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Fourier Transform

S H A H R E A R K H A N R A S E L S E N I O R L E C T U R E R D E P T . O F G E D

D A F F O D I L I N T E R N A T I O N A L U N I V E R S I T Y

▪ Virtually everything in the world can be described via a waveform - a function of time, space or some other variable.

▪ For instance, sound waves, electromagnetic fields, the elevation of a hill versus location, a plot of VSWR versus frequency, the price of your favorite stock versus time, etc.

▪ The Fourier Transform gives us a unique and

powerful way of viewing these waveforms.

Fourier Transform

Finite Transform Infinite Transform Complex Transform Finite

Fourier Sine Transform

Finite Fourier

Cosine Transform

Infinite Fourier

Sine Transform

Infinite Fourier

Cosine

Transform

Finite Transforms

Finite Fourier Sine transform: The finite Fourier sine transform of 𝐹 𝑥 , 0 < 𝑥 < 𝑙, is defined as 𝑓_𝑠 𝑛 = ׬₀^𝑙 𝐹(𝑥) sin^𝑛𝜋𝑥

𝑙 𝑑𝑥 where 𝑛 is an integer.

The function 𝐹(𝑥)is then called the inverse finite Fourier sine transform of 𝑓_𝑠 𝑛 , and is given by 𝐹 𝑥 = ²

𝑙 σ_𝑛=1^∞ 𝑓_𝑠 𝑛 sin^𝑛𝜋𝑥

𝑙

Finite Fourier Cosine transform: The finite Fourier cosine transform of 𝐹 𝑥 , 0 < 𝑥 < 𝑙, is defined as 𝑓_𝑐 𝑛 = ׬₀^𝑙𝐹(𝑥) cos^𝑛𝜋𝑥

𝑙 𝑑𝑥 where 𝑛 is an integer.

The function 𝐹(𝑥) is then called the inverse finite Fourier cosine transform of 𝑓_𝑐 𝑛 , and is given by 𝐹 𝑥 = ¹

𝑙 𝑓_𝑐 0 + ²

𝑙 σ_𝑛=1^∞ 𝑓_𝑐 𝑛 cos^𝑛𝜋𝑥

𝑙

Finite Fourier Transforms are used to solve partial differential equations.

Infinite Transforms

Infinite Finite Fourier Sine Transform: The Infinite Fourier Sine transform of 𝐹(𝑥), 0 < 𝑥 < ∞, is defined as 𝑓_𝑠 𝑛 = ׬₀^∞𝐹(𝑥) sin 𝑛𝑥 𝑑𝑥 ; where 𝑛 is an integer.

The function 𝐹(𝑥)is then called the inverse infinite Fourier sine transform of 𝑓_𝑠 𝑛 , and is given by 𝐹 𝑥 = ²

𝜋׬₀^∞𝑓_𝑠 𝑛 sin 𝑛𝑥 𝑑𝑛 ; where 𝑛 is an integer.

Infinite Finite Fourier Cosine Transform: The infinite Fourier cosine transform of 𝐹(𝑥), 0 < 𝑥 < ∞, is defined as 𝑓_𝑐 𝑛 = ׬₀^∞𝐹(𝑥) cos 𝑛𝑥 𝑑𝑥 where 𝑛 is an integer.

The function 𝐹(𝑥)is then called the inverse infinite Fourier cosine transform of 𝑓_𝑐 𝑛 , and is given by 𝐹 𝑥 = ²

𝜋׬₀^∞𝑓_𝑐 𝑛 cos 𝑛𝑥 𝑑𝑛 ; where 𝑛 is an integer

Example 5.7 Find the Fourier Sine Transform of 𝒔𝒊𝒏 𝒌𝒙, 𝟎 ≤ 𝒙 ≤ 𝝅, 𝒌 > 𝟎.

