EERI 222 – Signal Theory 1 Study Unit 2 – Continuous-Time Signals and Systems Prof ASJ Helberg 2026 Class Program July 14 Introduction to Course, Matlab 16 SU 1.1 - 1.3 17 SU 2.1 - 2.2 21 TUT SU1 23 SU 2.3-2.4 24 August Sept TUT SU2 30 Semester Test 2 July Date August Event Date Saturday 1 Sunday 2 Monday 3 Tuesday 4 RECESS September Event Date Event October Date Event November Date 2 TUTSU 3 + miniprac handout 3 6 SU 3.3 3 TUTSU5 + Hand out Prac Friday 3 7 SU 3.4 - 3.5 4 17 SU 6 Saturday 4 8 5 3 7 18 SU 6 Sunday 5 9 6 4 8 Monday 6 10 Public Holiday 7 5 Tuesday 7 11 TUTSU 3 8 Wednesday 8 Thursday 9 13 SU4 10 Semester test 1 6 TUTPrac SU 3.1 8 Tut Prac 31 SU 3.2 9 Semester Test 3 4 TUTSU 3 + mini-prac handout 13 Demo Practical 6 SU 3.3 15 Demo Practical 7 SU 3.4 - 3.5 16 Demo Practical 11 TUTSU 3 20 Demo Practical 13 SU4 22 Exam Review 14 Semester test 1 23 Demo Practical (ReDo) 27 Demo Practical (ReDo) 2 Mid Semester assessments 2 RECESS 5 1 Thursday TUTSU 6 12 9 Semester Test 2 Mid Semester assessments 4 1 2 6 RECESS 6 9 TUTPrac 7 10 11 8 Tut Prac 12 Semester Test 3 13 Friday 10 14 11 9 11 15 12 10 14 Sunday 12 16 13 11 15 Monday 13 RECESS 17 14 12 16 Introduction to Course, Matlab Matlab Mini-Prac Tuesday 14 Wednesday 15 Thursday 16 SU 1.1 - 1.3 20 SU4 17 SU 6 15 Demo Practical 19 Friday 17 SU 2.1 - 2.2 21 SU4 18 SU 6 16 Demo Practical 20 Saturday 18 22 19 17 21 22 19 15 TUTSU5 + Hand out Prac 16 13 Sunday 19 23 20 18 20 24 21 19 Tuesday 21 Wednesday 22 Thursday 23 SU 2.3-2.4 27 SU 5 24 Public Holiday 22 SU 2.5 28 SU 5 25 RECESS 23 TUT SU1 25 TUTSU 4 26 22 Demo Practical 14 Monday TUTSU 6 23 20 17 25 Exam Review Demo Practical (ReDo) SU4 Friday 24 21 SU4 Saturday 25 29 26 24 28 Sunday 26 30 27 25 29 31 28 26 27 SU 5 Tuesday 28 28 SU 5 Wednesday 29 Thursday 30 SU 3.1 29 SU 3.2 30 Friday 31 Saturday 1 Sunday 2 TUT SU2 29 30 RECESS 27 27 30 Demo Practical (ReDo) 2nd OPP 28 31 2nd OPP 26 20 27 2nd OPP 24 Matlab Mini-Prac Monday 1st OPP 23 Demo Practical 18 TUTSU 4 1st OPP 18 21 25 1st OPP 5 Saturday 18 Event 1 1 22 Oct Days Wednesday 15 SU 2.5 28 1 - 11 Months 1st OPP Agenda • SS 2.1 – Transformations of Signals • SS 2.2 – Signal Characteristics • SS 2.3 – Basic Signals • SS 2.4 – Block Diagrams • SS 2.5 – Characteristics of Systems • Resources: • Phillips et al, Chapter 2 SS 2.1 – Transformations of Signals Study Outcomes: • Formulate Continuous-Time Signals by Means of Mathematical Functions. • Apply the Following Time-Transformations to OneDimensional Continuous-Time Signals: Reversal; Scaling, and Translation (Shifting). • Apply the Following Amplitude-Transformations on One-Dimensional Continuous-Time Signals: Reversal; Scaling and Shifting. Continuous-Time Signals and Systems Transformations of Signals Original Signal ⇒ π₯(π‘) Transformed Signal ⇒ π¦ π‘ Two Types of Signal Transformations: 1. Transformations in Time 2. Transformations in Amplitude Transformations of Signals: Transformations in Time Three Sub-Types of Signal Transformations in Time: π¦ π‘ = π₯ ππ‘ + π 1. Time Reversal π¦(π‘) = π₯(−π‘) 2. Time Scaling π¦ π‘ = π₯(ππ‘) 3. Time Shifting π¦ π‘ = π₯ π‘ − π‘0 Transformations in Time Time Reversal Formula: π¦(π‘) = π₯(−π‘) Description: