I am evaluating the fourier transform of a function + a constant:...
0
$begingroup$
I am evaluating the fourier transform of a function plus a constant $c$:
$$frac{1}{2pi}int_{-infty}^{+infty}(f(x)+c)e^{-ikx}dx.$$
As a result, I should get the fourier transform of the function $f(x)$ plus a dirac delta:
$$frac{1}{2pi}int_{-infty}^{+infty}(f(x)+c)e^{-ikx}dx=hat{f}(k)+cdelta(k).$$
However, I cannot understand whether I am making some mistake or not and the meaning of such a result.
fourier-transform distribution-theory dirac-delta
edited Jan 13 at 23:19
Qmechanic
4,96711855
asked Jan 8 at 12:07
ARFARF
31
$endgroup$
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show 2 more comments
0
$begingroup$
I am evaluating the fourier transform of a function plus a constant $c$:
$$frac{1}{2pi}int_{-infty}^{+infty}(f(x)+c)e^{-ikx}dx.$$
As a result, I should get the fourier transform of the function $f(x)$ plus a dirac delta:
$$frac{1}{2pi}int_{-infty}^{+infty}(f(x)+c)e^{-ikx}dx=hat{f}(k)+cdelta(k).$$
However, I cannot understand whether I am making some mistake or not and the meaning of such a result.
fourier-transform distribution-theory dirac-delta
edited Jan 13 at 23:19
Qmechanic
4,96711855
asked Jan 8 at 12:07
ARFARF
31
$endgroup$
$begingroup$
the integral $int_{-infty}^infty ce^{-ikt}, dt$ doesnt exists. And the Dirac delta is not a function, so Im not sure what you mean by $delta(k)$
$endgroup$
– Masacroso
Jan 8 at 12:11
$begingroup$
Your result is almost correct, the Fourier transform of $1$ is $delta$ with your normalization, so you should have $c delta(k)$.
$endgroup$
– Botond
Jan 8 at 12:15
$begingroup$
@Botond I cannot understand the physical meaning. The Fourier transform can be associated to an energy spectrum. In this case do I have an infinite energy at zero frequency?
$endgroup$
– ARF
Jan 8 at 13:13
$begingroup$
I can't really say anything about it without context. But why don't you ask it on PSE?
$endgroup$
– Botond
Jan 8 at 13:35
$begingroup$
I can only assume you are talking about the physical meaning as I do not see a question otherwise. In that case superimposing $cos(omega_0 t)$ with low $omega_0$ to a signal $f(t)$ will result in a disturbed signal on top. In the frequency domain you will see spikes at $pm omega_0$. Think of a low constant frequency bass sound on top of a melody. Now lot $omega_0 rightarrow 0$.
$endgroup$
– Diger
Jan 8 at 14:32
|
show 2 more comments
0
0
0
$begingroup$
I am evaluating the fourier transform of a function plus a constant $c$:
$$frac{1}{2pi}int_{-infty}^{+infty}(f(x)+c)e^{-ikx}dx.$$
As a result, I should get the fourier transform of the function $f(x)$ plus a dirac delta:
$$frac{1}{2pi}int_{-infty}^{+infty}(f(x)+c)e^{-ikx}dx=hat{f}(k)+cdelta(k).$$
However, I cannot understand whether I am making some mistake or not and the meaning of such a result.
fourier-transform distribution-theory dirac-delta
edited Jan 13 at 23:19
Qmechanic
4,96711855
asked Jan 8 at 12:07
ARFARF
31
$endgroup$
I am evaluating the fourier transform of a function plus a constant $c$:
$$frac{1}{2pi}int_{-infty}^{+infty}(f(x)+c)e^{-ikx}dx.$$
As a result, I should get the fourier transform of the function $f(x)$ plus a dirac delta:
$$frac{1}{2pi}int_{-infty}^{+infty}(f(x)+c)e^{-ikx}dx=hat{f}(k)+cdelta(k).$$
However, I cannot understand whether I am making some mistake or not and the meaning of such a result.
fourier-transform distribution-theory dirac-delta
fourier-transform distribution-theory dirac-delta
edited Jan 13 at 23:19
Qmechanic
4,96711855
asked Jan 8 at 12:07
ARFARF
31
edited Jan 13 at 23:19
Qmechanic
4,96711855
asked Jan 8 at 12:07
ARFARF
31
edited Jan 13 at 23:19
Qmechanic
4,96711855
edited Jan 13 at 23:19
Qmechanic
4,96711855
edited Jan 13 at 23:19
Qmechanic
4,96711855
4,96711855
asked Jan 8 at 12:07
ARFARF
31
asked Jan 8 at 12:07
ARFARF
31
asked Jan 8 at 12:07
ARFARF
31
31
$begingroup$
the integral $int_{-infty}^infty ce^{-ikt}, dt$ doesnt exists. And the Dirac delta is not a function, so Im not sure what you mean by $delta(k)$
$endgroup$
– Masacroso
Jan 8 at 12:11
$begingroup$
Your result is almost correct, the Fourier transform of $1$ is $delta$ with your normalization, so you should have $c delta(k)$.
$endgroup$
– Botond
Jan 8 at 12:15
$begingroup$
@Botond I cannot understand the physical meaning. The Fourier transform can be associated to an energy spectrum. In this case do I have an infinite energy at zero frequency?
$endgroup$
– ARF
Jan 8 at 13:13
$begingroup$
I can't really say anything about it without context. But why don't you ask it on PSE?