Solution: By the definition of Fourier sine Transform of 𝐹(𝑥) for 0 ≤ 𝑥 ≤ 𝑙, we have 𝑓_𝑠 𝑛 = ׬₀^𝑙𝐹 𝑥 sin^𝑛𝜋𝑥

𝑙 𝑑𝑥

∴ 𝑓_𝑠 𝑛 = ׬₀^𝜋sin 𝑘𝑥 sin^𝑛𝜋𝑥

𝜋 𝑑𝑥 = ¹

2׬₀^𝜋2 sin 𝑘𝑥 sin 𝑛𝑥 𝑑𝑥 = ¹

2׬₀^𝜋 cos 𝑘 − 𝑛 𝑥 − cos 𝑘 + 𝑛 𝑥 𝑑𝑥

= ¹

2

sin 𝑘−𝑛 𝑥

𝑘−𝑛 − ^{sin 𝑘+𝑛 𝑥}

𝑘+𝑛 0

𝜋 = ¹

2

sin 𝑘−𝑛 𝜋

𝑘−𝑛 − ^{sin 𝑘+𝑛 𝜋}

𝑘+𝑛

= ¹

2

sin 𝑘𝜋 cos 𝑛𝜋−cos 𝑘𝜋 sin 𝑛𝜋

𝑘−𝑛 − sin 𝑘𝜋 cos 𝑛𝜋+cos 𝑘𝜋 sin 𝑛𝜋

𝑘+𝑛 ;

= ¹

2

sin 𝑘𝜋 cos 𝑛𝜋

𝑘−𝑛 − sin 𝑘𝜋 cos 𝑛𝜋

𝑘+𝑛 ; [∵ sin 𝑛𝜋 = 0]

= ¹

2 1

𝑘−𝑛 − ¹

𝑘+𝑛 sin 𝑘𝜋 cos 𝑛𝜋 = ^2𝑛

2 𝑘²−𝑛² −1 ^𝑛 sin 𝑘𝜋 [∵ cos 𝑛𝜋 = −1 ^𝑛]

= ^𝑛

𝑘²−𝑛² −1 ^𝑛sin 𝑘𝜋 ∴ 𝑓_𝑠 𝑛 = ^𝑛

𝑘²−𝑛² −1 ^𝑛 sin 𝑘𝜋

Example 5.8 Find the Fourier Cosine Transform of 𝒔𝒊𝒏 𝒌𝒙, 𝟎 ≤ 𝒙 ≤ 𝝅, 𝒌 > 𝟎.

Solution: By the definition of Fourier Cosine Transform of 𝐹(𝑥) for 0 ≤ 𝑥 ≤ 𝑙, we have 𝑓_𝑐 𝑛 = ׬₀^𝑙𝐹 𝑥 cos^𝑛𝜋𝑥

𝑙 𝑑𝑥

∴ 𝑓_𝑐 𝑛 = ׬₀^𝜋sin 𝑘𝑥 cos^𝑛𝜋𝑥

𝜋 𝑑𝑥 = ¹

2׬₀^𝜋2 sin 𝑘𝑥 cos 𝑛𝑥 𝑑𝑥 = ¹

2׬₀^𝜋 sin 𝑘 + 𝑛 𝑥 + s𝑖𝑛 𝑘 − 𝑛 𝑥 𝑑𝑥

= ¹

2

−cos 𝑘+𝑛 𝑥

𝑘+𝑛 − ^{cos 𝑘−𝑛 𝑥}

𝑘−𝑛 0

𝜋 = ¹

2

−cos 𝑘+𝑛 𝜋

𝑘+𝑛 − ^{cos 𝑘−𝑛 𝜋}

𝑘−𝑛 + ¹

𝑘+𝑛 + ¹

𝑘−𝑛

= ¹

2 −cos 𝑘𝜋 cos 𝑛𝜋−sin 𝑘𝜋 sin 𝑛𝜋

𝑘+𝑛 − cos 𝑘𝜋 cos 𝑛𝜋+sin 𝑘𝜋 sin 𝑛𝜋

𝑘−𝑛 + ^{𝑘−𝑛+𝑘+𝑛}

(𝑘+𝑛)(𝑘−𝑛)

