Foruier Series coefficients.












0














I am given a following signal and want to calculate it's Fourier series coefficients:



$$ x(t) = left{
begin{array}{ll}
0 & quad 0< t < 2 \
2 & quad 2< t < 3 \
0& quad 3< t < 4 \
-2 & quad 5< t < 6
end{array}
right. $$



Following is my method:



$$ a_k= frac{1}{6} ( int_{2}^{3} 2 e^{-jkw_0t}dt + int_{5}^{6} (-2) e^{-jkw_0t}dt) $$



$$ a_k= frac{2}{6} ( int_{2}^{3} e^{-jkw_0t}dt - int_{5}^{6} e^{-jkw_0t}dt) $$



After doing some simplification we get:



$$ a_k= frac{1}{jkpi}(e^{-2jfrac{kpi}{3}}-e^{-jkpi}+e^{-2jkpi}-e^{-5jfrac{kpi}{3}}) $$



I am stuck after this last step.






integration







share|cite|improve this question






















  • I forgot to say you can use Euler's formula $$e^{jx}=cos x+jsin x$$
    – Nosrati
    Nov 20 '18 at 19:28
















0














I am given a following signal and want to calculate it's Fourier series coefficients:



$$ x(t) = left{
begin{array}{ll}
0 & quad 0< t < 2 \
2 & quad 2< t < 3 \
0& quad 3< t < 4 \
-2 & quad 5< t < 6
end{array}
right. $$



Following is my method:



$$ a_k= frac{1}{6} ( int_{2}^{3} 2 e^{-jkw_0t}dt + int_{5}^{6} (-2) e^{-jkw_0t}dt) $$



$$ a_k= frac{2}{6} ( int_{2}^{3} e^{-jkw_0t}dt - int_{5}^{6} e^{-jkw_0t}dt) $$



After doing some simplification we get:



$$ a_k= frac{1}{jkpi}(e^{-2jfrac{kpi}{3}}-e^{-jkpi}+e^{-2jkpi}-e^{-5jfrac{kpi}{3}}) $$



I am stuck after this last step.






integration







share|cite|improve this question






















  • I forgot to say you can use Euler's formula $$e^{jx}=cos x+jsin x$$
    – Nosrati
    Nov 20 '18 at 19:28














0












0








0







I am given a following signal and want to calculate it's Fourier series coefficients:



$$ x(t) = left{
begin{array}{ll}
0 & quad 0< t < 2 \
2 & quad 2< t < 3 \
0& quad 3< t < 4 \
-2 & quad 5< t < 6
end{array}
right. $$



Following is my method:



$$ a_k= frac{1}{6} ( int_{2}^{3} 2 e^{-jkw_0t}dt + int_{5}^{6} (-2) e^{-jkw_0t}dt) $$



$$ a_k= frac{2}{6} ( int_{2}^{3} e^{-jkw_0t}dt - int_{5}^{6} e^{-jkw_0t}dt) $$



After doing some simplification we get:



$$ a_k= frac{1}{jkpi}(e^{-2jfrac{kpi}{3}}-e^{-jkpi}+e^{-2jkpi}-e^{-5jfrac{kpi}{3}}) $$



I am stuck after this last step.






integration







share|cite|improve this question













I am given a following signal and want to calculate it's Fourier series coefficients:



$$ x(t) = left{
begin{array}{ll}
0 & quad 0< t < 2 \
2 & quad 2< t < 3 \
0& quad 3< t < 4 \
-2 & quad 5< t < 6
end{array}
right. $$



Following is my method:



$$ a_k= frac{1}{6} ( int_{2}^{3} 2 e^{-jkw_0t}dt + int_{5}^{6} (-2) e^{-jkw_0t}dt) $$



$$ a_k= frac{2}{6} ( int_{2}^{3} e^{-jkw_0t}dt - int_{5}^{6} e^{-jkw_0t}dt) $$



After doing some simplification we get:



$$ a_k= frac{1}{jkpi}(e^{-2jfrac{kpi}{3}}-e^{-jkpi}+e^{-2jkpi}-e^{-5jfrac{kpi}{3}}) $$



I am stuck after this last step.






integration




integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 '18 at 16:08









Ahmad Qayyum

555




555












  • I forgot to say you can use Euler's formula $$e^{jx}=cos x+jsin x$$
    – Nosrati
    Nov 20 '18 at 19:28


















