Mixing
Fourier transform of general complex signal
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I am trying to understand a sinusoid and its fourier transform.
Given an example sinusoid
$Ze^{iwt}$, how do i calculate the fourier transform of it?
How do i even represent this?
In euler form?
like $Zcost(wt) + iZsin(wt)$ ?
Been looking at online materials but still cannot understand
fourier-series fourier-transform
asked Jan 23 at 2:56
acemineraceminer
230111
$endgroup$
add a comment |
$begingroup$
I am trying to understand a sinusoid and its fourier transform.
Given an example sinusoid
$Ze^{iwt}$, how do i calculate the fourier transform of it?
How do i even represent this?
In euler form?
like $Zcost(wt) + iZsin(wt)$ ?
Been looking at online materials but still cannot understand
fourier-series fourier-transform
asked Jan 23 at 2:56
acemineraceminer
230111
$endgroup$
$begingroup$
What is it that you don't understand, exactly? Any online resource would tell you, at least from the engineering/physics perspective where you do some hand-waving, that it's a shifted impulse (scaled by some constant).
$endgroup$
– Metric
Jan 23 at 3:33
$begingroup$
@Metric like do i integrate the euler form?
$endgroup$
– aceminer
Jan 23 at 3:35
$begingroup$
No, you integrate it directly, but this is probably from an engineering/physics class, so there's going to be some handwaving.
$endgroup$
– Metric
Jan 23 at 3:38
$begingroup$
To offer some assitance towards the classical approach of finding it, I suggest you look at the inverse fourier transform of the dirac delta.
$endgroup$
– Metric
Jan 23 at 3:39
add a comment |
$begingroup$
I am trying to understand a sinusoid and its fourier transform.
Given an example sinusoid
$Ze^{iwt}$, how do i calculate the fourier transform of it?
How do i even represent this?
In euler form?
like $Zcost(wt) + iZsin(wt)$ ?
Been looking at online materials but still cannot understand
fourier-series fourier-transform
asked Jan 23 at 2:56
acemineraceminer
230111
$endgroup$
I am trying to understand a sinusoid and its fourier transform.
Given an example sinusoid
$Ze^{iwt}$, how do i calculate the fourier transform of it?
How do i even represent this?
In euler form?
like $Zcost(wt) + iZsin(wt)$ ?
Been looking at online materials but still cannot understand
fourier-series fourier-transform
fourier-series fourier-transform
asked Jan 23 at 2:56
acemineraceminer
230111
asked Jan 23 at 2:56
acemineraceminer
230111
asked Jan 23 at 2:56
acemineraceminer
230111
asked Jan 23 at 2:56
acemineraceminer
230111
asked Jan 23 at 2:56
acemineraceminer
230111
230111
$begingroup$
What is it that you don't understand, exactly? Any online resource would tell you, at least from the engineering/physics perspective where you do some hand-waving, that it's a shifted impulse (scaled by some constant).
$endgroup$
– Metric
Jan 23 at 3:33
$begingroup$
@Metric like do i integrate the euler form?
$endgroup$
– aceminer
Jan 23 at 3:35
$begingroup$
No, you integrate it directly, but this is probably from an engineering/physics class, so there's going to be some handwaving.
$endgroup$
– Metric
Jan 23 at 3:38
$begingroup$
To offer some assitance towards the classical approach of finding it, I suggest you look at the inverse fourier transform of the dirac delta.
$endgroup$
– Metric
Jan 23 at 3:39
add a comment |
$begingroup$
What is it that you don't understand, exactly? Any online resource would tell you, at least from the engineering/physics perspective where you do some hand-waving, that it's a shifted impulse (scaled by some constant).
$endgroup$
– Metric
Jan 23 at 3:33
$begingroup$
@Metric like do i integrate the euler form?
$endgroup$
– aceminer
Jan 23 at 3:35
$begingroup$
No, you integrate it directly, but this is probably from an engineering/physics class, so there's going to be some handwaving.
$endgroup$
– Metric
Jan 23 at 3:38
$begingroup$
To offer some assitance towards the classical approach of finding it, I suggest you look at the inverse fourier transform of the dirac delta.
$endgroup$
– Metric
Jan 23 at 3:39
$begingroup$
What is it that you don't understand, exactly? Any online resource would tell you, at least from the engineering/physics perspective where you do some hand-waving, that it's a shifted impulse (scaled by some constant).
