Fourier Transform of Manchester (twinned-binary) function

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Given the following function used in binary telecommunications:



enter image description here



I'd like to know how I would obtain $P(f)$ the Fourier transform of this function. The function can be written as:



$$
p(t) = left{
beginarrayll
+1 & quad 0 leq fracTb2 \
-1 & quad fracTb2 lt Tb
endarray
right.
$$



I thought of maybe expressing it as some sort of shifted and truncated $sgn(x)$ function but I'm looking for the easiest/simplest mathematical solution.




fourier-transform




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  • The Fourier transform of the Heaviside step function is a bit troublesome; the direct approach yields $$ hat H(omega) = frac1sqrt2pi int_0^infty e^-iomega t mathsf dt = lim_xtoinfty frac1sqrt2picdot frac1-e^-iomega xiomega, $$ but this limit does not exist. Rather, the correct answer is $$ hat H(omega) = frac1sqrt2pifrac1iomega + sqrtfracpi 2delta(omega), $$ where $delta(cdot)$ is the Dirac delta function. See here: cs.uaf.edu/~bueler/M611heaviside.pdf
    – Math1000
    Jul 25 at 11:11







  • 1




    Applying this logic to $p(t) = H(t)-2H(t-T_b/2)+H(t-T_b)$ we have $$ hat p(omega) = frac1sqrt2pifrac1iomegaleft(-1+2 e^iT_bomega/2-e^iT_bomegaright). $$
    – Math1000
    Jul 25 at 11:16















up vote
0
down vote

favorite












Given the following function used in binary telecommunications:



enter image description here



I'd like to know how I would obtain $P(f)$ the Fourier transform of this function. The function can be written as:



$$
p(t) = left{
beginarrayll
+1 & quad 0 leq fracTb2 \
-1 & quad fracTb2 lt Tb
endarray
right.
$$



I thought of maybe expressing it as some sort of shifted and truncated $sgn(x)$ function but I'm looking for the easiest/simplest mathematical solution.




fourier-transform




share|cite|improve this question



















  • The Fourier transform of the Heaviside step function is a bit troublesome; the direct approach yields $$ hat H(omega) = frac1sqrt2pi int_0^infty e^-iomega t mathsf dt = lim_xtoinfty frac1sqrt2picdot frac1-e^-iomega xiomega, $$ but this limit does not exist. Rather, the correct answer is $$ hat H(omega) = frac1sqrt2pifrac1iomega + sqrtfracpi 2delta(omega), $$ where $delta(cdot)$ is the Dirac delta function. See here: cs.uaf.edu/~bueler/M611heaviside.pdf
    – Math1000
    Jul 25 at 11:11







  • 1




    Applying this logic to $p(t) = H(t)-2H(t-T_b/2)+H(t-T_b)$ we have $$ hat p(omega) = frac1sqrt2pifrac1iomegaleft(-1+2 e^iT_bomega/2-e^iT_bomegaright). $$
    – Math1000
    Jul 25 at 11:16













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given the following function used in binary telecommunications:



enter image description here



I'd like to know how I would obtain $P(f)$ the Fourier transform of this function. The function can be written as:



$$
p(t) = left{
beginarrayll
+1 & quad 0 leq fracTb2 \
-1 & quad fracTb2 lt Tb
endarray
right.
$$



I thought of maybe expressing it as some sort of shifted and truncated $sgn(x)$ function but I'm looking for the easiest/simplest mathematical solution.




fourier-transform




share|cite|improve this question











Given the following function used in binary telecommunications:



enter image description here



I'd like to know how I would obtain $P(f)$ the Fourier transform of this function. The function can be written as:



$$
p(t) = left{
beginarrayll
+1 & quad 0 leq fracTb2 \
-1 & quad fracTb2 lt Tb
endarray
right.
$$



I thought of maybe expressing it as some sort of shifted and truncated $sgn(x)$ function but I'm looking for the easiest/simplest mathematical solution.





