Inverse fourier transform of this phase shift system
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$$H(Omega)=begin{cases}exp(-j pi/2) ,;&Omega >0 \
exp(jpi/2) ,;&Omega<0.end{cases}$$
How can I find the inverse fourier transform of this $(h(t))$ using fourier properties?
Fourier Transform Used:
$$X(Omega)equivint_{-infty}^{infty}x(t),e^{-jOmega t},dt.$$
fourier-transform signal-processing
edited Nov 19 at 16:16
Adrian Keister
4,58251933
asked Nov 19 at 15:57
Prestyy
434
|
show 3 more comments
up vote
0
down vote
favorite
$$H(Omega)=begin{cases}exp(-j pi/2) ,;&Omega >0 \
exp(jpi/2) ,;&Omega<0.end{cases}$$
How can I find the inverse fourier transform of this $(h(t))$ using fourier properties?
Fourier Transform Used:
$$X(Omega)equivint_{-infty}^{infty}x(t),e^{-jOmega t},dt.$$
fourier-transform signal-processing
edited Nov 19 at 16:16
Adrian Keister
4,58251933
asked Nov 19 at 15:57
Prestyy
434
1
Is this a piece-wise defined function?
– Adrian Keister
Nov 19 at 16:00
@AdrianKeister yeah i had an issue with the format
– Prestyy
Nov 19 at 16:01
1
Ok, there's a better way to typeset: use thecasesenvironment.
– Adrian Keister
Nov 19 at 16:03
Which definition of the Fourier Transform and Fourier Inverse Transform are you using? (There are at least three, maybe more.)
– Adrian Keister
Nov 19 at 16:04
1
@Prestyy: I'm sorry, but that's not what I was asking. You need to type up the exact formula for the Fourier Transform that you're using, and add that to your question body. There are at least three different ways of handling the constants that multiply the integrals in the definitions.
– Adrian Keister
Nov 19 at 16:12
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$H(Omega)=begin{cases}exp(-j pi/2) ,;&Omega >0 \
exp(jpi/2) ,;&Omega<0.end{cases}$$
How can I find the inverse fourier transform of this $(h(t))$ using fourier properties?
Fourier Transform Used:
$$X(Omega)equivint_{-infty}^{infty}x(t),e^{-jOmega t},dt.$$
fourier-transform signal-processing
edited Nov 19 at 16:16
Adrian Keister
4,58251933
asked Nov 19 at 15:57
Prestyy
434
$$H(Omega)=begin{cases}exp(-j pi/2) ,;&Omega >0 \
exp(jpi/2) ,;&Omega<0.end{cases}$$
How can I find the inverse fourier transform of this $(h(t))$ using fourier properties?
Fourier Transform Used:
$$X(Omega)equivint_{-infty}^{infty}x(t),e^{-jOmega t},dt.$$
fourier-transform signal-processing
fourier-transform signal-processing
edited Nov 19 at 16:16
Adrian Keister
4,58251933
asked Nov 19 at 15:57
Prestyy
434
edited Nov 19 at 16:16
Adrian Keister
4,58251933
asked Nov 19 at 15:57
Prestyy
434
edited Nov 19 at 16:16
Adrian Keister
4,58251933
edited Nov 19 at 16:16
Adrian Keister
4,58251933
edited Nov 19 at 16:16
Adrian Keister
4,58251933
4,58251933
asked Nov 19 at 15:57
Prestyy
434
asked Nov 19 at 15:57
Prestyy
434
asked Nov 19 at 15:57
Prestyy
434
434
1
Is this a piece-wise defined function?
– Adrian Keister
Nov 19 at 16:00
@AdrianKeister yeah i had an issue with the format
– Prestyy
Nov 19 at 16:01
1
Ok, there's a better way to typeset: use thecasesenvironment.
– Adrian Keister
Nov 19 at 16:03
Which definition of the Fourier Transform and Fourier Inverse Transform are you using? (There are at least three, maybe more.)
– Adrian Keister
Nov 19 at 16:04
1
@Prestyy: I'm sorry, but that's not what I was asking. You need to type up the exact formula for the Fourier Transform that you're using, and add that to your question body. There are at least three different ways of handling the constants that multiply the integrals in the definitions.
