Download Trigonometric FS - The University of Jordan

Lecture 10:
EE 221: Signals Analysis and Systems
Instructor: Dr. Ghazi Al Sukkar
Dept. of Electrical Engineering
The University of Jordan
Email: [email protected]
Compact (Combined) trigonometric Form

For a periodic signal 𝑥(𝑡) with period 𝑇𝑜 the compact
trigonometric form of Fourier series is:
∞
𝑥 𝑡 = 𝐴0 +
𝐴𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝜃𝑛
𝑛=1

How?
∗ or 𝑐 ∗ = 𝑐
For real 𝑥(𝑡): 𝑐𝑛 = 𝑐−𝑛
𝑛
−𝑛
But
𝑐𝑛 = 𝑐𝑛 𝑒 𝑗𝜃𝑛 = 𝑐𝑛 𝑒 𝑗∡𝑐𝑛
⟹ 𝑐−𝑛 = 𝑐−𝑛 𝑒 𝑗𝜃−𝑛 = 𝑐𝑛∗ = 𝑐𝑛 𝑒 −𝑗𝜃𝑛
Hence,
𝑐−𝑛 = 𝑐𝑛 (The Magnitude spectrum is an even function of 𝑛)
𝜃−𝑛 = −𝜃𝑛 (The phase spectrum is an odd function of 𝑛)
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Cont..

From the exponential form
∞
𝑐𝑛 𝑒 𝑗𝑛𝜔𝑜 𝑡
𝑥 𝑡 =
𝑛=−∞
−𝑗𝜔𝑜 𝑡
+ 𝑐−1 𝑒
+ 𝑐0 + 𝑐1 𝑒 𝑗𝜔𝑜 𝑡 + 𝑐2 𝑒 𝑗2𝜔𝑜𝑡 + ⋯
⟹ 𝑐−𝑛 𝑒 −𝑗𝑛𝜔𝑜𝑡 + 𝑐𝑛 𝑒 𝑗𝑛𝜔𝑜𝑡
= 𝑐−𝑛 𝑒 𝑗𝜃−𝑛 𝑒 −𝑗𝑛𝜔𝑜 𝑡 + 𝑐𝑛 𝑒 𝑗𝜃𝑛 𝑒 𝑗𝑛𝜔𝑜𝑡
= 𝑐𝑛 𝑒 −𝑗𝜃𝑛 𝑒 −𝑗𝑛𝜔𝑜 𝑡 + 𝑐𝑛 𝑒 𝑗𝜃𝑛 𝑒 𝑗𝑛𝜔𝑜𝑡
2
−𝑗
𝑛𝜔
𝑡+𝜃
𝑗
𝑛𝜔
𝑡+𝜃
𝑜
𝑛
𝑜
𝑛
= 𝑐𝑛 𝑒
+𝑒
×
2
= 2 𝑐𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝜃𝑛
𝑥 𝑡 = ⋯ + 𝑐−2 𝑒
−𝑗2𝜔𝑜 𝑡
∞
⟹ 𝑥 𝑡 = 𝑐0 +
2 𝑐𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝜃𝑛
𝑛=1
The compact form is:
∞
𝑥 𝑡 = 𝐴0 +
𝐴𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝜃𝑛
𝑛=1
Hence: 𝐴0 = 𝑐0 (The average value)
𝐴𝑛 = 2 𝑐𝑛
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Cont..
𝐴𝑛 represents the single-sided magnitude spectrum.
2 𝑐𝑛 =
4𝑘
𝜋
4𝑘
, 𝑛 𝑜𝑑𝑑
𝑛𝜋
0, 𝑛 𝑒𝑣𝑒𝑛
4𝑘
3𝜋
4𝑘
5𝜋
𝜔𝑜 2𝜔𝑜 3𝜔𝑜 4𝜔𝑜 5𝜔𝑜
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…
𝑛𝜔𝑜
4
Trigonometric form of Fourier Series

The trigonometric form of a periodic signal 𝑥(𝑡) with period 𝑇𝑜
is:
∞
𝑥 𝑡 = 𝐴0 +
𝑎𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝑏𝑛 sin 𝑛𝜔𝑜 𝑡
𝑛=1
 How?
Return back to the compact trigonometric form:
∞
𝑥 𝑡 = 𝑐0 +
2 𝑐𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝜃𝑛
𝑛=1
From the trigonometric identity:
cos 𝛼 + 𝛽 = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽
∞
⟹ 𝑥 𝑡 = 𝑐0 +
∞
𝑥 𝑡 = 𝑐0 +
2 𝑐𝑛 cos 𝑛𝜔𝑜 𝑡 cos 𝜃𝑛 − sin 𝑛𝜔𝑜 𝑡 sin 𝜃𝑛
𝑛=1
2 𝑐𝑛 cos 𝜃𝑛 cos 𝑛𝜔𝑜 𝑡 − 2 𝑐𝑛 sin 𝜃𝑛 sin 𝑛𝜔𝑜 𝑡
𝑛=1
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Cont..

