- #1
atrus_ovis
- 101
- 0
finding the FT of x(t)=sin(πt) sin(50πt) :
( '*' is the convolution operator)
its FT X(Ω)=(1/2π) F(sin(πt)) * F(sin(50πt))
= (1/2π) jπ(δ(Ω+π)-δ(Ω-π)) * (jπ (δ(Ω+π)-δ(Ω-π)) ) (a)
from my professor's solution it next goes:
= (π/2) (-δ(Ω+π)+δ(Ω-π)) * ((-δ(Ω+π)+δ(Ω-π)) ) (b)
1)How did the signs change?
2)Where did the j's go? ( i suspect 1,2 are related =P )
also,
3) doesn't the 1/π cancel out the π coefficients in each quantity to be convoluted?
from wiki:
a (f * g) = (a f) * g = f * (a g)
why is that? i mean, why isn't it:
a (f * g) = (a f) *(a g)
?
( '*' is the convolution operator)
its FT X(Ω)=(1/2π) F(sin(πt)) * F(sin(50πt))
= (1/2π) jπ(δ(Ω+π)-δ(Ω-π)) * (jπ (δ(Ω+π)-δ(Ω-π)) ) (a)
from my professor's solution it next goes:
= (π/2) (-δ(Ω+π)+δ(Ω-π)) * ((-δ(Ω+π)+δ(Ω-π)) ) (b)
1)How did the signs change?
2)Where did the j's go? ( i suspect 1,2 are related =P )
also,
3) doesn't the 1/π cancel out the π coefficients in each quantity to be convoluted?
from wiki:
a (f * g) = (a f) * g = f * (a g)
why is that? i mean, why isn't it:
a (f * g) = (a f) *(a g)
?