1

Fourier Series and Transform

ENGIN 211, Engineering Math

Periodic Functions and Harmonics

2

a a+T t

f(t) Period: 𝑇

Frequency: 𝑓 = 1𝑇

Angular velocity (or angular frequency):

𝜔 = 2𝜋𝑓 = 2𝜋𝑇

Such a periodic function can be expressed in terms of a series of cosine and sine functions, known as the Fourier series:

𝑓 𝑡 =𝑎02 + � 𝑎𝑛 cos𝑛𝜔𝑡 + 𝑏𝑛 sin𝑛𝜔𝑡

∞

𝑛=1

=𝑎02 + � 𝑎𝑛 cos

2𝑛𝜋𝑡𝑇 + 𝑏𝑛 sin

2𝑛𝜋𝑡𝑇

∞

𝑛=1

where cos𝑛𝜔𝑡 and sin𝑛𝜔𝑡 are the harmonics.

1st (or fundamental) Harmonic: cos𝜔𝑡 and sin𝜔𝑡, (𝑛 = 1)

2nd Harmonic: cos2𝜔𝑡 and sin 2𝜔𝑡, (𝑛 = 2)

3rd Harmonic: cos3𝜔𝑡 and sin 3𝜔𝑡, (𝑛 = 3)

….

Significance of the Harmonics

3

𝑛 = 1

𝑛 = 1,2

𝑛 = 1,2,3

𝑛 = 1,2,3,4

Fourier series of a square wave containing various orders of harmonics

Orthogonality

4

𝑓 𝑡 and 𝑔 𝑡 are orthogonal to each other over the interval 𝑎 ≤ 𝑡 ≤ 𝑏 if

� 𝑓 𝑡𝑏

𝑎𝑔 𝑡 𝑑𝑡 = 0

In fact, the harmonics cos𝑛𝜔𝑡 and sin𝑛𝜔𝑡 (𝑛 = 0,1,2,⋯) form an infinite collection of periodic functions that are mutually orthogonal on the interval − 𝑇/2 ≤ 𝑡 ≤ 𝑇/2 because

� cos 𝑚𝜔𝑡𝑇/2

−𝑇/2cos 𝑛𝜔𝑡 𝑑𝑡 = 0, for 𝑚 ≠ 𝑛

� sin 𝑚𝜔𝑡𝑇/2

−𝑇/2sin 𝑛𝜔𝑡 𝑑𝑡 = 0, for 𝑚 ≠ 𝑛

� cos 𝑚𝜔𝑡𝑇/2

−𝑇/2sin 𝑛𝜔𝑡 𝑑𝑡 = 0

Fourier Coefficients

5

𝑎𝑛 =2𝑇� 𝑓 𝑡𝑇/2

−𝑇/2cos 𝑛𝜔𝑡 𝑑𝑡, for 𝑛 = 0,1,2,⋯

𝑏𝑛 =2𝑇� 𝑓 𝑡 sin 𝑛𝜔𝑡 𝑑𝑡

𝑇2

−𝑇2

, for 𝑛 = 1,2,3,⋯

Please note 𝑎0 has been included in the first set of integrals.

𝑓 𝑡 =𝑎02 + � 𝑎𝑛 cos𝑛𝜔𝑡 + 𝑏𝑛 sin𝑛𝜔𝑡

∞

𝑛=1

Example: Determine the Fourier series for the function

𝑓 𝑡 = �2 1 + 𝑡 −1 < 𝑡 < 00 0 < 𝑡 < 1

𝑓 𝑡 + 2 = 𝑓 𝑡 −1 1 −2 2

2

𝑓 𝑡

𝑡 0

Example (Cont’d)

