But we can still find valid Laplace transforms of f (t) = t and g (t) = (t^2). If we multiply their Laplace transforms, and then inverse Laplace transform the result, shouldn't the result be a convolution of f and g?
The question that is still popping in my head is: The convolution as a concept appeared after we knew the LTI systems (The convolution is a concept that is dependent on LTI systems in the first place)? or it was an independent concept and was adopted by the LTI systems transfer function?
3 The definition of convolution is known as the integral of the product of two functions $$ (f*g) (t):=\int_ {-\infty}^ {\infty} f (t -\tau)g (\tau),\mathrm d\tau$$ But what does the product of the functions give? Why are is it being integrated on negative infinity to infinity? What is the physical significance of the convolution?
I am currently learning about the concept of convolution between two functions in my university course. The course notes are vague about what convolution is, so I was wondering if anyone could giv...
Proving commutativity of convolution $ (f \ast g) (x) = (g \ast f) (x)$
On this math.MO post, "What is convolution intuitively?", Terence Tao's answer (in the case where one function is a bump function) involves "blurring" and "fuzz." Could someone clarify his
The term inside the parentheses is the discrete convolution of the coefficients. The fact that convolution shows up when doing products of polynomials is pretty closely tied to group theory and is actually very important for the theory of locally compact abelian groups.