could someone help me with fourier?
I dont understand the parseval's identity... for example this question: Consider the periodic block function f(t) with period pi and for some a∈R with 0 < a< pi, and the period block function g(t) with period pi and some b∈R with 0<b<pi. Assume a≤b Use parseval's identity to show that: ∞ Σ (sin(na)*sin(nb) )/n^2 n=1 is the same as (a( pi - b))/2
\[\sum_{n=1}^\infty {\frac{\sin(n*a) * \sin(n*b)}{n^2}}\]\[ = \frac{a(\pi - b)}{2}\]
yeahhh i have typed that.. but could you explain it?
This expression on the right \[ a(\pi - b)/2 \] looks like a convolution and then the expression on the left looks like the product of two Fourier series as you might expect from a convolution.
Ah ... and now I see what the functions are you are convoluting; they're given to you. So the convolution formula applied to those two functions gives you the infinite sum; explicitly integrating the convolution of the two functions gives you the RHS expression. Then use Parseval's theorem to set them equal.
JamesJ could you be more specific? So i have use the convolution integral?
Can you tell me exactly what are the two functions given to you in the problem? As written they're not well defined.
I think on the left side with the Σ thats just a sommation of a and b. And i dont know that the other is, im afraid of...
I mean these functions: Consider the periodic block function f(t) with period pi and for some a∈R with 0 < a< pi, and the period block function g(t) with period pi and some b∈R with 0<b<pi. Assume a≤b
owhh ... there are 2 periodic signals a and b, both signals are block functions
Can you write them out for me explicitly?
the period is \[\omega0 = 2\] fn = sin(na)/(n*pi) f0 = a/pi gn = sin(nb)/(n*pi) g0 = b/pi
that what i calculated
Sorry, I'm still not getting it. What is a periodic block function?
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