$endgroup$
– Metric
Jan 23 at 3:33
$begingroup$
What is it that you don't understand, exactly? Any online resource would tell you, at least from the engineering/physics perspective where you do some hand-waving, that it's a shifted impulse (scaled by some constant).
$endgroup$
– Metric
Jan 23 at 3:33
$begingroup$
@Metric like do i integrate the euler form?
$endgroup$
– aceminer
Jan 23 at 3:35
$begingroup$
@Metric like do i integrate the euler form?
$endgroup$
– aceminer
Jan 23 at 3:35
$begingroup$
No, you integrate it directly, but this is probably from an engineering/physics class, so there's going to be some handwaving.
$endgroup$
– Metric
Jan 23 at 3:38
$begingroup$
No, you integrate it directly, but this is probably from an engineering/physics class, so there's going to be some handwaving.
$endgroup$
– Metric
Jan 23 at 3:38
$begingroup$
To offer some assitance towards the classical approach of finding it, I suggest you look at the inverse fourier transform of the dirac delta.
$endgroup$
– Metric
Jan 23 at 3:39
$begingroup$
To offer some assitance towards the classical approach of finding it, I suggest you look at the inverse fourier transform of the dirac delta.
$endgroup$
– Metric
Jan 23 at 3:39
add a comment |
1 Answer
1
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oldest
votes
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Since I think I know what you're asking for, I'm going to hand-wave a little bit here.
Suppose the fourier transform $F$ of a function $f$ is defined as
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} dt$$
and the inverse-fourier transform $F^{-1}$ of a function $g$ is defined as
$$F^{-1}(g)(t) =frac{1}{2pi}int_{-infty}^infty g(omega) e^{i omega t} d omega$$
Since
$$ F(delta)(omega) = int_{-infty}^infty delta(t)e^{-iomega t} d t = int_{-infty}^infty delta(t)e^{0} d t = int_{-infty}^infty delta(t) d t = 1$$
we have that
$$F^{-1}(1)(t) = frac{1}{2pi}int_{-infty}^infty e^{iomega t} domega = delta(t) $$
That is,
$$ int_{-infty}^infty e^{i omega t} d omega = 2 pi delta(t) tag 1$$
Now, if your function is $f(t) = ze^{i omega_0 t}$, then using $(1)$ we have that its fourier transform is
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} d t = zint_{-infty}^infty e^{i (omega_0-omega) t} dt = 2 pi zdelta(omega_0 - omega) = 2 pi zdelta(omega - omega_0)$$
answered Jan 23 at 3:57
MetricMetric
1,25159
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since I think I know what you're asking for, I'm going to hand-wave a little bit here.
Suppose the fourier transform $F$ of a function $f$ is defined as
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} dt$$
and the inverse-fourier transform $F^{-1}$ of a function $g$ is defined as
$$F^{-1}(g)(t) =frac{1}{2pi}int_{-infty}^infty g(omega) e^{i omega t} d omega$$
Since
$$ F(delta)(omega) = int_{-infty}^infty delta(t)e^{-iomega t} d t = int_{-infty}^infty delta(t)e^{0} d t = int_{-infty}^infty delta(t) d t = 1$$
we have that
$$F^{-1}(1)(t) = frac{1}{2pi}int_{-infty}^infty e^{iomega t} domega = delta(t) $$
That is,
$$ int_{-infty}^infty e^{i omega t} d omega = 2 pi delta(t) tag 1$$
Now, if your function is $f(t) = ze^{i omega_0 t}$, then using $(1)$ we have that its fourier transform is
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} d t = zint_{-infty}^infty e^{i (omega_0-omega) t} dt = 2 pi zdelta(omega_0 - omega) = 2 pi zdelta(omega - omega_0)$$
answered Jan 23 at 3:57
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
Since I think I know what you're asking for, I'm going to hand-wave a little bit here.
