City University of Hong Kong,
EE3210 Signals & Systems
Lecture note Instructor: Dr. Young Jin Chun.
Semester B, 2020/21
EE3210 Section: CA1 & TA1 – Course Syllabus
• Lecturer Time: Thursday, 1 PM – 2:50 PM, YEUNG, LT-6 – Tutorial: Monday, 4 PM – 4:50 PM, YEUNG, LT-6
• Lecturer: Dr. Young Jin Chun (Email: yjchun@cityu.edu.hk) – Office: G6525, Yeung Kin Man Academic Building (Tel: 7709)
– Office Hour: Monday, 2 PM – 3 PM.
• Course Support (2 TAs and 2 Student Tutors(ST)) – TA1: WANG Chao
– TA2: YU Wing Yin – ST1:
– ST2:
• Textbook: Lecture Note
– Ref): Signals and Systems by Alan V. Oppenheim et. al., 2nd edition, Prentice Hall – Ref): Schaum’s Outline of Signals and Systems, 2nd Edition, 2010, McGraw-Hill
• Assessment: 60% by Continuous Assessment and 40% by Final exam – Assignment (3 times): 20 %
– Mid-term tests (2 times) : 40 %
– Final Exam: 40 %
• Remark: To pass the course, students are required to achieve at least 30% in course work (hw + mid-term tests) and 30% in the examination.
• Course Content
– Part 1. Signals and Systems
– Part 2. Linear Time-Invariant Systems
– Part 3. Fourier Series for Periodic Signals
– Part 4. Continuous-time Fourier Transform – Part 5. Discrete-time Fourier Transform
– Part 6. Laplace Transform
– Part 7. Z-Transform
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(Email: cwang224-c@my.cityu.edu.hk) (Email: wingyinyu8-c@my.cityu.edu.hk) (Email: ) (Email: )
1 Signals and Systems
Major References:
• Chapter 1, Signals and Systems by Alan V. Oppenheim et. al., 2nd edition, Prentice Hall • Chapter 1, Schaum’s Outline of Signals and Systems, 2nd Edition, 2010, McGraw-Hill
Introduction
• Objective: Describe a system (physical, mathematical, or computational) by the way it transforms an input signal into an output signal.
• System level approach
– Abstraction
∗ Identify the system input and output signals → characterize the signal types
∗ Write input-output relation of the system → Operational transformations (e.g.,
Fourier analysis, Laplace Transformation).
∗ Characterize the system types by the input-output relation → system types
– Modular design
∗ Break down the system into a number of interconnected subsystems (module) ∗ Each module performs some specific task.
∗ Focus on the flow of signal, abstract everything else away
– Composite system
∗ Determine the input-output relationship between each modules
∗ Combine the components (module) to composite the overall system
∗ Component and composite systems have the same form, and are analyzed with same methods.
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1 Signals and Systems
1 Signals and Systems
Example 1.1
Let’s consider a typical mobile communication between the transmitter and receiver. Abstract the system input and output signals, then determine the input-output relationship.
Sol) We first describe each module as a cascade of component systems.
Then we combine the modules into a composite system.
Example 1.2
Determine the input-output relationship of the following electrical circuit.
Sol) We first characterize the input and output signals, then perform modular design by breaking the circuit into several modular circuits
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1 Signals and Systems
Then we combine the modules into a composite system
1.1 Classification of Signal Types
• [Def] Signal 𝑥(𝑡) is a function of an independent variable 𝑡 representing a physical behavior of the phenomenon.
1. Continuous-Time (CT) and Discrete-Time (DT) Signals
A signal 𝑥 (𝑡 ) is a continuous-time signal if 𝑡 is a continuous variable. If 𝑡 is a discrete variable, i.e., 𝑥 (𝑡 ) is defined at discrete times, then 𝑥 (𝑡 ) is a discrete-time signal.
Fig. 1 (a) Continuous-time and (b) discrete-time signals DT signal 𝑥 [𝑛] may be obtained by sampling a CT signal 𝑥 (𝑡 ) such as
𝑥𝑛 = 𝑥 [𝑛] = 𝑥 (𝑡𝑛)
5
Two representations of the DT signals • Functional form. For example,
1𝑛, 𝑛≥0 𝑥=𝑥[𝑛]= 2
1 Signals and Systems
𝑛 0,𝑛<0
{𝑥𝑛} = {...,0,0,1,2,2,1,0,1,0,2,0,0,...}
2. Analog and Digital Signals
• Sequence form. For instance,
A CT signal 𝑥 (𝑡 ) is an analog signal if 𝑥 (𝑡 ) can take on any value in the continuous interval (𝑎,𝑏). A DT signal 𝑥 [𝑛] is a digital signal if 𝑥 [𝑛] can take on only a finite number of distinct values.
3. Periodic and Aperiodic (or nonperiodic) Signals
Asignal𝑥(𝑡)(or𝑥[𝑛])isperiodicwithperiod𝑇 (or𝑁)ifthereisapositivenon-zero value of 𝑇 (or 𝑁 ) for which the following equality holds
𝑥(𝑡+𝑇)=𝑥(𝑡), forCTsignal𝑥(𝑡), 𝑥[𝑛+𝑁]=𝑥[𝑛], forDTsignal𝑥[𝑛]
(1.1) Any signal that is not periodic is called a nonperiodic (or aperiodic) signal.
• Property 1. For a given periodic signal, {𝑇,2𝑇,3𝑇,...,𝑚𝑇,...} are all available period at all 𝑡 and any integer 𝑚.
• Property 2. Fundamental period 𝑇0 is the smallest positive value of the period 𝑇 for which (1.1) holds.
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4. Even and Odd Signals
A signal is referred to as an even signal if
𝑥 (−𝑡) = 𝑥 (𝑡) for CT signal 𝑥(𝑡), 𝑥 [−𝑛] = 𝑥 [𝑛]
A signal is referred to as an odd signal if
𝑥 (−𝑡) = −𝑥 (𝑡) for CT signal 𝑥(𝑡), 𝑥 [−𝑛] = −𝑥 [𝑛]
for DT signal 𝑥[𝑛] (1.2) for DT signal 𝑥[𝑛] (1.3)
(1.4) where 𝑥𝑒 (𝑡) (or 𝑥𝑒 [𝑛]) is the even part, 𝑥𝑜 (𝑡) (or 𝑥𝑜 [𝑛]) is the odd part, and these
components are related to the original signal 𝑥(𝑡) (or 𝑥[𝑛]) as follows 𝑥 (𝑡)=1[𝑥(𝑡)+𝑥(−𝑡)],
𝑒2
𝑥 (𝑡)=1[𝑥(𝑡)−𝑥(−𝑡)].
• Property 2. Product of signals
– Even signal × Even signal = Even signal, Odd × odd = Even signal – Even signal × odd signal = odd signal
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(1.5)
1 Signals and Systems
• Property 1. Any signal 𝑥 (𝑡 ) or 𝑥 [𝑛] can be expressed as a sum of two signals
𝑥(𝑡)=𝑥𝑒 (𝑡)+𝑥𝑜 (𝑡) forCTsignal, 𝑥[𝑛]=𝑥𝑒 [𝑛]+𝑥𝑜 [𝑛] forDTsignal,
𝑜2
5. Energy and Power Signals
Energy 𝐸 of a signal 𝑥 (𝑡) (or 𝑥 [𝑛]) is defined as
∫∞2 ∞2
for DT signal,
E =
whereas the Power 𝑃 of a signal is defined as follows
−∞
𝑛=−∞
1𝑁2
lim |𝑥[𝑛]|
𝑁→∞ 2𝑁 +1
𝑛=−𝑁
P=
lim
𝑇→∞𝑇 −𝑇/2
1∫𝑇/2
|𝑥(𝑡)|2𝑑𝑡
for CT signal,
forDTsignal • Energy signal has finite energy and zero power
1 Signals and Systems
|𝑥(𝑡)| 𝑑𝑡 for CT signal, E =
|𝑥 [𝑛] |
(1.6)
(1.7)
0
𝑠𝑔𝑛 (𝑡) = 2𝑢(𝑡) − 1 = 0, 𝑡 = 0 −1, 𝑡<0
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• Properties
– Aperiodic & odd signal – Power signal P = 1
– Infinite Energy E = ∞
b) Ramp Function • Definition
• Properties
𝑡, 𝑡≥0 𝑟(𝑡)= 0, 𝑡<0,
∫𝑡 −∞
𝑢(𝜏)𝑑𝜏=𝑟(𝑡)
2
2
– Aperiodic & Even signal – ZeroPowerP=0
– Energy Signal E = 𝜏
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𝑟𝑒𝑐𝑡(𝑡/𝜏)=𝑢 𝑡+𝜏 −𝑢 𝑡−𝜏 (1.9) 2 2
1 Signals and Systems
– Aperiodic
– Infinite Power P = ∞ – Infinite Energy E = ∞
c) Rectangular Pulse • Definition
1,|𝑡|<𝜏
𝑟𝑒𝑐𝑡(𝑡/𝜏)= , 0, |𝑡|>𝜏
• Properties
2. Unit Impulse Function (also referred as Direc delta function) • Definition
– Sifting Property
amplitude.
∫∞ −∞
∫𝑏
𝑎
𝑥(𝑡0), if𝑎<𝑡0<𝑏 𝑥(𝑡)𝛿(𝑡−𝑡0)𝑑𝑡= 0, otherwise
– Impulsefunctionisthebuildingblockofanysignal,i.e.,arbitrarysignal can be respresented as an infinite sum of impulse function and signal
𝑥 (𝑡) =
𝑥 (𝜏)𝛿 (𝑡 −𝜏)𝑑𝜏
(1.11)
1 Signals and Systems
0, 𝑡≠0 𝛿(𝑡)= ∞, 𝑡=0,
• Properties
– Sampling Property
∫∞ −∞
𝛿(𝑡)𝑑𝑡=1
(1.10)
𝑥(𝑡)𝛿 (𝑡 −𝑡0) = 𝑥 (𝑡0)𝛿 (𝑡 −𝑡0)
Relationship between Rectangular Pulse and Impulse Function
1𝑡 • 𝛿 (𝑡) = 𝑟𝑒𝑐𝑡
𝜖
• 𝛿𝜖 (0) = 1 𝜖
• 𝛿𝜖 (𝑡) = 0, |𝑡| > 𝜖 2
•∫∞𝛿 (𝑡)𝑑𝑡=1 −∞ 𝜖
𝜖→0 •𝛿(0)→∞
𝜖𝜖
•𝛿(𝑡)=lim𝛿 (𝑡) 𝜖
•𝛿(𝑡)=0,𝑡≠0
•∫∞𝛿(𝑡)𝑑𝑡=1 −∞
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Additional Properties of Unit impulse function
• Scaling Property: • Even Function:
• Derivative and Integral:
𝛿 (𝑎𝑡) = 1 𝛿 (𝑡) |𝑎|
𝛿 (−𝑡) = 𝛿 (𝑡) ∫𝑡 𝑑𝑢(𝑡)
3. Complex Exponential Function
• Definition • Properties
𝑒𝑗𝑤0𝑡 =cos(𝑤0𝑡)+𝑗sin(𝑤0𝑡) – Periodic with 𝑇 = 2𝜋𝑛 where 𝑛 is an integer
|𝑤0| – Infinite Energy E = ∞
1 ∫ 𝑇0
– FinitepowerP= lim 0 0 |𝑒𝑗𝑤0𝑡|2𝑑𝑡 = lim 0 0 1·𝑑𝑡 =1
4. Sinusoidal Function
𝐴cos(𝑤0𝑡 +𝜃) or 𝐴sin(𝑤0𝑡 +𝜃),
where 𝐴 is the amplitude, 𝜃 is the phase angle, 𝑤0 is the radian frequency with
Fundamental period 𝑇0 = 2𝜋 (sec), Fundamental frequency 𝑓0 = 1 hertz (Hz)
𝑤0
𝑇0
1 ∫ 𝑇0 𝑇0→∞ 𝑇 𝑇0→∞ 𝑇
−∞
𝑑𝑡
1 Signals and Systems
𝑢(𝑡) =
𝛿 (𝜏)𝑑𝜏,
𝛿 (𝑡) =
𝑤0
– Fundamental period 𝑇0 = 2𝜋
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1.4 Classification of System Types
• [Def] A system is a mathematical model of a physical process that relates the input signal to the output signal in the form 𝑦 = 𝑇𝑥.
1. Invertible and Noninvertible System
A system is said to be invertible if distinct inputs lead to distinct outputs. Otherwise, the system is said to be noninvertible.
1 Signals and Systems
[Examples] Invertible System
• 𝑦(𝑡) = 2𝑥(𝑡) ↔ 𝑤(𝑡) = 1𝑦(𝑡) 2
• 𝑦[𝑛]=𝑛 𝑥[𝑘]↔𝑤[𝑛]=𝑦[𝑛]−𝑦[𝑛−1] 𝑘 =−∞
2. Memory and Memoryless System
Noninvertible System
• 𝑦[𝑛] = 0
• 𝑦(𝑡)=𝑥2(𝑡)
A system is said to be memoryless if the output at any time depends only on the input at that same time. Otherwise, the system is said to have memory.
[Examples] Memoryless System
System with Memory
• 𝑦[𝑛]=𝑛𝑘=−∞𝑥[𝑘]
• 𝑦[𝑛]=𝑥[𝑛−1]
•𝑦(𝑡)=1∫𝑡 𝑥(𝜏)𝑑𝜏 𝑐 −∞
• • •
𝑦(𝑡) = 𝑅𝑥(𝑡)
𝑦[𝑛] = 2𝑥[𝑛] −𝑥2[𝑛]2
3. Causal and Noncausal System
A system is said to be causal if its output at the present time depends on only the present and/or past values of the input. If its output at the present time depends on future values of the input, the system is known as noncausal.
[Examples] Causal System
• 𝑦[𝑛]=𝑛𝑘=−∞𝑥[𝑘] • 𝑦(𝑡)=𝑥2(𝑡)
Noncausal System
• 𝑦[𝑛]=𝑥[𝑛]+𝑥[𝑛+2]
• 𝑦[𝑛]=𝑥[−𝑛]or𝑦(𝑡)=𝑥(𝑡+1)
∗ Note) All memoryless systems are causal, but not vice versa.
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4. Linear and Nonlinear System
A system is said to be linear if the following superposition property (1.12) holds for a given operator T. If the system does not satisfy (1.12), it is a nonlinear system.
[Examples] Linear System
• 𝑦[𝑛]=𝑛𝑘=−∞𝑥[𝑘] • 𝑦(𝑡)=𝑡𝑥(𝑡)
T {𝛼1𝑥1 + 𝛼2𝑥2} = 𝛼1T {𝑥1} + 𝛼2T {𝑥2}
Nonlinear System
• 𝑦(𝑡)=𝑥2(𝑡)
• 𝑦[𝑛]=2𝑥[𝑛]+3
(1.12)
1 Signals and Systems
∗ Note) For a linear system, zero input always yields a zero output. 5. Time-invariant and Time-Varying System
A system is time-invariant if a time-shift of the input causes a corresponding shift in the output. In other words, the system response is independent of time.
[Examples]
Time invariant System
• 𝑦[𝑛]=𝑛𝑘=−∞𝑥[𝑘] • 𝑦[𝑛]=𝑥[𝑛−𝑛0]
LTI System
Time varying System
• 𝑦(𝑡)=𝑥(2𝑡) • 𝑦[𝑛]=𝑛𝑥[𝑛]
If𝑦(𝑡) = T{𝑥(𝑡)}, then𝑦 (𝑡 −𝑡0) = T{𝑥 (𝑡 −𝑡0)}
(1.13)
Linear time-invariant (LTI) system: A system that is linear and also time-invariant.
6. Stable and Unstable System
A system is stable if every bounded input produces a bounded output for all time.
[Examples] Stable System
If|𝑥(𝑡)|<𝐴, then|𝑦(𝑡)|<𝐵where|𝐴|<∞,|𝐵|<∞ Unstable System
(1.14)
•𝑦(𝑡)=𝑥2(𝑡)
• 𝑦[𝑛]=𝑥[𝑛]+𝑥[𝑛+2]
•𝑦[𝑛]= 1 𝑥[𝑛]
• 𝑦[𝑛]=𝑛𝑥[𝑛]
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1.5 Examples
[Example 1-1] Determine whether or not each of the following signals is periodic. If a signal is periodic, determine its fundamental period.
a) 𝑥(𝑡)=cos𝑡+𝜋 b) 𝑥(𝑡)=sin2𝜋𝑡 43
√ d) 𝑥(𝑡)=cos(𝑡)+sin 2𝑡
2
f) 𝑥(𝑡)=𝑒𝑗[𝜋𝑡−1]
h) 𝑥(𝑡)=cos2(𝑡) j) 𝑥(𝑡)=𝑒𝑗𝜋𝑡
Solution)Tosolvethistypeofproblem,trytofindtheminimum𝑇 thatsatisfy𝑥 (𝑡 +𝑇) =𝑥(𝑡). For instance, in (a), if the following equality holds with a nonzero constant 𝑇 , then it is periodic
cos 𝑡+𝜋 =cos 𝑡+𝑇+𝜋 → cos(𝑡′)=cos(𝑡′+𝑇), (1.15) 44
(1.16)
c) 𝑥(𝑡)=cos 𝜋𝑡 +sin 𝜋𝑡 34
1 Signals and Systems
e) 𝑥(𝑡)=sin2(𝑡)
g) 𝑥(𝑡)=cos2𝑡+𝜋
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i) 𝑥(𝑡) = (cos(2𝜋𝑡))𝑢(𝑡)
where we used a change of variable 𝑡 ′ = 𝑡 + 𝜋 in the second equality. Since the minimum 𝑇 that 4
satisfy (1.15) is 2𝜋 , (a) is a periodic signal with period 𝑇 = 2𝜋 . Similarly, for (b), 2𝜋𝑡 2𝜋𝑡 2𝜋𝑇 2𝜋𝑇
sin 3 =sin 3 + 3 → 3 =2𝜋, and by denoting 𝑡 ′ = 2𝜋 𝑡 , the minimum 𝑇 that satisfy (1.16) is 3.
3
For (c) and (d), we can use (1.18); The period𝑇1 for cos 𝜋𝑡 in (c) is𝑇1 = 6 and𝑇2 for sin 𝜋𝑡 34
is 𝑇2 = 8. Since 𝑇1/𝑇2 = 3/4, (c) is a periodic signal with period 𝑇 = 24. In (d), the period 𝑇1 for √√ √
cos (𝑡) is 𝑇1 = 2𝜋 and 𝑇2 for sin 2𝑡 is 𝑇2 = 2𝜋. Since 𝑇1/𝑇2 = 2, (d) is aperiodic signal. For (e) and (h), convert 𝑥 (𝑡 ) as follows, then apply similar approach as (a).
cos2 (𝑡) = 1 (1+cos(2𝑡)), sin2 (𝑡) = 1 (1−cos(2𝑡)), (1.17) 22
and the remaining can be solved using similar method. The solutions are summarized below.
a) Periodic with 𝑇 = 2𝜋 d) Aperiodic
g) Periodic with 𝑇 = 𝜋 j) Periodic with 𝑇 = 2
b) Periodic with 𝑇 = 3 e) Periodic with 𝑇 = 𝜋 h) Periodic with 𝑇 = 𝜋
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c) Periodic with 𝑇 = 24 f ) Periodic with 𝑇 = 4 i) Aperiodic
Sum of Periodic Signals
• Let 𝑥1(𝑡) and 𝑥2(𝑡) be periodic signals with fundamental periods 𝑇1 and 𝑇2, re- spectively. The sum 𝑥(𝑡) = 𝑥1(𝑡) + 𝑥2(𝑡) is periodic if and only if the following
condition holds
where the fundamental period 𝑇 is the least common multiple of 𝑇1 and 𝑇2.
𝑇1 = 𝑘 = rational number (1.18) 𝑇2 𝑚
• Let 𝑥1 [𝑛] and 𝑥2 [𝑛] be periodic sequence with fundamental periods 𝑁1 and 𝑁2, respectively. The sum 𝑥 [𝑛] = 𝑥1 [𝑛]+𝑥2 [𝑛] is periodic given the following condition
𝑚𝑁1 = 𝑘𝑁2 = 𝑁 (1.19) where the fundamental period 𝑁 is the least common multiple of 𝑁1 and 𝑁2.