= ¹

2 −cos 𝑘𝜋 cos 𝑛𝜋(𝑘−𝑛+𝑘+𝑛)

𝑘²−𝑛² + ^2𝑘

𝑘²−𝑛² ; [sin 𝑛𝜋 = 0]

= ¹

2 −^{2𝑘 −1 𝑛cos 𝑘𝜋}

𝑘²−𝑛² + ^2𝑘

𝑘²−𝑛²

= ^𝑘

𝑘²−𝑛² 1 − −1 ^𝑛cos 𝑘𝜋 = ^𝑘

𝑛²−𝑘² −1 ^𝑛cos 𝑘𝜋 − 1

∴ 𝑓_𝑐 𝑛 = ^𝑘

𝑛²−𝑘² −1 ^𝑛 cos 𝑘𝜋 − 1

Also we can show that, 𝐹 𝑥 = sin 𝑘𝑥 , 0 ≤ 𝑥 ≤ 𝜋, 𝑘 > 0, 𝑓_𝑠 𝑛 + 𝑓_𝑐 𝑛 = 1

𝑘² − 𝑛² −1 ^𝑛 𝑛 sin 𝑘𝜋 + 𝑘 cos 𝑘𝜋 − 𝑘

Example 5.9 For 𝐹 𝑥 = 𝑒

^𝑚𝑥

, 0 ≤ 𝑥 ≤ 𝜋, 𝑚 > 0, Show that 𝑓

_𝑠

𝑛 + 𝑓

_𝑐

𝑛 = (𝑚 − 𝑛)

𝑚

²

+ 𝑛

²

𝑒

^𝑚𝜋

cos 𝑛𝜋 − 1

Solution: By the definition of Fourier sine Transform of 𝐹(𝑥) for 0 ≤ 𝑥 ≤ 𝑙, we have 𝑓

