Hanning Function

An Apodization Function, also called the Hann Function, frequently used to reduce Aliasing in Fourier Transforms. The illustrations above show the Hanning function, its Instrument Function, and a blowup of the Instrument Function sidelobes. The Hanning function is given by

$$f(x)=\cos^2\left({\pi x\over 2a}\right)= {\textstyle{1\over 2}}-{\textstyle{1\over 2}}\cos\left({\pi x\over a}\right).$$ (1)

The Instrument Function for Hanning apodization can also be written

$$a[\mathop{\rm sinc}\nolimits (2\pi ka)+{\textstyle{1\over 2}... ...textstyle{1\over 2}}\mathop{\rm sinc}\nolimits (2\pi ka+\pi)].$$ (2)

Its Full Width at Half Maximum is $a$. It has Apparatus Function

$$A(x) = \int_{-a}^a \left[{{\textstyle{1\over 2}}-{\textstyle{1\over 2}}\cos\left({\pi x\over a}\right)}\right]e^{-2\pi ikx}\,dx$$

$$= {\textstyle{1\over 2}}\int_{-a}^a e^{-2\pi ikx}\,dx-{\textstyle{1\over 2}}\int_{-a}^a e^{-2\pi ikx}\,dx$$

$$\equiv {\textstyle{1\over 2}}(A_1+A_2).$$ (3)

The first integral is

$$I_1=\int_{-a}^a e^{-2\pi ikx}\,dx ={\sin(2\pi ka)\over \pi k}=2a\mathop{\rm sinc}\nolimits (2\pi ka).$$ (4)

The second integral can be rewritten

$$I_2 = \int_{-a}^0 \cos\left({\pi x\over a}\right)e^{-2\pi ikx}\,dx+ \int_0^a \cos\left({\pi x\over a}\right)e^{-2\pi ikx}\,dx$$

$$= \int_0^a \cos\left({\pi x\over a}\right)(e^{2\pi ikx}+e^{-2\pi ikx})\,dx$$

$$= 2\int_0^a \cos\left({\pi x\over a}\right)\cos(2\pi kx)\,dx$$

$$= 2\left\{{{\sin\left({{\pi\over a}-2\pi k}\right)x\over 2\left({{\... ...\over a}+2\pi k}\right)x\over 2\left({{\pi\over a}+2\pi k}\right)}}\right\}_0^a$$

$$= a\left[{{\sin(\pi-2\pi ka)\over \pi-2\pi ka}+{\sin(\pi+2\pi ka)\over \pi+2\pi ka}}\right]$$

$$= {a\over\pi}\left[{{\sin(2\pi ka)\over 1-2ka}-{\sin(2\pi ka)\over 1+2ka}}\right]$$

$$= a[\mathop{\rm sinc}\nolimits (\pi-2\pi ka)+\mathop{\rm sinc}\nolimits (\pi+2\pi ka)].$$ (5)

Combining (4) and (5) gives

$$A(x)=a[\mathop{\rm sinc}\nolimits (2\pi ka)+{\textstyle{1\ov... ...textstyle{1\over 2}}\mathop{\rm sinc}\nolimits (\pi+2\pi ka)].$$ (6)

To find the extrema, define $x\equiv 2\pi ka$ and rewrite (6) as

$$A(x)=a[\sin x+{\textstyle{1\over 2}}\mathop{\rm sinc}\nolimi... ...pi)+{\textstyle{1\over 2}}\mathop{\rm sinc}\nolimits (x+\pi)].$$ (7)

Then solve

$${dA\over dx}={\pi^2(-x^3\cos x+3x^2\sin x+\pi^2 x\cos x-\pi^2\sin x)\over x^2(\pi^2-x^2)^2}=0$$ (8)

to find the extrema. The roots are $x=7.42023$ and 10.7061, giving a peak Negative sidelobe of $-0.026708$ and a peak Positive sidelobe (in units of $a$) of 0.00843441. The peak in units of $a$ is 1, and the full-width at half maximum is given by setting (7) equal to 1/2 and solving for $x$, yielding

$$x_{1/2}=2\pi k_{1/2} a=\pi.$$ (9)

Therefore, with $L\equiv 2a$, the Full Width at Half Maximum is

$${\rm FWHM}=2k_{1/2} = {1\over a} ={2\over L}.$$ (10)

See also Apodization Function, Hamming Function