Find Fourier transform of $delta(e^{ipi t}-i)$












1












$begingroup$


Find Fourier transform of $delta(e^{ipi t}-i)$
I used distribution $langledelta(t),phi(t)rangle=phi(0)$ but it didn't work.






fourier-transform







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you mean inverse fourier transform maybe ?
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 19:58










  • $begingroup$
    @AhmadBazzi i wrote wrong, it was edited
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 20:08










  • $begingroup$
    i have left you a detailed answer .. please check below.
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 20:37
















1












$begingroup$


Find Fourier transform of $delta(e^{ipi t}-i)$
I used distribution $langledelta(t),phi(t)rangle=phi(0)$ but it didn't work.






fourier-transform







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you mean inverse fourier transform maybe ?
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 19:58










  • $begingroup$
    @AhmadBazzi i wrote wrong, it was edited
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 20:08










  • $begingroup$
    i have left you a detailed answer .. please check below.
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 20:37














1












1








1





$begingroup$


Find Fourier transform of $delta(e^{ipi t}-i)$
I used distribution $langledelta(t),phi(t)rangle=phi(0)$ but it didn't work.






fourier-transform







share|cite|improve this question











$endgroup$




Find Fourier transform of $delta(e^{ipi t}-i)$
I used distribution $langledelta(t),phi(t)rangle=phi(0)$ but it didn't work.






fourier-transform




fourier-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 21:07







Thien Hoang Dang

















asked Dec 18 '18 at 19:46









Thien Hoang DangThien Hoang Dang

64




64












  • $begingroup$
    do you mean inverse fourier transform maybe ?
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 19:58










  • $begingroup$
    @AhmadBazzi i wrote wrong, it was edited
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 20:08










  • $begingroup$
    i have left you a detailed answer .. please check below.
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 20:37


















  • $begingroup$
    do you mean inverse fourier transform maybe ?
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 19:58










  • $begingroup$
    @AhmadBazzi i wrote wrong, it was edited
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 20:08










  • $begingroup$
    i have left you a detailed answer .. please check below.
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 20:37
















$begingroup$
do you mean inverse fourier transform maybe ?
$endgroup$
– Ahmad Bazzi
Dec 18 '18 at 19:58




$begingroup$
do you mean inverse fourier transform maybe ?
$endgroup$
– Ahmad Bazzi
Dec 18 '18 at 19:58












$begingroup$
@AhmadBazzi i wrote wrong, it was edited
$endgroup$
– Thien Hoang Dang
Dec 18 '18 at 20:08




$begingroup$
@AhmadBazzi i wrote wrong, it was edited
$endgroup$
– Thien Hoang Dang
Dec 18 '18 at 20:08












$begingroup$
i have left you a detailed answer .. please check below.
$endgroup$
– Ahmad Bazzi
Dec 18 '18 at 20:37




$begingroup$
i have left you a detailed answer .. please check below.
$endgroup$
– Ahmad Bazzi
Dec 18 '18 at 20:37










1 Answer
1






active

oldest

votes


















1












$begingroup$

Note: The OP was originally to find the inverse Fourier Transform of $delta(e^{ipi f}-i)$



Credits: Thanks @Skip for the comment below



Notice that the argument is zero only when $e^{i pi f} = i$, or in other words
begin{equation}
pi f = frac{pi}{2} + 2k pi qquad k in mathbb{Z}
end{equation}
hence for $f = frac{1}{2 } + 2k$ where $k in mathbb{Z}$, the Dirac function peaks at $1$, else it is $0$. This results in a "Dirac comb" or "train of impulses" as
begin{equation}
delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} delta(f - f_k)
end{equation}
where $f_k = frac{1}{2 } + 2k$. Applying inverse Fourier transforms on both sides, we have
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} mathcal{F}^{-1}delta(f - f_k) tag{1}
end{equation}