6

𝑓 𝑡 =𝑎02 + � 𝑎𝑛 cos

2𝑛𝜋𝑡𝑇

+ 𝑏𝑛 sin2𝑛𝜋𝑡𝑇

∞

𝑛=1

=𝑎02 + � 𝑎𝑛 cos𝑛𝜋𝑡 + 𝑏𝑛 sin𝑛𝜋𝑡

∞

𝑛=1

The coefficients: 𝑎𝑛 = ∫ 𝑓 𝑡1−1 cos 𝑛𝜋𝑡 𝑑𝑡 = ∫ 2 1 + 𝑡0

−1 cos 𝑛𝜋𝑡 𝑑𝑡, 𝑛 = 0,1,2,⋯

For 𝑛 = 0

𝑎0 = � 2 1 + 𝑡0

−1𝑑𝑡 = 2𝑡 + 𝑡2 −1

0 = 1

For 𝑛 ≠ 0, we can use integral by parts

𝑎𝑛 = � 2 1 + 𝑡0

−1cos 𝑛𝜋𝑡 𝑑𝑡 =

2𝑛𝜋 1 + 𝑡 sin 𝑛𝜋𝑡 −1

0 − � sin 𝑛𝜋𝑡 𝑑𝑡0

−1

=2𝑛𝜋 2 1 − cos 𝑛𝜋 = �

0, 𝑛 even4𝑛𝜋 2 , 𝑛 odd

Example (Cont’d)

7

Similarly

𝑏𝑛 = � 𝑓 𝑡1

−1sin 𝑛𝜋𝑡 𝑑𝑡 = � 2 1 + 𝑡

0

−1sin 𝑛𝜋𝑡 𝑑𝑡 = −

2𝑛𝜋

, 𝑛 = 1,2,3,⋯

The first few terms of the Fourier series:

𝑓 𝑡 =12

+4𝜋2

cos𝜋𝑡 +19

cos3𝜋𝑡 +1

25cos5𝜋𝑡 + ⋯

−2𝜋

sin𝜋𝑡 +12

sin2𝜋𝑡 +13

sin3𝜋𝑡 +14

sin4𝜋𝑡 + ⋯

Odd and Even Functions

8

Odd function: 𝒇 −𝒕 = −𝒇 𝒕 , symmetrical about the origin.

Sine function is odd.

Fourier series of an odd function contains only sine terms (𝑎𝑛 = 0,𝑛 = 0,1,2, …), because

� 𝑓 𝑡 cos 𝑛𝜔𝑡 𝑑𝑡𝑇/2

−𝑇/2

= 0 if 𝑓 𝑡 is odd.

Even function: 𝒇 −𝒕 = 𝒇 𝒕 , symmetrical about the 𝑦-axis

Cosine function is even.

Fourier series of an even function contains only cosine terms (𝑏𝑛 = 0,𝑛 = 1,2,3, …),

� 𝑓 𝑡 sin 𝑛𝜔𝑡 𝑑𝑡𝑇/2

−𝑇/2

= 0 if 𝑓 𝑡 is even.

Example (Even Function)

9

𝑥

𝑦

𝜋/2 3𝜋/2 −𝜋/2 −3𝜋/2

4

The function is symmetrical about the 𝑦-axis

𝑓 𝑥 =𝑎02 + � 𝑎𝑛 cos𝑛𝑥

∞

𝑛=1

𝑎0 =1𝜋� 𝑓 𝑥 𝑑𝑥𝜋

−𝜋=

2𝜋� 𝑓 𝑥 𝑑𝑥𝜋

0

=2𝜋� 4𝑑𝑥

𝜋/2

0= 4

𝑎𝑛 =1𝜋� 𝑓 𝑥 cos𝑛𝑥 𝑑𝑥

𝜋

−𝜋=

2𝜋� 𝑓 𝑥 cos𝑛𝑥 𝑑𝑥

𝜋

0=

2𝜋� 4 cos𝑛𝑥 𝑑𝑥

𝜋/2

0

=8𝑛𝜋

sin𝑛𝜋2 =

0, 𝑛 = 2𝑘, (𝑛 = 2,4,6,⋯ )8𝑛𝜋 , 𝑛 = 4𝑘 − 3, (n = 1,5,9,⋯ )

