Ambiguity function

Remarks

The substitution $t = s - x/2$ gives $Af(x, \omega) = e^{\pi i x \omega} V_f f(x, \omega)$, where $V_f f$ is the short-time Fourier transform of $f$ with window $f$. Hence $\abs{Af(x, \omega)} \le Af(0, 0) = \norm{f}_2^2$, by the Cauchy–Schwarz inequality. In radar texts the ambiguity function is often written as $\chi(\tau, \nu) = \int s(t) \overline{s(t - \tau)} e^{2\pi i \nu t} \, dt$. It differs from $Af$ by a sign convention and a phase factor, and $\abs{\chi(\tau, \nu)} = \abs{Af(\tau, -\nu)}$ for $s = f$.

Example: let $f(t) = 2^{1/4} e^{-\pi t^2}$. Then $f(t + x/2) \overline{f(t - x/2)} = \sqrt2 \, e^{-\pi x^2/2} e^{-2\pi t^2}$, and the Fourier transform of $e^{-2\pi t^2}$ is $2^{-1/2} e^{-\pi \omega^2/2}$. So $Af(x, \omega) = e^{-\pi (x^2 + \omega^2)/2}$, a single peak of height $1 = \norm{f}_2^2$ at the origin.

Non-example: the autocorrelation $R(x) = \int_\R f(t) \overline{f(t - x)} \, dt$ is not the ambiguity function. The substitution $t = s + x/2$ shows $R(x) = Af(x, 0)$, so $R$ is only the slice $\omega = 0$ of $Af$ and does not see frequency shifts.