3
$\begingroup$

I have read many excellent answers to similar questions, but never one this specific. Here is another way to ask it.

  • Why is the modulation transfer function (MTF) of $\textrm{rect}(x/5) = \textrm{sinc}(5x)$ not equal to the DFT of a vector of zeros with 5 ones in the center?

Relative MATLAB picture and code: picture

x = -512:511; x = x / 1024;
sq = zeros(1,1024); sq(511:515) = 1;
sqMTF = abs(sinc(x*5));               %'Modeled'
sqDFT = 1/5* abs(fftshift(fft(sq)));  %'Measured'

I can make the two curves overlay each other by multiplying the DFT result by $\textrm{sinc}(x)$, which I intuit should be related to windowing of the original signal and related spectral leakage.

  • But shouldn't this be a convolution in the frequency domain, as opposed to a multiplication?

Jack.

$\endgroup$
  • $\begingroup$ MTF, what is that ? $\endgroup$ – Gilles May 22 '16 at 9:00
  • $\begingroup$ Modulation Transfer Function = the modulus of the Fourier Tansform of the signal normalized to 1 at the origin. $\endgroup$ – Jack Hogan May 22 '16 at 11:28
1
$\begingroup$

Computing a DFT requires an input consisting of a finite length of samples instead of a infinite continuous function.

Because the full spectrum (FT) of a rect function is not bandlimited to below half the sample rate, aliased images will appear in the DFT of any finite window of samples of this non-bandlimited rect. The sum of all these aliased images folded (or wrapped around) into a finite DFT result window is a Dirichlet function, or periodic Sinc, which differs from a single Sinc by a denominator of sin(W) instead of W.

The aliasing will be due to the sampling of a non-bandlimited signal, no matter what the DFT length. However a longer DFT window will spread out the train of aliased images in terms of DFT result bins, thus changing the denominator to where sin(W) is closer to W.

A bandlimited version of a rect (allowing sampling without aliases) will not be a sequence of ones surrounded by zeros.

$\endgroup$
  • $\begingroup$ Thanks hotpaw2. From what I understand the asinc is the ratio of my sinc(5x) to sinc(x). Could the sinc(x) term be related to the finite window over a non-periodic signal in the spatial domain fed to a DFT? If so how? $\endgroup$ – Jack Hogan May 22 '16 at 13:35
2
$\begingroup$

The difference is between $\mbox{sinc}$ and the periodic version obtained using the DFT.

See this answer for a comparison.


It strikes me that asinc = ratio of two sincs.

The $\mbox{asinc}$ function is a ratio of two $\sin$ functions: $$ \frac{\sin(N\omega/2)}{\sin(\omega/2)} = \frac{\sin(N\omega/2)}{\omega/2} \frac{\omega/2}{\sin(\omega/2)} $$ so, yes, you could see it as the ratio of two $\mbox{sinc}$ functions.

Underlying question: can I model the sampling of a continuous signal (by, say, a digital camera) in the frequency domain as the DFT of the signal convolved with a Dirac delta train times the DFT of the pixel footprint (a rect)? If so, why is the DFT of the pixel footprint an asinc instead of a sinc?

So you seem to be trying to model taking a picture with a digital camera? Camera modeling is a little tricky. Even Gonzales and Woods seem to gloss over it.

The fundamental issue is the DFT of a $\mbox{rect}$ ($\Pi$) is a $\mbox{asinc}$.

If you're doing a discrete-time Fourier transform (DTFT), then it's not, but usually when dealing with computed FTs, you want the DFT.

$\endgroup$
  • $\begingroup$ Thanks Peter. So I gather that sampling continuous rect(x/5) produces an asinc function via DTFT in the frequency domain. Is that what it would also yield via DFT? (Indulge me, I am a photographer, not a scientist;-) It strikes me that asinc = ratio of two sincs. Underlying question: can I model the sampling of a continuous signal (by, say, a digital camera) in the frequency domain as the DFT of the signal convolved with a Dirac delta train times the DFT of the pixel footprint (a rect)? If so, why is the DFT of the pixel footprint an asinc instead of a sinc? $\endgroup$ – Jack Hogan May 22 '16 at 13:19
  • $\begingroup$ @JackHogan I have a busy day today, so will try to answer tonight. $\endgroup$ – Peter K. May 22 '16 at 15:24
  • $\begingroup$ 'Preciate it Peter. $\endgroup$ – Jack Hogan May 24 '16 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.