−8𝑛𝜋 , 𝑛 = 4𝑘 − 1, (𝑛 = 3,7,11,⋯ )

𝑘 = 1,2,3,⋯

Example (Odd Function)

10

𝑥

𝑦

𝜋 2𝜋 0 −𝜋

2

−2

A shift of 𝜋/2 in x-axis and shift of 2 in y-axis in the previous example change it into an odd function

𝑔 𝑥 = � 𝑏𝑛 sin𝑛𝑥∞

𝑛=1

𝑏𝑛 =1𝜋� 𝑓 𝑥 sin𝑛𝑥 𝑑𝑥

𝜋

−𝜋=

2𝜋� 𝑓 𝑥 sin𝑛𝑥 𝑑𝑥

𝜋

0

=4𝜋� sin𝑛𝑥 𝑑𝑥

𝜋

0=

4𝑛𝜋 1 − −1 𝑛 = �

0, 𝑛 even8𝑛𝜋

, 𝑛 odd

𝑓 𝑥 −𝜋2

− 2 =8𝑛𝜋

cos 𝑥 −𝜋2

−13

cos 3 𝑥 −𝜋2

+15

cos 5 𝑥 −𝜋2

−17

cos 7 𝑥 −𝜋2

+ ⋯

𝑔 𝑥 =8𝑛𝜋

sin 𝑥 +13

sin 3𝑥 +15

sin 5𝑥 −17

sin7𝑥 + ⋯

Connection between the two examples

Complex Fourier Series

11

Euler’s Identity 𝑒𝑗𝑗 = cos𝜃 + 𝑗 sin𝜃 and 𝑒−𝑗𝑗 = cos𝜃 − 𝑗 sin𝜃

Thus, cos𝜃 = 𝑒𝑗𝑗+𝑒−𝑗𝑗

2 and sin𝜃 = 𝑒𝑗𝑗−𝑒−𝑗𝑗

2𝑗

Now the Fourier series

𝑓 𝑡 =𝑎02

+ � 𝑎𝑛 cos𝑛𝜔𝑡 + 𝑏𝑛 sin𝑛𝜔𝑡∞

𝑛=1

=𝑎02

+ �𝑎𝑛2

𝑒𝑗𝑛𝑗𝑗 + 𝑒−𝑗𝑛𝑗𝑗 +𝑏𝑛2𝑗

𝑒𝑗𝑛𝑗𝑗 − 𝑒−𝑗𝑛𝑗𝑗∞

𝑛=1

=𝑎02

+ �𝑎𝑛2

+ 𝑗𝑏𝑛2

𝑒−𝑗𝑛𝑗𝑗 +𝑎𝑛2− 𝑗

𝑏𝑛2

𝑒𝑗𝑛𝑗𝑗∞

𝑛=1

= � 𝑐𝑛

∞

𝑛=−∞

𝑒𝑗𝑛𝑗𝑗

where the complex coefficients

𝑐𝑛 =𝑎𝑛2− 𝑗

𝑏𝑛2

=1𝑇� 𝑓 𝑡 cos 𝑛𝜔𝑡 − 𝑗 sin 𝑛𝜔𝑡 𝑑𝑡𝑇2

−𝑇2

=1𝑇� 𝑓 𝑡 𝑒−𝑗𝑛𝑗𝑗𝑑𝑡𝑇2

−𝑇2

Obviously, 𝑐−𝑛 = 𝑐𝑛∗ = 1𝑇 ∫ 𝑓 𝑡 𝑒𝑗𝑛𝑗𝑗𝑑𝑡

𝑇2−𝑇2

, and 𝑐0 = 1𝑇 ∫ 𝑓 𝑡 𝑑𝑡

𝑇2−𝑇2

= 𝑎02

since 𝑏0 = 0

Example 1 (Complex Fourier)

12

𝑡

𝑓(𝑡)