Mirror Reflection about the Vertical Axis Transformations in Time Time Scaling Formula: π¦ π‘ = π₯ ππ‘ Description: • If π > 1, Time is Sped-Up • If π < 1, Time is Slowed-Down Transformations in Time Time Shifting Formula: π¦ π‘ = π₯ π‘ − π‘0 , Where π‘0 is Constant Description: • If π‘0 > 0, Time is delayed or shifted to the right (Going back in time) • If π‘0 < 0, Time is advanced or shifted to the Left (Going forward in time) Transformations in Time General Approach to Formulating/Plotting Independent-Variable Transformations Formula: π¦ π‘ = π₯ ππ‘ + π Where π and π are Real Constants If π denotes the time variable in the original signal, then π−π π = ππ‘ + π ⇒ π‘ = π Transformations in Time Example: Plot the transformed signal π¦ π‘ =π₯ 1− π‘ 2 Answer: 1. Replace π‘ with π, since π denotes the time variable in the original signal 2. Identify π and π: If π¦ π‘ = π₯ ππ‘ + π , then 1 2 π=1 π=− Transformations in Time Example: Plot the transformed signal π¦ π‘ = π₯ 1− π‘ 2 Answer (Continues): 3. Solve π‘ for each timestamp from the original time axis π−π 1 π‘= , Where π = − and π = 1 π 2 π‘= For π = −1, 0, 1, 2 π‘ = 4 , 2, 0, −2 π− 1 1 − 2 Transformations in Time Example: Plot the transformed signal π¦ π‘ =π₯ 1− π‘ 2 Answer (Continues): 4. Plot π‘ on the x-axis (From small to large) 5. Match the values of π‘ with the corresponding π values Transformations of Signals: Transformations in Amplitude Three sub-types of signal transformations in amplitude: π¦(π‘) = π΄π₯(π‘) + π΅ 1. Amplitude Reversal π¦ π‘ = −π₯(π‘) 2. Amplitude Scaling π¦(π‘) = π΄π₯(π‘) 3. Amplitude Shifting π¦(π‘) = π₯(π‘) + π΅ Transformations in Amplitude Amplitude Reversal Formula: π¦ π‘ = −π₯(π‘) Description: Mirror reflection about the horizontal axis Transformations in Amplitude Amplitude Scaling Formula: π¦ π‘ = π΄π₯ π‘ Description: • If π΄ > 1, Signal is Amplified • If π΄ < 1, Signal is Attenuated Transformations in Amplitude Amplitude Shifting Formula: π¦ π‘ = π₯ π‘ + π΅, where π΅ is Constant Description: • If π΅ > 0, signal is shifted upward (Raising DC) • If π΅ < 0, signal is shifted downward (Lowering DC) Transformations in Amplitude Example: Plot the transformed signal π¦ π‘ = 3π₯ π‘ − 1 Answer: 1. Solve π¦(π‘) for each value from the original signal axis For π₯ π‘ = 0, 1, 1, 0, 1, 1, 0 π¦ π‘ = −1, 2, 2, −1, 2, 2, −1 2. Plot π¦(π‘) Transformations in Time & Amplitude Example: Plot the transformed signal π¦ π‘ = 3π₯ 1 − π‘ −1 2 Answer: 1. Start with the amplitude transformation 2. Perform the time transformation 3. Plot π¦(π‘) Summary: Signal Transformations Individual Activity 2.1 1. Tackle Problem 2·1(a) on p. 78 of Phillips. (Purely to Make Sure that You are Able to Execute TimeTransformation of Signals Independently. 