Mixing
Foruier Series coefficients.
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
asked Nov 20 '18 at 16:08
Ahmad Qayyum
555
add a comment |
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
asked Nov 20 '18 at 16:08
Ahmad Qayyum
555
I forgot to say you can use Euler's formula $$e^{jx}=cos x+jsin x$$
– Nosrati
Nov 20 '18 at 19:28
add a comment |
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
asked Nov 20 '18 at 16:08
Ahmad Qayyum
555
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
asked Nov 20 '18 at 16:08
Ahmad Qayyum
555
asked Nov 20 '18 at 16:08
Ahmad Qayyum
555
asked Nov 20 '18 at 16:08
Ahmad Qayyum
555
asked Nov 20 '18 at 16:08
Ahmad Qayyum
555
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$
answered Nov 20 '18 at 16:41
Nosrati
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$
answered Nov 20 '18 at 16:41
Nosrati
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
add a comment |
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
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$
answered Nov 20 '18 at 16:41
Nosrati
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
add a comment |
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
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$
answered Nov 20 '18 at 16:41
Nosrati
26.5k62353
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
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$
answered Nov 20 '18 at 16:41
Nosrati
26.5k62353
answered Nov 20 '18 at 16:41
Nosrati
26.5k62353
answered Nov 20 '18 at 16:41
Nosrati
26.5k62353
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
add a comment |
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
add a comment |
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I forgot to say you can use Euler's formula $$e^{jx}=cos x+jsin x$$
– Nosrati
Nov 20 '18 at 19:28