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I'm having a hard time understanding how the convolution integral works (for Laplace transforms of two functions multiplied together) and was hoping someone could clear the topic up or link to sources that easily explain it. Note: I don't understand math notation very well, so a watered-down explanation would do just fine. $$(f * g)(t) = \int_0^t f(\tau)g(t-\tau)\ \mathrm{d}\tau$$

This is what my textbook has written. What do those lowercase t-like symbols represent (I haven't seen them before).

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1  
That's a greek letter called tau. It's just a variable here. –  Pedro Dec 11 '12 at 1:53
    
Lowercase t-like symbol is a greek letter "tau". Here it represents an integration (dummy) variable, which "runs" from lower integration limit, "0", to upper integration limit, "t". So, the convolution is a function, which value for any value of argument (independent variable) "t" is expressed as an integral over dummy variable "tau". –  mbaitoff Dec 11 '12 at 1:55
    
... and it's common to choose related names for related variables. Some other examples of letters that are often paired are $(z, \zeta)$, $(x, \xi)$, $(r, \rho)$, $(s, \sigma)$, $(w, \omega)$, $(a, \alpha)$, $(b, \beta)$, $(d, \delta)$, $(e, \epsilon)$. Also, the letters $\iota, \kappa, \lambda, \mu, \nu$ are often used in places where one might also tend to use $i, k, l, m, n$ respectively. All of the "strange" letters I've used here are from the Greek alphabet. –  Hurkyl Dec 11 '12 at 1:57
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Maybe the animated gifs at Wikipedia will give you some ideas. –  dtldarek Dec 11 '12 at 2:05

3 Answers 3

up vote 3 down vote accepted

Intuitively speaking when you are given two signals/or functions $f$ and $g$. You time reverese one of the signals, it doesnt matter which one, and shift it by a value of $t$ then you simply sum the area under the intersection.

If you consider a function say function of $x$, then time reversal means inserting $-x$ wherever you see $x$ in this function.

Example:

Question1: Assume you have a function $f(x)$ that is $1$ if $x\in[0,1]$, and $0$ elsewhere then how should you plot $f(-x)$?

Question2: Assume you have $g(x)=f(x)$ is there any intersecting area between $f(x)$ and $g(x)$?

Question3: Now shift $g(x)$ by $0.5$, that is to find $g(-x+0.5)$. How does it look like when you plot it?

Question4: Where does the intersecting region lie in this case $x\in?$ what is the are of the intersecting region?

Question5: If you select the shifting parameter not $0.5$ but all reals in $[0,2]$ what function should you get at the output?

EDIT: You define convolution integral in $[0,t]$ for bounded signals. The integral limits depend on where your signal is non-zero.

If you have two signals as you suggested $f(t)=e^{at}$ and $g(t)=e^{bt}$ then the first question: what is the relation between $a$ and $b$? are they positive? where is the function defined? For example when $a$ and $b$ are some positive terms then we have the following integral

$$h(t)=\int f(\tau)g(t-\tau)d\tau= \int e^{a\tau}e^{b(t-\tau)}d\tau=e^{bt}\int e^{(a-b)\tau}d\tau=\Bigg]_{\tau\in\Omega}\frac{e^{(a-b)\tau}}{a-b}$$

clearly $\Omega=\mathbb{R}$ is not possible because the integral does not converge.

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Can you use a more concrete example? For example, how can we find the convolution of $f(t) = \sin2t$ and $g(t) = e^t$ or $f(t) = e^{at}$ and $g(t) = e^{bt}$ with $a \neq b$ in the second example? Or perhaps using the same theorem to find an inverse Laplace of $F(s) = \frac{4}{s(s^2+4)}$? –  Bailor Tow Dec 11 '12 at 2:36
    
please see the edit –  Seyhmus Güngören Dec 11 '12 at 3:10

I believe the convolution functions makes the most sense when you see it applied in probability theory.

Let X and Y be two random variables and f(X) and g(Y) be the probability distributions of the random variables.

Then the distribution of the sum of two random variables:

$$ (f*g)(t) = \int^t_0 f(-\tau)g(\tau - t )d\tau $$

Why is this? Let us visualize the simple case of rolling dices. and X be the outcome of the first roll and Y be the outcome of the second roll. What is the distribution of the sum?

Since our distributions are discrete, $$ (f*g)(t) = \sum^t_{i=0} f(t)g(i-t) \quad t\in[2,12] $$ This basically translates to, sum up all probabilities such that it has this probability.

i.e. $$(f*g)(4)= \sum^4_{i=0} f(t)g(i-t) =f(1)g(3) + f(2)g(2)+ f(3)g(1) = 1/12 $$

Which is the answer we expect. We can also look at the question from a more physics point of view where it is a time reversed signals but I find this much more intuitive.

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Consider the sequences $x_0, x_1, \dotsc$ and $y_0, y_1, \dotsc$.

Now, $$ \left( \sum_{j=0}^n x_j \right) \left( \sum_{j=0}^n y_j \right) = \sum_{z=0}^n z_j, $$ where $$ z_j = \sum_{k=0}^j x_k y_{j-k}. $$

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