Finding x(t) from CTFT using inverse transform and limit evaluation

hquang001 · Nov 27, 2021

Given the CTFT X(ω):

Equation: for all ω in R, X(ω) = e^(-ω/2) u(ω), where u(ω) is the unit step function.

Handwritten derivation of an inverse Fourier transform: x(t) = 1/(2π) integral from negative infinity to infinity of e^(-ω/2) v(ω) e^(jωt) dω, then = 1/(2π) integral from 0 to infinity of e^(ω(jt - 1/2)) dω = 1/(2π(jt - 1/2)) times the limit as n goes to infinity of e^(ω(jt - 1/2)) evaluated from 0 to n.

and here are my solution to find CT signal x(t) associated with the given CTFT , but i got stuck at the limit part

BvU · Nov 27, 2021

FInd any hints in a table of FTs ?

##\ ##

hquang001 · Nov 27, 2021

BvU said:

FInd any hints in a table of FTs ?

##\ ##

Oh, my professor never mention this, i will check this and see
Thank you

MathematicalPhysicist · Nov 27, 2021

##\exp(n(jt-1/2))-1\to -1## as ##n\to \infty## since ##\exp(jtn)=\cos(tn)+j\sin(tn)## and ##\exp(-n/2)\to 0## as ##n\to \infty##.

hquang001 · Nov 28, 2021

MathematicalPhysicist said:

##\exp(n(jt-1/2))-1\to -1## as ##n\to \infty## since ##\exp(jtn)=\cos(tn)+j\sin(tn)## and ##\exp(-n/2)\to 0## as ##n\to \infty##.

Yes i have tried this before, but it didn't match the solution that my professor gave, that's why i ask in here to check

Handwritten derivation evaluating a limit of e^(jωt) times 1/(2π(jt - 1/2)) as ω tends to infinity, giving 1/(π(1 - 2jt)) then (1 + 2jt)/(π(1 + 4t²)), next to a printed solution x(t) = (1 + j2t)/(π√(1 + 4t²)) for all real t.

Finding x(t) from CTFT using inverse transform and limit evaluation

Discussion

Similar threads