𝑎/2

1

−𝑎/2 𝑇/2 −𝑇/2

𝑓 𝑡 = �0, −𝑇/2 < 𝑡 < −𝑎/21, −𝑎/2 < 𝑡 < 𝑎/20, 𝑎/2 < 𝑡 < 𝑇/2

,

and 𝑓 𝑡 + 𝑇 = 𝑓(𝑡)

𝑓(𝑡) = ∑ 𝑐𝑛∞𝑛=−∞ 𝑒𝑗𝑛𝑗𝑗, 𝜔 = 2𝜋

𝑇

𝑐𝑛 =1𝑇� 𝑓(𝑡)𝑒𝑗𝑛𝑗𝑗𝑑𝑡

𝑇/2

−𝑇/2=

1𝑇� 𝑒𝑗𝑛𝑗𝑗𝑑𝑡

𝑎/2

−𝑎/2=

1𝑗𝑛𝜔𝑇 𝑒

𝑗𝑛𝑗𝑗�−𝑎/2

𝑎/2

=1

𝑗𝑛𝜔𝑇 𝑒𝑗𝑛𝑗𝑎/2 − 𝑒−𝑗𝑛𝑗𝑎/2

=2

𝑛𝜔𝑇 sin𝑛𝜔𝑎

2 =𝑎𝑇

sin 𝑛𝜔𝑎/2𝑛𝜔𝑎/2 =

𝑎𝑇

sin 𝑛𝜋𝑎/𝑇𝑛𝜋𝑎/𝑇 =

𝑎𝑇 sinc

𝑛𝜋𝑎𝑇 , for 𝑛 ≠ 0

sinc 𝑥 = sin 𝑥𝑥

is undefined at 𝑥 = 0, but lim𝑥→0

sinc 𝑥 = lim𝑥→0

sin 𝑥𝑥

= 1,

If we define sinc 𝑥 = �sin 𝑥𝑥

, for 𝑥 ≠ 01, for 𝑥 = 0

and notice 𝑐0 = 1𝑇 ∫ 𝑓(𝑡)𝑑𝑡𝑇/2

−𝑇/2 = 1𝑇 ∫ 1𝑑𝑡𝑎/2

−𝑎/2 = 𝑎𝑇

,

then 𝑐𝑛 = 𝑎𝑇

sinc 𝑛𝜋𝑎𝑇

, for all 𝑛, −∞ < 𝑛 < ∞

Example (Complex Fourier)

13

𝑡

𝑔(𝑡)

𝑎

1

0 𝑇

𝑔 𝑡 = �1, 0 < 𝑡 < 𝑎 0, 𝑎 < 𝑡 < 𝑇 ,

and 𝑓 𝑡 + 𝑇 = 𝑓(𝑡)

𝑔(𝑡) = ∑ 𝑔𝑛∞𝑛=−∞ 𝑒𝑗𝑛𝑗𝑗, 𝜔 = 2𝜋

𝑇

𝑔𝑛 =1𝑇� 𝑓(𝑡)𝑒𝑗𝑛𝑗𝑗𝑑𝑡

𝑇

0=

1𝑇� 𝑒𝑗𝑛𝑗𝑗𝑑𝑡

𝑎

0=

1𝑗𝑛𝜔𝑇 𝑒

𝑗𝑛𝑗𝑗�0

𝑎

=1

𝑗𝑛𝜔𝑇 𝑒𝑗𝑛𝑗𝑎 − 1 =𝑒𝑗𝑛𝑗𝑎/2

𝑗𝑛𝜔𝑇 𝑒𝑗𝑛𝑗𝑎/2 − 𝑒−𝑗𝑛𝑗𝑎/2

=2𝑒𝑗𝑛𝑗𝑎/2

𝑛𝜔𝑇 sin𝑛𝜔𝑎

2 =𝑎𝑇

sin 𝑛𝜔𝑎/2𝑛𝜔𝑎/2 𝑒𝑗𝑛𝑗𝑎/2 =

𝑎𝑇 sinc

𝑛𝜋𝑎𝑇 𝑒𝑗𝑛π𝑎/𝑇 , Complex coefficients

The same result can be obtained from Example 1 by shifting the time origin by half the pulse width,

