Inverse Fourier transform of $frac{jwL}{R+jwl}$
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I am trying to find the inverse Fourier transform of:
$$frac{jwL}{R+jwl}$$
My current attempt is
$$mathcal{F^{-1}(frac{jwL}{R+jwl})}$$
$$mathcal{F^{-1} ({jwL})} oplus mathcal{F^{-1}(frac{1}{R+jwL}}) $$
$$L.mathcal{F^{-1} ({jw}}) oplus frac{1}{L}.mathcal{F^{-1}(frac{1}{frac{R}{L}+jw}}) $$
$$ L.mathcal{F^{-1} ({jw}}) oplus frac{1}{L}.u(t).e^{frac{Rt}{L}} $$
I get stuck with $$ mathcal{F^{-1} ({jw}}) $$ as $$frac{1}{2 pi} int_{-infty}^infty jwe^{jwt} $$ doesn't converge
fourier-transform
asked Nov 20 at 20:18
C. Begley
105
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up vote
1
down vote
favorite
I am trying to find the inverse Fourier transform of:
$$frac{jwL}{R+jwl}$$
My current attempt is
$$mathcal{F^{-1}(frac{jwL}{R+jwl})}$$
$$mathcal{F^{-1} ({jwL})} oplus mathcal{F^{-1}(frac{1}{R+jwL}}) $$
$$L.mathcal{F^{-1} ({jw}}) oplus frac{1}{L}.mathcal{F^{-1}(frac{1}{frac{R}{L}+jw}}) $$
$$ L.mathcal{F^{-1} ({jw}}) oplus frac{1}{L}.u(t).e^{frac{Rt}{L}} $$
I get stuck with $$ mathcal{F^{-1} ({jw}}) $$ as $$frac{1}{2 pi} int_{-infty}^infty jwe^{jwt} $$ doesn't converge
fourier-transform
asked Nov 20 at 20:18
C. Begley
105
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to find the inverse Fourier transform of:
$$frac{jwL}{R+jwl}$$
My current attempt is
$$mathcal{F^{-1}(frac{jwL}{R+jwl})}$$
$$mathcal{F^{-1} ({jwL})} oplus mathcal{F^{-1}(frac{1}{R+jwL}}) $$
$$L.mathcal{F^{-1} ({jw}}) oplus frac{1}{L}.mathcal{F^{-1}(frac{1}{frac{R}{L}+jw}}) $$
$$ L.mathcal{F^{-1} ({jw}}) oplus frac{1}{L}.u(t).e^{frac{Rt}{L}} $$
I get stuck with $$ mathcal{F^{-1} ({jw}}) $$ as $$frac{1}{2 pi} int_{-infty}^infty jwe^{jwt} $$ doesn't converge
fourier-transform
asked Nov 20 at 20:18
C. Begley
105
I am trying to find the inverse Fourier transform of:
$$frac{jwL}{R+jwl}$$
My current attempt is
$$mathcal{F^{-1}(frac{jwL}{R+jwl})}$$
$$mathcal{F^{-1} ({jwL})} oplus mathcal{F^{-1}(frac{1}{R+jwL}}) $$
$$L.mathcal{F^{-1} ({jw}}) oplus frac{1}{L}.mathcal{F^{-1}(frac{1}{frac{R}{L}+jw}}) $$
$$ L.mathcal{F^{-1} ({jw}}) oplus frac{1}{L}.u(t).e^{frac{Rt}{L}} $$
I get stuck with $$ mathcal{F^{-1} ({jw}}) $$ as $$frac{1}{2 pi} int_{-infty}^infty jwe^{jwt} $$ doesn't converge
fourier-transform
fourier-transform
asked Nov 20 at 20:18
C. Begley
105
asked Nov 20 at 20:18
C. Begley
105
edited Nov 20 at 21:38
asked Nov 20 at 20:18
C. Begley
105
asked Nov 20 at 20:18
C. Begley
105
asked Nov 20 at 20:18
C. Begley
105
105
add a comment |
add a comment |
1 Answer
1
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1
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accepted
Recall that:
- $$mathcal{F}(e^{-alpha t}u(t)) = frac{1}{alpha + jomega};$$
- $$mathcal{F}(delta(t)) = 1.$$
Notice that:
$$frac{jomega L}{R+jomega L} = frac{jomega L + R - R}{R+jomega L} = 1- frac{R}{R+jomega L}.$$
The inverse Fourier transform of is:
$$delta(t) - frac{R}{L}e^{-frac{Rt}{L}}u(t).$$
answered Nov 20 at 20:26
the_candyman
8,55921944
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Recall that:
- $$mathcal{F}(e^{-alpha t}u(t)) = frac{1}{alpha + jomega};$$
- $$mathcal{F}(delta(t)) = 1.$$
Notice that:
$$frac{jomega L}{R+jomega L} = frac{jomega L + R - R}{R+jomega L} = 1- frac{R}{R+jomega L}.$$
The inverse Fourier transform of is:
$$delta(t) - frac{R}{L}e^{-frac{Rt}{L}}u(t).$$
answered Nov 20 at 20:26
the_candyman
8,55921944
add a comment |
up vote
1
down vote
accepted
Recall that:
- $$mathcal{F}(e^{-alpha t}u(t)) = frac{1}{alpha + jomega};$$
- $$mathcal{F}(delta(t)) = 1.$$
Notice that:
$$frac{jomega L}{R+jomega L} = frac{jomega L + R - R}{R+jomega L} = 1- frac{R}{R+jomega L}.$$
The inverse Fourier transform of is:
$$delta(t) - frac{R}{L}e^{-frac{Rt}{L}}u(t).$$
answered Nov 20 at 20:26
the_candyman
8,55921944
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Recall that:
- $$mathcal{F}(e^{-alpha t}u(t)) = frac{1}{alpha + jomega};$$
- $$mathcal{F}(delta(t)) = 1.$$
Notice that:
$$frac{jomega L}{R+jomega L} = frac{jomega L + R - R}{R+jomega L} = 1- frac{R}{R+jomega L}.$$
The inverse Fourier transform of is:
$$delta(t) - frac{R}{L}e^{-frac{Rt}{L}}u(t).$$
answered Nov 20 at 20:26
the_candyman
8,55921944
Recall that:
- $$mathcal{F}(e^{-alpha t}u(t)) = frac{1}{alpha + jomega};$$
- $$mathcal{F}(delta(t)) = 1.$$
Notice that:
$$frac{jomega L}{R+jomega L} = frac{jomega L + R - R}{R+jomega L} = 1- frac{R}{R+jomega L}.$$
The inverse Fourier transform of is:
$$delta(t) - frac{R}{L}e^{-frac{Rt}{L}}u(t).$$
answered Nov 20 at 20:26
the_candyman
8,55921944
edited Nov 20 at 20:39
answered Nov 20 at 20:26
the_candyman
8,55921944
answered Nov 20 at 20:26
the_candyman
8,55921944
answered Nov 20 at 20:26
the_candyman
8,55921944
8,55921944
add a comment |
add a comment |
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