Prove fourier coefficients of an odd discrete signal is $a_n = -a_{-n}$
I'm self studying signal and system. I've come across this property: fourier coefficients of an odd discrete signal is $a_n = -a_{-n}$, how can this be proved?
fourier-series
asked Nov 30 '18 at 20:00
drerDdrerD
1519
add a comment |
I'm self studying signal and system. I've come across this property: fourier coefficients of an odd discrete signal is $a_n = -a_{-n}$, how can this be proved?
fourier-series
asked Nov 30 '18 at 20:00
drerDdrerD
1519
add a comment |
I'm self studying signal and system. I've come across this property: fourier coefficients of an odd discrete signal is $a_n = -a_{-n}$, how can this be proved?
fourier-series
asked Nov 30 '18 at 20:00
drerDdrerD
1519
I'm self studying signal and system. I've come across this property: fourier coefficients of an odd discrete signal is $a_n = -a_{-n}$, how can this be proved?
fourier-series
fourier-series
asked Nov 30 '18 at 20:00
drerDdrerD
1519
asked Nov 30 '18 at 20:00
drerDdrerD
1519
asked Nov 30 '18 at 20:00
drerDdrerD
1519
asked Nov 30 '18 at 20:00
drerDdrerD
1519
asked Nov 30 '18 at 20:00
drerDdrerD
1519
1519
add a comment |
add a comment |
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First the signal must be periodic. Second if the signal is $x[n]$ we have $$x[n]=sum_{k=0}^{N-1}a_ne^{j2pi kn}$$and $$a_k={1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}$$if $x[n] $ is odd we have$$a_{-k}{={1over N}sum_{n=0}^{N-1}x[n]e^{i2pi kn}\={1over N}sum_{n=0}^{N-1}-x[-n]e^{i2pi kn}\=-{1over N}sum_{n=1-N}^{0}x[n]e^{-i2pi kn}\=-{1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}\=-a_{k}}$$hence the proof is complete $blacksquare$
answered Nov 30 '18 at 20:23
Mostafa AyazMostafa Ayaz
14.5k3937
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
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votes
First the signal must be periodic. Second if the signal is $x[n]$ we have $$x[n]=sum_{k=0}^{N-1}a_ne^{j2pi kn}$$and $$a_k={1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}$$if $x[n] $ is odd we have$$a_{-k}{={1over N}sum_{n=0}^{N-1}x[n]e^{i2pi kn}\={1over N}sum_{n=0}^{N-1}-x[-n]e^{i2pi kn}\=-{1over N}sum_{n=1-N}^{0}x[n]e^{-i2pi kn}\=-{1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}\=-a_{k}}$$hence the proof is complete $blacksquare$
answered Nov 30 '18 at 20:23
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
First the signal must be periodic. Second if the signal is $x[n]$ we have $$x[n]=sum_{k=0}^{N-1}a_ne^{j2pi kn}$$and $$a_k={1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}$$if $x[n] $ is odd we have$$a_{-k}{={1over N}sum_{n=0}^{N-1}x[n]e^{i2pi kn}\={1over N}sum_{n=0}^{N-1}-x[-n]e^{i2pi kn}\=-{1over N}sum_{n=1-N}^{0}x[n]e^{-i2pi kn}\=-{1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}\=-a_{k}}$$hence the proof is complete $blacksquare$
answered Nov 30 '18 at 20:23
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
First the signal must be periodic. Second if the signal is $x[n]$ we have $$x[n]=sum_{k=0}^{N-1}a_ne^{j2pi kn}$$and $$a_k={1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}$$if $x[n] $ is odd we have$$a_{-k}{={1over N}sum_{n=0}^{N-1}x[n]e^{i2pi kn}\={1over N}sum_{n=0}^{N-1}-x[-n]e^{i2pi kn}\=-{1over N}sum_{n=1-N}^{0}x[n]e^{-i2pi kn}\=-{1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}\=-a_{k}}$$hence the proof is complete $blacksquare$
answered Nov 30 '18 at 20:23
Mostafa AyazMostafa Ayaz
14.5k3937
First the signal must be periodic. Second if the signal is $x[n]$ we have $$x[n]=sum_{k=0}^{N-1}a_ne^{j2pi kn}$$and $$a_k={1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}$$if $x[n] $ is odd we have$$a_{-k}{={1over N}sum_{n=0}^{N-1}x[n]e^{i2pi kn}\={1over N}sum_{n=0}^{N-1}-x[-n]e^{i2pi kn}\=-{1over N}sum_{n=1-N}^{0}x[n]e^{-i2pi kn}\=-{1over N}sum_{n=0}^{N-1}x[n]e^{-i2pi kn}\=-a_{k}}$$hence the proof is complete $blacksquare$
answered Nov 30 '18 at 20:23
Mostafa AyazMostafa Ayaz
14.5k3937
answered Nov 30 '18 at 20:23
Mostafa AyazMostafa Ayaz
14.5k3937
answered Nov 30 '18 at 20:23
Mostafa AyazMostafa Ayaz
14.5k3937
answered Nov 30 '18 at 20:23
Mostafa AyazMostafa Ayaz
14.5k3937
14.5k3937
add a comment |
add a comment |
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