RIPHAH INTERNATIONAL
UNIVERSITY
Signals And Systems
Engr. M.S. Orakzai
Review of the Fourier Series
Review of Fourier Series
• Deal with continuous-time periodic signals.
A Periodic Signal
f(t)
t
T
2T
3T
Two Forms for Fourier Series
Sinusoidal
Form
∞
𝑥 𝑡 = 𝑎0 + ෍ 𝑎𝑛 cos 𝑛𝜔0 𝑡 + 𝑏𝑛 s𝑖𝑛 𝑛𝜔0 𝑡
𝑛=1
2 T /2
a0 = 
f (t )dt
−
T
/
2
T
Complex
Form:
∞
𝑥 𝑡 = ෍ 𝐶𝑛 𝑒
𝑛=−∞
𝑗𝑛𝜔0 𝑡
2 T /2
an = 
f (t ) cos n0tdt
−
T
/
2
T
2 T /2
bn = 
f (t ) sin n0tdt
−
T
/
2
T
1
cn =
T

T /2
−T / 2
f (t )e − jn0t dt
How to deal with Aperiodic
Signals?
How to Deal with Aperiodic Signal?
A Periodic Signal
f(t)
t
T
If T→, what happens?
• The Fourier series expansion reveals the frequency content of the
periodic signal.
• It is also possible to analyze the frequency content of nonperiodic
signals. The tool that enables us to do this is the Fourier transform
• The transform assumes that a nonperiodic (or aperiodic) signal is a
periodic signal with an infinite period
Continuous-Time Fourier Transform
Fourier Integral derivation
2
0 =
T
Let
1 0
=
T 2
2
 = 0 =
T
T →   d =   0
Fourier Transform Pair
Fourier Transform:
∞
𝑋 𝜔 = න 𝑥(𝑡)𝑒 −𝑗𝜔𝑡 𝑑𝑡
−∞
Inverse Fourier Transform:
1 ∞
න 𝑋(𝜔)𝑒 𝑗𝜔𝑡 𝑑𝜔
𝑥 𝑡 =
2𝜋 −∞
𝑥 𝑡 is time domain signal
𝑋 𝜔 is frequency domain signal
Fourier Transform and Inverse Fourier transform are used convert between time and
frequency domain
Note: some books write 𝑋 𝜔 as 𝑋 𝑗𝜔 , its just a notation, don’t be confused
Example-1: FT of Exponential Signal
• Find the Fourier transform of
𝑥 𝑡 = 𝑒 −𝑎𝑡 𝑢 𝑡
𝑎>0
Solution
∞
𝑋 𝜔 = න 𝑥(𝑡)𝑒 −𝑗𝜔𝑡 𝑑𝑡
−∞
∞
1
− 𝑎+𝑗𝜔 𝑡
𝑋 𝜔 =−
.𝑒
ቚ
𝑎 + 𝑗𝜔
0
∞
𝑋 𝜔 = න 𝑒 −𝑎𝑡 𝑢 𝑡 𝑒 −𝑗𝜔𝑡 𝑑𝑡
𝑋 𝜔 =−
−∞
1
. 𝑒 −∞ − 𝑒 0
𝑎 + 𝑗𝜔
∞
𝑋 𝜔 = න 𝑒 −𝑎𝑡 . 𝑒 −𝑗𝜔𝑡 𝑑𝑡
𝑋 𝜔 =
0
1
𝑎 + 𝑗𝜔
∞
𝑋 𝜔 = න 𝑒−
0
𝑎+𝑗𝜔 𝑡 𝑑𝑡
𝑒 −𝑎𝑡 𝑢 𝑡
𝐹𝑇
1
𝑎 + 𝑗𝜔
Example-2: FT of Rectangular Pulse
• Find the Fourier transform of
1 −𝑇 ≤ 𝑡 ≤ 𝑇
𝑥 𝑡 =ቊ
0
𝑡 >𝑇
Solution
∞
𝑋 𝜔 =න
𝑥(𝑡)𝑒 −𝑗𝜔𝑡 𝑑𝑡
−∞
𝑇
𝑋 𝜔 = න 𝑒 −𝑗𝜔𝑡 𝑑𝑡
−𝑇
1 −𝑗𝜔𝑡 𝑇
𝑋 𝜔 = − .𝑒
ቚ
𝑗𝜔
−𝑇
𝑋 𝜔 =−
1 −𝑗𝜔𝑇
𝑒
− 𝑒 𝑗𝜔𝑇
𝑗𝜔
2 𝑒 𝑗𝜔𝑇 − 𝑒 −𝑗𝜔𝑇
𝑋 𝜔 =
𝜔
2𝑗
2
𝑋 𝜔 = sin(𝜔𝑇)
𝜔
𝑋 𝜔 = 2𝑇.
