Two dimensional Fourier Transform and Convolution of a periodic function
I want to calculate the Fourier transform of the following function
$$
frac{1}{T}e^{iomega_n t}e^{-iomega_m t^prime}Theta(t-t^prime),
$$
where $omega_n=frac{2pi}{T}n$ ($nin mathbb{Z}$) and $Theta$ is the Heaviside step function
$$
Theta(t-t^prime)=
begin{cases}
1, ,, t-t^primegeq 0\
0, ,, t-t^prime< 0, .
end{cases}
$$
Since it depends on two variables the transformation reads as follows
$$
frac{1}{T^2}int_0^Tdt, dt^prime , e^{i(omega_n-omega_p) t}e^{i(omega_l-omega_m) t^prime}Theta(t-t^prime),
$$
the Fourier transform only depends on the frequency differences, so I define $Delta_{np}equivomega_n-omega_p$ and $Delta_{lm}equivomega_l-omega_m$.
The convolution of a periodic function is defined as
$$
(f*g)(t)=frac{1}{sqrt{T}}int_0^Tdt^prime f(t^prime)g(t-t^prime).
$$
Hence, I interpret the two dimensional Fourier transform as a one dimensional Fourier transform of a convolution where $f(t^prime)=e^{i(omega_l-omega_m) t^prime}$ and $g(t-t^prime)=Theta(t-t^prime)$. With the substitutions $q=t-t^prime$ ($dq=dt$) and $t=t^prime+q$, the integrals can be written as the product of Fourier transforms
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}.
$$
With this substitution, the step function can be omitted because we integrate from $0$ to the fundamental period $T$. Now, if one of the frequency differences are non zero, then the integral vanishes and if both are zero then the integral is rendered $1$. Hence,
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}=delta_{n,p}delta_{m,l}.
$$
However, if I go back to my original integral and insert $Delta_{np}=Delta_{lm}=0$, I get
$$
frac{1}{T^2}int_0^Tdtint_0^T dt^prime , Theta(t-t^prime)=frac{1}{T^2}int_0^Tdtint_0^t dt^prime=frac{1}{2}.
$$
Now which result is the correct one? I have been calculating all day now to figure a way out of this ambiguity. How do I calculate this Fourier Transform of the convolution??
Thank you in advance
calculus complex-analysis fourier-transform convolution
asked Nov 26 '18 at 18:49
mr. curious
909
|
show 1 more comment
I want to calculate the Fourier transform of the following function
$$
frac{1}{T}e^{iomega_n t}e^{-iomega_m t^prime}Theta(t-t^prime),
$$
where $omega_n=frac{2pi}{T}n$ ($nin mathbb{Z}$) and $Theta$ is the Heaviside step function
$$
Theta(t-t^prime)=
begin{cases}
1, ,, t-t^primegeq 0\
0, ,, t-t^prime< 0, .
end{cases}
$$
Since it depends on two variables the transformation reads as follows
$$
frac{1}{T^2}int_0^Tdt, dt^prime , e^{i(omega_n-omega_p) t}e^{i(omega_l-omega_m) t^prime}Theta(t-t^prime),
$$
the Fourier transform only depends on the frequency differences, so I define $Delta_{np}equivomega_n-omega_p$ and $Delta_{lm}equivomega_l-omega_m$.
The convolution of a periodic function is defined as
$$
(f*g)(t)=frac{1}{sqrt{T}}int_0^Tdt^prime f(t^prime)g(t-t^prime).
$$
Hence, I interpret the two dimensional Fourier transform as a one dimensional Fourier transform of a convolution where $f(t^prime)=e^{i(omega_l-omega_m) t^prime}$ and $g(t-t^prime)=Theta(t-t^prime)$. With the substitutions $q=t-t^prime$ ($dq=dt$) and $t=t^prime+q$, the integrals can be written as the product of Fourier transforms
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}.
$$
With this substitution, the step function can be omitted because we integrate from $0$ to the fundamental period $T$. Now, if one of the frequency differences are non zero, then the integral vanishes and if both are zero then the integral is rendered $1$. Hence,
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}=delta_{n,p}delta_{m,l}.
$$
However, if I go back to my original integral and insert $Delta_{np}=Delta_{lm}=0$, I get
$$
frac{1}{T^2}int_0^Tdtint_0^T dt^prime , Theta(t-t^prime)=frac{1}{T^2}int_0^Tdtint_0^t dt^prime=frac{1}{2}.
$$
Now which result is the correct one? I have been calculating all day now to figure a way out of this ambiguity. How do I calculate this Fourier Transform of the convolution??
