Let $$g(x) = \sqrt{1+x^2}$$ and $$h(x) = g^{-3/2}(x) \exp(-i2\pi g(x)).$$

I can observe that the Fourier transform $|H(f)|$ is almost flat if $|f|<1$ and $H(f)\approx 0, \; |f|>1.$ This observation is important for my work, but I cannot understand why it happens. I tried for quite some time to figure out why, but failed. It basically means that $h(x)$ is a $sin(x)/x$ function.

Let $$g(x) := \sqrt{1+x^2},$$ and $$h(x) := g^{-3/2}(x) \exp(-i2\pi g(x)).$$

I can observe that the Fourier transform $|H(f)|$ is almost flat if $|f|<1$, and $H(f)\approx 0, \; |f|>1$. This observation is important for my work, but I cannot understand why it happens. I tried for quite some time to figure out why, but failed. It basically means that $h(x)$ is a $\sin(x)/x$ function.