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![Example 1
Example: Find CK for signal x(t)
Solutio
Ck =
𝟏
𝑻𝟎
𝑻𝟎
𝒙 𝒌 𝒆−jkω0t 𝒅𝒕
=
𝟏
𝑻𝟎
−𝑻𝟎/𝟐
𝑻𝟎/𝟐
𝒙(𝒕)𝒆−jkω0t 𝒅𝒕
=
𝟏
𝑻𝟎
[ −𝑻𝟎/𝟐
−𝑻/𝟐
𝟎. 𝒆−jkω0t 𝒅𝒕 + −𝑻/𝟐
𝑻/𝟐
𝑨𝟎 𝒆−jkω0t 𝒅𝒕 + 𝑻/𝟐
𝑻𝟎/𝟐
𝟎. 𝒆−jkω0t 𝒅𝒕
=
𝑨𝟎
𝑻𝟎
[ −𝑻/𝟐
𝑻/𝟐
𝒆−jkω0t 𝒅𝒕]
Ck =
𝑨𝟎
𝑻𝟎
[
jkω0 𝐓
𝟐
−jkω0 𝐓
𝟐
𝒆Ө (
−𝟏
𝐣𝐤ω𝟎
dӨ)
=
−𝑨𝟎
jkω0𝑻𝟎
[
jkω0 𝑻
𝟐
−jkω0 𝑻
𝟐
𝒆Ө dӨ =
−𝑨𝟎
jkω0𝑻𝟎
[ e Ө ] =
−𝑨𝟎
jkω0𝑻𝟎
[ e -jkω0
𝑻
𝟐 - e jkω0
𝑻
𝟐 ]
𝑻
𝟐
−𝑻
𝟐
−𝑻𝟎
𝟐
𝑻𝟎
𝟐
jkω0
T
2
-jkω0
T
2
Let -jkω0t = Ө
-jkω0 dt = dӨ dt =
−𝟏
𝐣𝐤ω𝟎
dӨ
When t = -T/2 Ө= -jkω0 (
−𝐓
𝟐
) = jkω0
𝑻
𝟐
When t = T/2 Ө= -jkω0 (
𝑻
𝟐
) = - jkω0
𝑻
𝟐](https://image.slidesharecdn.com/lect6-complex-exponential-fourier-series-221223060401-29cabebc/85/Lect6-Complex-Exponential-Fourier-Series-pdf-5-320.jpg)
![Example 1 cont.,
Ck =
−𝑨𝟎
jkω0𝑻𝟎
[ e -jkω0
𝑻
𝟐 - e jkω0
𝑻
𝟐 ]
Let x =
kω0𝑻
𝟐
Ck =
−𝑨𝟎
jkω0𝑻𝟎
[ e -𝒋𝒙 - e 𝒋𝒙]
e 𝒋𝒙 = cos x + j sin x
Ck =
−𝑨𝟎
jkω0𝑻𝟎
[ cos (-x) +j sin (-x) – [ cos x + j sin x] ]
=
−𝑨𝟎
jkω0𝑻𝟎
[cos x – j sin x – cos x – j sin x]
=
−𝑨𝟎
jkω0𝑻𝟎
[ -2 j sin x]
=
−𝑨𝟎
jkω0𝑻𝟎
[ -2 j sin
kω0𝑻
𝟐
]
=
−𝑨𝟎
jkω0𝑻𝟎
−2 j sin kω0𝑻
𝟐
kω0𝑻
𝟐
*
kω0𝑻
𝟐
=
𝑨𝟎𝑻
𝑻𝟎
𝑺𝒂(
kω0𝑻
𝟐
)
𝑻
𝟐
−𝑻
𝟐
−𝑻𝟎
𝟐
𝑻𝟎
𝟐
Sampling Function
Sa (t) = sin t /t](https://image.slidesharecdn.com/lect6-complex-exponential-fourier-series-221223060401-29cabebc/85/Lect6-Complex-Exponential-Fourier-Series-pdf-6-320.jpg)
![Example 1
Example: Find CK for signal x(t)
Solution:
T=2 ω0 =2 𝜋/T= 2𝜋/2= 𝜋
Ck =
1
𝑇 𝑇
𝑥 𝑘 𝑒−jkω0t 𝑑𝑡
=
1
2
[ −1
0
− 𝑒−jkω0t 𝑑𝑡 + 0
1
𝑒−jkω0t 𝑑𝑡]
=
1
2
[ -
1
−𝑗𝑘ω0
𝑒−jkω0t +
1
−𝑗𝑘ω0
𝑒−jkω0t ]
=
1
2
[
1
𝑗𝑘ω0
(𝑒0 −𝑒jkω0) -
1
𝑗𝑘ω0
(𝑒 −jkω0 - 𝑒0)]
=
1
2𝑗𝑘ω0