𝑔 𝑡 = 𝑓 𝑡 +𝑎2

= �𝑎𝑇

sinc𝑛𝜋𝑎𝑇

∞

𝑛=−∞

𝑒𝑗𝑛𝑗 𝑗+𝑎2 = �𝑎𝑇

sinc𝑛𝜋𝑎𝑇

𝑒𝑗𝑛π𝑎/𝑇∞

𝑛=−∞

𝑒𝑗𝑛𝑗𝑗

Complex Spectra

14

The Fourier coefficients of square-pulse wave all have

sinc 𝑛𝜋𝑎𝑇

, what does it look like?

In example 2, we have

𝑐𝑛 = 𝑎𝑇

sinc 𝑛𝜋𝑎𝑇

𝑒𝑗𝑛𝜋𝑎/𝑇 = 𝑐𝑛 𝑒𝑗𝜙𝑛

In both examples, their amplitudes

𝑐𝑛 = 𝑎𝑇

sinc 𝑛𝜋𝑎𝑇

for 𝑛 ≠ 0, and 𝑐0 = 𝑎𝑇

𝑐𝑛 describes the spectrum of 𝑓 𝑡 in the frequency domain

𝑐0 𝑐1 𝑐−1

𝑐2 𝑐−2

Power in Time and Frequency Domains

15

Power content of a periodic function

𝑓(𝑡) = � 𝑐𝑛

∞

𝑛=−∞

𝑒𝑗𝑛𝑗𝑗

is defined as the mean square value in time domain, and is related to the spectrum in the frequency domain

𝑃 =1𝑇� 𝑓2 𝑡 𝑑𝑡𝑇/2

−𝑇/2= � 𝑐𝑛 2

∞

𝑛=−∞

Because,

𝑃 =1𝑇� � 𝑐𝑛

∞

𝑛=−∞

𝑒𝑗𝑛𝑗𝑗𝑓(𝑡)𝑑𝑡𝑇/2

−𝑇/2= �

𝑐𝑛𝑇

∞

𝑛=−∞

� 𝑒𝑗𝑛𝑗𝑗𝑓(𝑡)𝑑𝑡𝑇/2

−𝑇/2

= �𝑐𝑛𝑇

∞

𝑛=−∞

𝑐−𝑛𝑇 = � 𝑐𝑛

∞

𝑛=−∞

𝑐−𝑛 = � 𝑐𝑛𝑐𝑛∗∞

𝑛=−∞

= � 𝑐𝑛 2∞

𝑛=−∞

Continuous Spectrum

16

In the example of the periodic square pulse function,

𝑓 𝑡 = ∑ 𝑐𝑛∞𝑛=−∞ 𝑒𝑗𝑛𝑗𝑗 = ∑ 𝑐𝑛∞

𝑛=−∞ 𝑒𝑗2𝑛𝜋𝑗/𝑇, and 𝑐𝑛 = 𝑎𝑇

sinc 𝑛𝜋𝑎𝑇

,

the spacing between two neighboring harmonics is 𝛿𝑥 = 𝜋𝑎𝑇

. If we increase the period 𝑇, this spacing 𝛿𝑥 decreases, and when 𝑇 → ∞, the function has a single pulse and no longer periodic, the spacing 𝛿𝑥 → 0, the spectrum becomes continuous.