But:
𝑐𝑛 = 𝑐𝑛 𝑒 𝑗𝜃𝑛 = 𝑐𝑛 cos 𝜃𝑛 + 𝑗 𝑐𝑛 sin 𝜃𝑛
2𝑐𝑛 = 2 𝑐𝑛 𝑒 𝑗𝜃𝑛 = 2 𝑐𝑛 cos 𝜃𝑛 + 𝑗2 𝑐𝑛 sin 𝜃𝑛
Let 𝑎𝑛 = 2 𝑐𝑛 cos 𝜃𝑛 and 𝑏𝑛 = −2 𝑐𝑛 sin 𝜃𝑛
⟹ 2𝑐𝑛 = 𝑎𝑛 − 𝑗𝑏𝑛
∞
𝑥 𝑡 = 𝑐0 +
𝑎𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝑏𝑛 sin 𝑛𝜔𝑜 𝑡
𝑛=1
∞
𝑥 𝑡 = 𝐴0 +
𝑎𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝑏𝑛 sin 𝑛𝜔𝑜 𝑡
𝑛=1
⟹ 𝐴0 = 𝑐0
𝑎𝑛 𝑗𝑏𝑛
𝑐𝑛 =
−
2
2
−𝑏𝑛
−1
𝜃𝑛 = tan
𝑎𝑛
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Cont..
1
𝑐𝑛 = 2 𝑎𝑛2 + 𝑏𝑛2


𝑎𝑛 = 𝐴𝑛 cos 𝜃𝑛 and 𝑏𝑛 = −𝐴𝑛 sin 𝜃𝑛

𝐴𝑛 =
𝑐𝑛 =
1
𝑇𝑜
𝑎𝑛2 + 𝑏𝑛2
𝑥(𝑡)𝑒 −𝑗𝑛𝜔𝑜𝑡 𝑑𝑡 = 𝑐𝑛 =
𝑇𝑜
𝑐𝑛 =
1
𝑇𝑜
1
𝑇𝑜
𝑥(𝑡) cos 𝑛𝜔𝑜 𝑡 − 𝑗 sin 𝑛𝜔𝑜 𝑡
𝑇𝑜
𝑥(𝑡) cos 𝑛𝜔𝑜 𝑡 𝑑𝑡 − 𝑗
𝑇𝑜
𝑑𝑡
1
𝑇𝑜
𝑥(𝑡) sin 𝑛𝜔𝑜 𝑡 𝑑𝑡
𝑇𝑜
But:
𝑎𝑛 𝑗𝑏𝑛
𝑐𝑛 =
−
2
2
2
⟹ 𝑎𝑛 =
𝑥(𝑡) cos 𝑛𝜔𝑜 𝑡 𝑑𝑡
𝑇𝑜
𝑇𝑜
2
𝑏𝑛 =
𝑇𝑜
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𝑥(𝑡) sin 𝑛𝜔𝑜 𝑡 𝑑𝑡
𝑇𝑜
7
Cont..