(1-𝑒jkω0−𝑒−jkω0 + 1) =
1
2𝑗𝑘ω0
(2-𝑒jkω0−𝑒−jkω0)
=
1
2𝑗𝑘ω0
(2- 2(𝑒jkω0+𝑒−jkω0) /2) =
1
2𝑗𝑘ω0
(2 – 2 cos k ω0)
=
1
𝑗𝑘ω0
( 1 – cos k ω0) k≠0
Ck =
1
𝑗𝑘ω0
( 1 – cos k ω0) k≠0
0 k=0
1
-1
1
-1 0
0
-1
1
0 𝒆at 𝒅𝒕 =
𝟏
𝒂
𝒆at
Cos a =
𝒆ja+ 𝒆−ja
2](https://image.slidesharecdn.com/lect6-complex-exponential-fourier-series-221223060401-29cabebc/85/Lect6-Complex-Exponential-Fourier-Series-pdf-7-320.jpg)
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- 5. Example 1 Example: FindCK for signal x(t) Solutio Ck = 𝟏 𝑻𝟎 𝑻𝟎 𝒙 𝒌 𝒆−jkω0t 𝒅𝒕 = 𝟏 𝑻𝟎 −𝑻𝟎/𝟐 𝑻𝟎/𝟐 𝒙(𝒕)𝒆−jkω0t 𝒅𝒕 = 𝟏 𝑻𝟎 [ −𝑻𝟎/𝟐 −𝑻/𝟐 𝟎. 𝒆−jkω0t 𝒅𝒕 + −𝑻/𝟐 𝑻/𝟐 𝑨𝟎 𝒆−jkω0t 𝒅𝒕 + 𝑻/𝟐 𝑻𝟎/𝟐 𝟎. 𝒆−jkω0t 𝒅𝒕 = 𝑨𝟎 𝑻𝟎 [ −𝑻/𝟐 𝑻/𝟐 𝒆−jkω0t 𝒅𝒕] Ck = 𝑨𝟎 𝑻𝟎 [ jkω0 𝐓 𝟐 −jkω0 𝐓 𝟐 𝒆Ө ( −𝟏 𝐣𝐤ω𝟎 dӨ) = −𝑨𝟎 jkω0𝑻𝟎 [ jkω0 𝑻 𝟐 −jkω0 𝑻 𝟐 𝒆Ө dӨ = −𝑨𝟎 jkω0𝑻𝟎 [ e Ө ] = −𝑨𝟎 jkω0𝑻𝟎 [ e -jkω0 𝑻 𝟐 - e jkω0 𝑻 𝟐 ] 𝑻 𝟐 −𝑻 𝟐 −𝑻𝟎 𝟐 𝑻𝟎 𝟐 jkω0 T 2 -jkω0 T 2 Let -jkω0t = Ө -jkω0 dt = dӨ dt = −𝟏 𝐣𝐤ω𝟎 dӨ When t = -T/2 Ө= -jkω0 ( −𝐓 𝟐 ) = jkω0 𝑻 𝟐 When t = T/2 Ө= -jkω0 ( 𝑻 𝟐 ) = - jkω0 𝑻 𝟐
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- 7. Example 1 Example: FindCK for signal x(t) Solution: T=2 ω0 =2 𝜋/T= 2𝜋/2= 𝜋 Ck = 1 𝑇 𝑇 𝑥 𝑘 𝑒−jkω0t 𝑑𝑡 = 1 2 [ −1 0 − 𝑒−jkω0t 𝑑𝑡 + 0 1 𝑒−jkω0t 𝑑𝑡] = 1 2 [ - 1 −𝑗𝑘ω0 𝑒−jkω0t + 1 −𝑗𝑘ω0 𝑒−jkω0t ] = 1 2 [ 1 𝑗𝑘ω0 (𝑒0 −𝑒jkω0) - 1 𝑗𝑘ω0 (𝑒 −jkω0 - 𝑒0)] = 1 2𝑗𝑘ω0 (1-𝑒jkω0−𝑒−jkω0 + 1) = 1 2𝑗𝑘ω0 (2-𝑒jkω0−𝑒−jkω0) = 1 2𝑗𝑘ω0 (2- 2(𝑒jkω0+𝑒−jkω0) /2) = 1 2𝑗𝑘ω0 (2 – 2 cos k ω0) = 1 𝑗𝑘ω0 ( 1 – cos k ω0) k≠0 Ck = 1 𝑗𝑘ω0 ( 1 – cos k ω0) k≠0 0 k=0 1 -1 1 -1 0 0 -1 1 0 𝒆at 𝒅𝒕 = 𝟏 𝒂 𝒆at Cos a = 𝒆ja+ 𝒆−ja 2
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