Why Does Deconvolution Result in a Peak Followed by a Valley?

[PaulPaul]

Sorry, but I don't know where the topic about signal analysis should go..?
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I'm doing research about quantifying blood flow in the brain. Basically I need to know how to perform deconvolution (I think.)

I have two functions that describe the concentration of contrast solution:

f(t) = a(t-to)^b exp[ -(t-to)/c ]
g(t) = d(t-t1)^k exp[ -(t-t1)/m ]

where a,b,c,d,k,m,to, and t1 are known parameters (found by fitting the function to measurements.)

now say f = h * g (* - convolution)
How can I find h(t)? What type of result can I expect?

I have tried using the discrete Fourier transformation. But I often find h to be a peak followed by a small valley. I don't understand why this would be the deconvolution.

Please help. I'm stuck.
(http://s153.photobucket.com/albums/s235/s1020099/)

 

[rbj]

Sorry, but I don't know where the topic about signal analysis should go..?

try the USENET newsgroup comp.dsp .

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I'm doing research about quantifying blood flow in the brain. Basically I need to know how to perform deconvolution (I think.)

I have two functions that describe the concentration of contrast solution:

f(t) = a(t-to)^b exp[ -(t-to)/c ]
g(t) = d(t-t1)^k exp[ -(t-t1)/m ]

where a,b,c,d,k,m,to, and t1 are known parameters (found by fitting the function to measurements.)

now say f = h * g (* - convolution)
How can I find h(t)? What type of result can I expect?

if c is not the same number as m, i don't think you can do it. well, it might depend a little on what if f() and g() have unit step functions applied to them.

but the basic idea is to compute the Fourier Transform of f(t) and g(t), divide F(\omega) by G(\omega) to get H(\omega) and then inverse Fourier transform that result.

I have tried using the discrete Fourier transformation. But I often find h to be a peak followed by a small valley. I don't understand why this would be the deconvolution.

Please help. I'm stuck.
(http://s153.photobucket.com/albums/s235/s1020099/)