The Fourier Transform of the signal x(t) = 5cos(2π1000t) results in X(ω) = 5√(π/2) [δ(ω - 2000π) + δ(ω + 2000π)]. The signal g(t) is defined as a train of delta functions, represented mathematically as g(t) = ∑ from n=-∞ to ∞ δ(t - n/10000), which has a Fourier Transform G(ω) that is not equal to 1, but rather represents a periodic function. The product of the Fourier Transforms X(ω) and G(ω) simplifies to X(ω), confirming that the product of a signal and a train of delta functions retains the original signal's Fourier Transform.
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##G(\omega)## isn't equal to 1. That's the Fourier transform of a single delta function. You have a train of delta functions.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) ?