  • I forgot to say you can use Euler's formula $$e^{jx}=cos x+jsin x$$
    – Nosrati
    Nov 20 '18 at 19:28
















I forgot to say you can use Euler's formula $$e^{jx}=cos x+jsin x$$
– Nosrati
Nov 20 '18 at 19:28




I forgot to say you can use Euler's formula $$e^{jx}=cos x+jsin x$$
– Nosrati
Nov 20 '18 at 19:28










1 Answer
1






active

oldest

votes


















1














Hint:
for these two terms
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}$$
let
$$alpha-x=-2jfrac{kpi}{3}~~~;~~~alpha+x=-5jfrac{kpi}{3}$$
then
$$alpha=-7jfrac{kpi}{6}~~~;~~~x=-jfrac{kpi}{2}$$
and therefore
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$






share|cite|improve this answer





















  • can you explain a bit more.
    – Ahmad Qayyum
    Nov 20 '18 at 17:03










  • We want to make an expression like $$e^{alpha-x}pm e^{alpha+x}=e^{alpha}left(e^{-x}pm e^{x}right)$$
    – Nosrati
    Nov 20 '18 at 17:07











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









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oldest

votes






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oldest

votes









1














Hint:
for these two terms
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}$$
let
$$alpha-x=-2jfrac{kpi}{3}~~~;~~~alpha+x=-5jfrac{kpi}{3}$$
then
$$alpha=-7jfrac{kpi}{6}~~~;~~~x=-jfrac{kpi}{2}$$
and therefore
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$






share|cite|improve this answer





















  • can you explain a bit more.
    – Ahmad Qayyum
    Nov 20 '18 at 17:03










  • We want to make an expression like $$e^{alpha-x}pm e^{alpha+x}=e^{alpha}left(e^{-x}pm e^{x}right)$$
    – Nosrati
    Nov 20 '18 at 17:07
















1














Hint:
for these two terms
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}$$
let
$$alpha-x=-2jfrac{kpi}{3}~~~;~~~alpha+x=-5jfrac{kpi}{3}$$
then
$$alpha=-7jfrac{kpi}{6}~~~;~~~x=-jfrac{kpi}{2}$$
and therefore
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$






share|cite|improve this answer





















  • can you explain a bit more.
    – Ahmad Qayyum
    Nov 20 '18 at 17:03










  • We want to make an expression like $$e^{alpha-x}pm e^{alpha+x}=e^{alpha}left(e^{-x}pm e^{x}right)$$
    – Nosrati
    Nov 20 '18 at 17:07














1












1








1






Hint:
for these two terms
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}$$
let
$$alpha-x=-2jfrac{kpi}{3}~~~;~~~alpha+x=-5jfrac{kpi}{3}$$
then
$$alpha=-7jfrac{kpi}{6}~~~;~~~x=-jfrac{kpi}{2}$$
and therefore
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$






share|cite|improve this answer












Hint:
for these two terms
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}$$
let
$$alpha-x=-2jfrac{kpi}{3}~~~;~~~alpha+x=-5jfrac{kpi}{3}$$
then
$$alpha=-7jfrac{kpi}{6}~~~;~~~x=-jfrac{kpi}{2}$$
and therefore
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 '18 at 16:41









Nosrati

26.5k62353




26.5k62353












  • can you explain a bit more.
    – Ahmad Qayyum
    Nov 20 '18 at 17:03










  • We want to make an expression like $$e^{alpha-x}pm e^{alpha+x}=e^{alpha}left(e^{-x}pm e^{x}right)$$
    – Nosrati
    Nov 20 '18 at 17:07


















  • can you explain a bit more.
    – Ahmad Qayyum
    Nov 20 '18 at 17:03










  • We want to make an expression like $$e^{alpha-x}pm e^{alpha+x}=e^{alpha}left(e^{-x}pm e^{x}right)$$
    – Nosrati
    Nov 20 '18 at 17:07
















can you explain a bit more.
– Ahmad Qayyum
Nov 20 '18 at 17:03




can you explain a bit more.
– Ahmad Qayyum
Nov 20 '18 at 17:03












We want to make an expression like $$e^{alpha-x}pm e^{alpha+x}=e^{alpha}left(e^{-x}pm e^{x}right)$$
– Nosrati
Nov 20 '18 at 17:07




We want to make an expression like $$e^{alpha-x}pm e^{alpha+x}=e^{alpha}left(e^{-x}pm e^{x}right)$$
– Nosrati
Nov 20 '18 at 17:07


















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