What does U2(t) mean in this context? I think I know, but I want to be sure that you do, as well.Dusty912 said:Homework Statement
Compute the convolution f*g for the given function f and g.
f(t)=cost g(t)=U2(t)Homework Equations
f*g=∫f(t-u)g(u)du
The Attempt at a Solution
So I pretty much only know how to plug in the functions into the integral for convolutions. Not really sure how to evaluate it with the U2(t)
f*g=∫cos(t-u)U2(u)du
so do I evaluate thi integral directly or are there other methods to evaluate this.?
What effect does this have on your integral as well as on the limits of integration?Dusty912 said:well it is a heaviside function. The U2(t) has a discontinuity at 2. 2 is when the function gets "turned on" so to speak.
OK, so what does your convolution integral look like then?Dusty912 said:Well it causes for the integral to be evaluated from 2 to t. Because the function would be 0 when t is less than 2
But what is U2(u) on that interval? What's the value of any Heaviside function?Dusty912 said:∫t2cos(t-u)U2(u)du
You're not sure?Dusty912 said:1?
Convolutions in differential equations refer to a mathematical operation that combines two functions to produce a third function. It is commonly used in the study of differential equations to solve problems involving linear differential equations with constant coefficients.
Convolutions are widely used in various fields such as physics, engineering, and economics to model and analyze complex systems. They are particularly useful in solving problems involving signals and systems, such as in image and audio processing.
The convolution theorem states that the convolution of two functions in the time domain is equal to the product of their Fourier transforms in the frequency domain. This theorem is particularly useful in solving differential equations as it allows for the simplification of complex calculations.
The convolution of two functions f(x) and g(x) is given by the integral of the product of the two functions, where one of them is reversed and shifted. This can be represented mathematically as: (f*g)(x) = ∫f(t)g(x-t)dt. In simpler terms, it involves sliding one function over the other and multiplying the overlapping portions, then integrating the result.
One of the main advantages of using convolutions in solving differential equations is that it allows for the simplification of complex calculations. It also provides a systematic and efficient approach to solving problems involving linear differential equations with constant coefficients. Additionally, convolutions have a wide range of real-world applications, making them a powerful tool in various fields of science and engineering.