Using the Fourier transform table, we know that
begin{equation}
mathcal{F}(e^{j 2 pi f_0 t})
=
2 pi delta(f - f_0)
end{equation}
which means that
begin{equation}
mathcal{F}^{-1}(delta(f - f_0)) = frac{1}{2pi}e^{j 2 pi f_0 t}tag{2}
end{equation}
Using equation $(2)$ in $(1)$ we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi f_k t}
end{equation}
Replacing $f_k =frac{1}{2 } + 2k$, we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi (frac{1}{2 } + 2k) t}
= frac{1}{2pi}e^{jpi t}sum_{k = -infty}^{infty} e^{4 k pi t }
end{equation}






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I believe you meant $pi f = pi/2 + 2pi k$.
    $endgroup$
    – AEngineer
    Dec 18 '18 at 20:29






  • 1




    $begingroup$
    yes you are right .. thanks for drawing my attention @Skip
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 20:30










  • $begingroup$
    nice solution @AhmadBazzi, but i want to find Fourier transform of $delta(e^{ipi t}-1)$
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:10












  • $begingroup$
    but you just edited right now @ThienHoangDang
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 21:11










  • $begingroup$
    sorry for wrong please bro @AhmadBazzi
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:12











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1 Answer
1






active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

Note: The OP was originally to find the inverse Fourier Transform of $delta(e^{ipi f}-i)$



Credits: Thanks @Skip for the comment below



Notice that the argument is zero only when $e^{i pi f} = i$, or in other words
begin{equation}
pi f = frac{pi}{2} + 2k pi qquad k in mathbb{Z}
end{equation}
hence for $f = frac{1}{2 } + 2k$ where $k in mathbb{Z}$, the Dirac function peaks at $1$, else it is $0$. This results in a "Dirac comb" or "train of impulses" as
begin{equation}
delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} delta(f - f_k)
end{equation}
where $f_k = frac{1}{2 } + 2k$. Applying inverse Fourier transforms on both sides, we have
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} mathcal{F}^{-1}delta(f - f_k) tag{1}
end{equation}



Using the Fourier transform table, we know that
begin{equation}
mathcal{F}(e^{j 2 pi f_0 t})
=
2 pi delta(f - f_0)
end{equation}
which means that
begin{equation}
mathcal{F}^{-1}(delta(f - f_0)) = frac{1}{2pi}e^{j 2 pi f_0 t}tag{2}
end{equation}
Using equation $(2)$ in $(1)$ we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi f_k t}
end{equation}
Replacing $f_k =frac{1}{2 } + 2k$, we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi (frac{1}{2 } + 2k) t}
= frac{1}{2pi}e^{jpi t}sum_{k = -infty}^{infty} e^{4 k pi t }
end{equation}






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I believe you meant $pi f = pi/2 + 2pi k$.
    $endgroup$
    – AEngineer
    Dec 18 '18 at 20:29






  • 1




    $begingroup$
    yes you are right .. thanks for drawing my attention @Skip
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 20:30










  • $begingroup$
    nice solution @AhmadBazzi, but i want to find Fourier transform of $delta(e^{ipi t}-1)$
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:10












  • $begingroup$
    but you just edited right now @ThienHoangDang
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 21:11










  • $begingroup$
    sorry for wrong please bro @AhmadBazzi
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:12
















1












$begingroup$

Note: The OP was originally to find the inverse Fourier Transform of $delta(e^{ipi f}-i)$



Credits: Thanks @Skip for the comment below



Notice that the argument is zero only when $e^{i pi f} = i$, or in other words
begin{equation}
pi f = frac{pi}{2} + 2k pi qquad k in mathbb{Z}
end{equation}
hence for $f = frac{1}{2 } + 2k$ where $k in mathbb{Z}$, the Dirac function peaks at $1$, else it is $0$. This results in a "Dirac comb" or "train of impulses" as
begin{equation}
delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} delta(f - f_k)
end{equation}
where $f_k = frac{1}{2 } + 2k$. Applying inverse Fourier transforms on both sides, we have
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} mathcal{F}^{-1}delta(f - f_k) tag{1}
end{equation}