Refer [Schaum’s text, Problem 1.14 & 1.15]
[Example 1-2] Determine whether the following signals are energy signals, power signals, or neither.
a) 𝑥(𝑡)=𝑒−𝑎𝑡𝑢(𝑡),𝑎>0 b) 𝑥(𝑡)=𝐴cos(𝜔0𝑡+𝜃)
Solution) To solve this type of problem, (Step 1.) you need to calculate the energy E first. If E is finite, the signal is a Energy signal. Otherwise, (Step 2.) if E is infinite, you need to calculate the power P as well. If P is finite, the signal is a Power signal. Otherwise, if P is infinite, then it is neither a energy nor a power signal. For example, in (a),
1 Signals and Systems
∫∞ ∫∞
E = 𝑒−2𝑎𝑡𝑢(𝑡)𝑑𝑡 = 𝑒−2𝑎𝑡𝑑𝑡 =
1 2𝑎
,
(1.20)
−∞ 0
where we used the definition of the step function in the second equality. Since 1
is finite, 𝑥 (𝑡 ) in (a) is a energy signal. For a periodic signal, the integration interval 𝑇 in (1.7) is equal to the
2𝑎 period. In (b), the period is 𝑇 = 2𝜋 and the signal power can be calculated as follows
1 ∫ 𝑇 𝑇→∞𝑇 0
𝜔0
2 2 𝐴2 ∫ 2𝜋+𝜃
2
2
cos (𝑙)𝑑𝑙
𝜔021 𝐴2
then applied the Cosine rule cos (𝑡) = 2 (1 + cos (2𝑡)) in the third equality. Since 2 is finite,
𝑥 (𝑡 ) in (b) is a power signal. In summary, the solutions are
P= lim
= lim 𝐴
𝐴 cos (𝜔0𝑡+𝜃)𝑑𝑡= lim 𝑇→∞2𝜋 𝜃
(1.21) where we used 𝑇 = 2𝜋 and a change of variable, 𝑙 = 𝜔0𝑡 + 𝜃 or 𝜔0𝑑𝑡 = 𝑑𝑙, in the second equality,
2 ∫ 2𝜋+𝜃 𝑇→∞4𝜋 𝜃
[1+cos(2𝑙)]𝑑𝑙 = 𝐴 , 2
a) Energy signal
b) Power signal
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Definition of Energy and Power Signals
• Energy signal has finite energy and zero power, i.e., 0 < E < ∞, P = 0
• Power signal has finite power and infinite energy, i.e., 0 < P < ∞, E = ∞, where
∫ ∞ E =
−∞
Properties of Periodic Signals
2
1 ∫ 𝑇/2 2
P = lim |𝑥(𝑡)| 𝑑𝑡
𝑇→∞𝑇 −𝑇/2
|𝑥(𝑡)| 𝑑𝑡,
1 Signals and Systems
The following equalities hold for a periodic signal 𝑥 (𝑡 + 𝑇 ) = 𝑥 (𝑡 )
∫ 𝛽 ∫ 𝛽+𝑇 𝑥(𝑡)𝑑𝑡 =
∫ 𝑇 ∫ 𝑎+𝑇
𝑥(𝑡)𝑑𝑡,
𝛼 𝛼+𝑇 0 𝑎
𝑥(𝑡)𝑑𝑡,
1 ∫ 𝑇/2 𝑇→∞𝑇 −𝑇/2
2
𝑥(𝑡)𝑑𝑡 = 1 ∫ 𝑇0 2
P = lim
where 𝑇0 is the fundamental period and 𝛼, 𝛽, 𝑎 are arbitrary real valued constants.
Refer [Schaum’s text, Problem 1.17 & 1.18]
[Example 1-3] Determine the even and odd component of the following signals a) 𝑥(𝑡)=𝑢(𝑡) b) 𝑥(𝑡)=sin𝜔0𝑡+𝜋
Solution) To solve this type of problem, you need to apply (1.22). In (a), 𝑥(−𝑡) = 𝑢(−𝑡) = 1 for 𝑡 < 0 and 𝑢(−𝑡) = 0 for 𝑡 > 0. Then, the following results can be derived
𝑥 (𝑡)=1[𝑢(𝑡)+𝑢(−𝑡)]=1, 𝑒22
1 1 0.5, 𝑡 > 0, 𝑥𝑜 (𝑡) = 2 [𝑢 (𝑡) −𝑢 (−𝑡)] = 2𝑠𝑔𝑛 (𝑡) = −0.5, 𝑡 < 0
17
|𝑥(𝑡)| 𝑑𝑡 =
𝑇0 0
|𝑥(𝑡)| 𝑑𝑡,
4
In (b), we first use Sine rule to expand the Sine function, then the following reults can be derived. 𝜋 𝜋 𝜋 1
sin 𝜔0𝑡+4 =sin(𝜔0𝑡)cos 4 +cos(𝜔0𝑡)sin 4 =√2(sin(𝜔0𝑡)+cos(𝜔0𝑡)). 11
1 Signals and Systems
𝑥𝑒(𝑡)=√2cos(𝜔0𝑡), 𝑥𝑜(𝑡)=√2sin(𝜔0𝑡). Similarly, the even and odd component of (c) and (d) can be found as follows
Even and Odd Component
Any signal 𝑥 (𝑡 ) can be expressed as a sum of two signals 𝑥(𝑡)=𝑥𝑒 (𝑡)+𝑥𝑜 (𝑡),
where 𝑥𝑒 (𝑡 ) and 𝑥𝑜 (𝑡 ) are related to the original signal 𝑥 (𝑡 ) as follows 𝑥 (𝑡)=1[𝑥(𝑡)+𝑥(−𝑡)], 𝑥 (𝑡)=1[𝑥(𝑡)−𝑥(−𝑡)].
(1.22)
𝑒2 𝑜2
[Example 1-4] [Part 1] Sketch the following signals.
a) 𝑥1(𝑡)=𝑢(𝑡)+5𝑢(𝑡−1)−2𝑢(𝑡−2) c) 𝑥3(𝑡)=𝑢(𝑡)𝑢(𝑎−𝑡),𝑎>0
b) 𝑥2(𝑡)=𝑟(𝑡)−𝑟(𝑡−1)−𝑢(𝑡−2) d) 𝑥4(𝑡)=𝑥0(𝑡)𝑢(1−𝑡)
18
e) 𝑥5(𝑡)=𝑥0(𝑡)[𝑢(𝑡)−𝑢(𝑡−1)] where the signal 𝑥0(𝑡) is plotted below.
[Part 2] For each of the signals plotted below, write an expression in terms of unit step and unit ramp functions.
Solution)[Part 1]
[Part 2]
(f)𝑥6(𝑡)=𝑟(𝑡)−𝑟(𝑡−𝑎)−𝑎𝑢(𝑡−𝑏), (g)𝑥7(𝑡)=(𝑟(𝑡)−𝑟(−𝑡))(𝑢(𝑡+𝑎)−𝑢(𝑡−𝑎))
19
1 Signals and Systems
• Time Reversal: Flip the signal around the vertical axis • Time Shifts: Shift the signal to left or right
– Right-shift if 𝑡0 > 0, Left-shift if 𝑡0 < 0.
• Time Scaling: Linearly stretch or compress the signal
– Compression if |𝑐| > 1, Expansion if |𝑐| < 1.
𝑥(𝑡) → 𝑥(−𝑡) 𝑥(𝑡) → 𝑥 (𝑡 −𝑡0)
𝑥(𝑡) → 𝑥 (𝑐𝑡)
1 Signals and Systems
[Example 1-5] Evaluate the following integrals.
a) ∫𝑡 cos(𝜏)𝑢(𝜏)𝑑𝜏 b) ∫𝑡 cos(𝜏)𝛿(𝜏)𝑑𝜏
−∞ −∞ ∞ 2𝜋
c) ∫ cos(𝑡)𝑢(𝑡−1)𝛿(𝑡)𝑑𝑡 d) ∫ 𝑡sin𝑡𝛿(𝑡−𝜋) −∞ 02
e) ∫∞ 2𝑡−3𝛿(𝑡−1)𝑑𝑡 f) ∫2 exp(1−𝑡)+sin2𝜋𝑡𝛿𝑡−3𝑑𝑡 −∞32 −332
Solution)
(a)
(b)
(c)
(d)
(e)
(f)
−3 exp(1−𝑡)+sin 3 𝛿 𝑡−2 𝑑𝑡=exp −2 +sin(𝜋)=exp(−0.5)
20
∫𝑡 −∞
∫𝑡
If𝑡>0, cos(𝜏)𝑑𝜏=sin(𝑡)
cos(𝜏)𝑢(𝜏)𝑑𝜏 =
∫𝑡 −∞
−∞
If 𝑡 < 0, 0
0
=𝑢(𝑡)sin(𝑡)
If𝑡>0,cos0=1 If𝑡 < 0,0
cos(𝜏)𝛿(𝜏)𝑑𝜏 =
∫ ∞cos(𝑡)𝑢(𝑡−1)𝛿(𝑡)𝑑𝑡=cos(0)𝑢(−1)=0
∫2𝜋
𝑡sin 𝑡 𝛿(𝑡−𝜋)=𝜋sin 𝜋 =𝜋 22
=𝑢(𝑡)
0
∫∞2 3 2 3 5
3𝑡−2 𝛿(𝑡−1)𝑑𝑡=3−2=−6 −∞
∫2
2𝜋𝑡 3 1
Properties of Unit impulse function
𝑥(𝑡), if𝑎<𝑡 <𝑏 𝑏00
•∫ 𝑥(𝑡)𝛿(𝑡−𝑡0)𝑑𝑡= 0, otherwise 𝑎
• 𝑥(𝑡)=∫∞ 𝑥(𝜏)𝛿(𝑡−𝜏)𝑑𝜏 −∞
• 𝛿(𝑎𝑡)= 1 𝛿(𝑡), 𝑢(𝑡)=∫𝑡 𝛿(𝜏)𝑑𝜏, 𝛿(𝑡)=𝑑𝑢(𝑡) |𝑎| −∞ 𝑑𝑡
[Example 1-6] Determine whether the following system is (i) memoryless, (ii) causal, (iii) linear, (iv) time-invariant, or (v) stable. Refer [Schaum’s text, Problem 1.33, 1.34, 1.36, 1.38]
a) 𝑦(𝑡)= 1 ∫𝑡 𝑥(𝜏)𝑑𝜏 b) 𝑦(𝑡)=𝑥(𝑡)cos(𝜔0𝑡) 𝐶 −∞
c) 𝑦[𝑛]=𝑥[𝑛−1] d) 𝑦[𝑛]=𝑛𝑥[𝑛]
Solution) In (a), the output depends on the past input, so it is not memoryless system. The output depends on the present and past values of the input, so it is a Causal system. To test linearlity, substitute 𝑥(𝑡) ← 𝛼1𝑥1(𝑡) + 𝛼2𝑥2(𝑡) as the input, where 𝑦1(𝑡) and 𝑦2(𝑡) is the corresponding output of 𝑥1(𝑡) and 𝑥2(𝑡), respectively. Then,
𝐶 −∞ 𝐶 −∞
so the superposition property holds, which indicates a linear system. To test time-invariance, input time shifted signal 𝑥(𝑡 − 𝑡0). If the corresponding output is 𝑦(𝑡 − 𝑡0), then it is a time invariant system.
1 Signals and Systems
1∫𝑡 𝐶 −∞
[𝛼1𝑥1(𝜏) +𝛼2𝑥2(𝜏)]𝑑𝜏
1∫𝑡 1∫𝑡
𝑦(𝑡) =
=𝛼1 𝑥1(𝜏)𝑑𝜏 +𝛼2 𝑥2(𝜏)𝑑𝜏 =𝛼1𝑦1(𝑡)+𝛼2𝑦2(𝑡),
1 ∫ 𝑡 𝐶 −∞
𝑥 (𝜏 − 𝑡0) 𝑑𝜏 =
1 ∫ 𝑡−𝑡0 𝐶 −∞
𝑥 (𝑙) 𝑑𝑙 = 𝑦 (𝑡 − 𝑡0) ,
by using a change of variable 𝑙 = 𝜏 − 𝑡0 in the first equality. Hence, it is a time-invariant system. For stability, (a)
can be easily proved to be a unstable by substituting a unit step function 𝑥 (𝑡 ) = 𝑢 (𝑡 ) as the input, which achieves
unbounded𝑦(𝑡) = 𝑡𝑢(𝑡) . The remaining can be proved using similar method. The solutions are summarized below. 𝐶
a) c)
memory, causal, linear, time-invariant, unstable b) memoryless, causal, linear, time-variant, stable memory, causal, linear, time-invariant, stable d) memoryless, causal, linear, time-variant, unstable.
System Characterization
1. MemorylessSystem;outputatanytimedependsonlyontheinputatthatsametime
2. CausalSystem;outputatthepresenttimedependsonlyonthepresentand/orpastinputvalues 3. LinearSystem;thesuperpositionpropertyholds,i.e.,T{𝛼1𝑥1+𝛼2𝑥2}=𝛼1T{𝑥1}+𝛼2T{𝑥2}
4. Time-invariantSystem;time-shiftoftheinputcausesasameamountofshiftingintheoutput 5. StableSystem;If|𝑥(𝑡)|<𝐴, then|𝑦(𝑡)|<𝐵where|𝐴|<∞,|𝐵|<∞
6. LTISystem;Asystemthatislinearandalsotime-invariant
21
2 Linear Time-Invariant Systems
Major References:
• Chapter2,SignalsandSystemsbyAlanV.Oppenheimet.al.,2ndedition,PrenticeHall • Chapter2,Schaum’sOutlineofSignalsandSystems,2ndEdition,2010,McGraw-Hill
2.1 Convolution
2.1.1 Convolution Integral of CT Signal
1. Definition
Convolution Integral of two continuous-time signals 𝑥 (𝑡 ) and 𝑦 (𝑡 ) is defined by
∫∞ −∞
(2.1)
a) Commutative b) Associative c) Distributive
𝑥 (𝑡 ) ∗ 𝑦 (𝑡 ) = 𝑦 (𝑡 ) ∗ 𝑥 (𝑡 )
{𝑥(𝑡) ∗𝑦1(𝑡)} ∗𝑦2(𝑡) = 𝑥(𝑡) ∗ {𝑦1(𝑡) ∗𝑦2(𝑡)} 𝑥(𝑡) ∗ {𝑦1(𝑡) +𝑦2(𝑡)} = 𝑥(𝑡) ∗𝑦1(𝑡) +𝑥(𝑡) ∗𝑦2(𝑡)
𝑧(𝑡) = 𝑥(𝑡) ∗𝑦(𝑡) =
𝑥 (𝜏)𝑦 (𝑡 −𝜏)𝑑𝜏.
Convolution 𝑥 (𝑡 ) ∗ 𝑦 (𝑡 ) represents the degree to which 𝑥 &𝑦 overlap at 𝑡 as 𝑦 sweeps across the domain 𝑡 . Step. 1) 𝑦(𝜏) is time-reversed, then shifted by 𝑡; 𝑦 (𝜏) → 𝑦 (−𝜏) → 𝑦 (𝑡 − 𝜏)
Step.2) 𝑥(𝜏)and𝑦(𝑡−𝜏)aremultiplied,thenintegratedover𝜏
Step.3) Convolutionwillremainzeroaslongas𝑥&𝑦donotoverlap
Step.4) Sweep𝑦(𝑡−𝜏)from𝑡=−∞to𝑡=∞toproducetheentireoutput
2. PropertiesoftheConvolutionIntegral
The convolution integral has the following properties. Refer [Schaum’s text, Problem 2.1] for the proof.
3. AdditionalProperties
Refer [Schaum’s text, Problem 2.2, 2.8] for the proof.
a) 𝑥(𝑡)∗𝛿(𝑡)=𝑥(𝑡)
b) 𝑥(𝑡)∗𝛿(𝑡−𝑡0)=𝑥(𝑡−𝑡0)
c) 𝑥(𝑡)∗𝑢(𝑡)=∫𝑡 𝑥(𝜏)𝑑𝜏 −∞
e) If𝑥(𝑡)and𝑦(𝑡)areperiodicsignalswithacommonperiod𝑇,theconvolutionin(2.1)doesnotconverge. Instead, we define the periodic convolution 𝑓 (𝑡) = 𝑥(𝑡) 𝑦(𝑡), where 𝑓 (𝑡) is periodic with period𝑇.
d) 𝑥(𝑡)∗𝑢(𝑡−𝑡0)=∫𝑡−𝑡0𝑥(𝜏)𝑑𝜏 −∞
∫𝑇 𝑓(𝑡)=𝑥(𝑡)𝑦(𝑡)= 0 𝑥(𝜏)𝑦(𝑡−𝜏)𝑑𝜏
∫ 𝑎+𝑇 𝑎
22
(2.2)
=
𝑥(𝜏)𝑦(𝑡−𝜏)𝑑𝜏 forarbitrary𝑎
2 Linear Time-Invariant Systems
2.1.2 Convolution Sum of DT Signal
1. Definition
Convolution Sum of two discrete-time sequence 𝑥 [𝑛] and 𝑦 [𝑛] is defined by
∞ 𝑧[𝑛]=𝑥[𝑛]∗𝑦[𝑛]= 𝑥[𝑘]𝑦[𝑛−𝑘]
𝑘 =−∞
Step. 1) 𝑦[𝑘] is time-reversed, then shifted by 𝑛; 𝑦 [𝑘] → 𝑦 [−𝑘] → 𝑦 [𝑛 − 𝑘] Step.2) 𝑥[𝑘]and𝑦[𝑛−𝑘]aremultiplied,thensummedoverall𝑘
Step.3) Convolutionwillremainzeroaslongas𝑥&𝑦donotoverlap
Step.4) Sweep𝑦[𝑛−𝑘]from𝑛=−∞to𝑛=∞toproducetheentireoutput
2. PropertiesoftheConvolutionSum
The convolution sum has the following properties. Refer [Schaum’s text, Problem 2.26] for the proof.
(2.3)
a) Commutative b) Associative c) Distributive
𝑥 [𝑛] ∗ 𝑦 [𝑛] = 𝑦 [𝑛] ∗ 𝑥 [𝑛]
{𝑥[𝑛] ∗𝑦1[𝑛]} ∗𝑦2[𝑛] = 𝑥[𝑛] ∗ {𝑦1[𝑛] ∗𝑦2[𝑛]}
𝑥 [𝑛] ∗ {𝑦1 [𝑛] + 𝑦2 [𝑛]} = 𝑥 [𝑛] ∗ 𝑦1 [𝑛] + 𝑥 [𝑛] ∗ 𝑦2 [𝑛]
3. AdditionalProperties
Refer [Schaum’s text, Problem 2.27, 2.31] for the proof.
a) 𝑥[𝑛]∗𝛿[𝑛]=𝑥[𝑛]
b) 𝑥[𝑛]∗𝛿[𝑛−𝑛0]=𝑥[𝑛−𝑛0]
𝑛
c) 𝑥[𝑛]∗𝑢[𝑛]= 𝑥[𝑘] 𝑘 =−∞
𝑛−𝑛0
d) 𝑥[𝑛]∗𝑢[𝑛−𝑛0]= 𝑥[𝑘]
𝑘 =−∞
e) If𝑥[𝑛]and𝑦[𝑛]areperiodicsequencewithacommonperiod𝑁,theconvolutionin(2.3)doesnot
converge. Instead, we define the periodic convolution 𝑓 [𝑛] = 𝑥[𝑛] 𝑦[𝑛], where 𝑓 [𝑛] is periodic with period 𝑁 .
𝑁−1 𝑓[𝑛]=𝑥[𝑛]𝑦[𝑛]= 𝑥[𝑘]𝑦[𝑛−𝑘]
𝑘=0
[Example 2-1] Evaluate the following convolutions 1. 𝑢(𝑡+𝑎)∗𝑢(𝑡+𝑏)
2. 𝑟𝑒𝑐𝑡(𝑡/𝜏)∗𝑟𝑒𝑐𝑡(𝑡/𝜏)
3. 𝑟𝑒𝑐𝑡(𝑡/𝜏)∗𝑢(𝑡)
𝑇𝑇
(2.4)
2 Linear Time-Invariant Systems
1 for0<𝑡 <2
5. 𝑟𝑒𝑐𝑡(𝑡/𝜏)∗𝛿 (𝑡)where𝜏<𝑇and𝛿 (𝑡)= 𝛿(𝑡−𝑛𝑇)istheunitimpulsetrain
1 for0<𝑡 <3
4. 𝑥(𝑡)∗𝑦(𝑡)where𝑥(𝑡)= 0 otherwise and𝑦(𝑡)= 0 otherwise
Refer [Schaum’s text, Problem 2.6, 2.7, 2.8]
∞ 𝑛=−∞
23
Solution) Example 2-1. 1) The convolution integral is given by ∫∞
𝑢 (𝑡 + 𝑎) ∗ 𝑢 (𝑡 + 𝑏) = 𝑢 (𝜏 +𝑎)𝑢 (−𝜏 +𝑏 +𝑡)𝑑𝜏 = (𝑡 +𝑎 +𝑏)𝑢 (𝑡 +𝑎 +𝑏) −∞
if𝑡+𝑎+𝑏>0
if 𝑡 + 𝑎 + 𝑏 < 0
(2.5)
∫ 𝑏+𝑡1·𝑑𝜏=(𝑡+𝑎+𝑏)
= − 𝑎 0
where the product of two step functions 𝑢 (𝜏 + 𝑎) 𝑢 (−𝜏 + 𝑏 + 𝑡) has a non-zero value at 𝜏 + 𝑎 > 0 and −𝜏 + 𝑏 + 𝑡 > 0. As shown in Fig. 2.1, if −𝑎 < 𝑏 + 𝑡 , then the product of two step functions overlap each other within the interval −𝑎 < 𝜏 < 𝑏 + 𝑡. If 𝑏 + 𝑡 < −𝑎, there is no overlap and the integral in (2.5) becomes zero.
