Fourier Transform of x(t) and g(t) with Product Calculation

nikki92
Messages
37
Reaction score
0

Homework Statement


x(t) = 5cos(2*pi*1000*t) and g(t) = ∑ from n=-infinity to infinity delta(t-n/10000)

find Fourier transform of x(t) and g(t) and the product of the two


The Attempt at a Solution



x(w) = 5*sqrt(pi/2) [delta(w-2000pi)+delta(w+2000pi)]

g(w) = 1

so would the product of the two just be the x(w) ?
 
Physics news on Phys.org
nikki92 said:

Homework Statement


x(t) = 5cos(2*pi*1000*t) and g(t) = ∑ from n=-infinity to infinity delta(t-n/10000)

find Fourier transform of x(t) and g(t) and the product of the two


The Attempt at a Solution



x(w) = 5*sqrt(pi/2) [delta(w-2000pi)+delta(w+2000pi)]

g(w) = 1

so would the product of the two just be the x(w) ?
##G(\omega)## isn't equal to 1. That's the Fourier transform of a single delta function. You have a train of delta functions.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top