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/)