Fourier transform of impulse response $f(t)$ produces conjugate?

Question

Let $$F(f) = \int\limits_{-\infty}^{\infty} f(\tau) e^{-j2\pi f\tau} d\tau$$

where $(f, \tau)$ are Fourier pairs representing frequency, time..the usual stuff. One may take $f(\tau)$ as the impulse response of some filter in signal processing.

In my textbook, there is a step which contains the expression

$$ \int\limits_{-\infty}^{\infty} f(\tau) e^{j2\pi f\tau} d\tau$$

But instead of equating it with $F(-f)$, which is what I expected, it was written as

$$ F^*(f) = \int\limits_{-\infty}^{\infty} f(\tau) e^{j2\pi f\tau} d\tau$$ where $*$ denotes the complex conjugate of $F$.

What is the rationale here? Why is it true that $F(-f) = F^*(f)$?

Answer 1

Your textbook is assuming that $f(\tau)$ is real.

First,

$F(-f)=\int_{-\infty}^{\infty} f(\tau) e^{-j2\pi(-f)\tau} d\tau$.

Canceling the two minus signs in the exponent gives

$F(-f)=\int_{-\infty}^{\infty} f(\tau) e^{j2\pi f\tau} d\tau$.

Also,

$F^{*}(f)=\left( \int_{-\infty}^{\infty} f(\tau) e^{-j2\pi f\tau} d\tau \right)^{*}$

We can take the complex conjugate inside the integral and use the fact that the conjugate of a product is the product of the conjugates to get

$F^{*}(f)= \int_{-\infty}^{\infty} f(\tau)^{*} \left( e^{-j2\pi f\tau} \right)^{*} d\tau . $

Since $f(\tau)$ is real,

$f(\tau)^{*}=f(\tau)$.

The conjugate of $e^{jx}$ is $e^{-jx}$. Thus

$\left( e^{-j2\pi f\tau} \right)^{*}=e^{j2\pi f\tau}$.

Finally,

$F^{*}(f)=\int_{-\infty}^{\infty} f(\tau) e^{j2\pi f \tau} d\tau$.