evaluate this, but I want to use the definition of the Fourier Transform:

$$ X(w) = \int_{-\infty}^\infty x(t) e^{-jwt} dt $$

The first thing I do is plug in $-2t+4$ for $t$, let's call the transform of this new function $Y(w)$: $$ Y(w) = \int_{-\infty}^\infty x(-2t+4) e^{-jw(-2t+4)} dt $$

I notice that an exponential can be factored out:

$$ Y(w) = \int_\infty^\infty x(-2t+4) e^{j2wt-j4w)} dt $$ $$ Y(w) =e^{-j4w} \int_\infty^\infty x(-2t+4) e^{j2wt} dt $$

I'm not sure how to proceed from here?

$\endgroup$
1
$\begingroup$

We have $$X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-i\omega t}dt $$ We are going to calculate the following integral $$X_1(\omega) = \int_{-\infty}^{\infty} x(-2t+4)e^{-i\omega t}dt $$ changing the variable $-2t+4 = u$ we have $$X_1(\omega) = \int_{-\infty}^{\infty} x(u)e^{-i\omega \frac{4-u}{2}}du (-\frac12) $$ consequently $$X_1(\omega) = e^{-2i\omega}\int_{-\infty}^{\infty} x(u)e^{-i\frac{-\omega}{2} {u}}du (-\frac12)$$ and therefore using the definition of $X(\omega)$ we have $$X_1(\omega) = e^{-2i\omega} \frac12 X(\frac{-\omega}{2}) $$.

$\endgroup$

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.