_𝑠

𝑛 = ׬

₀^𝑙

𝐹 𝑥 sin

^𝑛𝜋𝑥

𝑙

𝑑𝑥

⇒ 𝑓

_𝑠

𝑛 = ׬

₀^𝜋

𝑒

^𝑚𝑥

sin 𝑛𝑥 𝑑𝑥 =

^𝑒𝑚𝑥 𝑚 sin 𝑛𝑥−𝑛 cos 𝑛𝑥 𝑚²+𝑛² 0

𝜋

=

¹

𝑚²+𝑛²

𝑒

^𝑚𝜋

𝑚 sin 𝑛𝜋 − 𝑛 cos 𝑛𝜋 − 𝑒

⁰

𝑚 sin 0 − 𝑛 cos 0

=

¹

𝑚²+𝑛²

𝑒

^𝑚𝜋

0 − 𝑛 cos 𝑛𝜋 − 1 0 − 𝑛. 1

=

^{𝑛(1−𝑒𝑚𝜋cos 𝑛𝜋)}

𝑚²+𝑛²

∴ 𝑓

_𝑠

𝑛 =

^{𝑛(1−𝑒𝑚𝜋 cos 𝑛𝜋)}

𝑚²+𝑛²

By the definition of Fourier Cosine Transform of 𝐹(𝑥) for 0 ≤ 𝑥 ≤ 𝑙 , we have 𝑓

_𝑐

𝑛 = ׬

₀^𝑙

𝐹 𝑥 cos

^𝑛𝜋𝑥

𝑙

𝑑𝑥

⇒ 𝑓

_𝑐

𝑛 = ׬

₀^𝜋

𝑒

^𝑚𝑥

cos 𝑛𝑥 𝑑𝑥 =

^𝑒𝑚𝑥 𝑚 cos 𝑛𝑥+𝑛 sin 𝑛𝑥 𝑚²+𝑛² 0

𝜋

=

¹

𝑚²+𝑛²

𝑒

^𝑚𝜋

𝑚 cos 𝑛𝜋 + 𝑛 sin 𝑛𝜋 − 𝑒

⁰

𝑚 cos 0 + 𝑛 sin 0

=

¹

𝑚²+𝑛²

𝑒

^𝑚𝜋

𝑚 cos 𝑛𝜋 − 0 − 1 𝑚 + 0

=

^{−𝑚(1−𝑒𝑚𝜋 cos 𝑛𝜋)}

𝑚²+𝑛²

∴ 𝑓

_𝑐

𝑛 =

^{−𝑚(1−𝑒𝑚𝜋 cos 𝑛𝜋)}

𝑚²+𝑛²

Therefore, 𝑓 𝑛 + 𝑓 𝑛 =

^{𝑛(1−𝑒𝑚𝜋 cos 𝑛𝜋)}

+

^{−𝑚(1−𝑒𝑚𝜋 cos 𝑛𝜋)}

=

^{(𝑚−𝑛)}

𝑒

^𝑚𝜋

cos 𝑛𝜋 − 1 .

Example 5.10 Find the Fourier Sine and Cosine Transform of 𝒆

^−𝒎𝒙

, 𝒙 ≥ 𝟎. Hence show that 𝒇

_𝒔

𝒏 + 𝒇

_𝒄

𝒏 =

^𝒎+𝒏

𝒎^𝟐+𝒏^𝟐

.

Solution: By the definition of Infinite Fourier Sine Transform of 𝐹(𝑥) for 0 ≤ 𝑥 < ∞, we have 𝑓

_𝑠

𝑛 = ׬

₀^∞

𝐹 𝑥 sin 𝑛𝑥 𝑑𝑥

∴ 𝑓

_𝑠

𝑛 = ׬

₀^∞

𝑒

^−𝑚𝑥

sin 𝑛𝑥 𝑑𝑥

=

^{𝑒−𝑚𝑥} − 𝑚 sin 𝑛𝑥−𝑛 cos 𝑛𝑥

𝑚²+𝑛² 𝑥=0

∞

= 0 −

^{𝑒−𝑚.0} − 𝑚 sin 0−𝑛 cos 0

𝑚²+𝑛²

; [∵ 𝑒

^−∞

= 0]

⇒ 𝑓

_𝑠

𝑛 =

^𝑛

𝑚²+𝑛²

.

By the definition of Infinite Fourier Cosine Transform of 𝐹(𝑥) for 0 ≤ 𝑥 < ∞, we have 𝑓

_𝑐

𝑛 = ׬

₀^∞

𝐹 𝑥 cos 𝑛𝑥 𝑑𝑥

∴ 𝑓

_𝑐

𝑛 = ׬

₀^∞

𝑒

^−𝑚𝑥

cos 𝑛𝑥 𝑑𝑥

=

^{𝑒−𝑚𝑥} − mcos 𝑛𝑥+𝑛 sin 𝑛𝑥

𝑚²+𝑛² 𝑥=0

∞

= 0 −

^{𝑒−𝑚.0} − mcos 0+𝑛 sin 0

𝑚²+𝑛²

; [∵ 𝑒

^−∞

= 0]

=

^𝑚

𝑚²+𝑛²

⇒ 𝑓

_𝑐

𝑛 =

^𝑚

𝑚²+𝑛²

.

Therefore, 𝑓

_𝑠

𝑛 + 𝑓

_𝑐

𝑛 =

^𝑚+𝑛

𝑚²+𝑛²

.

Complex Fourier Transform:

The complex Fourier transform of 𝐹(𝑥) is defined by 𝑓 𝑛 = ׬_−∞^∞ 𝐹 𝑥 𝑒^{−𝑖𝑛𝑥} 𝑑𝑥.

Example 5.13 Find the complex Fourier transform of 𝑭 𝒙 = 𝒆^{−𝒂|𝒙|} where 𝒂 > 𝟎.