Example: Find the trigonometric form of Fourier series for 𝑥(𝑡).
𝑥(𝑡)
1
…
…
−0.5
0.5
0
1
1.5
𝑡
−1
∞
𝑥 𝑡 = 𝐴0 +
𝑎𝑛 cos 𝑛𝜔𝑜 𝑡 + 𝑏𝑛 sin 𝑛𝜔𝑜 𝑡
𝑛=1
𝑇𝑜 = 2 ⟹ 𝜔𝑜 =
𝐴0 = 𝐶0 =
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1
𝑇𝑜
2𝜋
=𝜋
2
𝑥 𝑡 𝑑𝑡 = 0
𝑇𝑜
8
Cont..
𝑎𝑛 =
2
2
1.5
0.5
𝑥(𝑡) cos 𝑛𝜋𝑡 𝑑𝑡 =
−0.5
1.5
2𝑡 cos 𝑛𝜋𝑡 𝑑𝑡 +
−0.5
−2 𝑡 − 1 cos 𝑛𝜋𝑡 𝑑𝑡
0.5
Integration by parts or from tables:
𝑏
𝑎
1
𝑡 cos 𝑘𝑡 𝑑𝑡 = 2 cos 𝑘𝑡 + 𝑘𝑡 sin 𝑘𝑡 |𝑏𝑎
𝑘
⟹ 𝑎𝑛 = 0
𝑏𝑛 =
2
2
1.5
0.5
𝑥(𝑡) sin 𝑛𝜋𝑡 𝑑𝑡 =
−0.5
1.5
2𝑡 sin 𝑛𝜋𝑡 𝑑𝑡 +
−0.5
−2 𝑡 − 1 sin 𝑛𝜋𝑡 𝑑𝑡
0.5
Again integration by parts of from tables:
𝑏
𝑎
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1
𝑡 sin 𝑘𝑡 𝑑𝑡 = 2 sin 𝑘𝑡 + 𝑘𝑡 cos 𝑘𝑡 |𝑏𝑎
𝑘
9
Cont..
0, 𝑛 𝑒𝑣𝑒𝑛
8
, 𝑛 = 1,5,9,13, …
𝑛𝜋 2
⟹ 𝑏𝑛 =
8
−
, 𝑛 = 3,7,11,15, …
𝑛𝜋 2
8
1
1
1
⟹ 𝑥 𝑡 = 2 sin 𝜋𝑡 − sin 3𝜋𝑡 +
sin 5𝜋𝑡 −
sin 7𝜋𝑡 + ⋯
𝜋
9
25
49
 To find the combined trigonometric form, we use the identity:
𝜋
± sin 𝑥 = cos 𝑥 ∓
2
⟹𝑥 𝑡
8
𝜋
1
𝜋
1
𝜋
1
𝜋
= 2 cos 𝜋𝑡 −
+ cos 3𝜋𝑡 +
+
cos 5𝜋𝑡 −
+
cos 7𝜋𝑡 +
𝜋
2
9
2
25
2
49
2
∞
8
1
𝜋
𝑚+1
𝑥 𝑡 = 2
cos
2𝑚
+
1
𝜋𝑡
+
−1
𝜋
2𝑚 + 1 2
2
𝑚=0
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Existence of Fourier Series (Dirichlet Conditions)

A periodic signal 𝑥(𝑡) can be expanded into a Fourier series if it
satisfies Dirichlet conditions (sufficient but are not necessary):


𝑥(𝑡) has at most a finite number of discontinuities in one period.
𝑥(𝑡) has at most a finite number of maxima and minima in one
period.

𝑥(𝑡) is bounded or
𝑇𝑜
𝑥(𝑡) 𝑑𝑡 < ∞ (absolutely integrable) this will
include the singularity functions.

If 𝑥(𝑡) satisfies the Dirichlet conditions then the corresponding
Fourier series is convergent, i.e.,
∞
𝑐𝑛 𝑒 𝑗𝑛𝜔𝑜 𝜏
𝑥 𝜏 =
𝑛=−∞
 At points of discontinuity, the series converges to the mean of the
limits approached by 𝑥(𝑡) from the right and from the left, i.e.,
∞
−
+
𝑥
𝜏
+
𝑥(𝜏
)
𝑥(𝑡)
𝑐𝑛 𝑒 𝑗𝑛𝜔𝑜𝜏 =
2
𝑛=−∞
Where: 𝑥 𝜏 − = lim− 𝑥(𝑡) and 𝑥 𝜏 + = lim+ 𝑥(𝑡)
𝑡→𝜏
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𝑡→𝜏
𝑡
𝜏
11
Cont..

Example:
𝑥(𝑡)
1
…
…
−2𝜋

−𝜋
−
1
5
1
7
𝜋
2
𝜋
2
𝜋
2𝜋
𝑡
Using Fourier series
1 2
1
1
1
𝑥(𝑡) = + cos 𝑡 − cos 3𝑡 + cos 5𝑡 − cos 7𝑡 + ⋯
2 𝜋
3
5
7

1
2
𝑥 0 = +
1
3
1
5
2
𝜋
1
3
1− + − +⋯
1
7
But 1 − + − + ⋯ =
𝜋
4
⟹𝑥 0 =

𝑥
𝜋
2
1
2
= +
2
𝜋
0+0+⋯ =
1 2 𝜋
+ ∙ =1
2 𝜋 4
1
2
lim
𝑥(𝑡) = 1, lim + 𝑥(𝑡) = 0
𝜋 −
𝑡→
𝑡→
2
⟹𝑥
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𝜋
2
𝜋
1+0 1
=
=
2
2
2
12