fourier-transform





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 25 at 9:37









Blargian

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  • The Fourier transform of the Heaviside step function is a bit troublesome; the direct approach yields $$ hat H(omega) = frac1sqrt2pi int_0^infty e^-iomega t mathsf dt = lim_xtoinfty frac1sqrt2picdot frac1-e^-iomega xiomega, $$ but this limit does not exist. Rather, the correct answer is $$ hat H(omega) = frac1sqrt2pifrac1iomega + sqrtfracpi 2delta(omega), $$ where $delta(cdot)$ is the Dirac delta function. See here: cs.uaf.edu/~bueler/M611heaviside.pdf
    – Math1000
    Jul 25 at 11:11







  • 1




    Applying this logic to $p(t) = H(t)-2H(t-T_b/2)+H(t-T_b)$ we have $$ hat p(omega) = frac1sqrt2pifrac1iomegaleft(-1+2 e^iT_bomega/2-e^iT_bomegaright). $$
    – Math1000
    Jul 25 at 11:16

















  • The Fourier transform of the Heaviside step function is a bit troublesome; the direct approach yields $$ hat H(omega) = frac1sqrt2pi int_0^infty e^-iomega t mathsf dt = lim_xtoinfty frac1sqrt2picdot frac1-e^-iomega xiomega, $$ but this limit does not exist. Rather, the correct answer is $$ hat H(omega) = frac1sqrt2pifrac1iomega + sqrtfracpi 2delta(omega), $$ where $delta(cdot)$ is the Dirac delta function. See here: cs.uaf.edu/~bueler/M611heaviside.pdf
    – Math1000
    Jul 25 at 11:11







  • 1




    Applying this logic to $p(t) = H(t)-2H(t-T_b/2)+H(t-T_b)$ we have $$ hat p(omega) = frac1sqrt2pifrac1iomegaleft(-1+2 e^iT_bomega/2-e^iT_bomegaright). $$
    – Math1000
    Jul 25 at 11:16
















The Fourier transform of the Heaviside step function is a bit troublesome; the direct approach yields $$ hat H(omega) = frac1sqrt2pi int_0^infty e^-iomega t mathsf dt = lim_xtoinfty frac1sqrt2picdot frac1-e^-iomega xiomega, $$ but this limit does not exist. Rather, the correct answer is $$ hat H(omega) = frac1sqrt2pifrac1iomega + sqrtfracpi 2delta(omega), $$ where $delta(cdot)$ is the Dirac delta function. See here: cs.uaf.edu/~bueler/M611heaviside.pdf
– Math1000
Jul 25 at 11:11





The Fourier transform of the Heaviside step function is a bit troublesome; the direct approach yields $$ hat H(omega) = frac1sqrt2pi int_0^infty e^-iomega t mathsf dt = lim_xtoinfty frac1sqrt2picdot frac1-e^-iomega xiomega, $$ but this limit does not exist. Rather, the correct answer is $$ hat H(omega) = frac1sqrt2pifrac1iomega + sqrtfracpi 2delta(omega), $$ where $delta(cdot)$ is the Dirac delta function. See here: cs.uaf.edu/~bueler/M611heaviside.pdf
– Math1000
Jul 25 at 11:11





1




1




Applying this logic to $p(t) = H(t)-2H(t-T_b/2)+H(t-T_b)$ we have $$ hat p(omega) = frac1sqrt2pifrac1iomegaleft(-1+2 e^iT_bomega/2-e^iT_bomegaright). $$
– Math1000
Jul 25 at 11:16





Applying this logic to $p(t) = H(t)-2H(t-T_b/2)+H(t-T_b)$ we have $$ hat p(omega) = frac1sqrt2pifrac1iomegaleft(-1+2 e^iT_bomega/2-e^iT_bomegaright). $$
– Math1000
Jul 25 at 11:16
















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