Figure 2.1:
Example 2-1. 2) The convolution can be expanded by expressing the rectangular pulse signal via the step fucntion 𝜏 𝜏 𝜏 𝜏
2 Linear Time-Invariant Systems
𝑟𝑒𝑐𝑡(𝑡/𝜏)∗𝑟𝑒𝑐𝑡(𝑡/𝜏)= 𝑢 𝑡+2 −𝑢 𝑡−2 ∗ 𝑢 𝑡+2 −𝑢 𝑡−2 = (𝑡 + 𝜏) 𝑢 (𝑡 + 𝜏) − 2𝑡𝑢 (𝑡) + (𝑡 − 𝜏) 𝑢 (𝑡 − 𝜏) ,
(2.6)
where we used (1.9) in the first equality and the result from [Example 2-1. 1] in the second equality. The last expression has four separate intervals with different values, which is a triangular pulse signal with maximum magnitude 𝜏 at 𝑡 = 0 and width 2𝜏 (two times larger than the rectangular pulse signal 𝑟𝑒𝑐𝑡 (𝑡/𝜏)).
0
𝑡 + 𝜏
𝑡 + 𝜏 − 2𝑡 = 𝜏 − 𝑡
𝜏 − 𝑡 + 𝑡 − 𝜏 = 0
if 𝑡 < −𝜏
if − 𝜏 < 𝑡 < 0 if 0 < 𝑡 < 𝜏
if 𝑡 > 𝜏
(b) can also be solved using the direct definition of the convolution where we multiply one original signal to another signal that is time reversed, then time-shifted by t, which is plotted below.
24
2 Linear Time-Invariant Systems
• Forsub-figure(a)and(b),thereisnooverlapbetweentwosignals,hencetheconvolutioniszero.Thesecases
correspondtothecondition𝑡+𝜏 <−𝜏 and𝑡−𝜏 >𝜏,i.e.,𝑡<−𝜏forsub-figure(a)and𝑡>𝜏for(b). 2222
• Forsub-figure(c),if𝑡+𝜏 >−𝜏 and𝑡−𝜏 <−𝜏,theoverlapareais𝑡+𝜏−−𝜏=𝑡,whichistheconvolution 2222 22
in the interval −𝜏 < 𝑡 < 0.
• For sub-figure (d), if 𝑡 − 𝜏 < 𝜏 and 𝑡 + 𝜏 > 𝜏 , the overlap area is 𝜏 − 𝑡 − 𝜏 = 𝜏 − 𝑡 , which is the convolution
222222 intheinterval0<𝑡 <𝜏.
Example 2-1. 3) The convolution can be expanded in terms of the step fucntion as follows
𝜏 𝜏 𝜏 𝜏 𝜏 𝜏
𝑟𝑒𝑐𝑡(𝑡/𝜏)∗𝑢(𝑡)= 𝑢 𝑡+2 −𝑢 𝑡−2 ∗𝑢(𝑡)= 𝑡+2 𝑢 𝑡+2 − 𝑡−2 𝑢 𝑡−2 , (2.7) where we applied [Example 2-1. 1] in the second equality. The last expression has three separate intervals with
different values as follows
0
𝑥 (𝑡 ) ∗ 𝑦 (𝑡 ) = {𝑢 (𝑡 ) − 𝑢 (𝑡 − 3) } ∗ {𝑢 (𝑡 ) − 𝑢 (𝑡 − 2) }
= 𝑡𝑢 (𝑡 ) − (𝑡 − 3) 𝑢 (𝑡 − 3) − (𝑡 − 2) 𝑢 (𝑡 − 2) + (𝑡 − 5) 𝑢 (𝑡 − 5) ,
where we applied [Example 2-1. 1] in the second equality. The last expression has five separate intervals with different values as follows
if 𝑡 < −𝜏 2
if − 𝜏 < 𝑡 < 𝜏 222
𝑡 + 𝜏
𝜏𝜏𝜏
𝑡+ −𝑡− =𝜏 if𝑡>
222
Example 2-1. 4) The convolution can be expanded in terms of the step fucntion as follows
Example 2-1. 5)
0
𝑡
if 𝑡 < 0
if 0 < 𝑡 < 2 if 2 < 𝑡 < 3 if 3 < 𝑡 < 5 if 𝑡 > 5
𝑡 − (𝑡 − 2) = 2
2 − (𝑡 − 3) = 5 − 𝑡 5 − 𝑡 − (𝑡 − 5) = 0
∞∞∞ 𝑟𝑒𝑐𝑡(𝑡/𝜏)∗ 𝛿(𝑡−𝑛𝑇) = 𝑟𝑒𝑐𝑡(𝑡/𝜏)∗𝛿(𝑡−𝑛𝑇)= 𝑟𝑒𝑐𝑡
𝑡 − 𝑛𝑇 𝜏
(2.9) where we used distributive property in the second equality and 𝑥(𝑡) ∗ 𝛿 (𝑡 − 𝑡0) = 𝑥 (𝑡 − 𝑡0) in the last equality.
(2.8)
𝑛=−∞ 𝑛=−∞ 𝑛=−∞
25
Figure 2.2:
As shown in Fig. 2.2, the time-shifted rectangular pulse signal do not overlap each other as far as 𝜏 < 𝑇 is satisfied.
However, if 𝜏 > 𝑇 , then the time-shifted rectangular pulse signals will overlap each other. Important Convolution Pairs
1. 𝑢(𝑡+𝑎)∗𝑢(𝑡+𝑏)=(𝑡+𝑎+𝑏)𝑢(𝑡+𝑎+𝑏)
2 Linear Time-Invariant Systems
2. 𝑟𝑒𝑐𝑡(𝑡/𝜏)∗𝑟𝑒𝑐𝑡(𝑡/𝜏)=
0
𝜏 + 𝑡
𝜏 − 𝑡
0
if 𝑡 < −𝜏
if − 𝜏 < 𝑡 < 0 if 0 < 𝑡 < 𝜏
if 𝑡 > 𝜏
[Example 2-2] Evaluate the following convolutions
1. 𝑥(𝑡)∗𝑦(𝑡)where𝑥(𝑡)=𝑢(𝑡)and𝑦(𝑡)=𝑒−𝛼𝑡𝑢(𝑡),𝛼>0
2. 𝑥(𝑡)∗𝑦(𝑡)where𝑥(𝑡)=𝑒−𝛼𝑡𝑢(𝑡)and𝑦(𝑡)=𝑒𝛼𝑡𝑢(−𝑡),𝛼>0
Refer [Schaum’s text, Problem 2.4, 2.5]
Solution) Example 2-2. 1) 𝑥(𝑡) ∗𝑦(𝑡) =
∫ ∫𝑡1
if 𝑡 > 0 if 𝑡 < 0
∞ −∞
−𝛼𝜏
𝑢 (𝜏)𝑢 (𝑡 −𝜏)𝑑𝜏 =
0 0,
𝑒−𝛼𝜏𝑑𝜏 =
1 − 𝑒−𝛼𝑡 ,
𝑒
𝛼
(2.10)
Example 2-2. 2)
= 1 1−𝑒−𝛼𝑡𝑢(𝑡) 𝛼
𝑒−𝛼𝑡
𝑥(𝑡) ∗𝑦(𝑡) =
𝑒𝛼𝜏𝑢 (−𝜏)𝑒−𝛼(𝑡−𝜏)𝑢 (𝑡 −𝜏)𝑑𝜏
0
−𝛼𝑡∫2𝛼𝜏 1−𝛼𝑡
∫∞ −∞
(2.11) where the product of two step functions 𝑢 (−𝜏) 𝑢 (𝑡 − 𝜏) is determined by the magnitude of 𝑡 as shown in Fig. 2.3.
Then (2.11) can be derived as follows
𝑒 𝑒𝑑𝜏=𝑒,if𝑡>0
2𝛼 1
∫−∞ ⇒ 𝑒−𝛼 |𝑡 | (2.12) 𝑡1 2𝛼
𝑒2𝛼𝜏𝑑𝜏 = 𝑒𝛼𝑡, if𝑡 < 0
2𝛼
−∞
26
2.2 LTI System Response
𝑦(𝑡) = T{𝑥(𝑡)} = T
𝑥(𝜏)𝛿 (𝑡 −𝜏)𝑑𝜏 −∞ ∫
2 Linear Time-Invariant Systems
Figure 2.3:
Linear Time-Invariant (LTI) System, represented by T {·}, satisfy the following two attributes. • Linearity:T{𝛼1𝑥1(𝑡)+𝛼2𝑥2(𝑡)}=𝛼1T{𝑥1(𝑡)}+𝛼2T{𝑥2(𝑡)}
• Time-Invariance:T{𝑥(𝑡−𝑡0)}=𝑦(𝑡−𝑡0)
2.2.1 Response of a CT LTI System
1. ImpulseResponse
• ImpulseResponseisdefinedastheoutputofasystemwhentheinputisaimpulsesignal𝛿(𝑡).
h(𝑡) = T{𝛿(𝑡)}
𝑥(𝜏)𝛿 (𝑡 −𝜏)𝑑𝜏 (2.15) ∫ ∞
𝑥(𝑡) = 𝑥(𝑡) ∗𝛿(𝑡) =
USing the properties of an LTI system, the output to an arbitrary input 𝑥 (𝑡 ) can be expressed as
(2.16) by using (2.15) in the second equality, linearity in the third, and time-invariance in the fourth equality.
27
∫
=
−∞
𝑥(𝜏)T{𝛿 (𝑡 −𝜏)}𝑑𝜏 = 𝑥(𝜏)h (𝑡 −𝜏)𝑑𝜏 = 𝑥(𝑡) ∗h(𝑡), −∞
∞∞
−∞
(2.13)
Figure 2.4:
• TheoutputofanyCT-LTIsystemistheconvolutionoftheinput𝑥(𝑡)withtheimpulseresponseh(𝑡)
𝑦 (𝑡 ) = 𝑥 (𝑡 ) ∗ h (𝑡 ) (2.14) Proof ) Arbitrary input 𝑥 (𝑡 ) can be expressed in terms of the imulse signal 𝛿 (𝑡 ) using convolution
∫∞
2. StepResponse
• StepResponseisdefinedastheoutputofasystemwhentheinputisastepsignal𝑢(𝑡).
𝑠(𝑡) = T{𝑢(𝑡)} (2.17)
• Stepresponsecanbeobtainedbyintegratingtheimpulseresponseh(𝑡).Similarly,theimpulseresponse
h(𝑡) can be determined by differentiating the step response h(𝑡).
∫∞ ∫𝑡 𝑑𝑠(𝑡) 𝑠(𝑡) = h(𝑡) ∗𝑢(𝑡) = h(𝜏)𝑢 (𝑡 −𝜏)𝑑𝜏 = h(𝜏)𝑑𝜏 ⇔ h(𝑡) =
(2.18)
2 Linear Time-Invariant Systems
−∞ −∞
𝑑𝑡
3. PropertiesofCTLTISystem
• MemorylessorMemorySystem:Theoutputofamemorylesssystemdependsonlyonthepresent
input, so that the input-output relationship can be represented in the form 𝑦(𝑡) = 𝐾𝑥(𝑡). Since h(𝑡) is the system output for an impulse signal input 𝛿(𝑡), the impulse response h(𝑡) can be expressed as h(𝑡) = 𝐾𝛿(𝑡). Hence, the following statement holds
If h(𝑡0) ≠ 0 for 𝑡0 ≠ 0, then it is a LTI system with memory. (2.19)
• Causality:Foracausalsystem,theoutputatapresentinstantdonotanticipateinputfromfuture
instants. In other words, the input that occurs before 𝑡 < 𝑡0 can only determine the output with 𝑡 < 𝑡0.
𝑥(𝑡), 𝑡 <𝑡0 ⇔ 𝑦(𝑡), 𝑡 <𝑡0 (2.20)
Thus, in a causal system, it is impossible to obtain an output before an input is applied. Since the impulseresponseisdefinedash(𝑡)=T{𝛿(𝑡)},theimpulseresponseh(𝑡)iszerofor𝑡 <0
𝛿(𝑡)=0,𝑡<0 ⇔ h(𝑡)=0,𝑡<0
Due to (2.21), the output of a causal LTI system can be expressed as follows
(2.21) (2.22) (2.23) (2.24)
(2.25)
∫∞ ∫𝑡 Causal LTI System, Arbitrary Input: 𝑦(𝑡) = h(𝜏)𝑥 (𝑡 −𝜏)𝑑𝜏 =
𝑥(𝜏)h (𝑡 −𝜏)𝑑𝜏
𝑥(𝑡)=0, 𝑡<0,
the output of causal signal input on a causal LTI system can be expressed as follows
∫𝑡 ∫𝑡
Causal LTI System, Causal Input: 𝑦(𝑡) = 0 h(𝜏)𝑥 (𝑡 −𝜏)𝑑𝜏 = 0 𝑥(𝜏)h (𝑡 −𝜏)𝑑𝜏
• Stability:AnLTIsystemisstableifitsimpulseresponseh(𝑡)isabsolutelyintegrable. ∫∞
|h(𝜏)|𝑑𝜏 < ∞
0 −∞ Furthermore, if we define a causal signal to achieve the following condition
−∞
28
2 Linear Time-Invariant Systems Proof) If |𝑥(𝑡)| ≤ 𝑘1 < ∞ for any 𝑡, then the output can be bounded as follows
∫∞ ∫∞
|𝑦(𝑡)| = | h (𝜏)𝑥 (𝑡 −𝜏)𝑑𝜏| ≤ |h (𝜏)𝑥 (𝑡 −𝜏)|𝑑𝜏
∫
−∞ −∞
∞∞
(2.26)
|h (𝜏)||𝑥 (𝑡 − 𝜏)|𝑑𝜏 ≤ 𝑘1
If∫∞ |h(𝜏)|𝑑𝜏≤𝑘2<∞,then|𝑦(𝑡)|≤𝑘1×𝑘2<∞andthesystemisstable.
−∞
2.2.2 Response of a DT LTI System
1. ImpulseResponse
• ImpulseResponseistheoutputofasystemwhentheinputisaimpulsesignal𝛿[𝑛]
=
−∞
|h (𝜏)|𝑑𝜏
∫
−∞
(2.27) 𝑦[𝑛]=𝑥[𝑛]∗h[𝑛]= 𝑥[𝑘]h[𝑛−𝑘] (2.28)
h[𝑛] = T{𝛿[𝑛]}
• TheoutputofanyDT-LTIsystemistheconvolutionoftheinput𝑥[𝑛]withtheimpulseresponseh[𝑛]
2. StepResponse
• StepResponseisthesystemoutputforastepsignalinput𝑢[𝑛],i.e.,𝑠[𝑛]=T{𝑢[𝑛]}
• Stepresponsecanbeobtainedbycalculatingthecumulativesumoftheimpulseresponseh[𝑛]. Similarly,
the impulse response h[𝑛] can be determined by differentiating the step response h[𝑛].
∞𝑛 𝑠[𝑛]=h[𝑛]∗𝑢[𝑛]= h[𝑘]𝑢[𝑛−𝑘]= h[𝑘],
𝑘 =−∞ 𝑘 =−∞ h [𝑛] = 𝑠 [𝑛] − 𝑠 [𝑛 − 1]
3. PropertiesofDTLTISystem
• MemorySystem
If h[𝑛0] ≠ 0 for 𝑛0 ≠ 0, then it is a LTI system with memory.
• Causality:ThecausalityconditionforaDT-LTIsystemisgivenby h[𝑛] = 0, 𝑛 < 0
(2.29)
(2.30) (2.31)
Due to (2.31) and the definition of causal signal, the output of a causal LTI system can be expressed as
∞𝑛 CausalLTISystem,ArbitraryInput: 𝑦[𝑛]=h[𝑘]𝑥[𝑛−𝑘]= 𝑥[𝑘]h[𝑛−𝑘],
𝑘=0 𝑘=−∞ 𝑛𝑛
Causal LTI System, Causal Input: 𝑦[𝑛] = h[𝑘]𝑥 [𝑛 −𝑘] = 𝑥[𝑘]h [𝑛 −𝑘] 𝑘=0 𝑘=0
• Stability:AnLTIsystemisstableifitsimpulseresponseh(𝑡)isabsolutelysummable.
∞
|h[𝑘]| < ∞
𝑘 =−∞
29
(2.32)
(2.33)
∞
𝑘 =−∞
[Example 2-3] Consider a CT-LTI system with step response 𝑠(𝑡) = 𝑒−𝑡𝑢(𝑡). Determine the output of the system for input signal 𝑥(𝑡) = 𝑢 (𝑡 − 1) −𝑢 (𝑡 − 3).
Refer [Schaum’s text, Problem 2.10]
Solution) Based on the definition of a step response 𝑠(𝑡) = T {𝑢(𝑡)}, the output of the LTI system is given by
𝑦(𝑡) = 𝑠 (𝑡 − 1) − 𝑠 (𝑡 − 3) = 𝑒−(𝑡−1)𝑢 (𝑡 − 1) − 𝑒−(𝑡−3)𝑢 (𝑡 − 3) , where we used linearity and time-invariance of the LTI system in the second equality.
[Example 2-4] Consider a CT-LTI system described by
∫ 𝑡+𝑇
𝑦(𝑡) =
Find the impulse response h (𝑡 ) of the system and answer whether this system is causal or not.
Refer [Schaum’s text, Problem 2.11]
(2.34)
(2.35)
2 Linear Time-Invariant Systems
12
𝑥 (𝜏)𝑑𝜏.
𝑇 𝑡−𝑇 2
Solution) Since the impulse response is the output for the impulse signal input, the following equality holds
∫𝑇1𝑇𝑇
> 0,
where we applied the sifting property of 𝛿 (𝑡 ) in the second equality. The impulse response h (𝑡 ) is plotted in Fig. 2.5.
Sinceh(𝑡)≠0for−𝑇 <𝑡<0,thissystemisnotcausal. 2
Figure 2.5:
[Example 2-5] Consider CT-LTI systems composed of two component blocks where the impulse response of each block is given by h1(𝑡) = 𝑒−2𝑡𝑢(𝑡) and h2(𝑡) = 2𝑒−𝑡𝑢(𝑡). For (a) cascade and (b) parallel connection case, find the impulse response h (𝑡 ) of the overall system and answer whether the overall system is stable or not.
Refer [Schaum’s text, Problem 2.14, 2.53]
Figure 2.6: 30
1 𝑡+ 2 h(𝑡)=
𝑇 𝑡−𝑇 2
if 𝑡 − < 0 and 𝑡 + 𝛿(𝜏)𝑑𝜏= 𝑇 2
2
(2.36)
0 otherwise
Solution) For (a) cascaded connection case, the impulse response can be derived as follows
h(𝑡)=h1(𝑡)∗h2(𝑡)=
∫∞ −𝜏 −2(𝑡−𝜏)
2𝑒 𝑢(𝜏)𝑒 𝑢(𝑡−𝜏)𝑑𝜏
−∞
∞ 2𝑒 𝑒𝜏𝑢 (𝜏)𝑢 (𝑡 −𝜏)𝑑𝜏 =
𝑒𝑑𝜏=2𝑒 −𝑒
if𝑡>0, otherwise
∫
(2.37)
(2.38) (2.39) (2.40)
= 2𝑒−2𝑡
−𝑡 −2𝑡
0
0
−∞ =2𝑒−𝑒 𝑢(𝑡)
To check stability, we need to verify whether h (𝑡 ) is absolutely integrable.
∫∞ ∫∞ ∫∞ ∫∞
|h(𝜏)|𝑑𝜏= 2 𝑒−𝑡−𝑒−2𝑡 𝑑𝜏=2 𝑒−𝜏𝑑𝜏− −∞ 0 0 0
𝑒−2𝜏𝑑𝜏 =1<∞
For (b) parallel connection case, the overall impulse response is given by
−2𝑡 h(𝑡)=h1(𝑡)+h2(𝑡)= 𝑒
and the stability test can be performed as below
+2𝑒
−𝑡
𝑢(𝑡),
∫ ∞
−∞ 0
1
𝑑𝜏 = 2 + 2 < ∞,
−𝜏 which indicates that the parallel connection system in (b) is stable.
|h (𝜏) |𝑑𝜏 =
∫𝑡
−2𝑡𝜏 −𝑡−2𝑡
∫ ∞ −2𝜏
𝑒 + 2𝑒
2 Linear Time-Invariant Systems
[Example 2-6] For the following impulse responses, determine whether the given LTI system is causal and stable. a) h(𝑡)=𝑒−3𝑡sin(𝑡)𝑢(𝑡) b) h(𝑡)=𝛿(𝑡)+𝑒−3𝑡𝑢(𝑡)
1𝑛
c) h[𝑛]=𝛿[𝑛+1] d) h[𝑛]= −2 𝑢[𝑛−1]
Solution) For causality, we need to check whether the impulse response has a non-zero value on 𝑡 < 0 (or 𝑛 < 0). Then, it is clear that, except (c), the other systems are all causal system. For stability test, we need to check whether the given impulse response is absolutely integrable (or absolutely summable).
