What is the Inverse Fourier Transform [closed]
$begingroup$
what is the inverse fourier transform for:
$(6w^2 + 20 )/( w^4 + 7w^2 + 12)$
I tried to simplify it and look it up from the table but it didn't work, can someone give me a hint?
fourier-transform
edited Dec 12 '18 at 14:05
Key Flex
8,28261233
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closed as off-topic by Saad, Did, Dando18, Chris Custer, Brahadeesh Dec 13 '18 at 6:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, Dando18, Chris Custer, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
what is the inverse fourier transform for:
$(6w^2 + 20 )/( w^4 + 7w^2 + 12)$
I tried to simplify it and look it up from the table but it didn't work, can someone give me a hint?
fourier-transform
edited Dec 12 '18 at 14:05
Key Flex
8,28261233
$endgroup$
closed as off-topic by Saad, Did, Dando18, Chris Custer, Brahadeesh Dec 13 '18 at 6:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, Dando18, Chris Custer, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
what is the inverse fourier transform for:
$(6w^2 + 20 )/( w^4 + 7w^2 + 12)$
I tried to simplify it and look it up from the table but it didn't work, can someone give me a hint?
fourier-transform
edited Dec 12 '18 at 14:05
Key Flex
8,28261233
$endgroup$
what is the inverse fourier transform for:
$(6w^2 + 20 )/( w^4 + 7w^2 + 12)$
I tried to simplify it and look it up from the table but it didn't work, can someone give me a hint?
fourier-transform
fourier-transform
edited Dec 12 '18 at 14:05
Key Flex
8,28261233
edited Dec 12 '18 at 14:05
Key Flex
8,28261233
edited Dec 12 '18 at 14:05
Key Flex
8,28261233
edited Dec 12 '18 at 14:05
Key Flex
8,28261233
edited Dec 12 '18 at 14:05
Key Flex
8,28261233
8,28261233
asked Dec 12 '18 at 14:00
user616723
closed as off-topic by Saad, Did, Dando18, Chris Custer, Brahadeesh Dec 13 '18 at 6:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, Dando18, Chris Custer, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Did, Dando18, Chris Custer, Brahadeesh Dec 13 '18 at 6:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, Dando18, Chris Custer, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Use Partial Fraction Decomposition
Writing $w^2=y$
$$dfrac{6w^2+20}{(w^2+3)(w^2+4)}=dfrac{6y+20}{(y+3)(y+4)}=dfrac a{y+3}+dfrac b{y+4}$$
$implies6y+20=y(a+b)+3b+4a$
$a=?,b=?$
answered Dec 12 '18 at 14:05
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Finally, $$F^{-1}dfrac{2a}{a^2+w^2}=e^{-a|t|}$$
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 14:21
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Use Partial Fraction Decomposition
Writing $w^2=y$
$$dfrac{6w^2+20}{(w^2+3)(w^2+4)}=dfrac{6y+20}{(y+3)(y+4)}=dfrac a{y+3}+dfrac b{y+4}$$
$implies6y+20=y(a+b)+3b+4a$
$a=?,b=?$
answered Dec 12 '18 at 14:05
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Finally, $$F^{-1}dfrac{2a}{a^2+w^2}=e^{-a|t|}$$
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 14:21
add a comment |
$begingroup$
Hint:
Use Partial Fraction Decomposition
Writing $w^2=y$
$$dfrac{6w^2+20}{(w^2+3)(w^2+4)}=dfrac{6y+20}{(y+3)(y+4)}=dfrac a{y+3}+dfrac b{y+4}$$
$implies6y+20=y(a+b)+3b+4a$
$a=?,b=?$
answered Dec 12 '18 at 14:05
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Finally, $$F^{-1}dfrac{2a}{a^2+w^2}=e^{-a|t|}$$
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 14:21
add a comment |
$begingroup$
Hint:
Use Partial Fraction Decomposition
Writing $w^2=y$
$$dfrac{6w^2+20}{(w^2+3)(w^2+4)}=dfrac{6y+20}{(y+3)(y+4)}=dfrac a{y+3}+dfrac b{y+4}$$
$implies6y+20=y(a+b)+3b+4a$
$a=?,b=?$
answered Dec 12 '18 at 14:05
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint:
Use Partial Fraction Decomposition
Writing $w^2=y$
$$dfrac{6w^2+20}{(w^2+3)(w^2+4)}=dfrac{6y+20}{(y+3)(y+4)}=dfrac a{y+3}+dfrac b{y+4}$$
$implies6y+20=y(a+b)+3b+4a$
$a=?,b=?$
answered Dec 12 '18 at 14:05
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 12 '18 at 14:05
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 12 '18 at 14:05
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 12 '18 at 14:05
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
Finally, $$F^{-1}dfrac{2a}{a^2+w^2}=e^{-a|t|}$$
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 14:21
add a comment |
$begingroup$
Finally, $$F^{-1}dfrac{2a}{a^2+w^2}=e^{-a|t|}$$
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 14:21
$begingroup$
Finally, $$F^{-1}dfrac{2a}{a^2+w^2}=e^{-a|t|}$$
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 14:21
$begingroup$
Finally, $$F^{-1}dfrac{2a}{a^2+w^2}=e^{-a|t|}$$
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 14:21
add a comment |