Using the Fourier transform table, we know that
begin{equation}
mathcal{F}(e^{j 2 pi f_0 t})
=
2 pi delta(f - f_0)
end{equation}
which means that
begin{equation}
mathcal{F}^{-1}(delta(f - f_0)) = frac{1}{2pi}e^{j 2 pi f_0 t}tag{2}
end{equation}
Using equation $(2)$ in $(1)$ we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi f_k t}
end{equation}
Replacing $f_k =frac{1}{2 } + 2k$, we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi (frac{1}{2 } + 2k) t}
= frac{1}{2pi}e^{jpi t}sum_{k = -infty}^{infty} e^{4 k pi t }
end{equation}






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I believe you meant $pi f = pi/2 + 2pi k$.
    $endgroup$
    – AEngineer
    Dec 18 '18 at 20:29






  • 1




    $begingroup$
    yes you are right .. thanks for drawing my attention @Skip
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 20:30










  • $begingroup$
    nice solution @AhmadBazzi, but i want to find Fourier transform of $delta(e^{ipi t}-1)$
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:10












  • $begingroup$
    but you just edited right now @ThienHoangDang
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 21:11










  • $begingroup$
    sorry for wrong please bro @AhmadBazzi
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:12














1












1








1





$begingroup$

Note: The OP was originally to find the inverse Fourier Transform of $delta(e^{ipi f}-i)$



Credits: Thanks @Skip for the comment below



Notice that the argument is zero only when $e^{i pi f} = i$, or in other words
begin{equation}
pi f = frac{pi}{2} + 2k pi qquad k in mathbb{Z}
end{equation}
hence for $f = frac{1}{2 } + 2k$ where $k in mathbb{Z}$, the Dirac function peaks at $1$, else it is $0$. This results in a "Dirac comb" or "train of impulses" as
begin{equation}
delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} delta(f - f_k)
end{equation}
where $f_k = frac{1}{2 } + 2k$. Applying inverse Fourier transforms on both sides, we have
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} mathcal{F}^{-1}delta(f - f_k) tag{1}
end{equation}



Using the Fourier transform table, we know that
begin{equation}
mathcal{F}(e^{j 2 pi f_0 t})
=
2 pi delta(f - f_0)
end{equation}
which means that
begin{equation}
mathcal{F}^{-1}(delta(f - f_0)) = frac{1}{2pi}e^{j 2 pi f_0 t}tag{2}
end{equation}
Using equation $(2)$ in $(1)$ we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi f_k t}
end{equation}
Replacing $f_k =frac{1}{2 } + 2k$, we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi (frac{1}{2 } + 2k) t}
= frac{1}{2pi}e^{jpi t}sum_{k = -infty}^{infty} e^{4 k pi t }
end{equation}






share|cite|improve this answer











$endgroup$



Note: The OP was originally to find the inverse Fourier Transform of $delta(e^{ipi f}-i)$



Credits: Thanks @Skip for the comment below



Notice that the argument is zero only when $e^{i pi f} = i$, or in other words
begin{equation}
pi f = frac{pi}{2} + 2k pi qquad k in mathbb{Z}
end{equation}
hence for $f = frac{1}{2 } + 2k$ where $k in mathbb{Z}$, the Dirac function peaks at $1$, else it is $0$. This results in a "Dirac comb" or "train of impulses" as
begin{equation}
delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} delta(f - f_k)
end{equation}
where $f_k = frac{1}{2 } + 2k$. Applying inverse Fourier transforms on both sides, we have
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = sum_{k = -infty}^{infty} mathcal{F}^{-1}delta(f - f_k) tag{1}
end{equation}