– Adrian Keister
Nov 19 at 16:12
|
show 3 more comments
1
Is this a piece-wise defined function?
– Adrian Keister
Nov 19 at 16:00
@AdrianKeister yeah i had an issue with the format
– Prestyy
Nov 19 at 16:01
1
Ok, there's a better way to typeset: use thecasesenvironment.
– Adrian Keister
Nov 19 at 16:03
Which definition of the Fourier Transform and Fourier Inverse Transform are you using? (There are at least three, maybe more.)
– Adrian Keister
Nov 19 at 16:04
1
@Prestyy: I'm sorry, but that's not what I was asking. You need to type up the exact formula for the Fourier Transform that you're using, and add that to your question body. There are at least three different ways of handling the constants that multiply the integrals in the definitions.
– Adrian Keister
Nov 19 at 16:12
1
1
Is this a piece-wise defined function?
– Adrian Keister
Nov 19 at 16:00
Is this a piece-wise defined function?
– Adrian Keister
Nov 19 at 16:00
@AdrianKeister yeah i had an issue with the format
– Prestyy
Nov 19 at 16:01
@AdrianKeister yeah i had an issue with the format
– Prestyy
Nov 19 at 16:01
1
1
Ok, there's a better way to typeset: use the
cases environment.– Adrian Keister
Nov 19 at 16:03
Ok, there's a better way to typeset: use the
cases environment.– Adrian Keister
Nov 19 at 16:03
Which definition of the Fourier Transform and Fourier Inverse Transform are you using? (There are at least three, maybe more.)
– Adrian Keister
Nov 19 at 16:04
Which definition of the Fourier Transform and Fourier Inverse Transform are you using? (There are at least three, maybe more.)
– Adrian Keister
Nov 19 at 16:04
1
1
@Prestyy: I'm sorry, but that's not what I was asking. You need to type up the exact formula for the Fourier Transform that you're using, and add that to your question body. There are at least three different ways of handling the constants that multiply the integrals in the definitions.
– Adrian Keister
Nov 19 at 16:12
@Prestyy: I'm sorry, but that's not what I was asking. You need to type up the exact formula for the Fourier Transform that you're using, and add that to your question body. There are at least three different ways of handling the constants that multiply the integrals in the definitions.
– Adrian Keister
Nov 19 at 16:12
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
First, to simplify notation a bit, we notice that, since $e^{jpi/2}=j,$ it follows that
$$H(Omega)=-joperatorname{sgn}(Omega). $$
In looking at the wiki table of Fourier Transforms, in the non-unitary, angular frequency column (which corresponds to the definition you're using), we see that the tricky part here is to note that
$$mathcal{F}^{-1}(-jpioperatorname{sgn}(Omega))=frac1t,$$
so that
$$mathcal{F}^{-1}(-joperatorname{sgn}(Omega))=frac{1}{pi t},$$
since the Inverse Fourier Transform is linear.
However, this does not take into account the situation in which $t=0$. Technically, your $H(Omega)$ is undefined there, unless you made a typo in your definition and one of the inequalities is non-strict.
answered Nov 19 at 17:34
Adrian Keister
4,58251933
This is a well known-system, called Hilbert transformer.
– Matt L.
Nov 21 at 8:45
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
First, to simplify notation a bit, we notice that, since $e^{jpi/2}=j,$ it follows that
$$H(Omega)=-joperatorname{sgn}(Omega). $$
In looking at the wiki table of Fourier Transforms, in the non-unitary, angular frequency column (which corresponds to the definition you're using), we see that the tricky part here is to note that
$$mathcal{F}^{-1}(-jpioperatorname{sgn}(Omega))=frac1t,$$
so that
$$mathcal{F}^{-1}(-joperatorname{sgn}(Omega))=frac{1}{pi t},$$
since the Inverse Fourier Transform is linear.
However, this does not take into account the situation in which $t=0$. Technically, your $H(Omega)$ is undefined there, unless you made a typo in your definition and one of the inequalities is non-strict.
answered Nov 19 at 17:34
Adrian Keister
4,58251933
This is a well known-system, called Hilbert transformer.
