How do I find the convolution of two functions with different domains?

In summary, the convolution is given by:In summary, to find the convolution \beta * L, you need to integrate over the two-dimensional region A_a = { (x,y) : -1 \leq y \leq 1, |x-y| \leq a }. This region will look different depending on whether a < 1, a = 1, or a > 1. To find the limits of integration for the y-integral, you need to consider the vertical lines through x and determine if they intersect A_a in a line segment or not.
  • #1
bobred
173
0

Homework Statement


I have the two functions below and have to find the convolution [itex]\beta * L[/itex]

Homework Equations


Assume a<1
[tex]
\beta(x)=\begin{cases}
\frac{\pi}{4a}\cos\left(\frac{\pi x}{2a}\right) & \left|x\right|<a\\
0 & \left|x\right|\geq a
\end{cases}
[/tex]
[tex]
L(x)=\begin{cases}
1 & \left|x\right|<1\\
0 & \left|x\right|\geq 1
\end{cases}
[/tex]

The Attempt at a Solution


[tex]\beta * L=\int^\infty_{-\infty}\beta(y)L(x-y)dy=\int^\infty_{-\infty}\beta(x-y)L(y)dy[/tex]
What I am unsure about is the limits and which integral to use. Any hints would be greatly appreciated.
 
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  • #2
bobred said:

Homework Statement


I have the two functions below and have to find the convolution [itex]\beta * L[/itex]

Homework Equations


Assume a<1
[tex]
\beta(x)=\begin{cases}
\frac{\pi}{4a}\cos\left(\frac{\pi x}{2a}\right) & \left|x\right|<a\\
0 & \left|x\right|\geq a
\end{cases}
[/tex]
[tex]
L(x)=\begin{cases}
1 & \left|x\right|<1\\
0 & \left|x\right|\geq 1
\end{cases}
[/tex]

The Attempt at a Solution


[tex]\beta * L=\int^\infty_{-\infty}\beta(y)L(x-y)dy=\int^\infty_{-\infty}\beta(x-y)L(y)dy[/tex]
What I am unsure about is the limits and which integral to use. Any hints would be greatly appreciated.

Start by drawing the two-dimensional ##A## region for the second form:
$$A_a = \{ (x,y) : -1 \leq y \leq 1, |x-y| \leq a \}.$$
There are three cases: (1) ##a < 1##; (2) ##a = 1##; and (3) ##a > 1##. The region will look slightly different in cases (1) and (3).For some values of ##x## the line vertical line through ##x## will miss region ##A_a## completely, so the ##y##integral will equal 0. For other values of ##x## the vertical line through ##x## will intersect region ##A_a## in a vertical line segment, so will have the form ##y_1(x) \leq y \leq y_2(x)##, and your ##y##-integration will be from ##y_1(x)## to ##y_2(x)##.
 

What is the definition of convolution of two functions?

The convolution of two functions, f and g, is a mathematical operation that produces a third function, h, which represents the amount of overlap of f and a flipped and shifted version of g. It is denoted by f * g or f ⨂ g.

What is the significance of convolution in signal processing?

In signal processing, convolution is used to represent the output of a linear time-invariant system in terms of its input and impulse response. It is also used to filter signals and remove noise, as well as for correlation and deconvolution.

How is convolution related to the Fourier transform?

The convolution theorem states that the convolution of two functions in the time domain is equal to the multiplication of their Fourier transforms in the frequency domain. This relationship is used in many applications of convolution, particularly in signal and image processing.

What are some real-world applications of convolution?

Convolution has numerous applications in a variety of fields, including physics, engineering, economics, and biology. Some examples include image and signal processing, convolutional neural networks in machine learning, and modeling of physical systems such as heat transfer and diffusion.

Can any two functions be convolved with each other?

No, not all functions can be convolved. Both functions must be integrable over their common domain for the convolution to exist. Additionally, certain conditions must be met, such as the existence of a finite integral or the functions approaching zero at infinity.

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