sin 𝜔𝑇
𝜔𝑇
𝑋 𝜔 = 2𝑇. sin𝑐 𝜔𝑇
Example-2: Cont…
𝑟𝑒𝑐𝑡
𝑡
2𝑇
𝑟𝑒𝑐𝑡
𝑡
𝑇
𝐹𝑇
2𝑇𝑠𝑖𝑛𝑐(𝜔𝑇)
or
𝐹𝑇
𝑇𝑠𝑖𝑛𝑐(𝜔 𝑇/2)
Example-2: Cont…
𝑡
𝑟𝑒𝑐𝑡
𝑇
𝐹𝑇
𝑇𝑠𝑖𝑛𝑐(𝜔 𝑇/2)
X a ()
2
−2
𝑟𝑒𝑐𝑡
𝑡
2
𝐹𝑇
2 𝑠𝑖𝑛𝑐(𝜔)
−


2
Example-2: Cont…
𝑡
𝑟𝑒𝑐𝑡
𝑇
𝑋 𝜔 = 4𝑠𝑖𝑛𝑐 2𝜔
𝐹𝑇
𝑇𝑠𝑖𝑛𝑐(𝜔 𝑇/2)
Example-3: IFT of Rectangular Pulse
• Find the Inverse FT of rectangular
spectrum
1 −𝑊 ≤ 𝜔 ≤ 𝑊
𝑋(𝜔) = ቊ
0
𝜔 >𝑊
Solution:
1 ∞
𝑥 𝑡 =
න 𝑋(𝑗𝜔)𝑒 𝑗𝜔𝑡 𝑑𝜔
2𝜋 −∞
1 𝑒 𝑗𝑊𝑡 − 𝑒 −𝑗𝑊𝑡
=
𝜋𝑡
2𝑗
1 𝑊 𝑗𝜔𝑡
𝑥 𝑡 =
න 𝑒 𝑑𝜔
2𝜋 −𝑊
1
=
sin(𝑊𝑡)
𝜋𝑡
𝑊
1
𝑗𝜔𝑡
=
.𝑒 ቚ
2𝑗𝜋𝑡
−𝑊
1
=
𝑒 𝑗𝑊𝑡 − 𝑒 −𝑗𝑊𝑡
2𝑗𝜋𝑡
𝑊 sin 𝑊𝑡
=
.
𝜋
𝑊𝑡
=
𝑊
𝑠𝑖𝑛𝑐(𝑊𝑡)
𝜋
Example-3: cont…
Example-4: FT of Impulse
Unit Impulse
𝑥 𝑡 = 𝛿(𝑡)
Solution:
∞
𝑋 𝜔 = න 𝛿(𝑡) 𝑒 −𝑗𝜔𝑡 𝑑𝑡
−∞
∞
𝑋 𝜔 = න 1. 𝑒 −𝑗𝜔𝑡 𝑑𝑡
−∞
𝑋 𝜔 = 𝑒 −𝑗𝜔𝑡 ቚ
𝛿(𝑡)
𝐹𝑇
𝑡=0
1
Example-5: IFT of Impulse
• Find the Inverse FT of 𝑋 𝜔 = 2𝜋𝛿(𝜔)
1 ∞
𝑥 𝑡 =
න 𝑋(𝑗𝜔)𝑒 𝑗𝜔𝑡 𝑑𝜔
2𝜋 −∞
1 ∞
𝑥 𝑡 =
න 2𝜋𝛿(𝜔)𝑒 𝑗𝜔𝑡 𝑑𝜔
2𝜋 −∞
𝑥 𝑡 =1
1
𝐹𝑇
2𝜋𝛿(𝜔)
Example-6: FT of 𝑒 𝑗𝜔0𝑡
• Find the Fourier Transform of 𝑥 𝑡 = 𝑒 𝑗𝜔0 𝑡
Solution:
We know that
1
𝐹𝑇
2𝜋𝛿(𝜔)
∞
න 𝑒 −𝑗𝜔𝑡 𝑑𝑡 = 2𝜋𝛿(𝜔)
−∞
∞
ℱ 𝑒 𝑗𝜔0 𝑡 = න 𝑒 𝑗𝜔0 𝑡 . 𝑒 −𝑗𝜔𝑡 𝑑𝑡
∞
−∞
= න 𝑒 −𝑗(𝜔−𝜔0 )𝑡 𝑑𝑡
−∞
= 2𝜋𝛿 𝜔 − 𝜔0
Example-6: FT of 𝑒 𝑗𝜔0𝑡
• Find the Fourier Transform of 𝑥 𝑡 = 𝑒 𝑗𝜔0 𝑡
Solution:
We know that
𝛿(𝑡)
𝛿 𝑡 =
∞
𝐹𝑇
ℱ −1
1
1
1 ∞
𝛿 𝑡 =
න 1. 𝑒 𝑗𝜔𝑡 𝑑𝜔
2𝜋 −∞
ℱ 𝑒 𝑗𝜔0 𝑡 = න 𝑒 𝑗𝜔0 𝑡 . 𝑒 −𝑗𝜔𝑡 𝑑𝑡
∞
−∞
= න 𝑒 𝑗(𝜔0 −𝜔)𝑡 𝑑𝑡
−∞
= 2𝜋𝛿 𝜔0 − 𝜔
or
∞
න 𝑒 𝑗𝜔𝑡 𝑑𝜔 = 2𝜋𝛿 𝑡
−∞
Since the impulse function is an even
function, 𝛿 𝜔0 − 𝜔 = 𝛿(𝜔 − 𝜔0 ),