Using the Fourier transform table, we know that
begin{equation}
mathcal{F}(e^{j 2 pi f_0 t})
=
2 pi delta(f - f_0)
end{equation}
which means that
begin{equation}
mathcal{F}^{-1}(delta(f - f_0)) = frac{1}{2pi}e^{j 2 pi f_0 t}tag{2}
end{equation}
Using equation $(2)$ in $(1)$ we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi f_k t}
end{equation}
Replacing $f_k =frac{1}{2 } + 2k$, we get
begin{equation}
mathcal{F}^{-1}delta(e^{ipi f}-i) = frac{1}{2pi}sum_{k = -infty}^{infty} e^{j 2 pi (frac{1}{2 } + 2k) t}
= frac{1}{2pi}e^{jpi t}sum_{k = -infty}^{infty} e^{4 k pi t }
end{equation}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 21:13

























answered Dec 18 '18 at 20:16









Ahmad BazziAhmad Bazzi

8,2612824




8,2612824








  • 1




    $begingroup$
    I believe you meant $pi f = pi/2 + 2pi k$.
    $endgroup$
    – AEngineer
    Dec 18 '18 at 20:29






  • 1




    $begingroup$
    yes you are right .. thanks for drawing my attention @Skip
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 20:30










  • $begingroup$
    nice solution @AhmadBazzi, but i want to find Fourier transform of $delta(e^{ipi t}-1)$
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:10












  • $begingroup$
    but you just edited right now @ThienHoangDang
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 21:11










  • $begingroup$
    sorry for wrong please bro @AhmadBazzi
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:12














  • 1




    $begingroup$
    I believe you meant $pi f = pi/2 + 2pi k$.
    $endgroup$
    – AEngineer
    Dec 18 '18 at 20:29






  • 1




    $begingroup$
    yes you are right .. thanks for drawing my attention @Skip
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 20:30










  • $begingroup$
    nice solution @AhmadBazzi, but i want to find Fourier transform of $delta(e^{ipi t}-1)$
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:10












  • $begingroup$
    but you just edited right now @ThienHoangDang
    $endgroup$
    – Ahmad Bazzi
    Dec 18 '18 at 21:11










  • $begingroup$
    sorry for wrong please bro @AhmadBazzi
    $endgroup$
    – Thien Hoang Dang
    Dec 18 '18 at 21:12








1




1




$begingroup$
I believe you meant $pi f = pi/2 + 2pi k$.
$endgroup$
– AEngineer
Dec 18 '18 at 20:29




$begingroup$
I believe you meant $pi f = pi/2 + 2pi k$.
$endgroup$
– AEngineer
Dec 18 '18 at 20:29




1




1




$begingroup$
yes you are right .. thanks for drawing my attention @Skip
$endgroup$
– Ahmad Bazzi
Dec 18 '18 at 20:30




$begingroup$
yes you are right .. thanks for drawing my attention @Skip
$endgroup$
– Ahmad Bazzi
Dec 18 '18 at 20:30












$begingroup$
nice solution @AhmadBazzi, but i want to find Fourier transform of $delta(e^{ipi t}-1)$
$endgroup$
– Thien Hoang Dang
Dec 18 '18 at 21:10






$begingroup$
nice solution @AhmadBazzi, but i want to find Fourier transform of $delta(e^{ipi t}-1)$
$endgroup$
– Thien Hoang Dang
Dec 18 '18 at 21:10














$begingroup$
but you just edited right now @ThienHoangDang
$endgroup$
– Ahmad Bazzi
Dec 18 '18 at 21:11




$begingroup$
but you just edited right now @ThienHoangDang
$endgroup$
– Ahmad Bazzi
Dec 18 '18 at 21:11












$begingroup$
sorry for wrong please bro @AhmadBazzi
$endgroup$
– Thien Hoang Dang
Dec 18 '18 at 21:12




$begingroup$
sorry for wrong please bro @AhmadBazzi
$endgroup$
– Thien Hoang Dang
Dec 18 '18 at 21:12


















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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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