– Matt L.
Nov 21 at 8:45
add a comment |
up vote
1
down vote
accepted
First, to simplify notation a bit, we notice that, since $e^{jpi/2}=j,$ it follows that
$$H(Omega)=-joperatorname{sgn}(Omega). $$
In looking at the wiki table of Fourier Transforms, in the non-unitary, angular frequency column (which corresponds to the definition you're using), we see that the tricky part here is to note that
$$mathcal{F}^{-1}(-jpioperatorname{sgn}(Omega))=frac1t,$$
so that
$$mathcal{F}^{-1}(-joperatorname{sgn}(Omega))=frac{1}{pi t},$$
since the Inverse Fourier Transform is linear.
However, this does not take into account the situation in which $t=0$. Technically, your $H(Omega)$ is undefined there, unless you made a typo in your definition and one of the inequalities is non-strict.
answered Nov 19 at 17:34
Adrian Keister
4,58251933
This is a well known-system, called Hilbert transformer.
– Matt L.
Nov 21 at 8:45
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First, to simplify notation a bit, we notice that, since $e^{jpi/2}=j,$ it follows that
$$H(Omega)=-joperatorname{sgn}(Omega). $$
In looking at the wiki table of Fourier Transforms, in the non-unitary, angular frequency column (which corresponds to the definition you're using), we see that the tricky part here is to note that
$$mathcal{F}^{-1}(-jpioperatorname{sgn}(Omega))=frac1t,$$
so that
$$mathcal{F}^{-1}(-joperatorname{sgn}(Omega))=frac{1}{pi t},$$
since the Inverse Fourier Transform is linear.
However, this does not take into account the situation in which $t=0$. Technically, your $H(Omega)$ is undefined there, unless you made a typo in your definition and one of the inequalities is non-strict.
answered Nov 19 at 17:34
Adrian Keister
4,58251933
First, to simplify notation a bit, we notice that, since $e^{jpi/2}=j,$ it follows that
$$H(Omega)=-joperatorname{sgn}(Omega). $$
In looking at the wiki table of Fourier Transforms, in the non-unitary, angular frequency column (which corresponds to the definition you're using), we see that the tricky part here is to note that
$$mathcal{F}^{-1}(-jpioperatorname{sgn}(Omega))=frac1t,$$
so that
$$mathcal{F}^{-1}(-joperatorname{sgn}(Omega))=frac{1}{pi t},$$
since the Inverse Fourier Transform is linear.
However, this does not take into account the situation in which $t=0$. Technically, your $H(Omega)$ is undefined there, unless you made a typo in your definition and one of the inequalities is non-strict.
answered Nov 19 at 17:34
Adrian Keister
4,58251933
answered Nov 19 at 17:34
Adrian Keister
4,58251933
answered Nov 19 at 17:34
Adrian Keister
4,58251933
answered Nov 19 at 17:34
Adrian Keister
4,58251933
4,58251933
This is a well known-system, called Hilbert transformer.
– Matt L.
Nov 21 at 8:45
add a comment |
This is a well known-system, called Hilbert transformer.
– Matt L.
Nov 21 at 8:45
This is a well known-system, called Hilbert transformer.
– Matt L.
Nov 21 at 8:45
This is a well known-system, called Hilbert transformer.
– Matt L.
Nov 21 at 8:45
add a comment |
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1
Is this a piece-wise defined function?
– Adrian Keister
Nov 19 at 16:00
@AdrianKeister yeah i had an issue with the format
– Prestyy
Nov 19 at 16:01
1
Ok, there's a better way to typeset: use the
casesenvironment.– Adrian Keister
Nov 19 at 16:03
Which definition of the Fourier Transform and Fourier Inverse Transform are you using? (There are at least three, maybe more.)
– Adrian Keister
Nov 19 at 16:04
1
@Prestyy: I'm sorry, but that's not what I was asking. You need to type up the exact formula for the Fourier Transform that you're using, and add that to your question body. There are at least three different ways of handling the constants that multiply the integrals in the definitions.
– Adrian Keister
Nov 19 at 16:12