∫ ∞ −3𝑡 ∫ ∞ −3𝑡 1
(a) 0 𝑒 |sin(𝑡)|𝑑𝑡≤ 0 𝑒 𝑑𝑡=3<∞
(b)
(c)
∫ ∞ ∫ ∞ −3𝑡
4
𝑑𝑡 = 3 < ∞
|h (𝜏) |𝑑𝜏 = 1 + −∞ 0 ∞
|h[𝑛]|=1<∞, 𝑛=−∞
𝑒
(2.41)
(d)
∞ 1𝑛
| −2 |=
𝑛=1 1−2
1
2 1 =1<∞
Hence, the solutions are summarized below. a) Causal, Stable
c) Noncausal, Stable
b) Causal, Stable d) Causal, Stable
[Example 2-7] Compute the output 𝑦 [𝑛] of a DT-LTI system for the given impulse response and the input signals
a) 𝑥[𝑛]=𝛼𝑛𝑢[𝑛]andh[𝑛]=𝛽𝑛𝑢[𝑛] Refer [Schaum’s text, Problem 2.28, 2.29]
b) 𝑥[𝑛]=𝑢[𝑛]andh[𝑛]=𝛼𝑛𝑢[𝑛],where0<𝛼<1
31
Solution) In (a), the convolution sum between the input signal and the impulse response is given by
∞∞
𝑦[𝑛]= 𝑥[𝑘]h[𝑛−𝑘]= 𝛼𝑘𝛽𝑛−𝑘𝑢[𝑘]𝑢[𝑛−𝑘]
𝑘 =−∞
𝑛 𝛼𝑘
𝑘 =−∞
1−(𝛼/𝛽)𝑛+1
(2.42) For (b), we can apply [Example 2-7. a] by substituting 𝛼 ← 1 and 𝛽 ← 𝛼 into (2.42). Then, the output is given by
(2.43)
= 𝛽 𝑛 𝑘=0
𝛽𝑛
1 − 𝛼 / 𝛽
𝑢[𝑛]=
𝑢[𝑛]
𝛽𝑛+1−𝛼𝑛+1 𝛽 − 𝛼
𝑢[𝑛] if𝛼≠𝛽 if𝛼=𝛽
2 Linear Time-Invariant Systems
𝛽
𝑢 [ 𝑛 ] =
𝑦[𝑛] =
𝑛
𝛽 (𝑛+1)𝑢[𝑛]
1−𝛼𝑛+1 1−𝛼
32
3 Fourier Series
Major References:
• Chapter3,SignalsandSystemsbyAlanV.Oppenheimet.al.,2ndedition,PrenticeHall
• Chapter5.2&6.2,Schaum’sOutlineofSignalsandSystems,2ndEdition,2010,McGraw-Hill
3.1 Introduction
1st Metaphor of the Fourier Analysis (Source: https://betterexplained.com)
2nd Metaphor of the Fourier Analysis
• Arbitrary vector can be expressed via unit vectors and the magnitude
3 Fourier Series
Jean-Baptiste Joseph Fourier (1768-1830)
What is Fourier Analysis?
1. WhatdoestheFourierTransformdo?
Given a smoothie, it finds the recipe.
2. How?
Run the smoothie through filters to extract each ingredient.
3. Why?Recipesareeasiertoanalyze,compare, and modify than the smoothie itself.
4. Howdowegetthesmoothieback? Blend the ingredients.
Important Points to consider
1. Filtersmustbeindependent 2. Filtersmustbecomplete
3. Ingredients must be combineable. The ingredients must make the same re- sult when separated and combined in any order.
a=𝑎𝑥e𝑥 +𝑎𝑦e𝑦 +𝑎𝑧e𝑧
𝑛𝑛𝑛 𝑛=−∞
toward each unit vector.
• Can we break a function into it simple functions (referred to as base
functions) just like vector case?
• Canwecombinethebasefunctionstorepresentarbitrarysignals?
∞
⇒ 𝑥(𝑡) = 𝑎 𝜓 (𝑡), where𝜓 (𝑡) is the base function.
33
fundamental angular frequency:𝑤0 = 2𝜋 𝑓0 = 2𝜋
Then the CT-Fourier series can be expressed into the following two representations. All of the proof for Chapter 3.2
𝑐
𝑘
= 𝑥(𝑡)𝑒−𝑗𝑘𝑤0𝑡𝑑𝑡 = 𝑇0 T0
𝑥(𝑡)𝜓∗(𝑡)𝑑𝑡, 𝑘
1∫ 1∫
3 Fourier Series
1. Fourieranalysisisthestudyofthewaygeneralfunctionsmayberepresentedorapproximatedby sums of simpler trigonometric functions.
• Analysis:breakingupasignalintosimplerconstituentparts
• Synthesis:reassemblingasignalfromitsconstituentparts
⇒ Fourier analysis is all about breaking & reassembling a function
2. FourierSeries:fourieranalysisforperiodicsignals
3. FourierTransform:fourieranalysisfornon-periodicsignals
3.2 Continuous Time Fourier Series
For a periodic signal 𝑥 (𝑡 ) with fundamental period 𝑇0 , we adopt sinusoidal signals as the base function fundamental period:𝑇0, fundamental frequency:𝑓0 = 1 ,
∞∞
𝑥(𝑡)= 𝑐 𝜓 (𝑡)= 𝑐 𝑒𝑗𝑘𝑤0𝑡
𝑇0 𝑇0
are summarized at the end of the section.
1. Fourier Series (Complex Exponential Series Form)
Thebasefunctionforthisformis𝜓𝑘(𝑡) =𝑒𝑗𝑘𝑤0𝑡 =𝑒𝑗2𝜋𝑘𝑓0𝑡 1. Synthesis:
2. Analysis
𝑘𝑘𝑘 𝑘 =−∞ 𝑘 =−∞
(3.1)
(3.2)
(3.3)
𝑇0 T0
where the integration interval T0 is any period with length 𝑇0, e.g., [0,𝑇0] or
[Properties]
−𝑇0 , 𝑇0 22
1. Thesetofbasefunctions{𝜓𝑘(𝑡)}isorthogonalonanyintervaloveraperiod𝑇0,(𝛼,𝛼+𝑇0)
34
2. If 𝑥(𝑡) is a real funcction, then 𝑐
−𝑘
= 𝑐∗ 𝑘
∫ 𝛼+𝑇0
𝛼 𝑘 𝑇0,𝑚=𝑘
𝜓 (𝑡)𝜓∗(𝑡)𝑑𝑡 = 𝑚
0, 𝑚≠𝑘
By using the Euler’s Formula, 𝑒𝑗𝑘𝑤0𝑡 = cos (𝑘𝑤0𝑡) + sin (𝑘𝑤0𝑡), the Fourier series in the complex exponential series form can be converted to a trigonometric series form as follows.
2. Fourier Series (Trigonometric Series Form)
1. Synthesis
2. Analysis
𝑥(𝑡)cos(𝑘𝑤0𝑡)𝑑𝑡, 1. Theconvertionbetweentworepresentations
0∞
𝑥(𝑡)=𝑎 +(𝑎 cos(𝑘𝑤𝑡)+𝑏 sin(𝑘𝑤𝑡))
(3.4)
(3.5)
(3.6)
A periodic signal 𝑥(𝑡) has a Fourier series representation if it satisfies the Dirichlet conditions. In other words, Dirichlet conditions are the sufficient conditions (but not necessary condition) for the Fourier series to converge.
Peter Gustav Lejeune Dirichlet (1805-1859)
3. Dirichlet Condition (Sufficient conditions for FS to exist)
∫
nuities is finite.
If 𝑥 (𝑡 ) satisfies the Dirichlet condition, then the corresponding Fourier series is convergent and its sum is
𝑥(𝑡), except at any point 𝑡0 at which 𝑥(𝑡) is discontinuous.
𝑥 ( 𝑡 0 ) = 1 𝑥 𝑡 0+ + 𝑥 𝑡 0−
35
3 Fourier Series
𝑎𝑘 =
2
2∫ 2∫
𝑥(𝑡)sin(𝑘𝑤0𝑡)𝑑𝑡
[Properties]
𝑎0
0
1
𝑐 =2(𝑎 −𝑗𝑏), 𝑘 𝑘𝑘
𝑇0 T0
𝑏𝑘 =
𝑇0 T0
𝑘0𝑘0 𝑘=1
2=𝑐, 1
𝑎𝑘=𝑐𝑘+𝑐−𝑘, 𝑏𝑘=𝑗(𝑐𝑘−𝑐−𝑘) 𝑐 = (𝑎+𝑗𝑏), −𝑘 𝑘 𝑘
2 2. If𝑥(𝑡)isarealfuncction,then𝑎𝑘 =2Re[𝑐𝑘],𝑏𝑘 =−2Im[𝑐𝑘].
⇔
1. 𝑥(𝑡) is absolutely integrable over any period
2. 𝑥(𝑡)hasafinitenumberofmaximaandminimawithinanyfiniteintervaloft.
T0
3. 𝑥(𝑡)hasafinitenumberofdiscontinuitieswithinanyfiniteintervaloft,andeachofthesedisconti-
2
|𝑥(𝑡)|𝑑𝑡 < ∞
4. Parseval’s Theorem
Figure 3.1: Examples of the Dirichlet Condition
Average signal power can be calculated by integral over time domain or infinite sum over frequency domain.
P=
𝑇0 T0
|𝑥(𝑡)| 𝑑𝑡 =
|𝑐𝑘| (3.7)
1∫ 2 ∞ 2
3 Fourier Series
[Proof of Chapter 3.2]
1. Provethattheset{𝜓 (𝑡)}={𝑒𝑗𝑘𝑤0𝑡}ofbasefunctionsisorthogonal. 𝑘
Proof) Refer [Schaum’s text, Problem 5.1]
∫ 𝛼+𝑇0 𝛼𝛼
𝑒𝑗(𝑚−𝑘)𝑤0𝑇𝑑𝑡
If 𝑚 = 𝑘, then the complex exponential in the integral will be one (𝑒0 = 1) and the integration results will be
𝑒𝑗𝑚𝑤0𝑇𝑒−𝑗𝑘𝑤0𝑇𝑑𝑡 =
𝑇0. If 𝑚 ≠ 𝑘, then denote 𝑚 − 𝑘 = 𝑙, which is a nonzero integer, and the integral can be further expressed as
∫ 𝛼+𝑇0 𝛼
𝑒𝑗𝑙𝑤0𝑇𝑑𝑡=
1 𝑗𝑙𝑤0
𝛼+𝑇0 𝑒𝑗𝑙𝑤0𝑡
=
𝑒𝛼𝑙𝑤0𝑙 𝑒𝑗2𝜋𝑙−1 =0,
(3.8)
(3.9)
𝑘 =−∞
∫ 𝛼+𝑇0
where the second equality follows by 𝑤0 = 2𝜋/𝑇0. 1∫
(𝑡)}. ∞
2. Derive 𝑐 = 𝑥(𝑡)𝜓∗(𝑡)𝑑𝑡 using the orthogonality condition of {𝜓 𝑘𝑘
𝛼
𝑗𝑙𝑤0
𝑇0 T0
Proof) Refer [Schaum’s text, Problem 5.2]
𝑥(𝑡)𝜓 (𝑡)𝑑𝑡 =
= 1 𝑐𝑚 𝜓𝑚(𝑡)𝜓 (𝑡)𝑑𝑡=𝑐𝑘,
𝑘
=𝑇0𝛿𝑚,𝑘
𝑘
∫ 𝛼+𝑇0 ∗ 11
∫ 𝛼+𝑇0 𝑇0𝛼𝑘𝑇0𝛼 𝑘
∗ 𝑐𝑚𝜓𝑚(𝑡) 𝜓 (𝑡)𝑑𝑡
𝑚=−∞ ∞∫𝛼+𝑇0 ∗
where𝛿𝑚,𝑘 =1if𝑚=𝑘and𝛿𝑚,𝑘 =0if𝑚≠𝑘.
𝑇0 𝛼
𝑚=−∞
36
3. DerivethetrigonometricseriesrepresentationofFSfromthecomplexexponentialform.
Proof) Apply the Euler Formula 𝑒𝑗𝑤𝑡 = cos (𝑤𝑡) + 𝑗 sin (𝑤𝑡) into the complex exponential FS formula as
follows. Refer [Schaum’s text, Problem 5.3].
∞∞
=𝑐0 +
=𝑐 +[(𝑐 +𝑐 )cos(𝑘𝑤 𝑡)+𝑗(𝑐 −𝑐 )sin(𝑘𝑤 𝑡)]
where𝑎0 =𝑐0,𝑎𝑘=𝑐𝑘+𝑐−𝑘,and𝑏𝑘=𝑗(𝑐𝑘−𝑐−𝑘). 2
4. DerivetheParseval’stheoremin(3.7).
Proof) Let’s evaluate the power of a periodic signal 𝑥(𝑡). Refer [Schaum’s text, Problem 5.14].
P= =
=
𝑥(𝑡) = 𝑘=−∞
𝑐𝑘𝑒 ∞
2
𝑘0𝑘0 𝑘=1
𝑘=1 0∞
(3.10)
=𝑎 +(𝑎 cos(𝑘𝑤𝑡)+𝑏 sin(𝑘𝑤𝑡)),
𝑗𝑘𝑤0𝑡 −𝑗𝑘𝑤0𝑡 𝑐𝑘𝑒 +𝑐−𝑘𝑒
𝑗𝑘𝑤0𝑡
𝑘=1
0 𝑘−𝑘 0 𝑘−𝑘 0
3 Fourier Series
1∫21∫∗
|𝑥(𝑡)|𝑑𝑡= 𝑥(𝑡)𝑥 (𝑡)𝑑𝑡 𝑇0 T0
𝑇0 T0 1∫∞∞∞
𝑐𝑒𝑗𝑘𝑤0𝑡× 𝑐∗𝑒−𝑗𝑚𝑤0𝑡 𝑑𝑡 Applied𝑥(𝑡)= 𝑐𝑒𝑗𝑘𝑤0𝑡 𝑘𝑚𝑘
𝑇0 T0 𝑘=−∞ 𝑚=−∞ 1∞∞∫∞
𝑘=−∞
𝑘=−∞
(3.11)
𝑐𝑘𝑐𝑚∗ 𝑒𝑗𝑘𝑤0𝑡𝑒−𝑗𝑚𝑤0𝑡𝑑𝑡 = |𝑐𝑘|2
𝑇0 T 𝑘=−∞𝑚=−∞ 0
=𝑇0𝛿𝑘,𝑚
[Example 3-1] Derive the complex exponential FS representation of the following signals Refer [Schaum’s text, Problem 5.4]
a) 𝑥(𝑡)=cos(𝑤0𝑡) b) 𝑥(𝑡)=sin(𝑤0𝑡) c) 𝑥(𝑡) = cos (4𝑡) + sin (6𝑡) d) 𝑥(𝑡) = sin2(𝑡)
Solution) For sinusoidal functions, we can get the FS by simply expanding it in terms of the complex exponentials. 11
cos(𝑤0𝑡) = 𝑒𝑗𝑤0𝑡 +𝑒−𝑗𝑤0𝑡 , sin(𝑤0𝑡) =
2 2𝑗
cos(4𝑡)+sin(6𝑡)=1𝑒𝑗4𝑡+𝑒−𝑗4𝑡+ 1 𝑒𝑗6𝑡−𝑒−𝑗6𝑡, 2 2𝑗
sin2(𝑡) = 1 (1−cos2𝑡) = 1 − 1𝑒𝑗2𝑡 − 1𝑒−𝑗2𝑡 2244
𝑒𝑗𝑤0𝑡 −𝑒−𝑗𝑤0𝑡 ,
(3.12)
Hence, the complex Fourier coefficients for each signals are derived as follows.
(a) 𝑐1 = 1,𝑐−1 = 1,and𝑐𝑘 =0forother𝑘index. 22
(b) 𝑐1= 1,𝑐−1=−1,and𝑐𝑘 =0forother𝑘index. 2𝑗 2𝑗
(c) 𝑐−3=−1,𝑐−2=1,𝑐2=1,𝑐3= 1,and𝑐𝑘 =0forother𝑘index. 2𝑗 2 2 2𝑗
(d) 𝑐1 =−1,𝑐−1 =−1,𝑐0 = 1,and𝑐𝑘 =0forother𝑘index. 442
37
[Example 3-2] Determine the complex and trigonometric FS representation of the following signals. Refer [Schaum’s text, Problem 5.5, 5.6, 5.7]
a) Case (a) c) Case (c)
Case (a)
Case (b)
b) Case (b)
3 Fourier Series
Case (c)
Solution) For (a), the Complex Fourier coefficients can be derived as follows
𝑐0 =
∫ 𝑇0 𝐴2𝐴
𝑑𝑡 = 𝑇00 2
𝑐
𝑘
=
𝑇0 0
0
𝑒−𝑗𝑘𝑤0𝑡𝑑𝑡 = 𝑒−𝑗𝑘𝑤0𝑡 𝑗𝑘𝑤0𝑇0
𝑇0 2
=
𝑗𝑘𝑤0𝑇0
1−𝑒−𝑗𝑘𝜋 =
𝑗2𝜋𝑘
(3.13)
∫𝑇0 0
𝐴2𝐴𝐴𝐴
1−(−1)𝑘
= 𝐴
for even 𝑘 = 2𝑚, for odd 𝑘 = 2𝑚 + 1
𝑗𝜋 (2𝑚 + 1)
The trigonometric Fourier coefficients can be derived as follows.
𝑎0=2𝑐0=𝐴, 𝑎𝑘=2Re[𝑐𝑘]=0,
𝑏 =−2Im[𝑐]= 𝑘𝑘
2𝐴
The signals 𝑥 (𝑡) for case (b) is a time-shifted version of the case (a), i.e., 𝑥 (𝑡) = 𝑥 𝑡 + 𝑇0
𝑏 𝑏𝑎4 coefficient of the time-shifted signal is derived as follows
0
(2𝑚 + 1) 𝜋
for even 𝑘 = 2𝑚, for odd 𝑘 = 2𝑚 + 1
(3.14) . Then, the Fourier
1∫ 𝑇0−𝑗𝑘𝑤𝑡 1𝑗𝑘𝑤𝑇0∫ 𝑥𝑡+ 𝑒 0𝑑𝑡= 𝑒 04
𝑥(𝜏)𝑒
−𝑗𝑘𝑤𝜏 𝑗𝑘𝜋 𝑘 0𝑑𝑡=𝑐𝑒2=𝑗𝑐,
(3.15)
𝑇4𝑇
0T0 0T0
𝑘𝑘
38
3 Fourier Series where we used a change of variable, i.e., 𝜏 = 𝑡 + 𝑇0 . Hence, the Fourier coefficients of case (b) are given by
𝑐0 = 𝐴 × 𝑗0 = 𝐴 22
2𝑚+1
4
for even 𝑘 = 2𝑚, for odd 𝑘 = 2𝑚 + 1
𝜋 (2𝑚 + 1) for odd 𝑘 = 2𝑚 + 1
The signals 𝑥𝑐 (𝑡 ) for case (c) is a scaled version of the case (a), i.e., 𝑥𝑐 (𝑡 ) = 2𝑥𝑎 (𝑡 ) − 𝐴. Then, the Fourier coefficient
0
𝐴(−1)
𝜋 (2𝑚 + 1)
𝑐𝑘= 𝐴×𝑗
𝑚
=
𝑎0=2𝑐0=𝐴, 𝑏𝑘=−2Im[𝑐𝑘]=0,
𝑗𝜋 (2𝑚 + 1)
0 for even 𝑘 = 2𝑚,
(3.16)
of the case (c) is derived as follows
𝑎𝑘 =2Re[𝑐𝑘]=2𝐴(−1)𝑚
1∫
(2𝑥𝑎 (𝑡)−𝐴)𝑑𝑡 =2[𝑐0 forcase(a)]−𝐴 𝑘𝑎𝑘
𝑐0 =
𝑐 = (2𝑥 (𝑡)−𝐴)𝑒−𝑗𝑘𝑤0𝑡𝑑𝑡=2[𝑐 forcase(a)],
for even 𝑘 = 2𝑚, for odd 𝑘 = 2𝑚 + 1
for even 𝑘 = 2𝑚, for odd 𝑘 = 2𝑚 + 1
𝑇0 T0 1 ∫
(3.17)
(3.18)
where ∫ T0
𝑇0 T0
𝑒−𝑗𝑘𝑤0𝑡𝑑𝑡 = 0. Hence, the Fourier coefficients of case (c) are given by
0 𝑐=2×𝐴−𝐴=0,𝑐= 2𝐴
0𝑘
2
,
𝑗 𝜋 ( 2 𝑚 + 1 ) 𝑎0=2𝑐0=0, 𝑎𝑘=2Re[𝑐𝑘]=0,
0
𝑏=−2Im[𝑐]= 4𝐴 𝑘𝑘
(2𝑚 + 1) 𝜋
[Example 3-3] Determine the complex and trigonometric FS representation of the periodic impulse trains 𝛿𝑇0 (𝑡 ) ∞
signals, which is defined by 𝛿 (𝑡) = 𝛿 (𝑡 − 𝑘𝑇 ). Refer [Schaum’s text, Problem 5.8] 𝑇0 0
𝑘 =−∞
Solution) The Complex Fourier coefficients can be derived as follows
𝛿(𝑡)𝑑𝑡= , 𝑐𝑘= 𝑇0
∫ 𝑇0 ∫ 𝑇0 12112∞1
𝑐0=
2
𝛿(𝑡−𝑘𝑇0)𝑒−𝑗𝑘𝑤0𝑡𝑑𝑡= (3.19) 𝑇0 −𝑇0 𝑇0
𝑇0 −𝑇0
2 𝑘=−∞
39
The trigonometric Fourier coefficients can be derived as follows.