$endgroup$
– Botond
Jan 8 at 13:35
$begingroup$
I can only assume you are talking about the physical meaning as I do not see a question otherwise. In that case superimposing $cos(omega_0 t)$ with low $omega_0$ to a signal $f(t)$ will result in a disturbed signal on top. In the frequency domain you will see spikes at $pm omega_0$. Think of a low constant frequency bass sound on top of a melody. Now lot $omega_0 rightarrow 0$.
$endgroup$
– Diger
Jan 8 at 14:32
|
show 2 more comments
$begingroup$
the integral $int_{-infty}^infty ce^{-ikt}, dt$ doesnt exists. And the Dirac delta is not a function, so Im not sure what you mean by $delta(k)$
$endgroup$
– Masacroso
Jan 8 at 12:11
$begingroup$
Your result is almost correct, the Fourier transform of $1$ is $delta$ with your normalization, so you should have $c delta(k)$.
$endgroup$
– Botond
Jan 8 at 12:15
$begingroup$
@Botond I cannot understand the physical meaning. The Fourier transform can be associated to an energy spectrum. In this case do I have an infinite energy at zero frequency?
$endgroup$
– ARF
Jan 8 at 13:13
$begingroup$
I can't really say anything about it without context. But why don't you ask it on PSE?
$endgroup$
– Botond
Jan 8 at 13:35
$begingroup$
I can only assume you are talking about the physical meaning as I do not see a question otherwise. In that case superimposing $cos(omega_0 t)$ with low $omega_0$ to a signal $f(t)$ will result in a disturbed signal on top. In the frequency domain you will see spikes at $pm omega_0$. Think of a low constant frequency bass sound on top of a melody. Now lot $omega_0 rightarrow 0$.
$endgroup$
– Diger
Jan 8 at 14:32
$begingroup$
the integral $int_{-infty}^infty ce^{-ikt}, dt$ doesnt exists. And the Dirac delta is not a function, so Im not sure what you mean by $delta(k)$
$endgroup$
– Masacroso
Jan 8 at 12:11
$begingroup$
the integral $int_{-infty}^infty ce^{-ikt}, dt$ doesnt exists. And the Dirac delta is not a function, so Im not sure what you mean by $delta(k)$
$endgroup$
– Masacroso
Jan 8 at 12:11
$begingroup$
Your result is almost correct, the Fourier transform of $1$ is $delta$ with your normalization, so you should have $c delta(k)$.
$endgroup$
– Botond
Jan 8 at 12:15
$begingroup$
Your result is almost correct, the Fourier transform of $1$ is $delta$ with your normalization, so you should have $c delta(k)$.
$endgroup$
– Botond
Jan 8 at 12:15
$begingroup$
@Botond I cannot understand the physical meaning. The Fourier transform can be associated to an energy spectrum. In this case do I have an infinite energy at zero frequency?
$endgroup$
– ARF
Jan 8 at 13:13
$begingroup$
@Botond I cannot understand the physical meaning. The Fourier transform can be associated to an energy spectrum. In this case do I have an infinite energy at zero frequency?
$endgroup$
– ARF
Jan 8 at 13:13
$begingroup$
I can't really say anything about it without context. But why don't you ask it on PSE?
$endgroup$
– Botond
Jan 8 at 13:35
$begingroup$
I can't really say anything about it without context. But why don't you ask it on PSE?
$endgroup$
– Botond
Jan 8 at 13:35
$begingroup$
I can only assume you are talking about the physical meaning as I do not see a question otherwise. In that case superimposing $cos(omega_0 t)$ with low $omega_0$ to a signal $f(t)$ will result in a disturbed signal on top. In the frequency domain you will see spikes at $pm omega_0$. Think of a low constant frequency bass sound on top of a melody. Now lot $omega_0 rightarrow 0$.
$endgroup$
– Diger
Jan 8 at 14:32
$begingroup$
I can only assume you are talking about the physical meaning as I do not see a question otherwise. In that case superimposing $cos(omega_0 t)$ with low $omega_0$ to a signal $f(t)$ will result in a disturbed signal on top. In the frequency domain you will see spikes at $pm omega_0$. Think of a low constant frequency bass sound on top of a melody. Now lot $omega_0 rightarrow 0$.
$endgroup$
– Diger
Jan 8 at 14:32
|
show 2 more comments
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$begingroup$
the integral $int_{-infty}^infty ce^{-ikt}, dt$ doesnt exists. And the Dirac delta is not a function, so Im not sure what you mean by $delta(k)$
$endgroup$
– Masacroso
Jan 8 at 12:11
$begingroup$
Your result is almost correct, the Fourier transform of $1$ is $delta$ with your normalization, so you should have $c delta(k)$.
$endgroup$
– Botond
Jan 8 at 12:15
$begingroup$
@Botond I cannot understand the physical meaning. The Fourier transform can be associated to an energy spectrum. In this case do I have an infinite energy at zero frequency?
$endgroup$
– ARF
Jan 8 at 13:13
$begingroup$
I can't really say anything about it without context. But why don't you ask it on PSE?
$endgroup$
– Botond
Jan 8 at 13:35
$begingroup$
I can only assume you are talking about the physical meaning as I do not see a question otherwise. In that case superimposing $cos(omega_0 t)$ with low $omega_0$ to a signal $f(t)$ will result in a disturbed signal on top. In the frequency domain you will see spikes at $pm omega_0$. Think of a low constant frequency bass sound on top of a melody. Now lot $omega_0 rightarrow 0$.
$endgroup$
– Diger
Jan 8 at 14:32