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
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active

oldest

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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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