𝑎0=2𝑐0= 2, 𝑎𝑘=2Re[𝑐𝑘]= 2, 𝑏𝑘=−2Im[𝑐𝑘]=0
(3.20)
3 Fourier Series
Hence, we get
𝑇0 𝑇0 ∞1∞12∞
𝑇0
𝛿𝑇0(𝑡)= 𝛿(𝑡−𝑘𝑇0)= 𝑘=−∞
𝑒𝑗𝑘𝑤0𝑡 = 𝑘=−∞
+ cos(𝑘𝑤0𝑡)
𝑇0 𝑇0
𝑘=1
[Example 3-4] Determine the complex and trigonometric FS representation of the periodic signal 𝑥 (𝑡 ) defined by 𝑥(𝑡)=𝑡2, −𝜋<𝑡<𝜋 and 𝑥(𝑡+2𝜋)=𝑥(𝑡).
Refer [Schaum’s text, Problem 5.62]
Solution) For the given signal, 𝑇0 = 2𝜋 and 𝑤0 = 2𝜋 = 1. Then, the Fourier coefficients can be derived as 𝑇0
0 1∫𝜋2 𝜋2 𝑐= 𝑡𝑑𝑡=,
2𝜋−𝜋 3 1 ∫ 𝜋
1 𝑡2𝑒−𝑗𝑘𝑡 −𝜋
𝑐
𝑘 2𝜋
+
−𝜋 𝜋−𝜋
= 𝑡2𝑒−𝑗𝑘𝑡𝑑𝑡 =
2 ∫ 𝜋 2𝜋 𝑗𝑘 𝑗𝑘
𝑡𝑒−𝑗𝑘𝑡𝑑𝑡
= sin(𝜋𝑘𝑡)+
𝑘 𝑗𝑘𝜋𝑗𝑘
2cos(𝑘𝜋𝑡) 2(−1)𝑘 =2=2,
𝑘𝑘 2𝜋2
𝜋 1 𝑡𝑒−𝑗𝑘𝑡−𝜋 1 −𝜋
+ 𝑒−𝑗𝑘𝑡 2
(3.21)
𝜋 (𝑗𝑘𝜋) 𝜋
𝑎0=2𝑐0= 3, 𝑏𝑘=−2Im[𝑐𝑘]=0,
𝑎𝑘 =2Re[𝑐𝑘]=
𝑘
2
4 (−1)𝑘
[Example 3-5] Determine the complex and trigonometric FS representation of the periodic signal 𝑥 (𝑡 ) defined by 𝑥(𝑡)=𝐴𝑡, 0<𝑡<𝑇0 and 𝑥(𝑡+𝑇0)=𝑥(𝑡).
𝑇0 Refer [Schaum’s text, Problem 5.63]
40
Solution) The Complex Fourier coefficients can be derived as follows
𝐴 ∫ 𝑇0 𝐴 𝑐0= 2 𝑡𝑑𝑡= ,
𝑇00 2
𝐴 ∫ 𝑇0 −𝑗𝑘𝑤0𝑡 𝐴 𝑡𝑒−𝑗𝑘𝑤0𝑡 0 𝐴
∫ 𝑇0
𝑐=2 𝑡𝑒 𝑑𝑡=2 + 2 𝑒 𝑑𝑡
−𝑗𝑘𝑤0𝑡
(3.22)
(3.23)
3 Fourier Series
𝑘
=
𝑇0 0 𝑇0 𝑗𝑘𝑤0 $X$0
𝑗𝑘𝑤0𝑇0 0
𝑇0 𝐴𝑗 𝐴 $$ 𝐴𝑗
+ $1−𝑒−𝑗2𝑘𝜋 = $
2𝜋𝑘 $2$ 2𝜋𝑘 (𝑗$𝑘𝑤 𝑇 )
$00
The trigonometric Fourier coefficients can be derived as follows.
𝑎 =2𝑐 =𝐴, 𝑎 =2Re[𝑐 ]=0, 𝑏 =−2Im[𝑐 ]=−𝐴
00𝑘𝑘𝑘𝑘
Properties of Fourier Series. (Refer [Oppenheim], Chapter 3.5 for detailed proof ) 1. LinearProperty
−𝑗𝑘𝑤0𝑡
where we used a change of variable (𝑡 − 𝑡0 = 𝜏) in the first equality and (3.2) in the second equality.
𝑥(𝑡−𝑡0)𝑒 3. ConjugateProperty
𝑑𝑡= 𝑥(𝜏)𝑒 𝑇0 T0
𝑑𝜏 ×𝑒
=𝑒
−𝑗𝑘𝑤0𝑡0
𝑐𝑘,
If 𝑥 (𝑡 ) is a real-valued signal, then 𝑐
Proof) By taking the conjugate of (3.2), we get
𝑘
−𝑘
= 𝑐 ∗ follows by the conjugate property.
∗ 1∫ 𝑗𝑘𝑤0𝑡 ∗ 1∫ −𝑗𝑘𝑤0𝑡 𝑐= 𝑥(𝑡)𝑒 𝑑𝑡,⇒𝑐= 𝑥(𝑡)𝑒 𝑑𝑡,
−𝑗𝑘𝑤0𝑡0
𝜋𝑘
IftheFScoefficientsof𝑥1(𝑡)and𝑥2(𝑡)are𝑐1,𝑘 and𝑐2,𝑘,thentheFScoefficientsof𝛼1𝑥1(𝑡)+𝛼2𝑥2(𝑡)are 𝛼1𝑥1(𝑡) + 𝛼2𝑥2(𝑡) ↔ 𝛼1𝑐1,𝑘 + 𝛼2𝑐2,𝑘
2. TimeShifting
If𝑥(𝑡)isaCTperiodicsignalwithperiod𝑇0 andFScoefficients𝑐𝑘,theFScoefficientsof𝑥(𝑡−𝑡0)are
−𝑗𝑘𝑤0𝑡0 𝑥(𝑡−𝑡0) ↔ 𝑒 𝑐𝑘
Proof) The FS coefficients of the delayed signal are given by 1∫1∫
−𝑗𝑘𝑤0𝜏
𝑇0 T0
If𝑥(𝑡)isaCTperiodicsignalwithperiod𝑇 andFScoefficients𝑐 ,theFScoefficientsof𝑥∗(𝑡)aregivenby
−𝑘
0𝑘 𝑥∗(𝑡) ↔ 𝑐∗
𝑘 𝑇0 T0 −𝑘 𝑇0 T0
where the second expression follows by substituting 𝑘 to −𝑘. Then, by comparing (3.2) to the last expression,
itfollowsthat 𝑐∗ aretheFScoefficientsoftheconjugetsignal𝑥∗(𝑡). −𝑘
4. FrequencyShifting
If𝑥(𝑡)isaCTperiodicsignalwithperiod𝑇 andFScoefficients𝑐 ,theFScoefficientsof𝑒𝑗𝑘0𝑤0𝑡𝑥(𝑡)are
0𝑘 𝑒𝑗𝑘0𝑤0𝑡𝑥(𝑡) ↔ 𝑐𝑘−𝑘0
41
Proof) Based on (3.2), the Fourier series coefficients of 𝑒𝑗𝑘0𝑤0𝑡𝑥(𝑡) are given by
1 ∫ 𝑇0 T0
𝑗𝑘0𝑤0𝑡
𝑥 (𝑡) 𝑒
−𝑗𝑘𝑤0𝑡
1 ∫ −𝑗(𝑘−𝑘0)𝑤0𝑡
𝑇0 T0
𝑑𝑡 =
𝑥 (𝑡) 𝑒
𝑑𝑡 = 𝑐𝑘−𝑘0 If𝑥(𝑡)isaCTperiodicsignalwithperiod𝑇0 andFScoefficients𝑐𝑘,theFScoefficientsof𝑥(−𝑡)aregivenby
𝑒
3 Fourier Series
5. TimeReversalProperty
𝑥(−𝑡) ↔ 𝑐−𝑘 If𝑥(𝑡)isaCTperiodicsignalwithperiod𝑇0 andFScoefficients𝑐𝑘,theFScoefficientsof𝑥(𝛼𝑡),𝛼 >0,are
𝑥(𝛼𝑡) ↔ 𝑐 withperiod𝑇0 𝑘
𝛼
Proof) Since the period of 𝑥 (𝛼𝑡) reduces to 𝑇0 = 𝑇0 , the Fourier series coefficients of 𝑥 (𝛼𝑡) are 𝛼
6. TimeScalingProperty
0 𝑑𝑡=
where we used a change of variable (𝑡 = 𝛼𝜏) in the second equality.
0 𝑑𝜏=𝑐, 𝑘
1 ∫ 𝑇0𝑇
7. Differentiation
If𝑥(𝑡)isaCTperiodicsignalwithperiod𝑇 andFScoefficients𝑐 ,theFScoefficientsof𝑑𝑥(𝑡) are
−𝑗𝑘2𝜋𝛼𝑡 1 ∫ −𝑗𝑘2𝜋𝜏
−𝑗𝑘2𝜋𝑡 𝛼 ∫
𝑇 𝑇 𝑇
𝑥(𝛼𝑡)𝑒 0 𝑑𝑡= 𝑥(𝜏)𝑒
𝑥(𝛼𝑡)𝑒
𝑇0 𝑇0 0 𝛼 0 𝑇0
8. Integration
𝑥(𝑡) = 𝑐𝑘𝑒 𝑘 =−∞
𝑗𝑘𝑤0𝑡 𝑘
∫
𝑐 𝑗𝑘𝑤0
𝑇
0𝑘
𝑑𝑡
𝑑𝑥(𝑡) 𝑑𝑡
𝑗𝑘𝑤0𝑡
↔ 𝑗𝑘𝑤0𝑐𝑘
Proof) By differentiating (3.1), we get ∞
𝑑𝑥(𝑡) ∞ =
𝑗𝑘𝑤0𝑡
𝑗𝑘𝑤0𝑐𝑘𝑒 If𝑥(𝑡)isaCTperiodicsignalwithperiod𝑇 andFScoefficients𝑐 ,theFScoefficientsof∫ 𝑥(𝑡)𝑑𝑡are
𝑥(𝑡) = 𝑐𝑘𝑒 𝑘 =−∞
→
0𝑘
𝑥(𝑡)𝑑𝑡 ↔
𝑘
𝑑𝑡
𝑘 =−∞
Proof) By integrating (3.1), we get
∞ ∫∞𝑐
𝑗𝑘𝑤0𝑡
→
𝑥(𝑡)𝑑𝑡 =
0𝑒
9. PeriodicConvolution
If 𝑥1(𝑡) and 𝑥2(𝑡) are periodic signals with common period 𝑇0 and FS coefficients 𝑐1,𝑘 and 𝑐2,𝑘, the FS coefficients of the periodic convolution defined in (2.2) are given by
∫
T0
42
𝑥1(𝑡) 𝑥2(𝑡) =
𝑥1 (𝜏)𝑥2 (𝑡 −𝜏)𝑑𝜏 ↔ 𝑇0𝑐1,𝑘𝑐2,𝑘
𝑘 =−∞
𝑗𝑘𝑤
Proof) The FS coefficients of 𝑥1(𝑡) 𝑥2(𝑡) are given by
1 ∫ −𝑗𝑘𝑤0
1 ∫ ∫ −𝑗𝑘𝑤0𝑡
𝑥1(𝑡)𝑥2(𝑡)𝑒
𝑑𝑡= = =
𝑥1(𝜏)𝑥2(𝑡−𝜏)𝑑𝜏 𝑒
1∫∫1 21 −𝑗𝑘𝑤0𝜏−𝑗𝑘𝑤0𝑡1 1
3 Fourier Series
𝑇0 T0
𝑇0 T0 T0
𝑑𝑡
𝑇0 T0 T0
𝑥 (𝜏)𝑥 (𝑡 )𝑑𝜏𝑒 𝑒 𝑑𝜏𝑑𝑡
1 ∫ 𝑇0 T0
in the second equality, and employed (3.2) for 𝑐1,𝑘 and 𝑐2,𝑘 in the last equality
10. MultiplicationProperty
If 𝑥1(𝑡) and 𝑥2(𝑡) are periodic signals with common period 𝑇0 and FS coefficients 𝑐1,𝑘 and 𝑐2,𝑘, the FS coefficients of the product of two signals 𝑥1(𝑡)𝑥2(𝑡) are given by
𝑥1(𝑡)𝑥2(𝑡) ↔
Proof) The FS coefficients of 𝑥1(𝑡)𝑥2(𝑡) are given by
∞
𝑐1,𝑙𝑐2,𝑘−𝑙
1 ∫ 𝑇0 T0
𝑥1(𝑡)𝑥2(𝑡)𝑒
−𝑗𝑘𝑤0𝑡
𝑑𝑡 =
1 ∫ ∞
𝑇0
𝑐1,𝑙𝑒
𝑗𝑙𝑤0𝑡
𝑥2(𝑡)𝑒
−𝑗𝑘𝑤0𝑡
𝑑𝑡
1 −𝑗𝑘𝑤0𝜏 𝑥(𝜏)𝑒 𝑑𝜏·
𝑥(𝑡)𝑒 𝑑𝜏𝑑𝑡=𝑇𝑐 𝑐
∫ 2 1 −𝑗𝑘𝑤0𝑡1 1 0 1,𝑘 2,𝑘
T0
where we used the definition of periodic convoltion in the first equality, applied a change of variable (𝑡 −𝜏 = 𝑡1)
𝑙 =−∞
T0 𝑙=−∞
∞1∫ ∞
1. LinearProperty: 𝛼1𝑥1(𝑡)+𝛼2𝑥2(𝑡) ↔ 𝛼1𝑐1,𝑘 +𝛼2𝑐2,𝑘
2. Time Shifting: 𝑥 (𝑡 − 𝑡0) ↔ 𝑒−𝑗𝑘𝑤0𝑡0𝑐𝑘
𝑐1,𝑙
where we applied (3.1) for 𝑥1(𝑡) in the first equality, then used (3.2) in the last equality.
Properties of Fourier Series
=
𝑇0 T0
𝑥2(𝑡)𝑒−𝑗(𝑘−𝑙)𝑤0𝑡𝑑𝑡 = 𝑐1,𝑙𝑐2,𝑘−𝑙 𝑙=−∞
𝑙=−∞
3. Conjugate Property: 𝑥∗ (𝑡) ↔ 𝑐∗ −𝑘
4. Frequency Shifting: 𝑒𝑗𝑘0𝑤0𝑡𝑥(𝑡) ↔
𝑐
5. Time Reversal Property:
6. Time Scaling Property:
𝑥 (−𝑡) ↔
𝑘−𝑘0 𝑐−𝑘
with period 𝑇0 𝑘𝛼
𝑥 (𝛼𝑡)
↔
𝑗𝑘𝑤0𝑐𝑘 𝑐𝑘
10. Multiplication Property: 𝑥1(𝑡)𝑥2(𝑡) ↔ ∞ 𝑐1,𝑙𝑐2,𝑘−𝑙 𝑙 =−∞
𝑐
7. Differentiation: 𝑑𝑥(𝑡) ↔
𝑑𝑡
8. Integration: ∫ 𝑥(𝑡)𝑑𝑡 ↔
9. PeriodicConvolution: 𝑥1(𝑡)𝑥2(𝑡)=∫ 𝑥1(𝜏)𝑥2(𝑡−𝜏)𝑑𝜏 ↔ 𝑇0𝑐1,𝑘𝑐2,𝑘
𝑗𝑘𝑤0
T0
43
[Example 3-6] Derive the complex exponential FS coefficients of the following signal using the differentiation / or integration property. Refer [Schaum’s text, Problem 5.9]
Solution) The differentiation of the given signal is plotted below, which is similar to the signal in Example 3-2. Case (c). The amplitude of Ex 3-2. Case (c) was 𝐴, whereas the that of the illustrated signal is 2𝐴 .
𝑇0
3 Fourier Series
By using the integration property, the complex exponential FS coefficients of the given signal is 𝑐
where 𝑐𝑐,𝑘 represents the FS coefficients of Ex 3-2. Case (c). The constant 𝑐0 for 𝑘 = 0 can not determined by the integration property, so it should be derived through the following equation
0 1∫𝑇0 1 𝐴𝑇0 𝐴 𝑐= 𝑥(𝑡)𝑑𝑡=·=,
𝑇0 0
𝑇0 2 2 2𝐴
for even 𝑘 = 2𝑚, for odd 𝑘 = 2𝑚 + 1
(3.24)
𝑘
= 𝑐𝑐,𝑘 for 𝑘 ≠ 0, 𝑗𝑘𝑤0
0 𝑐= 4𝐴
𝑘
( 𝑗 𝑘 ) 2 𝑤 0 𝑇 0
= −
𝜋 2 ( 2 𝑚 + 1 ) 2
44
[Example 3-7] Derive the complex exponential FS coefficients of the following signal using the property of FS.
.Duetothetimeshifting
property, the FS coefficients are determined by 𝑒𝑗𝑘𝑤0𝑇0/2𝑐𝑘 = (−1)𝑘𝑐𝑘. 𝐴
𝑐0 = 2,
𝑘+1 𝑐𝑘= (−1) 2𝐴
2 (𝑘𝜋)
2 𝑒𝑥3−6 for even 𝑘 = 2𝑚,
3 Fourier Series
Solution)Thesignalisatime-shiftedversionofExample3-6,i.e.,𝑥(𝑡)=𝑥 𝑡+𝑇0
0
(3.25)
=
2𝐴
2 2
for odd 𝑘 = 2𝑚 + 1
𝜋 (2𝑚+1)
[Example 3-8] Consider a signal defined by 𝑥 (𝑡) = |𝐴 sin (𝜋𝑡)|. Refer [Schaum’s text, Problem 5.61] a) Sketch𝑥(𝑡)andfinditsfundamentalperiod𝑇0 andangularfrequency𝑤0
b) Find the complex exponential FS series and the trigonometric FS series of 𝑥 (𝑡 )
Solution)
The fundamental period and fundamental angular frequency of the given signal are 𝑇0 =1, 𝑤0 = 2𝜋 =2𝜋
(3.26)
𝑇0
45
The complex exponential FS coefficients can be derived as follows
∫1 𝐴02𝐴
𝑐0 =𝐴 0 sin(𝜋𝑡)𝑑𝑡 = cos(𝜋𝑡)1 = , 𝜋𝜋
(3.27)
(3.28) (3.29)
𝑐 =𝐴 sin(𝜋𝑡)𝑒𝑗𝑘2𝜋𝑡𝑑𝑡=
𝑘0 𝜋1−4𝑘
2 , wheretheintegral∫01sin(𝜋𝑡)𝑒𝑗𝑘2𝜋𝑡𝑑𝑡 canbederivedbyusingintegrationbypartstwotimes
3 Fourier Series
∫1 2𝐴 1
∫1 21
sin(𝜋𝑡)𝑒𝑗𝑘2𝜋𝑡𝑑𝑡 = 2 𝜋 1−4𝑘
.
0
The trigonometric Fourier coefficients can be derived as follows.
