Analyzing Convolution of Exponential Functions with Unit Step Function

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In summary, convolution is the operation of multiplying two functions together, and is commutative. Thanks.
  • #1
purplebird
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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.
 
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  • #2
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,
[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]
 
  • #3
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 [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]?
 
  • #4
quadraphonics said:
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.
 
  • #5


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

 
  • #6
what is MATLAB code for convolution in z domain?
 
  • #7
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
 

1. What is the purpose of convolving with exponentials?

The purpose of convolving with exponentials is to smooth out a signal or data set and highlight its underlying trends. This is particularly useful in time series analysis, where it can be used to remove noise and identify patterns or trends.

2. How does convolution with exponentials work?

In convolution with exponentials, the exponential function is used as a weighting function to combine neighboring data points in a signal. This process involves multiplying each data point by the corresponding value of the exponential function and summing the results to create a new, smoothed signal.

3. Can convolution with exponentials be used for data sets with irregular intervals?

Yes, convolution with exponentials can be used for data sets with irregular intervals. The exponential function can be scaled and shifted to fit the intervals of the data, allowing for a smooth convolved signal to be generated.

4. Is convolution with exponentials the same as exponential smoothing?

No, convolution with exponentials and exponential smoothing are not the same. Exponential smoothing is a specific type of smoothing technique that uses a weighted average of past data to forecast future values, while convolution with exponentials is a more general technique that can be used for smoothing and trend identification in any type of signal or data set.

5. What are the limitations of convolution with exponentials?

Convolution with exponentials is not suitable for all types of data. It works best on data with a relatively small amount of noise and a clear underlying trend. It also assumes that the data is stationary, meaning that the underlying trend does not change over time. Additionally, it may not be effective in identifying trends in data sets with irregular or non-linear patterns.

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