𝑇 → ∞

Fourier Transform

17

We start from a periodic function 𝑓 𝑡 = ∑ 𝑐𝑛∞𝑛=−∞ 𝑒𝑗𝑛2𝜋𝑗/𝑇 , where 𝑐𝑛 = 1

𝑇 ∫ 𝑓 𝜏 𝑒−𝑗𝑛2𝜋𝑗/𝑇𝑑𝜏𝑇2−𝑇2

We can plug in 𝑐𝑛 into the series and use 𝛿𝜔 = 2𝜋𝑇

𝑓 𝑡 = �1𝑇� 𝑓 𝜏 𝑒−𝑗𝑛2𝜋𝑗/𝑇𝑑𝜏𝑇2

−𝑇2

∞

𝑛=−∞

𝑒𝑗𝑛2𝜋𝑗/𝑇 =12𝜋

� 𝛿𝜔� 𝑓 𝜏 𝑒−𝑗𝑛𝑗𝑗𝑗𝑑𝜏𝑇2

−𝑇2

∞

𝑛=−∞

𝑒𝑗𝑛𝑗𝑗𝑗

Let 𝑇 → ∞, then 𝛿𝜔 → 0, thus 𝛿𝜔 = 𝑑𝜔, and 𝑛𝛿𝜔 = 𝜔,

𝑓 𝑡 = �12𝜋

� 𝑓 𝜏 𝑒−𝑗𝑗𝑗𝑑𝜏∞

−∞

∞

𝑛=−∞

𝑒𝑗𝑗𝑗𝑑𝜔 =12𝜋

�12𝜋

� 𝑓 𝜏 𝑒−𝑗𝑗𝑗𝑑𝜏∞

−∞

∞

−∞𝑒𝑗𝑗𝑗𝑑𝜔

So if we introduce 𝐹 𝜔 = 12𝜋 ∫ 𝑓 𝑡 𝑒−𝑗𝑗𝑗𝑑𝑡∞

−∞ - Fourier transform

Then 𝑓 𝑡 = 12𝜋 ∫ 𝐹 𝜔∞

−∞ 𝑒𝑗𝑗𝑗𝑑𝜔 - inverse Fourier Transform

As 𝑇 → ∞, the discrete harmonic values 𝑛𝜔0 = 2𝜋𝑛/𝑇 become a continuous value 𝜔, and the discrete spectrum 𝑐𝑛 = 𝑐𝑛 𝑒𝑗𝜙𝑛 becomes the continuous spectrum 𝐹 𝜔 = 𝐹 𝜔 𝑒𝑗𝜙 𝑗 .

Example

18

𝐹 𝜔 =12𝜋

� 𝑓 𝑡 𝑒−𝑗𝑗𝑗𝑑𝑡∞

−∞=

12𝜋

� 𝑒−𝑗𝑗𝑗𝑑𝑡𝑎

0= −

1𝑗 2𝜋𝜔

𝑒−𝑗𝑗𝑗�0

𝑎

=1

𝑗 2𝜋𝜔1 − 𝑒−𝑗𝑗𝑎 =

𝑒−𝑗𝑗𝑎/2

𝑗 2𝜋𝜔𝑒𝑗𝑗𝑎/2 − 𝑒−𝑗𝑗𝑎/2 =

𝑒−𝑗𝑗𝑎/2

𝑗 2𝜋𝜔2𝑗 sin

𝜔𝑎2

=𝑎𝑒−𝑗𝑗𝑎/2

2𝜋

sin𝜔𝑎2𝜔𝑎2

=𝑎𝑒−𝑗𝑗𝑎/2

2𝜋sinc

𝜔𝑎2

𝐹 𝜔

𝜔

This is a continuous spectrum!

𝑎 0

1

𝑓(𝑡)

𝑡

𝑓 𝑡 = �1, 0 < 𝑡 < 𝑎0, otherwise

Properties of Fourier Transform

19

Linearity ℱ 𝛼1𝑓1 𝑡 + 𝛼2𝑓2 𝑡 = 𝛼1𝐹1 𝜔 + 𝛼2𝐹2 𝜔

Time shifting ℱ 𝑓 𝑡 − 𝑡0 = 𝑒𝑗𝑗𝑗0𝐹 𝜔 Frequency shifting ℱ 𝑒𝑗𝑗0𝑗𝑓 𝑡 = 𝐹 𝜔 − 𝜔0

Time scaling ℱ 𝑓 𝑘𝑡 = 1𝑘𝐹 𝑗

𝑘

Symmetry ℱ 𝐹 𝑡 = 𝑓 −𝜔

Proof: We start with the inverse Fourier transform and then employ two variable substitutions