4𝐴
𝑎0=2𝑐0= , 𝑎𝑘=2Re[𝑐𝑘]=
4𝐴 1
2 , 𝑏𝑘=−2Im[𝑐𝑘]=0
𝜋
𝜋 1−4𝑘
46
4 Continuous Time Fourier Transform
Major References:
• Chapter4,SignalsandSystemsbyAlanV.Oppenheimet.al.,2ndedition,PrenticeHall • Chapter5,Schaum’sOutlineofSignalsandSystems,2ndEdition,2010,McGraw-Hill
Develope Fourier Transform from Fourier Series
Let’sconsideranonperiodicsignalwithfiniteduration𝑥(𝑡),where𝑥(𝑡)=0for|𝑡|>𝑇1andassumethat𝑇1 <𝑇0. We introduce a periodic signal 𝑥𝑇0 (𝑡 ) by repeating 𝑥 (𝑡 ) with fundamental period 𝑇0 as follows 2
𝑥 (𝑡+𝑇)=𝑥 (𝑡) where 𝑥 (𝑡)=𝑥(𝑡)for−𝑇0 ≤𝑡<𝑇0
𝑇00𝑇0 𝑇0
22
4 Continuous Time Fourier Transform
If we let 𝑇0 → ∞, we have the following results by using the complex exponential FS representation
∫ 𝑇0 ∞ 𝑗2𝜋𝑘𝑡 ∞12
−𝑗2𝜋𝑘𝜏𝑗2𝜋𝑘𝑡 𝑇0 𝑑𝜏 𝑒 𝑇0
𝑥(𝑡)= lim 𝑥 (𝑡)= lim 𝑐 𝑒 𝑇0 = lim 𝑥 (𝜏)𝑒
𝑇0→∞ =lim
𝑇0→∞ ∞ 1 ∫ ∞
𝑇0→∞
− 𝑗 2 𝜋 𝑘 𝜏 𝑗 2 𝜋 𝑘 𝑡
𝑇0 −𝑇0 2
(4.1)
𝑇0 𝑘 𝑇0
𝑘=−∞
𝑥(𝜏)𝑒 𝑇0𝑑𝜏𝑒𝑇0
𝑘=−∞
𝑇0 −∞ 222
𝑇0→∞
Since𝑥 (𝑡)=𝑥(𝑡)for−𝑇0 ≤𝑡<𝑇0 and𝑥(𝑡)=0for|𝑡|>𝑇0,thelastequalityof(4.1)follows
𝑇0
𝑥(𝑡) = lim 𝑇0→∞
= lim 𝑇0→∞
𝑇0𝑓=−∞
𝑘𝑘
−𝑗2𝜋𝑓𝜏 𝑗2𝜋𝑓𝑡 𝑑𝜏 𝑒
𝑘 =−∞
𝑇0→∞
∫ 𝑇0
2 −𝑗𝑘𝑤0𝜏
∫ ∞
𝑑𝜏= 𝑥(𝜏)𝑒
−𝑗𝑘𝑤0𝜏
𝑇0 𝑥𝑇0 (𝜏)𝑒
−2 −∞
𝑑𝜏
Let’sdenote𝑓=𝑘,then lim 𝑓=Δ𝑓=Δ𝑘.Since𝑘isaninteger,Δ𝑘=1.Byapplying𝑇0Δ𝑓=1to(4.1),
𝑇0
∞ 1 ∫ ∞
𝑇0
− 𝑗 2 𝜋 𝜏 𝑗 2 𝜋 𝑡
∞ ∫ ∞
lim 𝑥 (𝜏)𝑒
𝑇0→∞ −∞
𝑥 (𝜏)𝑒
𝑇0 𝑑𝜏 𝑒
𝑇0
· 1 =
· Δ𝑓
𝑋(𝑓) (4.2)
where we denoted 𝑋 (𝑓 ) ∫
𝑥 (𝑡) 𝑒
𝑇0 −∞
∞ ∫∞
𝑘 =−∞
𝑘 =−∞
𝑋 (𝑓)𝑒𝑗2𝜋𝑓𝑡 ·Δ𝑓 =
𝑋 (𝑓)𝑒𝑗2𝜋𝑓𝑡𝑑𝑓,
𝑑𝑡 and applied the definition of the Riemann Integral.
49
∞ −∞
−∞
−𝑗2𝜋𝑓𝑡
4.1 Fourier Transform Pair and Properties
Based on (4.1, 4.2), we derived the Fourier Transform pair of an aperiodic continuous time signal 𝑥 (𝑡 ) as follows. 1. CT – Fourier Transform Pair
For a non-periodic CT signal 𝑥(𝑡), the FT pair in terms of frequence 𝑓 are given by
• FourierTransformof𝑥(𝑡);
• InverseFourierTransformof𝑋(𝑓); • Notation;
F{𝑥(𝑡)}=𝑋(𝑓)
F−1{𝑋(𝑓)}=𝑥(𝑡)
𝑥(𝑡) F 𝑋(𝑓) F−1
∫∞F∫∞
𝑋 (𝑓)𝑒𝑗2𝜋𝑓𝑡𝑑𝑓 𝑋 (𝑓) = 𝑥 (𝑡)𝑒−𝑗2𝜋𝑓𝑡𝑑𝑡
(4.3)
(4.4)
𝑥(𝑡) =
The FT pair in terms of angular frequence 𝑤 = 2𝜋 𝑓 are given by
2𝜋 −∞ F−1
2. Dirichlet Condition (Sufficient conditions for FT to exist)
−∞
−∞
F−1 −∞
1∫∞F∫∞ 𝑥(𝑡) = 𝑋 (𝑤)𝑒𝑗𝑤𝑡𝑑𝑤 𝑋 (𝑤) =
𝑥 (𝑡)𝑒−𝑗𝑤𝑡𝑑𝑡
4 Continuous Time Fourier Transform
∫∞ −∞
1. 𝑥(𝑡) is absolutely integrable
2. 𝑥(𝑡)hasafinitenumberofmaximaandminimawithinanyfiniteintervaloft.
|𝑥(𝑡)|𝑑𝑡 < ∞
3. 𝑥(𝑡)hasafinitenumberofdiscontinuitieswithinanyfiniteintervaloft,andeachofthesedisconti- nuities is finite.
If 𝑥 (𝑡 ) satisfies the Dirichlet condition, then the corresponding Fourier transform is convergent. Properties of Fourier Transform
1. LinearProperty
If the FT of 𝑥1(𝑡) and 𝑥2(𝑡) are 𝑋1(𝑓 ) and 𝑋2(𝑓 ), then the FT of 𝛼1𝑥1(𝑡) + 𝛼2𝑥2(𝑡) is given by
𝛼1𝑥1(𝑡) + 𝛼2𝑥2(𝑡) ↔ 𝛼1𝑋1(𝑓 ) + 𝛼2𝑋2(𝑓 ) Proof) The FT of the linearly combined signals 𝛼1𝑥1(𝑡) + 𝛼2𝑥2(𝑡) is
∫ ∞ −𝑗2𝜋𝑓𝑡 [𝛼1𝑥1(𝑡) + 𝛼2𝑥2(𝑡)] 𝑒
∫ ∞
−∞ −∞ −∞
2. ParameterShifting
IftheFTof𝑥(𝑡) is𝑋(𝑓),thentheFTofthefollowingmappedsignalsaregivenby
0 −𝑗2𝜋𝑓𝑡0 ±𝑗2𝜋𝑓0𝑡 0 F[𝑥(𝑡−𝑡)]=𝑒 𝑋(𝑓), F 𝑥(𝑡)𝑒 =𝑋(𝑓∓𝑓)
Proof)TheFTof𝑥(𝑡−𝑡0)and𝑥(𝑡)𝑒±𝑗2𝜋𝑓0𝑡 arederivedasbelow
∫ ∞ −∞
∫ ∞ −∞
−𝑗2𝜋𝑓𝑡 ±𝑗2𝜋𝑓0𝑡 −𝑗2𝜋𝑓𝑡
𝑒
∫ ∞ −𝑗2𝜋𝑓𝜈 −𝑗2𝜋𝑓𝑡0 −𝑗2𝜋𝑓𝑡0
𝑥 (𝑡 − 𝑡0) 𝑒 𝑥 (𝑡) 𝑒
𝑑𝑡 = 𝑥 (𝜈) 𝑒 −∞
𝑑𝑡 · 𝑒 −𝑗2𝜋(𝑓∓𝑓0)𝑡
= 𝑋 (𝑓 ) 𝑒
𝑑𝑡 = 𝛼1
= 𝛼1𝑋1(𝑓 ) + 𝛼2𝑋2(𝑓 ).
𝑥2(𝑡)𝑒
𝑑𝑡 =
∫ ∞ −∞
50
𝑥 (𝑡) 𝑒
𝑑𝑡 = 𝑋 (𝑓 ∓ 𝑓 ) , 0
𝑥1(𝑡)𝑒
𝑑𝑡
,
−𝑗2𝜋𝑓𝑡 ∫ ∞ 𝑑𝑡 + 𝛼2
−𝑗2𝜋𝑓𝑡
where we used a change of variable (𝑡 − 𝑡0 = 𝜈) in the first equality of the first expression. 3. Scaling
IftheFTof𝑥(𝑡) is𝑋(𝑓),thentheFTofthefollowingscaledsignalisgivenby
1𝑓
F [𝑥 (𝑎𝑡)] = Proof) The FT of 𝑥 (𝑎𝑡) is given by
|𝑎|
𝑋
, F [𝑥 (−𝑡)] = 𝑋 (−𝑓 )
1∫∞ 𝜏
−𝑗2𝜋𝑓
𝑥(𝜏)𝑒 𝑎𝑑𝜏, if𝑎>0
4 Continuous Time Fourier Transform
∫∞ F [𝑥 (𝑎𝑡)] =
−∞
𝑥 (𝑎𝑡)𝑒
−𝑗2𝜋𝑓𝑡
𝑎
𝑑𝑡 =
𝑎 −∞ ∫
1∞ 𝜏 𝑎
− 𝑥(𝜏)𝑒−𝑗2𝜋𝑓 𝑑𝜏, if𝑎<0 𝑎 −∞
4. ConjugateProperty
IftheFTof𝑥(𝑡) is𝑋(𝑓),theFTof𝑥∗(𝑡) forarbitrarysignal𝑥(𝑡) isgivenby
F 𝑥∗ (𝑡) =𝑋∗ (−𝑓).
If 𝑥(𝑡) is a real-valued signal, then 𝑋∗ (−𝑓 ) = 𝑋 (𝑓 ) follows by the conjugate property. Proof) By taking the complex conjugage of (4.3), we get
∫∞ ∗ ∫∞ ∫∞ 𝑋∗(𝑓) = 𝑥(𝑡)𝑒−𝑗2𝜋𝑓𝑡𝑑𝑡 = 𝑥∗(𝑡)𝑒𝑗2𝜋𝑓𝑡𝑑𝑡 ⇒ 𝑋∗(−𝑓) =
−∞ −∞ −∞
5. Duality
IftheFTof𝑥(𝑡) is𝑋(𝑓),theFTof𝑋(𝑡) canbederivedasfollows
F [𝑋(𝑡)] =𝑥(−𝑓) Proof) By using a change of variable (𝑓 ′ ← −𝑡, 𝑡′ ← 𝑓 ), we get
𝑥∗(𝑡)𝑒−𝑗2𝜋𝑓𝑡𝑑𝑡
∫∞ ∫∞
𝑗2𝜋𝑓𝑡 ′ ′ −𝑗2𝜋𝑓′𝑡′ ′
𝑥(𝑡)=𝑋(𝑓)𝑒𝑑𝑓⇒𝑥−𝑓=𝑋𝑡𝑒 𝑑𝑡 −∞ −∞
6. ConvolutionandMultiplication
IftheFTof𝑥(𝑡) and𝑦(𝑡) are𝑋(𝑓) and𝑌(𝑓),respectively,theFTofthefollowingsignalsaregivenby
F[𝑥(𝑡)∗𝑦(𝑡)]=𝑋(𝑓)𝑌(𝑓), F[𝑥(𝑡)𝑦(𝑡)]=𝑋(𝑓)∗𝑌(𝑓) Proof) The FT of the convoltion signal can be derived as
∫ ∞∫ ∞ −∞ −∞
−𝑗2𝜋𝑓𝑡 𝑥(𝜏)𝑦(𝑡−𝜏)𝑑𝜏 𝑒 𝑑𝑡=
∫ ∞∫ ∞ −𝑗2𝜋𝑓𝜏 −𝑗2𝜋𝑓(𝑡−𝜏) 𝑥(𝜏)𝑒 𝑦(𝑡−𝜏)𝑒 𝑑𝑡𝑑𝜏
∫∞ =
∫∞ 𝑥(𝜏)𝑒−𝑗2𝜋𝑓𝜏𝑑𝜏 ×
−∞ −∞
𝑦(𝑙)𝑒−𝑗2𝜋𝑓𝑙𝑑𝑙 =𝑋(𝑓)𝑌(𝑓)
where we used a change of variable (𝑡 − 𝜏 = 𝑙 ) in the second equality. Similarly,
−∞
−∞
{𝑋(𝑓)∗𝑌(𝑓)}𝑒𝑗2𝜋𝑓𝑡𝑑𝑓 =
F−1[𝑋(𝑓)∗𝑌(𝑓)]= =
∫ ∞ −∞
∫ ∞ ∫ ∞ −∞ −∞
𝑋(𝑙)𝑌(𝑓 −𝑙)𝑑𝑙 𝑒𝑗2𝜋𝑓𝑡𝑑𝑓
∫∞ −∞
∫∞ 𝑋(𝑙)𝑒𝑗2𝜋𝑙𝑡𝑑𝑙 ×
𝑌(𝑓−𝑙)𝑒𝑗2𝜋(𝑓−𝑙)𝑡𝑑𝑓 =𝑥(𝑡)𝑦(𝑡) where we used a change of variable (𝑓 − 𝑙 = 𝑓 ′) in the last equality.
51
−∞
7. Differentiation
IftheFTof𝑥(𝑡) is𝑋(𝑓),theFTofthefollowingsignalsaregivenby
F
=(𝑗2𝜋𝑓)𝑛𝑋(𝑓), F (−𝑗2𝜋𝑡)𝑛𝑥(𝑡) = 𝑛𝑛
𝑑𝑛𝑥(𝑡) 𝑑𝑛𝑋(𝑓)
𝑑𝑓
(𝑗2𝜋𝑓)𝑛𝑋(𝑓)𝑒𝑗2𝜋𝑓𝑡𝑑𝑓 =F−1(𝑗2𝜋𝑓)𝑛𝑋(𝑓), (−𝑗2𝜋𝑡)𝑛𝑥(𝑡)𝑒−𝑗2𝜋𝑓𝑡𝑑𝑡=F(−𝑗2𝜋𝑡)𝑛𝑥(𝑡)
8. Integration
IftheFTof𝑥(𝑡) is𝑋(𝑓),theFTofthefollowingsignalisgivenby
F
Proof) The integral can be expressed in terms of convoltion as follows ∫𝑡
𝑥 (𝜏)𝑑𝜏 = 𝑥(𝑡) ∗𝑢 (𝑡) −∞
Then, by using the convoltion property of the FT, we get
Proof) By differentiating (4.3), we get 𝑛∫∞
𝑑 𝑥(𝑡) =
4 Continuous Time Fourier Transform
𝑑𝑡
𝑛 𝑛∫∞
𝑑𝑡
𝑑 𝑋(𝑓)=
−∞ −∞
𝑛
𝑑𝑓
F
∫𝑡
𝑥(𝜏)𝑑𝜏 =F[𝑥(𝑡)∗𝑢(𝑡)]=𝑋(𝑓)×F[𝑢(𝑡)]
−∞
−∞
where the FT pair of the unit step function is derived in the example problem F[𝑢(𝑡)]=1𝛿(𝑓)+ 1
2 𝑗2𝜋𝑓 Hence, we obtained the FT of integrated signals as follows
11 2 𝑗2𝜋𝑓
∫𝑡
𝑥(𝜏)𝑑𝜏 = 𝑋(0)𝛿(𝑓)+ 𝑋(𝑓)
F[𝑥(𝑡)∗𝑢(𝑡)]=𝑋(𝑓)×F[𝑢(𝑡)]=1𝑋(0)𝛿(𝑓)+ 1 𝑋(𝑓) 2 𝑗2𝜋𝑓
9. Parseval’sTheorem
IftheFTof𝑥(𝑡) is𝑋(𝑓),theParseval’stheoremis
∫∞2∫∞2
−∞
|𝑥(𝑡)| 𝑑𝑡 = |𝑋(𝑓)| 𝑑𝑓 −∞
52
4 Continuous Time Fourier Transform
Figure 4.1: Properties of the Fourier Transform
53
4 Continuous Time Fourier Transform
Figure 4.2: Table of the Fourier Transform 54
[Example 4-1. Functions related to 𝛿(𝑡)] Derive the FT of the following signals a) F[𝛿(𝑡−𝑡0)]= b) F[𝛿(𝑡)]=
𝑗2𝜋𝑡𝑓0 c) F[1]= d) F 𝑒 =
Solution) From (4.3), we can directly calculate (a) and (b) as
[Example 4-2. Functions related to Exponential] Derive the FT of the following signals where 𝛼 > 0
∫ ∞ −∞
−𝑗2𝜋𝑓𝑡0 𝑡0=0
−−−→ (b) F [𝛿 (𝑡)] = 1
−𝑗2𝜋𝑓𝑡
We can apply duality on (b) to calculate (c), then use parameter shifting (frequency shifting) on (c) to obtain (d).
(a) F [𝛿 (𝑡 − 𝑡0)] =
(c)F[1]=𝛿(−𝑓)=𝛿(𝑓), (d)F 𝑒 ×1 =𝛿(𝑓−𝑓0)
(4.5) (4.6)
𝛿 (𝑡 − 𝑡0) 𝑒
𝑑𝑡 = 𝑒
a) F 𝑒−𝛼𝑡𝑢 (𝑡) =
c) F[𝑠𝑔𝑛(𝑡)]=F[2𝑢(𝑡)−1]=
−𝛼|𝑡| e)F𝑒 =
Solution) The FT of the one sided decaying exponential function in (a) can be derived as −𝛼𝑡 ∫∞−𝛼𝑡−𝑗2𝜋𝑓𝑡 1
0
𝛼+𝑗2𝜋𝑓 𝛼−𝑗2𝜋𝑓 𝛼 +4(𝜋𝑓) unit step function 𝑢(𝑡) in (d) can be derived from the signum function as 𝑢(𝑡) = 1 + 1𝑠𝑔𝑛(𝑡).
4 Continuous Time Fourier Transform
𝑗2𝜋𝑡𝑓0
b) F 𝑒−𝛼𝑡𝑢 (𝑡) − 𝑒𝛼𝑡𝑢 (−𝑡) =
d) F[𝑢(𝑡)]=
1
f)F2 2= 𝛼 +4(𝜋𝑡)
(a)F 𝑒 𝑢(𝑡) =
The functions in (b) and (e) are referred as odd two-sided exponential and even two-sided exponential signal, respectively.
𝑒 𝑒 𝑑𝑡=
𝛼 + 𝑗2𝜋𝑓
(4.7)
∫∞ ∫0
(b) F 𝑒−𝛼𝑡𝑢 (𝑡) −𝑒𝛼𝑡𝑢 (−𝑡) = 𝑒−𝛼𝑡𝑒−𝑗2𝜋𝑓𝑡𝑑𝑡 − 𝑒𝛼𝑡𝑒−𝑗2𝜋𝑓𝑡𝑑𝑡
0 −∞
1 1 𝑗4𝜋𝑓 =−=−22,
𝛼 + 𝑗2𝜋𝑓 𝛼 − 𝑗2𝜋𝑓 𝛼 +4(𝜋𝑓) ∫∞ ∫0
𝑒−𝛼|𝑡| = 𝑒−𝛼𝑡𝑒−𝑗2𝜋𝑓𝑡𝑑𝑡 + 𝑒𝛼𝑡𝑒−𝑗2𝜋𝑓𝑡𝑑𝑡
(e) F
The signum function 𝑠𝑔𝑛(𝑡) in (c) can be approximated by odd two-sided exponential signal in (b) by 𝛼 → 0 and the
01 1−∞2𝛼 =+=22
(4.8)
22
𝑗4𝜋𝑓 1 (c)F[𝑠𝑔𝑛(𝑡)]= limF 𝑒−𝛼𝑡𝑢(𝑡)−𝑒𝛼𝑡𝑢(−𝑡) = lim − = ,
𝛼→0 𝛼→0 𝛼2 +4(𝜋𝑓)2 𝑗𝜋𝑓 1111 11 (4.9)
(d)F[𝑢(𝑡)]=F + 𝑠𝑔𝑛(𝑡) = F[1]+ F[𝑠𝑔𝑛(𝑡)]= 𝛿(𝑓)+ 22 2 2 2 𝑗2𝜋𝑓
The FT of (f) can be derived by using the duality property with (e).
1 1 Duality 1 1
(f) F 𝑒−𝛼 |𝑡 | = 2 2 −−−−−→ F 2 2 = 𝑒−𝛼 |𝑓 | (4.10)
2𝛼 𝛼 +4(𝜋𝑓) 𝛼 +4(𝜋𝑡) 2𝛼
55
[Example 4-3. Functions related to Rectangular Pulse] Derive the FT of the following signals where 𝜏 > 0
a) Frect𝑡= 𝜏
c) F𝜏tri𝑡= 𝜏
b) F[sinc(2𝐵𝑡)]= d) Fsinc2(2𝐵𝑡)=
f) F[sinc(2𝐵(𝑡−𝑡0))]=
where the rectangular pulse 𝑟𝑒𝑐𝑡(𝑡), Sinc function 𝑠𝑖𝑛𝑐(𝑡), and triangular function 𝑡𝑟𝑖(𝑡) are illustrated below.
