I'm trying to find the inverse Fourier transform of two impulse functions, which correspond to the Fourier transform of the function $h(t)=A\sin(2πf_0t)$.

The Fourier transform of the above sine function is as follows:

$$H(f)=j \frac{A}{2} \delta\left(f+f_0\right)-j \frac{A}{2} \delta\left(f-f_0\right)$$

When I try to find the inverse Fourier transform of $H(f)$, I keep getting a minus sign which is not in the original time-domain function. Here is my procedure:

\begin{align} h(t)&=j \frac{A}{2} \int_{-\infty}^\infty\delta(f+f_0 ) e^{j2πft} df-j \frac{A}{2}\int_{-\infty}^\infty\delta(f-f_0 ) e^{j2πft}df\\ &=j\frac{A}{2}e^{j2πf_0t}-j\frac{A}{2}e^{-j2πf_0t}\\ &=jA\left(\frac{e^{j2πf_0t}-e^{-j2πf_0t}}{2}\right)\\ &=-A\left(\frac{e^{j2πf_0t}-e^{-j2πf_0t}}{2j}\right)\\ &=-A\sin(2\pi f_0t) \end{align}

THAT is the minus sign that, so far, I've been unable to get rid of!

What is it that I am missing here? I hope you can help me!

Thank you very much!

  • 1
    $\begingroup$ I think that inverse transform of the product of the delta function at (t-t_0) and the exponent yields the value of the exponent at t_0, not at -t_0, so you lost the minus signs computing the inverse Fourier integrals. $\endgroup$
    – mbaitoff
    Nov 15, 2020 at 6:32

1 Answer 1


HINT: you should review the sifting property of the Dirac delta impulse:


if $f(t)$ is continuous at $t=t_0$.


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