# How can I get rid of this unexpected minus sign on my inverse Fourier transform of two impulse functions?

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!

• 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. – mbaitoff Nov 15 '20 at 6:32

$$\int_{-\infty}^{\infty}f(t)\delta(t-t_0)dt=f(t_0)\tag{1}$$
if $$f(t)$$ is continuous at $$t=t_0$$.