𝑓 𝑡 = 12𝜋 ∫ 𝐹 𝜔∞

−∞ 𝑒𝑗𝑗𝑗𝑑𝜔 = 12𝜋 ∫ 𝐹 𝑢∞

−∞ 𝑒𝑗𝑗𝑗𝑑𝑢

Let 𝑡 → −𝜔, then 𝑓 −𝜔 = 12𝜋 ∫ 𝐹 𝑢∞

−∞ 𝑒−𝑗𝑗𝑗𝑑𝑢 = 12𝜋 ∫ 𝐹 𝑡∞

−∞ 𝑒−𝑗𝑗𝑗𝑑𝑡 = ℱ 𝐹 𝑡

Example: ℱ 1 =? (hard to solve without using the symmetry)

Since ℱ 𝛿(𝑡) = 12𝜋

, then using symmetry: ℱ 12𝜋

= 𝛿 −𝜔 = 𝛿 𝜔 , thus ℱ 1 = 2𝜋𝛿 𝜔 .

Cosine and Sine Transforms

20

𝐹 𝜔 =12𝜋

� 𝑓 𝑡 𝑒−𝑗𝑗𝑗𝑑𝑡∞

−∞=

12𝜋

� 𝑓 𝑡 cos𝜔𝑡 + 𝑗 sin𝜔𝑡 𝑑𝑡∞

−∞

If 𝑓 𝑡 is even function, ∫ 𝑓 𝑡 sin𝜔𝑡 𝑑𝑡∞−∞ = 0, then the cosine transform:

𝐹𝑐 𝜔 =2𝜋� 𝑓 𝑡 cos𝜔𝑡 𝑑𝑡

∞

0

If 𝑓 𝑡 is odd function, ∫ 𝑓 𝑡 cos𝜔𝑡 𝑑𝑡∞−∞ = 0, then the sine transform:

𝐹𝑠 𝜔 =2𝜋� 𝑓 𝑡 sin𝜔𝑡 𝑑𝑡

∞

0

For those functions that are defined only for 𝑡 ≥ 0, and the extension into 𝑡 < 0 can make them either even or odd, where cosine and sine transforms can then be used, respectively.

Example

21

Example: 𝑓 𝑡 = � 1, 0 < 𝑡 < 𝑎 0, 𝑎 ≤ 𝑡 < ∞ and

undefined in −∞ < 𝑡 < 0

𝑎 0

1 𝑓(𝑡)

𝑡

𝑎 0

1

𝑡 −𝑎

𝑎 0

1

𝑡 −𝑎

−1

Sine transform:

𝐹𝑠 𝜔 =2𝜋� 𝑓 𝑡 sin𝜔𝑡 𝑑𝑡∞

0=

2𝜋� sin𝜔𝑡 𝑑𝑡𝑎

0 =

2𝜋

1 − cos𝜔𝑎𝜔

=2𝜋

2𝜔

sin2𝜔𝑎2

=2𝜋𝜔𝑎2

2sin2 𝜔𝑎2𝜔𝑎2

2 =𝜔𝑎2

2𝜋sinc2

𝜔𝑎2

𝐹𝑐 𝜔 =2𝜋� 𝑓 𝑡 cos𝜔𝑡 𝑑𝑡∞

0=

2𝜋� cos𝜔𝑡 𝑑𝑡𝑎

0

=2𝜋

sin𝜔𝑎𝜔

=2𝜋𝑎sinc 𝜔𝑎

Cosine transform:

Summary Key points:

Periodic functions

Fourier series and coefficients

Significance of harmonics

Sine and cosine for odd and even functions

Complex Fourier series and discrete spectrum

Fourier transform and continuous spectrum

Properties of Fourier transform

22