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From: mbaran on 2 Aug 2010 10:24
Say I have a signal x(t) and a window w(t). I multiply the two signals
together, y(t)=x(t)w(t) and take the Fourier transform to form Y(f). Now
say I apply a linear phase to Y(f) such that Y'(f) = exp(-jwD)Y(f) where D
is an unknown.

Now assume that x(t), D, and Y(f) are both unknown, while w(t) and Y'(f)
are known.

Any ideas on how I can calculate D, or recover x(t)?



More specifically,
I have an image in the Fourier domain that was formed by weighting (with a
hamming) each row, and forward transforming. However, when I back transform
it (in the row direction), I can detect (by looking at the column energy)
that the signal is shifted. In fact I am not 100% sure that the same linear
phase was applied to each row. However, for now... if I can solve the above
example for the 1D case, I think I'm good.

Thanks in advance!



From: dbd on 2 Aug 2010 15:49
The window function can be applied by a vector multiply in the time
domain or a small convolution in the frequency domain. The frequency
domain version can be represented as a vector-matrix multiply. If the
frequency domain windowing matrix is invertible, the inverse matrix
can be used to remove the windowing. For example see:

'Practical methods for rapid and accurate computation of
interferometric spectra for remote sensing applications'
Geoscience and Remote Sensing, IEEE Transactions on
Jan 2000, Vol 38, Issue 1, page 169 - 183

Dale B. Dalrymple
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