I know that the property of duality says:

$$x(t) \iff X(f)$$ $$X(t) \iff x(-f) "="x(t=-f)$$

and I know that:

$$\delta(t-t_0) \iff exp(-j2\pi ft_0)$$

If I apply the duality property, I get:

$$exp(-j2\pi f_0t) \iff \delta(-f-f_0)$$ $$exp(j2\pi f_0t) \iff \delta(f-f_0)$$

instead the correct result is:

$$exp(j2\pi f_0t) \iff \delta(f+f_0)$$


Thank you for your help.

  • 2
    $\begingroup$ I guess $exp(j2\pi f_0t) \iff \delta(f-f_0)$ is correct, while $exp(j2\pi f_0t) \iff \delta(f+f_0)$ is false. $\endgroup$ – the_candyman Jul 1 '17 at 10:10

I think you're getting confused between $t$ and $f$ , Duality says

$$x(t) \Leftrightarrow X(\omega) \implies X(t)\Leftrightarrow2\pi x(-\omega) $$ Or in terms of $f$ $$x(t) \Leftrightarrow X(f) \implies X(t)\Leftrightarrow x(-f) $$

So for $\delta(t-t_0)$ $$\delta(t-t_0) \Leftrightarrow \mathrm{exp}(-jt_0 2\pi f) \implies \mathrm{exp}(-jt_02\pi t) \Leftrightarrow \delta(-f-t_{0})=\delta(f+t_0)$$ And for $\delta(t+t_0)$ $$\delta(t+t_0) \Leftrightarrow \mathrm{exp}(jt_0 2\pi f) \implies \mathrm{exp}(jt_02\pi t) \Leftrightarrow \delta(-f+t_{0})=\delta(f-t_0)$$ Note that $\delta(t)$ is even function

Please let me know if i can improve my answer somehow , thanks !


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