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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?

Mamal
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Bob Shannon
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1 Answers1

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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 $$X_1(\omega) = e^{-2i\omega} \frac12 X(\frac{-\omega}{2}) $$.

Mamal
  • 386