Analyzing Convolution of Exponential Functions with Unit Step Function

  • Thread starter Thread starter purplebird
  • Start date Start date
  • Tags Tags
    Convolution
AI Thread Summary
The discussion focuses on the convolution of exponential functions multiplied by the unit step function, specifically exp(x(n))*u(x(n)) and exp(x(n-1))*u(x(n-1)). It highlights that convolution is commutative and provides a general result for the convolution of an exponential function with a unit step function. An example calculation is presented, demonstrating how to compute y(1) through integration, resulting in an approximate value of 0.233. There is also a mention of a related question regarding the convolution of an arbitrary input function with a piecewise function, emphasizing the need for analytical solutions over numerical methods. The conversation underscores the complexity of convolutions involving unit step functions and the importance of understanding the underlying principles.
purplebird
Messages
14
Reaction score
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.
 
Physics news on Phys.org
purplebird said:
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,
y(t) = u(t) * e^{t} = e^{t} * u(t)
therefore, using the general result
h(t)*u(t) = \int_{-\infty} ^{t} *e^{\tau} d\tau= e^{t} - e^{-\infty } = e^{t}

So, as an example given that

h(t) = e^{-t}*u(t)
f(t) = e^{-2t}*u(t)

and y(t) = h(t)* f(t) , determine y(1).

Solution,

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

for any value of t, it fallows that
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.

I hope this helps a little bit.


Reference. Convolution examples from Kudeki and Munson
 
purplebird said:
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 x(n) is inside the u(x)? That makes it difficult to do, unless you tell us something more about what x(n) looks like.

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

Are you sure it's not supposed to be just u(n)?

You are right, the other approach will be maniopulate the equation with the properties of convolution.
 


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

 
what is MATLAB code for convolution in z domain?
 
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
 

Similar threads

Replies
2
Views
2K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
Back
Top