Thank you in advance
calculus complex-analysis fourier-transform convolution
asked Nov 26 '18 at 18:49
mr. curious
909
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 '18 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 '18 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 '18 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 '18 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 '18 at 19:42
|
show 1 more comment
I want to calculate the Fourier transform of the following function
$$
frac{1}{T}e^{iomega_n t}e^{-iomega_m t^prime}Theta(t-t^prime),
$$
where $omega_n=frac{2pi}{T}n$ ($nin mathbb{Z}$) and $Theta$ is the Heaviside step function
$$
Theta(t-t^prime)=
begin{cases}
1, ,, t-t^primegeq 0\
0, ,, t-t^prime< 0, .
end{cases}
$$
Since it depends on two variables the transformation reads as follows
$$
frac{1}{T^2}int_0^Tdt, dt^prime , e^{i(omega_n-omega_p) t}e^{i(omega_l-omega_m) t^prime}Theta(t-t^prime),
$$
the Fourier transform only depends on the frequency differences, so I define $Delta_{np}equivomega_n-omega_p$ and $Delta_{lm}equivomega_l-omega_m$.
The convolution of a periodic function is defined as
$$
(f*g)(t)=frac{1}{sqrt{T}}int_0^Tdt^prime f(t^prime)g(t-t^prime).
$$
Hence, I interpret the two dimensional Fourier transform as a one dimensional Fourier transform of a convolution where $f(t^prime)=e^{i(omega_l-omega_m) t^prime}$ and $g(t-t^prime)=Theta(t-t^prime)$. With the substitutions $q=t-t^prime$ ($dq=dt$) and $t=t^prime+q$, the integrals can be written as the product of Fourier transforms
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}.
$$
With this substitution, the step function can be omitted because we integrate from $0$ to the fundamental period $T$. Now, if one of the frequency differences are non zero, then the integral vanishes and if both are zero then the integral is rendered $1$. Hence,
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}=delta_{n,p}delta_{m,l}.
$$
However, if I go back to my original integral and insert $Delta_{np}=Delta_{lm}=0$, I get
$$
frac{1}{T^2}int_0^Tdtint_0^T dt^prime , Theta(t-t^prime)=frac{1}{T^2}int_0^Tdtint_0^t dt^prime=frac{1}{2}.
$$
Now which result is the correct one? I have been calculating all day now to figure a way out of this ambiguity. How do I calculate this Fourier Transform of the convolution??
Thank you in advance
calculus complex-analysis fourier-transform convolution
asked Nov 26 '18 at 18:49
mr. curious
909
I want to calculate the Fourier transform of the following function
$$
frac{1}{T}e^{iomega_n t}e^{-iomega_m t^prime}Theta(t-t^prime),
$$
where $omega_n=frac{2pi}{T}n$ ($nin mathbb{Z}$) and $Theta$ is the Heaviside step function
$$
Theta(t-t^prime)=
begin{cases}
1, ,, t-t^primegeq 0\
0, ,, t-t^prime< 0, .
end{cases}
$$
Since it depends on two variables the transformation reads as follows
$$
frac{1}{T^2}int_0^Tdt, dt^prime , e^{i(omega_n-omega_p) t}e^{i(omega_l-omega_m) t^prime}Theta(t-t^prime),
$$
the Fourier transform only depends on the frequency differences, so I define $Delta_{np}equivomega_n-omega_p$ and $Delta_{lm}equivomega_l-omega_m$.