4 Continuous Time Fourier Transform
e) Frect𝑡−𝑡0= 𝜏
Solution) The FT of the rect(𝑡) in (a) can be directly derived using (4.3)
∫𝜏 𝑡 2
−𝜏
1 2 2𝑗sin(𝜋𝑓𝜏)
𝑒−𝑗2𝜋𝑓𝑡𝑑𝑡= 22
𝑒−𝑗2𝜋𝑓𝑡 =
𝑗2𝜋𝑓 𝜏 𝑗2𝜋𝑓
=𝜏·sinc(𝑓𝜏).
The y-intercept of the Sinc function is 1 and the zero-crossing occurs at every integer 𝑡 , so 𝜏 sinc ( 𝑓 𝜏 ) has a maximum
value 𝜏 at 𝑓 = 0 with zero-crossing at 𝑓 = 𝑛 for any nonzero integer 𝑛. 𝜋
sinc(𝑡) = sin (𝜋𝑡) ⇒ 0 for 𝑡 = arbitrary nonzero integer, lim [sinc(𝑡)] = 1 𝜋𝑡 𝑡→0
The FT of (b) can be derived by using the duality property with (a) as follows
(a)F rect =
𝜏 −𝜏
1 𝑡 F (b) rect 𝜏 𝜏 F−1
F 1 𝑓
Duality
======⇒
𝑡 𝜏 − |𝑡 | for |𝑡 | ≤ 𝜏 =
sinc (𝑓 𝜏) , wherewesubtituted𝜏=2𝐵andusedachangeofvariable(𝑡→−𝑓′,𝑓 →𝑡′).Hence,F[sinc(2𝐵𝑡)]= 1 rect𝑓 .
sinc (2𝐵𝑡)
In Example 2-1. (b) (pp26), we proved that the convoltion of two rectangular pulse signal is a triangluar function
rect F−1 2𝐵
2𝐵
2𝐵 2𝐵
𝑡 𝑡 rect ∗rect
=𝜏 ·tri
𝜏 𝜏 𝜏 0 for|𝑡|>𝜏
Then, the FT of (c) can be obtained by convoltion properties and (d) can be derived based on the duality with (c).
22 (c)F𝜏tri𝑡 =Frect𝑡 ∗rect𝑡 =Frect𝑡 ×Frect𝑡 =𝜏sinc(𝑓𝜏),
𝜏𝜏𝜏𝜏𝜏
1𝑡F 2 Duality 2 F1𝑓
2𝐵
(d) tri sinc (𝑓𝜏), ======⇒ sinc (2𝐵𝑡) tri
𝜏 𝜏 F−1
F−1 2𝐵
56
wherewesubtituted𝜏=2𝐵andusedachangeofvariable(𝑡→−𝑓′,𝑓→𝑡′).Hence,Fsinc2(2𝐵𝑡)= 1 tri𝑓 . 2𝐵 2𝐵
The FT of (e) and (f) can be obtained by using the time-shift property of (a) and frequency-shift of (b), respectively. (e)Frect𝑡−𝑡0=𝜏𝑒−𝑗2𝜋𝑓𝑡0sinc(𝑓𝜏), (f)F[sinc(2𝐵(𝑡−𝑡0))]= 1𝑒−𝑗2𝜋𝑓𝑡0 rect𝑡
𝜏 2𝐵2𝐵
4 Continuous Time Fourier Transform
[Example 4-4. Functions related to Modulation] Derive the FT of the following signals where 𝛼 > 0
a) F [cos(2𝜋𝑓0𝑡)] =
c) F [𝑥(𝑡) cos (2𝜋𝑓0𝑡)] =
e) F [cos(2𝜋𝑓0𝑡)𝑢(𝑡)] =
g) F 𝑒−𝛼𝑡 cos (2𝜋𝑓0𝑡)𝑢(𝑡) =
b) F [sin(2𝜋𝑓0𝑡)] =
d) F [𝑥(𝑡) sin (2𝜋𝑓0𝑡)] =
f) F [sin(2𝜋𝑓0𝑡)𝑢(𝑡)] =
h) F 𝑒−𝛼𝑡 sin (2𝜋𝑓0𝑡)𝑢(𝑡) =
Solution) The FT of (a) and (b) can be derived by using Example 4-1 (d) and linear property as follows
𝑗2𝜋𝑓0𝑡 −𝑗2𝜋𝑓0𝑡 (a)F[cos(2𝜋𝑓0𝑡)]=F 2 𝑒 +𝑒
=2[𝛿(𝑓−𝑓0)+𝛿(𝑓+𝑓0)],
11
1𝑗
𝑗2𝜋𝑓0𝑡 −𝑗2𝜋𝑓0𝑡 (b)F[sin(2𝜋𝑓0𝑡)]=F 𝑒 −𝑒
The FT of (c) and (d) can be obtained based on the multiplication property of the FT with (a) and (b).
=− [𝛿(𝑓−𝑓0)−𝛿(𝑓+𝑓0)] 2𝑗 2
(c) F [𝑥 (𝑡) cos (2𝜋 𝑓0𝑡)] = 𝑋 (𝑓 ) ∗ 1 [𝛿 (𝑓 − 𝑓0) + 𝛿 (𝑓 + 𝑓0)] = 1 [𝑋 (𝑓 − 𝑓0) + 𝑋 (𝑓 + 𝑓0)] , 22
(d)F[𝑥(𝑡)sin(2𝜋𝑓0𝑡)]=𝑋(𝑓)∗ 1 [𝛿(𝑓−𝑓0)−𝛿(𝑓+𝑓0)]=−𝑗[𝑋(𝑓−𝑓0)−𝑋(𝑓+𝑓0)] 2𝑗 2
The FT of (e) and (f) can be obtained by using Example 4-2, i.e., F [𝑢(𝑡)] = 1𝛿(𝑓 ) + 1 2 𝑗2𝜋𝑓
11111 (e)F[cos(2𝜋𝑓0𝑡)𝑢(𝑡)]= 𝛿(𝑓−𝑓0)+ + 𝛿(𝑓+𝑓0)+
2 2 𝑗2𝜋(𝑓−𝑓0) 2 𝑗2𝜋(𝑓+𝑓0) 10 0 𝑗2𝜋𝑓
= 4 [𝛿 (𝑓 − 𝑓 ) +𝛿 (𝑓 + 𝑓 )] + (2𝜋𝑓0)2 − (2𝜋𝑓)2,
11111
(f)F[sin(2𝜋𝑓0𝑡)𝑢(𝑡)]=
= 1 [𝛿(𝑓−𝑓0)−𝛿(𝑓+𝑓0)]+ 2𝜋𝑓0
𝛿(𝑓−𝑓0)+ − 𝛿(𝑓+𝑓0)−
𝑗2𝜋(𝑓 −𝑓0) 2 𝑗2𝜋(𝑓 +𝑓0)
2𝑗 2
4𝑗 (2𝜋𝑓0)2 −(2𝜋𝑓)2
The FT of (g) and (h) can be obtained by the multiplication property with F 𝑒−𝛼𝑡𝑢 (𝑡) = (𝛼 + 𝑗2𝜋𝑓 )−1 (g)F 𝑒−𝛼𝑡 cos(2𝜋𝑓0𝑡)𝑢(𝑡) = 1 ∗ 1 [𝛿(𝑓 −𝑓0)+𝛿(𝑓 +𝑓0)]
𝛼 + 𝑗2𝜋𝑓 2
11 1𝛼+𝑗2𝜋𝑓 =+=22,
2 𝛼+𝑗2𝜋(𝑓 −𝑓0) 𝛼+𝑗2𝜋(𝑓 +𝑓0) (𝛼+𝑗2𝜋𝑓) +4(𝜋𝑓0) (h)F𝑒−𝛼𝑡sin(2𝜋𝑓0𝑡)𝑢(𝑡)= 1 ∗ 1 [𝛿(𝑓−𝑓0)−𝛿(𝑓+𝑓0)]
𝛼 + 𝑗2𝜋𝑓 2𝑗
1 1 1 2𝜋𝑓0 =−=22
2𝑗 𝛼+𝑗2𝜋(𝑓 −𝑓0) 𝛼+𝑗2𝜋(𝑓 +𝑓0) (𝛼+𝑗2𝜋𝑓) +4(𝜋𝑓0)
57
[Example 4-5. etc] Derive the FT of the following signals where 𝛼 > 0 1. FTofaperiodicsignal𝑥(𝑡)withperiod𝑇0
∞ 2. FToftheperiodicimpulsetrainF 𝛿 (𝑡)= 𝛿(𝑡−𝑘𝑇 )
3. F 𝑡𝑛𝑒−𝛼𝑡𝑢 (𝑡)
Solution) In (a), we first express a periodic signal 𝑥 (𝑡 ) using the FS, then apply F
𝑗2𝜋𝑡𝑓0
𝑒 = 𝛿 ( 𝑓 − 𝑓0 ) as follows
𝑇0 0 𝑘 =−∞
∞∞∞
𝑘 𝑘 𝑘2𝜋
𝑘=−∞ 𝑘=−∞ 𝑘=−∞
which indicates that the Fourier transform of a periodic signal consists of a sequence of equidistant impulses located
at the harmonic frequencies. For (b), the FS of the periodic impulse train 𝛿𝑇0 (𝑡) is derived in Example 3-3, pp42 1 ∞ ∞ 1 ∞ 1 ∞ 𝑘𝑤
(a)F[𝑥(𝑡)]=F 𝑐𝑒𝑗𝑘𝑤0𝑡 =𝑐F𝑒𝑗𝑘𝑤0𝑡 =𝑐𝛿𝑓−𝑘𝑤0 ,
4 Continuous Time Fourier Transform
𝛿(𝑡)= 𝑒𝑗𝑘𝑤0𝑡 ⇒(b)F𝛿(𝑡−𝑘𝑇)= F𝑒𝑗𝑘𝑤0𝑡= 𝛿𝑓− 0
𝑇0𝑇0
The FT of (c) can be derived by using F 𝑒−𝛼𝑡𝑢 (𝑡) =
0𝑇0 𝑇0 2𝜋 𝑘=−∞ 𝑘=−∞
1 and the higher order differentiation of the FT 𝛼+𝑗2𝜋𝑓
1 𝑛!
𝑘=−∞ 𝑘=−∞
1𝑑𝑛𝑋(𝑓)
where𝑥(𝑡)=𝑒−𝛼𝑡𝑢(𝑡),𝑋(𝑓)= ⇒ (c)F 𝑡𝑛𝑒−𝛼𝑡𝑢(𝑡) = 𝛼+𝑗2𝜋𝑓
F 𝑡𝑛𝑥(𝑡) =
(−𝑗2𝜋)𝑛 𝑑𝑓𝑛
(𝛼 + 𝑗2𝜋𝑓)𝑛+1
4.2 Frequency Response
GiventheFTofaninput𝑥(𝑡) andtheimpulseresponseh(𝑡) as𝑋(𝑓) and𝐻(𝑓),theoutput𝑦(𝑡) ofanLTIsystemis
𝑦(𝑡) =h(𝑡) ∗𝑥(𝑡) = F−1 (𝑌(𝑓)) = F−1 (𝑋(𝑓)𝐻(𝑓)) ∫∞ ∫∞
(4.11) where we refer the FT of the impulse response 𝐻 ( 𝑓 ) as the Frequency Response, the magnitude of the frequency
= 𝑌(𝑓)𝑒𝑗2𝜋𝑓𝑡𝑑𝑓 = 𝑋(𝑓)𝐻(𝑓)𝑒𝑗2𝜋𝑓𝑡𝑑𝑓 forAperiodicinput −∞ −∞
response |𝐻 (𝑓 )| as the Magnitude Response, the phase of the frequency response 𝜃𝐻 (𝑓 ) as the Phase response. • FrequencyResponse:
𝑑𝑡 = |𝐻(𝑓)|𝑒
• Phaseresponse:𝜃𝐻(𝑓)
The output in (4.11) holds for a non-periodic input signal 𝑥(𝑡). Given a periodic input 𝑥(𝑡),
• If𝑥(𝑡)=𝑒𝑗𝑘𝑤0𝑡,thenF(𝑥(𝑡))=𝛿(𝑓−𝑘𝑓0),where𝑓0=𝑤0,and 2𝜋
𝑌(𝑓)=𝐻(𝑓)𝑋(𝑓)=𝐻(𝑓)𝛿(𝑓 −𝑘𝑓0)=𝐻(𝑘𝑓0)𝛿(𝑓 −𝑘𝑓0)
58
𝐻(𝑓) =
• MagnitudeResponse:|𝐻(𝑓)|
∫∞ −𝑗2𝜋𝑓𝑡 h(𝑡)𝑒
𝑗𝜃𝐻(𝑓)
−∞
• If𝑥(𝑡)=∞ 𝑐𝑘𝑒𝑗𝑘𝑤0𝑡,thenF(𝑥(𝑡))=∞ 𝑐𝑘𝛿(𝑓−𝑘𝑓0)and 𝑘 =−∞ 𝑘 =−∞
LTI Systems Characterized by Differential Equations
If the LTI system is described by differential equations, we can derive the Frequency response using the differentiation property of the Fourier transform as follows
∞∞
𝑌(𝑓)=𝐻(𝑓)𝑋(𝑓)= 𝑐 𝐻(𝑓)𝛿(𝑓 −𝑘𝑓 )= 𝑐 𝑐 𝐻(𝑘𝑓 )𝛿(𝑓 −𝑘𝑓 )
forperiodicinput
𝑘0𝑘𝑘00 𝑘 =−∞ 𝑘 =−∞
𝑁𝑀𝑁𝑀 𝑑𝑘𝑦(𝑡) 𝑑𝑘𝑥(𝑡) F 𝑑𝑘𝑦(𝑡) 𝑑𝑘𝑥(𝑡)
𝑎𝑘 =𝑏𝑘 =⇒𝑎𝑘F =𝑏𝑘F ,𝑀≤𝑁 𝑘𝑘𝑘𝑘
4 Continuous Time Fourier Transform
𝑘=0 𝑑𝑡 𝑘=0 𝑑𝑡 𝑘=0 𝑑𝑡 𝑘=0 𝑑𝑡
𝑀 𝑏 (𝑗2𝜋𝑓)𝑘
⇒ 𝐻(𝑓)=𝑌(𝑓)= 𝑘=0 𝑘 ⇔ h(𝑡)=F−1(𝐻(𝑓))
[Example 4-6.] Consider CT-LTI systems described by the following differential equations. Derive the frequency response 𝐻 (𝑓 ) and the corresponding impulse response h(𝑡). Furthermore, derive the system output when the input signal is 𝑥1(𝑡) = 𝑒−𝑡𝑢(𝑡), 𝑥2(𝑡) = 𝑡𝑒−2𝑡𝑢(𝑡), and 𝑥3(𝑡) = 𝑢(𝑡), respectively.
𝑋(𝑓)
𝑁 𝑎 (𝑗2𝜋𝑓)𝑘 𝑘=0 𝑘
a) 𝑑𝑦(𝑡)+2𝑦(𝑡)=𝑥(𝑡)+𝑑𝑥(𝑡) ⇒ 𝑑𝑡 𝑑𝑡
b) 𝑑𝑦(𝑡)+2𝑦(𝑡)=𝑥(𝑡) ⇒ 𝑑𝑡
𝐻(𝑓),h(𝑡),𝑦𝑖(𝑡)=T{𝑥𝑖(𝑡)},where𝑖={1,2,3}
𝐻(𝑓),h(𝑡),𝑦𝑖(𝑡)=T{𝑥𝑖(𝑡)},where𝑖={1,2,3}
c) 𝑑2𝑦(𝑡)+4𝑑𝑦(𝑡)+3𝑦(𝑡)=𝑑𝑥(𝑡)+2𝑥(𝑡) ⇒ 𝑑𝑡 𝑑𝑡 𝑑𝑡
d) 𝑑2𝑦(𝑡)+6𝑑𝑦(𝑡)+8𝑦(𝑡)=2𝑥(𝑡) ⇒ 𝐻(𝑓),h(𝑡),𝑦(𝑡)=T{𝑥(𝑡)},where𝑖={1,2,3} 2𝑖𝑖
𝑑𝑡 𝑑𝑡
𝐻(𝑓),h(𝑡),𝑦(𝑡)=T{𝑥(𝑡)},where𝑖={1,2,3} 2𝑖𝑖
Solution)SinceF 𝑑𝑛𝑥(𝑡)=(𝑗2𝜋𝑓)𝑛𝑋(𝑓),theFTofeachdifferentialequationscanbeexpressedas
(a)F
⇒ 𝐻(𝑓)=
+2𝑦(𝑡)=𝑥(𝑡)+
𝑛
𝑑𝑡
𝑑𝑦(𝑡) 𝑑𝑥(𝑡)
⇔ (𝑗2𝜋𝑓)𝑌(𝑓)+2𝑌(𝑓)=𝑋(𝑓)+(𝑗2𝜋𝑓)𝑋(𝑓), 1 F−1 −2𝑡
𝑑𝑡
𝑑𝑡
𝑌(𝑓) 𝑋 ( 𝑓 )
1+ 𝑗2𝜋𝑓 2 + 𝑗 2𝜋 𝑓
=1−
where we used the FT pairs in Table 4.2. The FT of each input signals are given by
=
===⇒ h(𝑡)=𝛿(𝑡)−𝑒 𝑢(𝑡),
1 + 𝑗 2𝜋 𝑓
𝑋1(𝑓) = F 𝑒−𝑡𝑢(𝑡) = 1 , 𝑋2(𝑓) = F 𝑡𝑒−2𝑡𝑢(𝑡) = 1 2,
1+ 𝑗2𝜋𝑓 (2+ 𝑗2𝜋𝑓) 𝑋3(𝑓)=F{𝑢(𝑡)}=1𝛿(𝑓)+ 1 .
2 𝑗2𝜋𝑓
Since the FT of the system output is given by 𝑌𝑖 (𝑓 ) = F (𝑥𝑖 (𝑡) ∗ h(𝑡)) = 𝑋𝑖 (𝑓 )𝐻 (𝑓 ), the output for each signals are
𝑌1(𝑓) = 1 ⇒ 2 + 𝑗 2𝜋 𝑓
𝑌2(𝑓 ) = 1 + 𝑗2𝜋𝑓 = (2+ 𝑗2𝜋𝑓)3
where we applied the Partial Fraction Expansion. Refer [Oppenheim, Appendix], [https://bit.ly/3aNDl2K], [https: //bit.ly/3cSXtCi].
