How to calculate this $sumlimits_{n=0}^{infty}frac{n}{2^n}e^{jwn}$
0
How to calculate this $sumlimits_{n=0}^{infty}frac{n}{2^n}e^{-jwn}$,because it is not geometric progression,so i can't know how to solve it,can anyone help me?
sequences-and-series summation fourier-transform
asked Dec 4 '18 at 10:38
electronic componentelectronic component
387
add a comment |
0
How to calculate this $sumlimits_{n=0}^{infty}frac{n}{2^n}e^{-jwn}$,because it is not geometric progression,so i can't know how to solve it,can anyone help me?
sequences-and-series summation fourier-transform
asked Dec 4 '18 at 10:38
electronic componentelectronic component
387
Still you can write your series as $$sum_{ngeq 0}frac{e^{-jwn}}{2^n}+sum_{ngeq 1}frac{e^{-jwn}}{2^n}+sum_{ngeq 2}frac{e^{-jwn}}{2^n}+ldots $$ which is a geometric series of geometric series.
– Jack D'Aurizio
Dec 4 '18 at 10:43
See math.stackexchange.com/questions/647587/…
– lab bhattacharjee
Dec 4 '18 at 10:43
add a comment |
0
0
0
How to calculate this $sumlimits_{n=0}^{infty}frac{n}{2^n}e^{-jwn}$,because it is not geometric progression,so i can't know how to solve it,can anyone help me?
sequences-and-series summation fourier-transform
asked Dec 4 '18 at 10:38
electronic componentelectronic component
387
How to calculate this $sumlimits_{n=0}^{infty}frac{n}{2^n}e^{-jwn}$,because it is not geometric progression,so i can't know how to solve it,can anyone help me?
sequences-and-series summation fourier-transform
sequences-and-series summation fourier-transform
asked Dec 4 '18 at 10:38
electronic componentelectronic component
387
asked Dec 4 '18 at 10:38
electronic componentelectronic component
387
asked Dec 4 '18 at 10:38
electronic componentelectronic component
387
asked Dec 4 '18 at 10:38
electronic componentelectronic component
387
asked Dec 4 '18 at 10:38
electronic componentelectronic component
387
387
Still you can write your series as $$sum_{ngeq 0}frac{e^{-jwn}}{2^n}+sum_{ngeq 1}frac{e^{-jwn}}{2^n}+sum_{ngeq 2}frac{e^{-jwn}}{2^n}+ldots $$ which is a geometric series of geometric series.
– Jack D'Aurizio
Dec 4 '18 at 10:43
See math.stackexchange.com/questions/647587/…
– lab bhattacharjee
Dec 4 '18 at 10:43
add a comment |
Still you can write your series as $$sum_{ngeq 0}frac{e^{-jwn}}{2^n}+sum_{ngeq 1}frac{e^{-jwn}}{2^n}+sum_{ngeq 2}frac{e^{-jwn}}{2^n}+ldots $$ which is a geometric series of geometric series.
– Jack D'Aurizio
Dec 4 '18 at 10:43
See math.stackexchange.com/questions/647587/…
– lab bhattacharjee
Dec 4 '18 at 10:43
Still you can write your series as $$sum_{ngeq 0}frac{e^{-jwn}}{2^n}+sum_{ngeq 1}frac{e^{-jwn}}{2^n}+sum_{ngeq 2}frac{e^{-jwn}}{2^n}+ldots $$ which is a geometric series of geometric series.
– Jack D'Aurizio
Dec 4 '18 at 10:43
Still you can write your series as $$sum_{ngeq 0}frac{e^{-jwn}}{2^n}+sum_{ngeq 1}frac{e^{-jwn}}{2^n}+sum_{ngeq 2}frac{e^{-jwn}}{2^n}+ldots $$ which is a geometric series of geometric series.
– Jack D'Aurizio
Dec 4 '18 at 10:43
See math.stackexchange.com/questions/647587/…
– lab bhattacharjee
Dec 4 '18 at 10:43
See math.stackexchange.com/questions/647587/…
– lab bhattacharjee
Dec 4 '18 at 10:43
add a comment |
1 Answer
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For $|z|<1$ we have $sum_{n=0}^{infty}n z^n= frac{z}{(1-z)^2}$.
Now let $z= frac{e^{-jw}}{2}$. Can you proceed ?
answered Dec 4 '18 at 10:44
FredFred
44.4k1845
thx,i will see this math.stackexchange.com/questions/647587/…
– electronic component
Dec 4 '18 at 10:47
add a comment |
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1 Answer
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1 Answer
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active
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For $|z|<1$ we have $sum_{n=0}^{infty}n z^n= frac{z}{(1-z)^2}$.
Now let $z= frac{e^{-jw}}{2}$. Can you proceed ?
answered Dec 4 '18 at 10:44
FredFred
44.4k1845
thx,i will see this math.stackexchange.com/questions/647587/…
– electronic component
Dec 4 '18 at 10:47
add a comment |
2
For $|z|<1$ we have $sum_{n=0}^{infty}n z^n= frac{z}{(1-z)^2}$.
Now let $z= frac{e^{-jw}}{2}$. Can you proceed ?
answered Dec 4 '18 at 10:44
FredFred
44.4k1845
thx,i will see this math.stackexchange.com/questions/647587/…
– electronic component
Dec 4 '18 at 10:47
add a comment |
2
2
2
For $|z|<1$ we have $sum_{n=0}^{infty}n z^n= frac{z}{(1-z)^2}$.
Now let $z= frac{e^{-jw}}{2}$. Can you proceed ?
answered Dec 4 '18 at 10:44
FredFred
44.4k1845
For $|z|<1$ we have $sum_{n=0}^{infty}n z^n= frac{z}{(1-z)^2}$.
Now let $z= frac{e^{-jw}}{2}$. Can you proceed ?
answered Dec 4 '18 at 10:44
FredFred
44.4k1845
answered Dec 4 '18 at 10:44
FredFred
44.4k1845
answered Dec 4 '18 at 10:44
FredFred
44.4k1845
answered Dec 4 '18 at 10:44
FredFred
44.4k1845
44.4k1845
thx,i will see this math.stackexchange.com/questions/647587/…
– electronic component
Dec 4 '18 at 10:47
add a comment |
thx,i will see this math.stackexchange.com/questions/647587/…
– electronic component
Dec 4 '18 at 10:47
thx,i will see this math.stackexchange.com/questions/647587/…
– electronic component
Dec 4 '18 at 10:47
thx,i will see this math.stackexchange.com/questions/647587/…
– electronic component
Dec 4 '18 at 10:47
add a comment |
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Still you can write your series as $$sum_{ngeq 0}frac{e^{-jwn}}{2^n}+sum_{ngeq 1}frac{e^{-jwn}}{2^n}+sum_{ngeq 2}frac{e^{-jwn}}{2^n}+ldots $$ which is a geometric series of geometric series.
– Jack D'Aurizio
Dec 4 '18 at 10:43
See math.stackexchange.com/questions/647587/…
– lab bhattacharjee
Dec 4 '18 at 10:43