The convolution of a periodic function is defined as
$$
(f*g)(t)=frac{1}{sqrt{T}}int_0^Tdt^prime f(t^prime)g(t-t^prime).
$$
Hence, I interpret the two dimensional Fourier transform as a one dimensional Fourier transform of a convolution where $f(t^prime)=e^{i(omega_l-omega_m) t^prime}$ and $g(t-t^prime)=Theta(t-t^prime)$. With the substitutions $q=t-t^prime$ ($dq=dt$) and $t=t^prime+q$, the integrals can be written as the product of Fourier transforms
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}.
$$
With this substitution, the step function can be omitted because we integrate from $0$ to the fundamental period $T$. Now, if one of the frequency differences are non zero, then the integral vanishes and if both are zero then the integral is rendered $1$. Hence,
$$
frac{1}{T^2}int_0^Tdq, Theta(q)e^{iDelta_{np} q}int_0^Tdt^prime, e^{i(Delta_{lm}+Delta_{np}) t^prime}=delta_{n,p}delta_{m,l}.
$$
However, if I go back to my original integral and insert $Delta_{np}=Delta_{lm}=0$, I get
$$
frac{1}{T^2}int_0^Tdtint_0^T dt^prime , Theta(t-t^prime)=frac{1}{T^2}int_0^Tdtint_0^t dt^prime=frac{1}{2}.
$$
Now which result is the correct one? I have been calculating all day now to figure a way out of this ambiguity. How do I calculate this Fourier Transform of the convolution??
Thank you in advance
calculus complex-analysis fourier-transform convolution
calculus complex-analysis fourier-transform convolution
asked Nov 26 '18 at 18:49
mr. curious
909
asked Nov 26 '18 at 18:49
mr. curious
909
asked Nov 26 '18 at 18:49
mr. curious
909
asked Nov 26 '18 at 18:49
mr. curious
909
asked Nov 26 '18 at 18:49
mr. curious
909
909
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 '18 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 '18 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 '18 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 '18 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 '18 at 19:42
|
show 1 more comment
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 '18 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 '18 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 '18 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 '18 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 '18 at 19:42
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 '18 at 18:58
Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 '18 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 '18 at 19:01
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 '18 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 '18 at 19:16
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 '18 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 '18 at 19:21
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 '18 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 '18 at 19:42
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 '18 at 19:42
|
show 1 more comment
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Ok, thank you for the quick reply. (I guess $h$ is the step function and $delta$ the $delta$ distribution.) How can I use the FT of the step function, when it is multiplied by another. Afterall, I am transforming a product of functions.
– mr. curious
Nov 26 '18 at 18:58
The Fourier transform of $H(x_1,x_2) = h(frac{x_1-x_2}{sqrt{2}})$ is $widehat{H}(xi_1,xi_2)=delta(frac{xi_1+xi_2}{sqrt{2}})widehat{h}(frac{xi_1-xi_2}{sqrt{2}})$
– reuns
Nov 26 '18 at 19:01
Ok, I see. One could consider the frequency differences simply as the Fourier exponentials. But, the function under consideration is a product of two periodic functions and the Heaviside step function. So my integral is calculate over one period, instead of $-infty$ to $infty$....Does that matter?
– mr. curious
Nov 26 '18 at 19:16
I don't understand where you see a convolution. Assume $G(x_1,x_2) = h_1(x_1)h_2(x_2)$, find its Fourier transform, use a change of variable to obtain the FT of $H(x_1,x_2) = h_1(frac{x_1-x_2}{sqrt{2}})h_2(frac{x_1+x_2}{sqrt{2}})$, extend to the case $h_1,h_2$ are distributions
– reuns
Nov 26 '18 at 19:21
The convolution (for periodic functions) is seen in the third and fourth equation $(e^{iDelta_{l,m}t})*(Theta(t))=1/sqrt{T}int_0^Te^{iDelta_{l,m}t^prime}Theta(t-t^prime)$.
– mr. curious
Nov 26 '18 at 19:42