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Convolution with exponentials

  • Thread starter purplebird
  • Start date
18
0
How do you do the convolution of

exp (x(n))*u(x(n)) and exp(x(n-1))*u(x(n-1))

where u(x) is the unit step function. Thanks.
 

Answers and Replies

65
0
How do you do the convolution of

exp (x(n))*u(x(n)) and exp(x(n-1))*u(x(n-1))

where u(x) is the unit step function. Thanks.
Because convolution is commutative,
[tex] y(t) = u(t) * e^{t} = e^{t} * u(t)[/tex]
therefore, using the general result
[tex] h(t)*u(t) = \int_{-\infty} ^{t} *e^{\tau} d\tau= e^{t} - e^{-\infty } = e^{t}[/tex]

So, as an example given that

[tex] h(t) = e^{-t}*u(t)[/tex]
[tex] f(t) = e^{-2t}*u(t) [/tex]

and [tex] y(t) = h(t)* f(t)[/tex] , determine [tex] y(1). [/tex]

Solution,

[tex]y(t) = \int_{-\infty} ^{\infty} h(\tau)f(t-\tau) d\tau = \int_{-\infty} ^{\infty} [e^{-\tau}u(\tau)] [e^{-2(t-\tau)} u(t - \tau)] d\tau [/tex]

for any value of t, it fallows that
[tex] y(1) = \int_{-\infty} ^{\infty} [e^{-\tau}u(\tau)] [e^{-2(1-\tau)} u(1 - \tau)] d\tau = \int_{0} ^{1} e^{-\tau}e^{-2(1-\tau) }d\tau = e^{-2}(e^1 - 1)=e^{-1}-e^{-2} \approx 0.233.[/tex]

I hope this helps a little bit.


[tex] Reference. Convolution examples from Kudeki and Munson [/tex]
 
quadraphonics
How do you do the convolution of

exp (x(n))*u(x(n)) and exp(x(n-1))*u(x(n-1))

where u(x) is the unit step function. Thanks.
Wait, the [itex]x(n)[/itex] is inside the [itex]u(x)[/itex]? That makes it difficult to do, unless you tell us something more about what [itex]x(n)[/itex] looks like.

Are you sure it's not supposed to be just [itex]u(n)[/itex]?
 
65
0
Wait, the [itex]x(n)[/itex] is inside the [itex]u(x)[/itex]? That makes it difficult to do, unless you tell us something more about what [itex]x(n)[/itex] looks like.

Are you sure it's not supposed to be just [itex]u(n)[/itex]?
You are right, the other approach will be maniopulate the equation with the properties of convolution.
 
12
0


Nice Job!

I am new here and I bring a similar question for my first post.....

I want to know how to do a convolution where the two functions are:

Ca(t) - arbitrary Input function
(it actually represents the time activity curve of a CT contrast bolus injection in the blood)

R(t) - a piecewise function defined as follows:

R(t) = 1, 0<t<Tm
and E*(exp)^(kt), t>Tm

(this R(t) is called the Impulse Residue Function for the Johnson WIlson model for capillary tracer exchange)

so therefore:

Ca(t)*R(t) = (from 0 to Tm){Ca(t) convolved with 1} + (from Tm to t)E*{Ca(t) convolved with (*exp)^(-kt)}

Can anyone shed some light on this please?!

If anyone is curious the context of this convolution is for determining the representation of CT tissue attenuation in tracer kinetics modelling, considering a distributive parameter model. A background link for those interested is below.

http://www.minervamedica.it/index2.t?show=R39Y2003N03A0171

 
7
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what is matlab code for convolution in z domain?
 
12
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Nevermind, I worked this thing out

I don't want Matlab code I just wanted to do it by hand, an analytical solution

Matlab is worthless for giving insight into a solution, but it will provide you with a numerical solution...if you don't understand how it works and why it works the solution is useless for understanding your problem
 

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