DTFT of odd function?
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I am confused about the result of an odd function being purely imaginary for the DTFT of an odd function. I was under the understanding that the result of a FT was amplitude and phase as a function of frequency, with phase information being imaginary and amplitude being real. The magnitude could then be calculated from this complex number.
However, in a purely odd function the DTFT is purely imaginary. Is there then no amplitude information? How is the magnitude found?
For instance, if $x[n] = delta(n - pi/2) - delta(n - 3pi/2)$, the DTFT could be found to be $X(jw) = exp(-jw(pi/2)) - exp(-jw(3pi/2))$. Even if this is simplified via symmetry I get $2jexp(-jwpi)sin(w*pi/2)$. This is purely imaginary and I am confused how the results are interpreted.
discrete-mathematics fourier-analysis
edited Jan 4 at 22:27
Gnumbertester
668113
asked Jan 4 at 21:49
ZeariaZearia
245
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add a comment |
$begingroup$
I am confused about the result of an odd function being purely imaginary for the DTFT of an odd function. I was under the understanding that the result of a FT was amplitude and phase as a function of frequency, with phase information being imaginary and amplitude being real. The magnitude could then be calculated from this complex number.
However, in a purely odd function the DTFT is purely imaginary. Is there then no amplitude information? How is the magnitude found?
For instance, if $x[n] = delta(n - pi/2) - delta(n - 3pi/2)$, the DTFT could be found to be $X(jw) = exp(-jw(pi/2)) - exp(-jw(3pi/2))$. Even if this is simplified via symmetry I get $2jexp(-jwpi)sin(w*pi/2)$. This is purely imaginary and I am confused how the results are interpreted.
discrete-mathematics fourier-analysis
edited Jan 4 at 22:27
Gnumbertester
668113
asked Jan 4 at 21:49
ZeariaZearia
245
$endgroup$
$begingroup$
Of course there is amplitude/magnitude for a purely imaginary quantity. What is the magnitude of $j$?
$endgroup$
– Math1000
Jan 4 at 22:48
add a comment |
$begingroup$
I am confused about the result of an odd function being purely imaginary for the DTFT of an odd function. I was under the understanding that the result of a FT was amplitude and phase as a function of frequency, with phase information being imaginary and amplitude being real. The magnitude could then be calculated from this complex number.
However, in a purely odd function the DTFT is purely imaginary. Is there then no amplitude information? How is the magnitude found?
For instance, if $x[n] = delta(n - pi/2) - delta(n - 3pi/2)$, the DTFT could be found to be $X(jw) = exp(-jw(pi/2)) - exp(-jw(3pi/2))$. Even if this is simplified via symmetry I get $2jexp(-jwpi)sin(w*pi/2)$. This is purely imaginary and I am confused how the results are interpreted.
discrete-mathematics fourier-analysis
edited Jan 4 at 22:27
Gnumbertester
668113
asked Jan 4 at 21:49
ZeariaZearia
245
$endgroup$
I am confused about the result of an odd function being purely imaginary for the DTFT of an odd function. I was under the understanding that the result of a FT was amplitude and phase as a function of frequency, with phase information being imaginary and amplitude being real. The magnitude could then be calculated from this complex number.
However, in a purely odd function the DTFT is purely imaginary. Is there then no amplitude information? How is the magnitude found?
For instance, if $x[n] = delta(n - pi/2) - delta(n - 3pi/2)$, the DTFT could be found to be $X(jw) = exp(-jw(pi/2)) - exp(-jw(3pi/2))$. Even if this is simplified via symmetry I get $2jexp(-jwpi)sin(w*pi/2)$. This is purely imaginary and I am confused how the results are interpreted.
discrete-mathematics fourier-analysis
discrete-mathematics fourier-analysis
edited Jan 4 at 22:27
Gnumbertester
668113
asked Jan 4 at 21:49
ZeariaZearia
245
edited Jan 4 at 22:27
Gnumbertester
668113
asked Jan 4 at 21:49
ZeariaZearia
245
edited Jan 4 at 22:27
Gnumbertester
668113
edited Jan 4 at 22:27
Gnumbertester
668113
edited Jan 4 at 22:27
Gnumbertester
668113
668113
asked Jan 4 at 21:49
ZeariaZearia
245
asked Jan 4 at 21:49
ZeariaZearia
245
asked Jan 4 at 21:49
ZeariaZearia
245
245
$begingroup$
Of course there is amplitude/magnitude for a purely imaginary quantity. What is the magnitude of $j$?
$endgroup$
– Math1000
Jan 4 at 22:48
add a comment |
$begingroup$
Of course there is amplitude/magnitude for a purely imaginary quantity. What is the magnitude of $j$?
$endgroup$
– Math1000
Jan 4 at 22:48
$begingroup$
Of course there is amplitude/magnitude for a purely imaginary quantity. What is the magnitude of $j$?