Suppose the fourier transform $F$ of a function $f$ is defined as
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} dt$$
and the inverse-fourier transform $F^{-1}$ of a function $g$ is defined as
$$F^{-1}(g)(t) =frac{1}{2pi}int_{-infty}^infty g(omega) e^{i omega t} d omega$$
Since
$$ F(delta)(omega) = int_{-infty}^infty delta(t)e^{-iomega t} d t = int_{-infty}^infty delta(t)e^{0} d t = int_{-infty}^infty delta(t) d t = 1$$
we have that
$$F^{-1}(1)(t) = frac{1}{2pi}int_{-infty}^infty e^{iomega t} domega = delta(t) $$
That is,
$$ int_{-infty}^infty e^{i omega t} d omega = 2 pi delta(t) tag 1$$
Now, if your function is $f(t) = ze^{i omega_0 t}$, then using $(1)$ we have that its fourier transform is
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} d t = zint_{-infty}^infty e^{i (omega_0-omega) t} dt = 2 pi zdelta(omega_0 - omega) = 2 pi zdelta(omega - omega_0)$$
answered Jan 23 at 3:57
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
Since I think I know what you're asking for, I'm going to hand-wave a little bit here.
Suppose the fourier transform $F$ of a function $f$ is defined as
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} dt$$
and the inverse-fourier transform $F^{-1}$ of a function $g$ is defined as
$$F^{-1}(g)(t) =frac{1}{2pi}int_{-infty}^infty g(omega) e^{i omega t} d omega$$
Since
$$ F(delta)(omega) = int_{-infty}^infty delta(t)e^{-iomega t} d t = int_{-infty}^infty delta(t)e^{0} d t = int_{-infty}^infty delta(t) d t = 1$$
we have that
$$F^{-1}(1)(t) = frac{1}{2pi}int_{-infty}^infty e^{iomega t} domega = delta(t) $$
That is,
$$ int_{-infty}^infty e^{i omega t} d omega = 2 pi delta(t) tag 1$$
Now, if your function is $f(t) = ze^{i omega_0 t}$, then using $(1)$ we have that its fourier transform is
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} d t = zint_{-infty}^infty e^{i (omega_0-omega) t} dt = 2 pi zdelta(omega_0 - omega) = 2 pi zdelta(omega - omega_0)$$
answered Jan 23 at 3:57
MetricMetric
1,25159
$endgroup$
Since I think I know what you're asking for, I'm going to hand-wave a little bit here.
Suppose the fourier transform $F$ of a function $f$ is defined as
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} dt$$
and the inverse-fourier transform $F^{-1}$ of a function $g$ is defined as
$$F^{-1}(g)(t) =frac{1}{2pi}int_{-infty}^infty g(omega) e^{i omega t} d omega$$
Since
$$ F(delta)(omega) = int_{-infty}^infty delta(t)e^{-iomega t} d t = int_{-infty}^infty delta(t)e^{0} d t = int_{-infty}^infty delta(t) d t = 1$$
we have that
$$F^{-1}(1)(t) = frac{1}{2pi}int_{-infty}^infty e^{iomega t} domega = delta(t) $$
That is,
$$ int_{-infty}^infty e^{i omega t} d omega = 2 pi delta(t) tag 1$$
Now, if your function is $f(t) = ze^{i omega_0 t}$, then using $(1)$ we have that its fourier transform is
$$F(f)(omega) = int_{-infty}^infty f(t) e^{-i omega t} d t = zint_{-infty}^infty e^{i (omega_0-omega) t} dt = 2 pi zdelta(omega_0 - omega) = 2 pi zdelta(omega - omega_0)$$
answered Jan 23 at 3:57
MetricMetric
1,25159
edited Jan 24 at 0:18
answered Jan 23 at 3:57
MetricMetric
1,25159
answered Jan 23 at 3:57
MetricMetric
1,25159
answered Jan 23 at 3:57
MetricMetric
1,25159
1,25159
add a comment |
add a comment |
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$begingroup$
What is it that you don't understand, exactly? Any online resource would tell you, at least from the engineering/physics perspective where you do some hand-waving, that it's a shifted impulse (scaled by some constant).
$endgroup$
– Metric
Jan 23 at 3:33
$begingroup$
@Metric like do i integrate the euler form?
$endgroup$
– aceminer
Jan 23 at 3:35
$begingroup$
No, you integrate it directly, but this is probably from an engineering/physics class, so there's going to be some handwaving.
$endgroup$
– Metric
Jan 23 at 3:38
$begingroup$
To offer some assitance towards the classical approach of finding it, I suggest you look at the inverse fourier transform of the dirac delta.
$endgroup$
– Metric
Jan 23 at 3:39