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𝑦1(𝑡) =𝑒−2𝑡𝑢(𝑡),
1 − 1 ⇒ 𝑦2(𝑡) = 𝑡𝑒−2𝑡𝑢(𝑡) − 1𝑡2𝑒−2𝑡𝑢(𝑡),
(2+ 𝑗2𝜋𝑓)2 (2+ 𝑗2𝜋𝑓)3 2
11+𝑗2𝜋𝑓11111 11−2𝑡 𝑌3(𝑓)= 𝛿(𝑓)+ = 𝛿(𝑓)+ + ⇒ 𝑦3(𝑡)= 𝑢(𝑡)+ 𝑒 𝑢(𝑡),
4 𝑗2𝜋𝑓(2+𝑗2𝜋𝑓) 2 2 𝑗2𝜋𝑓 2 2+𝑗2𝜋𝑓 2 2
For system (b), we can derive the Frequency response using similar approach
⇔ ( 𝑗 2𝜋 𝑓 ) 𝑌 ( 𝑓 ) + 2𝑌 ( 𝑓 ) = 𝑋 ( 𝑓 ),
(b) F
⇒ 𝐻(𝑓)=
+ 2𝑦 (𝑡 ) = 𝑥 (𝑡 )
𝑑𝑦(𝑡)
𝑑𝑡
𝑌(𝑓) 1 F−1 −2𝑡
= ===⇒ h(𝑡)=𝑒 𝑢(𝑡), 2 + 𝑗 2𝜋 𝑓
4 Continuous Time Fourier Transform
𝑋 ( 𝑓 )
where we used the FT pairs in Table 4.2. The system output for each input signals are derived as
1 = 1 − 1 ⇒ 𝑦1(𝑡) = 𝑒−𝑡𝑢(𝑡) − 𝑒−2𝑡𝑢(𝑡), (1 + 𝑗2𝜋𝑓 ) (2 + 𝑗2𝜋𝑓 ) 1 + 𝑗2𝜋𝑓 2 + 𝑗2𝜋𝑓
1 3 ⇒ 𝑦2(𝑡) = 1𝑡2𝑒−2𝑡𝑢(𝑡), (2+ 𝑗2𝜋𝑓) 2
1 111111 11−2𝑡
𝑑2𝑦(𝑡) 𝑑𝑦(𝑡) 𝑑𝑥(𝑡) 2
(c)F 2 +4 +3𝑦(𝑡)= +2𝑥(𝑡) ⇔ (𝑗2𝜋𝑓) +4(𝑗2𝜋𝑓)+3 𝑌(𝑓)={𝑗2𝜋𝑓 +2}𝑋(𝑓),
𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑌(𝑓) 2+𝑗2𝜋𝑓 1/2 1/2
𝑌1(𝑓 ) = 𝑌2(𝑓) =
𝑌3(𝑓 ) =
For system (c), the Frequency response and impulse response are derived as follows
4
𝛿(𝑓 ) + = 𝛿(𝑓 ) + − ⇒ 𝑦3(𝑡) = 𝑢(𝑡) − 𝑒 𝑢(𝑡). 𝑗2𝜋𝑓(2+𝑗2𝜋𝑓) 2 2 𝑗2𝜋𝑓 2 2+𝑗2𝜋𝑓 2 2
⇒ 𝐻(𝑓)= = 2 = +
𝑋(𝑓) (𝑗2𝜋𝑓) +4(𝑗2𝜋𝑓)+3 3+𝑗2𝜋𝑓 1+𝑗2𝜋𝑓
F−1 1 −3𝑡 1 −𝑡 ===⇒h(𝑡)= 𝑒 𝑢(𝑡)+ 𝑒 𝑢(𝑡),
2 2 where we used the FT pairs in Table 4.2. The system output for each input signals are derived as
F−1 ===⇒
𝑌1(𝑓)= 𝑦1(𝑡) =
2 + 𝑗 2𝜋 𝑓 = 1/4 + 1/2 + −1/4 (1+ 𝑗2𝜋𝑓)2 (3+ 𝑗2𝜋𝑓) 1+ 𝑗2𝜋𝑓 (1+ 𝑗2𝜋𝑓)2 3+ 𝑗2𝜋𝑓
1−𝑡 1−𝑡 1−3𝑡
4𝑒 𝑢(𝑡) + 2𝑡𝑒 𝑢(𝑡) − 4𝑒 𝑢(𝑡),
1
(1 + 𝑗2𝜋𝑓 ) (2 + 𝑗2𝜋𝑓 ) (3 + 𝑗2𝜋𝑓 )
= 1/2 + −1 + 1/2
1 + 𝑗2𝜋𝑓 2 + 𝑗2𝜋𝑓 3 + 𝑗2𝜋𝑓
= 1𝛿(𝑓)+ 2/3 + −1/2 + −1/6
𝑌2(𝑓)=
===⇒ 𝑦2(𝑡)=2𝑒 𝑢(𝑡)−𝑒 𝑢(𝑡)+2𝑒 𝑢(𝑡),
1 −𝑡 −2𝑡 1 −3𝑡 𝑌3(𝑓)= 1𝛿(𝑓)+ 2+𝑗2𝜋𝑓
21 1−1/2−1/6 = 𝛿(𝑓)+ + +
𝑦3(𝑡) = 3𝑢(𝑡) − 2𝑒 𝑢(𝑡) − 6𝑒 𝑢(𝑡).
For system (d), we can derive the Frequency response using similar approach
F−1
3 𝑗2𝜋𝑓 (1+𝑗2𝜋𝑓)(3+𝑗2𝜋𝑓)
3 𝑗2𝜋𝑓
1+𝑗2𝜋𝑓 3+𝑗2𝜋𝑓
3 2 𝑗 2𝜋 𝑓 1 + 𝑗 2𝜋 𝑓 3 + 𝑗 2𝜋 𝑓 2 1−𝑡 1−3𝑡
F−1 ===⇒
(d) F
⇒ 𝐻(𝑓)=𝑌(𝑓)=
𝑑2𝑦(𝑡) 𝑑𝑦(𝑡)
2 ⇔ ( 𝑗 2𝜋 𝑓 )
+ 6 ( 𝑗 2𝜋 𝑓 ) + 8
= 1 + −1
𝑑𝑡
𝑑𝑡
2 + 6
𝑋(𝑓) (𝑗2𝜋𝑓) +6(𝑗2𝜋𝑓)+8 (4+𝑗2𝜋𝑓)(2+𝑗2𝜋𝑓) 2+𝑗2𝜋𝑓 4+𝑗2𝜋𝑓
+ 8𝑦 (𝑡 ) = 2𝑥 (𝑡 )
2 2 = 2
𝑌 ( 𝑓 ) = 2𝑋 ( 𝑓 ),
F−1
===⇒h(𝑡)=𝑒 𝑢(𝑡)−𝑒 𝑢(𝑡)
−2𝑡 −4𝑡
60
where we used the FT pairs in Table 4.2. The system output for each input signals are derived as 𝑌1(𝑓)= 2 = 2/3 + −1 + 1/3
(1 + 𝑗2𝜋𝑓 ) (2 + 𝑗2𝜋𝑓 ) (4 + 𝑗2𝜋𝑓 ) 1 + 𝑗2𝜋𝑓 2 + 𝑗2𝜋𝑓 4 + 𝑗2𝜋𝑓 2 −𝑡 −2𝑡 1 −4𝑡
4 Continuous Time Fourier Transform
F−1
===⇒ 𝑦1(𝑡)=3𝑒 𝑢(𝑡)−𝑒 𝑢(𝑡)+3𝑒 𝑢(𝑡),
𝑌2(𝑓)= 2
(2+ 𝑗2𝜋𝑓)3 (4+ 𝑗2𝜋𝑓)
= 1/4 + −1/2 + 2+ 𝑗2𝜋𝑓 (2+ 𝑗2𝜋𝑓)2
1
(2+ 𝑗2𝜋𝑓)3
+ −1/4 4+ 𝑗2𝜋𝑓
1−2𝑡 1 −2𝑡
1 2 111−1/21/4
F−1
===⇒ 𝑦2(𝑡)=4𝑒 𝑢(𝑡)−2𝑡𝑒 𝑢(𝑡)+2𝑡𝑒 𝑢(𝑡)−4𝑒 𝑢(𝑡),
12−2𝑡 1−4𝑡
𝑌3(𝑓)= 𝛿(𝑓)+ 8
=
4 2
𝛿(𝑓)+ + + 𝑗2𝜋𝑓 2+𝑗2𝜋𝑓
4+𝑗2𝜋𝑓
𝑗2𝜋𝑓 (2+𝑗2𝜋𝑓)(4+𝑗2𝜋𝑓) 1 1−2𝑡 1−4𝑡
F−1 ===⇒
4.3 Nyquist Sampling
𝑦3(𝑡) = 4𝑢(𝑡) − 2𝑒 𝑢(𝑡) + 4𝑒 𝑢(𝑡).
GivenaCTsignal𝑥(𝑡),ifwemeasureitsvaluesevery𝑇 timeunits,thenwegetthefollowingsamples ··· ,𝑥 (−2𝑇),𝑥 (−𝑇),𝑥 (0),𝑥 (𝑇),𝑥 (2𝑇),···
This process is called Sampling.
Sampling period: 𝑇 , Sampling frequency: 𝑓𝑠 = 1
𝑇
In general, some information is lost in this process if the signal is not bandlimited (the bandwidth is not finite).
Sampling Theorem
Let 𝑥 (𝑡 ) be a band-limited signal with 𝑋 ( 𝑓 ) = 0 for | 𝑓 | > 𝑓𝑀 . Then 𝑥 (𝑡 ) is uniquely determined by its samples {𝑥 (−𝑘𝑇 ) , arbitrary integer𝑘 } if the following condition is satisfied
𝑓𝑠 ≥ 2𝑓𝑀, (4.12) where 𝑓𝑀 , 𝑓𝑠 , 2𝑓𝑀 are the signal bandwidth, sampling frequency, Nyquist rate (minimum 𝑓𝑠 ), respectively.
Intuitive Explanation: Sampling theorem in (4.12) indicates that the sampling rate should be higher than the frequency content of the signal. Otherwise, the recovered signal after sampling will suffer significant distortion.
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Formal Explanation: Analytically, Sampling process is multiplication of the bandlimited signal 𝑥 (𝑡 ) and the periodic impulse train 𝛿𝑇 (𝑡) = ∞ 𝛿 (𝑡 − 𝑘𝑇 ), which we will denote as 𝑥𝑇 (𝑡)
𝑘 =−∞
𝑥𝑇(𝑡)=𝑥(𝑡)𝛿𝑇(𝑡)= 𝑥(𝑘𝑇)𝛿(𝑡−𝑘𝑇).
By using the multiplication property, the FT of the sampled signal 𝑥𝑇 (𝑡 ) is obtained as follows 1∞∞1
𝑋 (𝑓)=F(𝑥(𝑡))∗F(𝛿 (𝑡))=𝑋(𝑓)∗ 𝛿(𝑓−𝑘𝑓)=𝑓 𝑋(𝑓−𝑘𝑓), 𝑓 𝑇𝑇𝑠𝑠𝑠𝑠
∞
𝑘 =−∞
4 Continuous Time Fourier Transform
𝑇𝑇
𝑘 =−∞ 𝑘 =−∞
The shifted replicas do not overlap if and only if 𝑓𝑠 ≥ 2𝑓𝑀. If (4.12) is satisfied, then 𝑥(𝑡) can be exactly recovered from 𝑥𝑇 (𝑡 ) by using an ideal lowpass filter with amplitude 1 and cutoff frequency 𝑓𝑐 where 𝑓𝑀 ≤ 𝑓𝑐 ≤ 𝑓𝑠 − 𝑓𝑀 . In
𝑓𝑠
other words, ideal lowpass filter (or equivalently, sinc interpolation) can exactly reproduce the original signal.
[Example 4-7] Find the Nyquist sampling rate for the following signals
a) 𝑥(𝑡)=2cos(200𝜋𝑡)+sin(500𝜋𝑡) b) 𝑥(𝑡)=2cos2(200𝜋𝑡)
Solution) In (a), the signal bandwidth for cos (200𝜋𝑡) and sin (500𝜋𝑡) are respectively given by 𝑓1 = 200𝜋 =100, 𝑓2 = 500𝜋 =250⇒𝑓𝑀 =250
2𝜋 2𝜋
and the bandwidth of the 𝑥(𝑡) is 𝑓𝑀 = 250. Hence the Nyquist sampling rate for (a) is 2𝑓𝑀 = 500. In (b), 2 cos2 (200𝜋𝑡) = 1 + cos (400𝜋𝑡) whose signal bandwidth is 𝑓 = 400𝜋 = 200. Hence the Nyquist sampling ratefor(b)is2𝑓𝑀 =400. 𝑀 2𝜋
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4.4 Filtering
Distortionless Transmission: In order to have a distortionless transmission, the output of an LTI system needs to have an identical shape to the input signal, although its amplitude may be different and it may be delayed in time.
Inputsignal𝑥(𝑡) ⇒ OutputSignal𝑦(𝑡)=𝐾𝑥(𝑡−𝑡𝑑),
where 𝑡𝑑 is the time delay, 𝐾 is a gain constant. By using the FT, we can obtain the Frequency response as follows
Hence,thesystemswithdistortionlesstransmissionshouldhaveFrequencyresponse𝐻(𝑓) =𝐾𝑒−𝑗2𝜋𝑓𝑡𝑑,constant magnituderesponse|𝐻(𝑓)|=𝐾,andlinearphaseresponse𝜃𝐻(𝑓)=−𝑗2𝜋𝑓𝑡𝑑 overtheentirefrequencyrange.
Filtering is a process that changes the amplitude (or phase) of some frequency components of an input signal.
1. Frequency-ShapingFilter
• amplifysomefrequencycomponentswhilesuppresssomeotherfrequencycomponents • e.g.differntiator,equalizerinaHi-Fisystem
2. Frequency-SelectiveFilter
• selectsomebandsoffrequenciesandrejectothers.
• e.g.low-passfilter,high-passfilter,band-passfilter,band-stopfilter
Ideal Frequency Selective Filter: An ideal frequency selective filter is one that passes signals at one set of frequencies (referred as the pass band) and completely rejects the rest (referred as the stop band).
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4 Continuous Time Fourier Transform
a) Ideal Low-Pass Filter b) Ideal High-Pass Filter
1, |𝑓|≤𝑓 0, |𝑓|≤𝑓 𝑐𝑐
𝐻(𝑓) =
where 𝑤𝑐 is the cutoff frequency.
𝐻(𝑓) =
d) Ideal Bandstop Filter
0, |𝑓|>𝑓𝑐 c) Ideal Bandpass Filter
1, |𝑓|>𝑓𝑐
0, 𝑓1<|𝑓|<𝑓2
1, 𝑓1<|𝑓|<𝑓2 𝐻(𝑓) = 0, otherwise
Bandwidth: There are many different definitions of filter bandwidth. 1. AbsoluteBandwidth𝑓𝐵
𝐻(𝑓) =
1,
otherwise
• absolute BW (Bandwidth) of an ideal low-pass filter: 𝑓𝐵 = 𝑓𝑐
• absolute BW of an ideal bandpass filter: 𝑓𝐵 = 𝑓2 − 𝑓1
• absoluteBSofanidealhigh-passorbandstopfilterarenotdefined.
2. 3-dBBandwidth(Half-PowerBandwidth)𝑓3dB
• 𝑓3 dB is the frequency where the peak magnitude spectrum |𝐻 (0)| drops to |𝐻 (0)|/ 2. Since the
√
power is proportional to the square of |𝐻 (𝑓 )|, i.e., 𝑃 ∝ |𝐻 (𝑓 )|, then the 3-dB bandwidth represent the
frequency where the peak power reduces to the half.
4 Continuous Time Fourier Transform
|𝐻(0)| 𝑃(𝑓3dB) |𝐻(𝑓3dB)|2 |𝐻(𝑓3dB)|= √2
10 log10 𝑃 (𝑓max) 𝑑𝐵 = 10 log10 |𝐻 (0)|2 𝑑𝐵 ===============⇒ 10 log10 (1/2) 𝑑𝐵 = −10log10 (2) 𝑑𝐵 ≃ −3𝑑𝐵
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3. EquivalentBandwidth𝑓eq.Refer[Schaum’stext,Problem5.55]
• 𝑓eqisdefinedasthebandwidthofanidealfilterwherethepoweroftheidealfilterisequaltothatof
the real filter given the same input signals as follows
0 𝐻max
where 𝐻max denotes the maximum value of the magnitude spectrum.
4. 𝑥 percent Energy containment Bandwidth 𝑓𝑥%. (e.g., 90 percent energy containment BW 𝑓90%)
∫ ∞𝐻(𝑓)2 𝑓eq = 𝑑𝑓
[Example 4-8] Consider the following filters with impulse response h(𝑡). For each filter, derive the 3-dB bandwidth 𝑓3 dB, equivalent bandwidth 𝑓eq, and 90 percent energy containment bandwidth 𝑓90%, respectively.
a) h(𝑡)=𝑤0𝑒−𝑤0𝑡𝑢(𝑡)where𝑤0=2𝜋𝑓0 b) h(𝑡)= 2 1 2 where𝛼>0 𝛼 +(2𝜋𝑡)
Solution) For (a), the frequency response 𝐻 ( 𝑓 ) is obtained as
(a) h(𝑡) = 𝑤0 𝑒−𝑤0𝑡𝑢 (𝑡) ⇔ 𝐻 (𝑓 ) = 𝑤0 = 1 where 𝑓0 = 𝑤0
4 Continuous Time Fourier Transform
∫𝑓𝑥% 2
|𝑋(𝑓)| 𝑑𝑓 =(𝑥%)𝐸𝑥
−𝑓𝑥%
Refer [Schaum’s text, Problem 5.57]
where
𝐸𝑥 =
∫∞ 2 |𝑋(𝑓)| 𝑑𝑓.
−∞
𝑤0 + 𝑗2𝜋𝑓 1+ 𝑗 (𝑓/𝑓0) 2𝜋
Since the maximum magnitude spectrum is |𝐻 (0)| = 1, the 3-dB bandwidth 𝑓3 dB, the equivalent bandwidth 𝑓eq, and
the 90 percent energy containment bandwidth 𝑓90% occurs when the following conditions are satisfied.
|𝐻(0)|
𝑓3dB; |𝐻(𝑓3dB)|= √ ,
2 ∞2
∫
|𝐻(𝑓)|
1
1 + (𝑓 /𝑓0)
𝑓eq; 𝑓eq=
0 |𝐻(0)|
2𝑑𝑓,
, where|𝐻(𝑓)|=
2
(4.13)
∫ ∫ 𝑓90% ∞
|𝐻(𝑓)|2𝑑𝑓 65
𝑓
90%
;
|𝐻(𝑓)|2𝑑𝑓 =0.9
−𝑓90%
−∞
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Since |𝐻 (𝑓3 dB) | = 1+(𝑓3 dB/𝑓0)2 = √2 when 𝑓3 dB = 𝑓0, the 3-dB bandwidth is equal to 𝑓0. By using (4.14), we obtain
4 Continuous Time Fourier Transform
eq∫∞ 1 0∫∞1 𝑓0𝜋 𝑓 = 0 1+(𝑓/𝑓0)2𝑑𝑓=𝑓 0 1+𝑥2𝑑𝑥= 2,
∫ 𝑓90% 2 LHSof𝑓90%in(1.13) |𝐻(𝑓)|𝑑𝑓=2
∫ 𝑓90%
1
1 + (𝑓 /𝑓0)2
−1 𝑓90%/𝑓0 𝑑𝑓=2𝑓0tan (𝑥)
−1𝑓90% =2𝑓0tan ,
0
2𝑑𝑓=0.9𝑓0𝜋 ⇒ 𝑓90%=𝑓0tan(0.45𝜋)
0 ∫∞1∫∞1
𝑓0
−𝑓90%
RHSof𝑓90%in(1.13)0.9
Hence, the three BWs 𝑓3 dB, 𝑓eq, and 𝑓90% are given by
(b)h(𝑡)= 1 ⇔ 𝐻(𝑓)= 1𝑒−𝛼|𝑓| 𝛼2 + (2𝜋𝑡)2 2𝛼
and the maximum magnitude spectrum is |𝐻 (0)| = 1 . Then, based on (4.13), we obtain 2𝛼
|𝐻(0)|1 1 1√ |𝐻(𝑓3dB)|=√ ⇒ 𝑒−𝛼|𝑓3dB|=√ ⇒𝑓3dB=ln2,
2𝑑𝑓=0.9∗2
(a)𝑓3dB=𝑓0, 𝑓eq=𝑓0𝜋≃1.57𝑓0, 𝑓90%=𝑓0tan(0.45𝜋)≃6.31𝑓0, ⇔ 𝑓3dB<𝑓eq<𝑓90%
−∞ 1+(𝑓/𝑓0)
0 1+(𝑓/𝑓0)
2
Similarly, for (b), the frequency response 𝐻 ( 𝑓 ) is obtained as
2 2𝛼 2𝛼2 𝛼 eq ∫∞−2𝛼𝑓 1
𝑓 = 𝑒 𝑑𝑓= ,
0
LHS of 𝑓90% in (1.13)
2𝛼
∫ 𝑓90%
2
|𝐻 ( 𝑓 ) | 𝑑 𝑓 = 2
∫ 𝑓90%
𝑒
1 𝛼
−2𝛼𝑓
∫ ∞ −2𝛼𝑓
−2𝛼𝑓90% ⇒ 𝑓90% =
1 − 𝑒
0
−∞ 0 𝛼 2𝛼
𝑑 𝑓 = RHSof𝑓90% in(1.13)0.9 𝑒 𝑑𝑓 =0.9∗2 𝑒 𝑑𝑓 =
,
−𝑓90%
∫ ∞ −2𝛼𝑓
0.9
ln(10)
Hence, the BWs 𝑓3 dB, 𝑓eq, and 𝑓90% are given by
(b)𝑓3dB=1ln√2≃0.35, 𝑓eq= 1 =0.5, 𝑓90%=ln(10)≃1.15, ⇔ 𝑓3dB<𝑓eq<𝑓90%
𝛼 𝛼 2𝛼 𝛼 2𝛼 𝛼 Useful Equality
Refer Wolfram Alpha [https://bit.ly/3d3RZoA]
∫1 tan−1𝑥∫∞1 𝜋
2 2𝑑𝑥= 𝑎, 2 2𝑑𝑥= forreal-valued𝑎 (4.14)
𝑎+𝑥 𝑎 0𝑎+𝑥 2|𝑎|
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5 Discrete Time Fourier Transform
Major References:
• Chapter3&5,SignalsandSystemsbyAlanV.Oppenheimet.al.,2ndedition,PrenticeHall • Chapter6,Schaum’sOutlineofSignalsandSystems,2ndEdition,2010,McGraw-Hill
Aperiodic Signal
Periodic Signal
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Leonhard Euler P. S. Laplace (1749-1827) (1707-1783)
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