$endgroup$
– Math1000
Jan 4 at 22:48
$begingroup$
Of course there is amplitude/magnitude for a purely imaginary quantity. What is the magnitude of $j$?
$endgroup$
– Math1000
Jan 4 at 22:48
add a comment |
1 Answer
1
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There are two ways to represent the Fourier transform - in terms of sines and cosines, or in terms of complex exponentials with positive and negative weights on the exponent. This is based on two choices of basis for the space of solutions to $y''+y=0$; either $y=sin t$ and $y=cos t$ or $y=e^{it}$ and $y=e^{-it}$.
The two versions are related by a simple linear transformation. A real-valued odd function will have a transform which is odd and purely imaginary in the complex form, or that is pure sine (and pure real) in the real form. A real-valued even function will have a transform which is even and pure real in the complex form, or real and pure cosine in the real form.
In your "amplitude/phase" framework, the amplitude is the absolute value of that complex number $z$ and the phase is its angle $arg z$.
answered Jan 4 at 22:09
jmerryjmerry
12.1k1628
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1 Answer
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1 Answer
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active
oldest
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active
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active
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$begingroup$
There are two ways to represent the Fourier transform - in terms of sines and cosines, or in terms of complex exponentials with positive and negative weights on the exponent. This is based on two choices of basis for the space of solutions to $y''+y=0$; either $y=sin t$ and $y=cos t$ or $y=e^{it}$ and $y=e^{-it}$.
The two versions are related by a simple linear transformation. A real-valued odd function will have a transform which is odd and purely imaginary in the complex form, or that is pure sine (and pure real) in the real form. A real-valued even function will have a transform which is even and pure real in the complex form, or real and pure cosine in the real form.
In your "amplitude/phase" framework, the amplitude is the absolute value of that complex number $z$ and the phase is its angle $arg z$.
answered Jan 4 at 22:09
jmerryjmerry
12.1k1628
$endgroup$
add a comment |
$begingroup$
There are two ways to represent the Fourier transform - in terms of sines and cosines, or in terms of complex exponentials with positive and negative weights on the exponent. This is based on two choices of basis for the space of solutions to $y''+y=0$; either $y=sin t$ and $y=cos t$ or $y=e^{it}$ and $y=e^{-it}$.
The two versions are related by a simple linear transformation. A real-valued odd function will have a transform which is odd and purely imaginary in the complex form, or that is pure sine (and pure real) in the real form. A real-valued even function will have a transform which is even and pure real in the complex form, or real and pure cosine in the real form.
In your "amplitude/phase" framework, the amplitude is the absolute value of that complex number $z$ and the phase is its angle $arg z$.
answered Jan 4 at 22:09
jmerryjmerry
12.1k1628
$endgroup$
add a comment |
$begingroup$
There are two ways to represent the Fourier transform - in terms of sines and cosines, or in terms of complex exponentials with positive and negative weights on the exponent. This is based on two choices of basis for the space of solutions to $y''+y=0$; either $y=sin t$ and $y=cos t$ or $y=e^{it}$ and $y=e^{-it}$.
The two versions are related by a simple linear transformation. A real-valued odd function will have a transform which is odd and purely imaginary in the complex form, or that is pure sine (and pure real) in the real form. A real-valued even function will have a transform which is even and pure real in the complex form, or real and pure cosine in the real form.
In your "amplitude/phase" framework, the amplitude is the absolute value of that complex number $z$ and the phase is its angle $arg z$.
answered Jan 4 at 22:09
jmerryjmerry
12.1k1628
$endgroup$
There are two ways to represent the Fourier transform - in terms of sines and cosines, or in terms of complex exponentials with positive and negative weights on the exponent. This is based on two choices of basis for the space of solutions to $y''+y=0$; either $y=sin t$ and $y=cos t$ or $y=e^{it}$ and $y=e^{-it}$.
The two versions are related by a simple linear transformation. A real-valued odd function will have a transform which is odd and purely imaginary in the complex form, or that is pure sine (and pure real) in the real form. A real-valued even function will have a transform which is even and pure real in the complex form, or real and pure cosine in the real form.
In your "amplitude/phase" framework, the amplitude is the absolute value of that complex number $z$ and the phase is its angle $arg z$.
answered Jan 4 at 22:09
jmerryjmerry
12.1k1628
answered Jan 4 at 22:09
jmerryjmerry
12.1k1628
answered Jan 4 at 22:09
jmerryjmerry
12.1k1628
answered Jan 4 at 22:09
jmerryjmerry
12.1k1628
12.1k1628
add a comment |
add a comment |
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$begingroup$
Of course there is amplitude/magnitude for a purely imaginary quantity. What is the magnitude of $j$?
$endgroup$
– Math1000
Jan 4 at 22:48