Interchanging variables t and ω results in
∞
න 𝑒 𝑗𝜔𝑡 𝑑𝑡 = 2𝜋𝛿 𝜔
−∞
ℱ 𝑒 𝑗𝜔0 𝑡 = 2𝜋𝛿 𝜔 − 𝜔0
Example-7: FT of 𝑒 −𝑗𝜔0𝑡
• Find the Fourier Transform of 𝑥 𝑡 = 𝑒 −𝑗𝜔0 𝑡
Solution:
∞
ℱ 𝑒 𝑗𝜔0 𝑡 = න 𝑒 −𝑗𝜔0 𝑡 . 𝑒 −𝑗𝜔𝑡 𝑑𝑡
−∞
∞
= න 𝑒 −𝑗(𝜔+𝜔0 )𝑡 𝑑𝑡
−∞
= 2𝜋𝛿 𝜔 + 𝜔0
Example-8: FT of cos signals
• Find the Fourier Transform of 𝑥 𝑡 = 𝑐𝑜𝑠 𝜔0 𝑡
Solution:
∞
𝑋 𝜔 = න cos 𝜔0 𝑡 𝑒 −𝑗𝜔𝑡 𝑑𝑡
−∞
∞
𝑋 𝜔 =න
−∞
𝑒 𝑗𝜔0𝑡 + 𝑒 −𝑗𝜔0 𝑡 −𝑗𝜔𝑡
.𝑒
𝑑𝑡
2
1 ∞ −𝑗(𝜔−𝜔 )𝑡
1 ∞ −𝑗(𝜔+𝜔 )𝑡
0 𝑑𝑡 +
0 𝑑𝑡
= න 𝑒
න 𝑒
2 −∞
2 −∞
=
1
2𝜋𝛿 𝜔 − 𝜔0
2
+
1
2𝜋𝛿 𝜔 + 𝜔0
2
= 𝜋𝛿 𝜔 − 𝜔0 + 𝜋𝛿(𝜔 + 𝜔0 )
Example-9: FT of sin signal
• Find the Fourier Transform of 𝑥 𝑡 = 𝑠𝑖𝑛 𝜔0 𝑡
Solution:
∞
𝑋 𝜔 = න 𝑠𝑖𝑛 𝜔0 𝑡 𝑒 −𝑗𝜔𝑡 𝑑𝑡
−∞
∞
𝑋 𝜔 =න
−∞
𝑒 𝑗𝜔0𝑡 − 𝑒 −𝑗𝜔0𝑡 −𝑗𝜔𝑡
.𝑒
𝑑𝑡
2𝑗
1 ∞ −𝑗(𝜔−𝜔 )𝑡
1 ∞ −𝑗(𝜔+𝜔 )𝑡
0 𝑑𝑡 +
0 𝑑𝑡
= න 𝑒
න 𝑒
2𝑗 −∞
2𝑗 −∞
=
1
2𝜋𝛿 𝜔 − 𝜔0
2𝑗
+
1
2𝜋𝛿 𝜔 + 𝜔0
2𝑗
= −𝑗𝜋𝛿 𝜔 − 𝜔0 + 𝑗𝜋𝛿(𝜔 + 𝜔0 )
= 𝑗𝜋 𝛿 𝜔 + 𝜔0 + 𝛿(𝜔 − 𝜔0 )
Example-10: Find the FT of given Signal
Obtain the Fourier transform of the signal shown in Figure
Solution:
Example-11: Find the FT of given Signal
Solution:
Practice Problem-1
• Find the Fourier Transform of
Practice Problem-2
Practice Problem-3
• Find the Fourier Transform of
𝑥 𝑡 = 𝑒 𝑎𝑡 𝑢 −𝑡
Answer:
𝑋 𝑗𝜔 =
1
𝑎−𝑗𝜔
Practice Problem-4
• Find the Fourier Transform of
Exam Question
• Determine the Fourier Transform of the following Signal
𝑥(𝑡)
−4
−1
0
1
4
Table of Fourier Transform
Fourier Transform Pairs
Fourier Transform Pairs