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kpou
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[itex]\geq[/itex]
Find the Fourier Transform of
y = exp([itex]^{}-at[/itex])sin([itex]\omega[/itex][itex]_{}0[/itex]t) for t ≥ 0
and = 0 for t < 0
Find the amplitudes C([itex]\omega[/itex], S([itex]\omega[/itex]), and energy spectrum [itex]\Phi[/itex]' for [itex]\omega[/itex] > 0 if the term that peaks at negative frequency can be disregarded for pos frequency.
Y([itex]\omega[/itex]) = C([itex]\omega[/itex]) - iS([itex]\omega[/itex])
[itex]\Phi[/itex]' = C^2([itex]\omega[/itex]) + iS^2([itex]\omega[/itex])
C([itex]\omega[/itex])= [itex]\int[/itex]( y(t)cosw([itex]\omega[/itex]t)dt from -∞ -> ∞
S([itex]\omega[/itex])= [itex]\int[/itex] ( y(t)sin([itex]\omega[/itex]t)dt from -∞ -> ∞
I have page after page of trying to simplify the algebra down with no luck. In my text it writes "y" in the equation with no function of ( t ) or ([itex]\omega[/itex]) for most every other equation I see contains either of those. Is there something different about y?
It looks like I have to take the Fourier of two functions exp([itex]^{}-at[/itex]) and sin([itex]\omega[/itex][itex]_{}0[/itex]t) over t = 0 -> [itex]\infty[/itex]
I try exp(-at)[/itex]sin[itex]\omega[/itex][itex]_{}0[/itex]tcos([itex]\omega[/itex]t) using sin(bx)=(exp(ibx)-exp(-ibx)/2i
Am I missing something? Are there any algebraic tricks I may be missing? Thanks !
C([itex]\omega[/itex])= [itex]\int[/itex]( y(t)cosw([itex]\omega[/itex]t)dt from 0 -> ∞ Since y(t) = 0 for negative t
= [itex]\int[/itex] exp(-at)sin(ω0)t)cos(ω0t)dt
= [itex]\int[/itex] exp(-at)[ (1/2i) ( exp(iω0t) - exp(-iω0t) ) (1/2) ( exp(iωt) + exp(-iωt))] dt
Homework Statement
Find the Fourier Transform of
y = exp([itex]^{}-at[/itex])sin([itex]\omega[/itex][itex]_{}0[/itex]t) for t ≥ 0
and = 0 for t < 0
Find the amplitudes C([itex]\omega[/itex], S([itex]\omega[/itex]), and energy spectrum [itex]\Phi[/itex]' for [itex]\omega[/itex] > 0 if the term that peaks at negative frequency can be disregarded for pos frequency.
Homework Equations
Y([itex]\omega[/itex]) = C([itex]\omega[/itex]) - iS([itex]\omega[/itex])
[itex]\Phi[/itex]' = C^2([itex]\omega[/itex]) + iS^2([itex]\omega[/itex])
C([itex]\omega[/itex])= [itex]\int[/itex]( y(t)cosw([itex]\omega[/itex]t)dt from -∞ -> ∞
S([itex]\omega[/itex])= [itex]\int[/itex] ( y(t)sin([itex]\omega[/itex]t)dt from -∞ -> ∞
The Attempt at a Solution
I have page after page of trying to simplify the algebra down with no luck. In my text it writes "y" in the equation with no function of ( t ) or ([itex]\omega[/itex]) for most every other equation I see contains either of those. Is there something different about y?
It looks like I have to take the Fourier of two functions exp([itex]^{}-at[/itex]) and sin([itex]\omega[/itex][itex]_{}0[/itex]t) over t = 0 -> [itex]\infty[/itex]
I try exp(-at)[/itex]sin[itex]\omega[/itex][itex]_{}0[/itex]tcos([itex]\omega[/itex]t) using sin(bx)=(exp(ibx)-exp(-ibx)/2i
Am I missing something? Are there any algebraic tricks I may be missing? Thanks !
C([itex]\omega[/itex])= [itex]\int[/itex]( y(t)cosw([itex]\omega[/itex]t)dt from 0 -> ∞ Since y(t) = 0 for negative t
= [itex]\int[/itex] exp(-at)sin(ω0)t)cos(ω0t)dt
= [itex]\int[/itex] exp(-at)[ (1/2i) ( exp(iω0t) - exp(-iω0t) ) (1/2) ( exp(iωt) + exp(-iωt))] dt
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