Solution. We have 𝑥 = ቊ−𝑥 𝑖𝑓 𝑥 < 0 𝑥 𝑖𝑓 𝑥 > 0

∴ 𝑓 𝑛 = ׬_−∞^∞ 𝑒^{−𝑎 𝑥} 𝑒^{−𝑖𝑛𝑥} 𝑑𝑥 = ׬_−∞⁰ 𝑒^{−𝑎(−𝑥)}𝑒^{−𝑖𝑛𝑥} 𝑑𝑥 + ׬₀^∞𝑒^−𝑎𝑥𝑒^{−𝑖𝑛𝑥} 𝑑𝑥

= ׬_−∞⁰ 𝑒^{𝑎𝑥−𝑖𝑛𝑥} 𝑑𝑥 + ׬₀^∞𝑒^{−𝑎𝑥−𝑖𝑛𝑥} 𝑑𝑥 = ׬_−∞⁰ 𝑒^{(𝑎−𝑖𝑛)𝑥} 𝑑𝑥 + ׬₀^∞𝑒^{−(𝑎+𝑖𝑛)𝑥} 𝑑𝑥

= ^{𝑒 𝑎−𝑖𝑛 𝑥}

𝑎−𝑖𝑛 −∞

0

− ^{𝑒− 𝑎+𝑖𝑛 𝑥}

𝑎+𝑖𝑛 0

∞

= ¹

𝑎−𝑖𝑛 𝑒⁰ − 𝑒^−∞ − ¹

𝑎+𝑖𝑛 𝑒^−∞ − 𝑒⁰

= ^2𝑎

𝑎²+𝑛²

1. Find the Fourier Sine and Cosine Transform of 𝐹 𝑥 = cos 𝑚𝑥 , 0 ≤ 𝑥 ≤ 𝜋, 𝑚 > 0, hence show that 𝑓_𝑠 𝑛 + 𝑓_𝑐 𝑛 = ¹

𝑚²−𝑛² −1 ^𝑛 𝑚 sin 𝑚𝜋 + 𝑛 cos 𝑚𝜋 − 𝑛 Ans. 𝑓_𝑠 𝑛 = ^𝑛

𝑚²−𝑛² −1 ^𝑛 cos 𝑚𝜋 − 1 , 𝑓_𝑐 𝑛 = ^𝑚

𝑚²−𝑛² −1 ^𝑛 sin 𝑚𝜋

2. Find the Fourier Sine and Cosine Transform of 𝐹 𝑥 = 2𝑥, 0 ≤ 𝑥 ≤ 4, 𝑚 > 0, hence show that 𝑓_𝑠 𝑛 + 𝑓_𝑐 𝑛 = ³²

𝑛²𝜋²[cos 𝑛𝜋 1 − 𝑛𝜋 − 1] ;Ans. 𝑓_𝑠 𝑛 = ^{−32 cos 𝑛𝜋}

𝑛𝜋 , 𝑓_𝑐 𝑛 = 32(cos 𝑛𝜋−1) 𝑛²𝜋²

3. Find the Fourier Sine and Cosine Transform of 𝐹 𝑥 = 𝑥², 0 ≤ 𝑥 ≤ 1 Ans. 𝑓_𝑠 𝑛 = ^{−2+(2−𝑛2𝜋2) cos 𝑛𝜋}

𝑛³𝜋³ , 𝑓_𝑐 𝑛 = ^{2 cos 𝑛𝜋}

𝑛²𝜋²

4. Find the Fourier Sine transform of 𝐹 𝑥 = 2𝑒^−5𝑥 + 5𝑒^−2𝑥, 0 ≤ 𝑥 < ∞ Ans. ^8𝑛

3+133𝑛 (𝑛²+25)(𝑛²+4)

5. Find the Fourier Cosine transform of 𝐹 𝑥 = 2𝑒^−5𝑥 + 5𝑒^−2𝑥, 0 ≤ 𝑥 < ∞ Ans. ^20𝑛

2+290 (𝑛²+25)(𝑛²+4)

Exercises