2. Do the Following Problems at the End of Chapter 2 in Phillips (4th and 5th edition) as Preparation for the Tutorial Session: 2·1(b); 2·2(c); 2·3(a) Questions? SS 2.2 – Signal Characteristics Study Outcomes: • Know the definitions of even and uneven functions and describe it in your own words. • Deduce expressions with which the even and/or uneven components of a signal can be calculated from the signal and a mirror image thereof. • Prove the characteristics of even and uneven functions. • Determine the even and uneven components of an arbitrary signal. • Determine the average value of a signal. • Write down the definition of periodic signals. • Determine whether an arbitrary signal is periodic or not (and calculate the period of the signal if it is a periodic signal). Signal with Even Symmetry Formula: π₯π (π‘) = π₯π −π‘ Description: Symmetry with respect to the vertical axis Example: π₯ π‘ = cos ππ‘ Signal with Uneven Symmetry Formula: π₯π π‘ = −π₯π −π‘ Description: Symmetry with respect to the origin Example: π₯ π‘ = sin ππ‘ Signal Characteristic Expressions Defining Any Signal π₯ π‘ = π₯π π‘ + π₯π (π‘) Finding the Even Part of a Signal π₯π π‘ = 1 π₯ π‘ + π₯(−π‘) 2 π₯π π‘ = 1 π₯ π‘ − π₯(−π‘) 2 Finding the Odd Part of a Signal Finding the Average Value of a Signal π΄π₯ = 0 For Uneven Signals 1 π 1 π π΄π₯ = lim π₯ π‘ ππ‘ = lim π₯π π‘ ππ‘ π→∞ 2π −π π→∞ 2π −π Characteristics of Even and Odd Functions 1. The sum of two even functions is even 2. The sum of two odd functions is odd 3. The sum of an even function and an odd function is neither even nor odd 4. The product of two even functions is even 5. The product of two odd functions is even 6. The product of an even function and an odd function is odd Periodic Signals Formula: π₯ π‘ = π₯ π‘ + ππ , π > 0 and π ∈ β€ The Fundamental Period (π»π in s): The minimum value of π > 0 that satisfies π₯ π‘ = π₯(π‘ + π) The Fundamental Frequency: ππ (in Hz) 1 π0 = π 0 or ππ (in rad/s) 2π π0 = 2ππ0 = π 0 Test Signals for Periodicity Sum of periodic signals is periodic βΊ Ratios of periods are ratios of integers (i.e., Rational - β) What is the fundamental period for a sum of periodic signals? Given a set of periods of relevant signals (π01 , π02 , … , π0π ), π01 ,2 ≤ π ≤ π π0π 1. Convert each period ratio to a ratio of integers (β€). 2. Eliminate common factors from the numerator and denominator for each ratio (Simplify). 3. Then the common fundamental period is: π0 = π0 π01 Where π0 is the least common multiple of the denominators. Test Signals for Periodicity Example: Given three periodical signals: 7π‘ 6 The sum of which equals π£ π‘ . To determine whether π£ π‘ is periodic, we must see if the ratios of the periods of π₯1 π‘ , π₯2 π‘ and π₯3 π‘ are ratios of integers: π₯1 π‘ = πππ 3.5π‘ , We use: π₯2 = π ππ 2π‘ , πππ π₯3 π‘ = 2 cos 1 2π 0 0 π0 = 2ππ0 ⇒ π0 = π = π To first determine all the periods: 2π 2π 2π 2π π01 = = , π02 = = , π1 3.5 π2 2 2π 2π π03 = = . π3 7 6 Then, the ratios of periods are: 2π π01 2 4 3.5 = = = 2π π02 3.5 7 2 and π01 = π03 Both are ratios of integers and therefore, π£ π‘ is periodic. 2π 2π 3.5 7 6 7 7 = 3.56 = 21 . Finding The Fundamental Frequency/Period The period of π£ π‘ is determined by first eliminating any common factors from the numerator and denominator of each ratio of periods. π 4 • In this case, we have no common factor in the ratio π01 = 7, π 7 But the ratio π01 = 21 has 7 as a common factor. 03 π π03 02 1 3 • After eliminating the common factor, we have 01 = . • The next step is to find the least common multiple of the denominators of the ratios. in this case, the least common multiple is π1 = 3 × 7 = 21. • Therefore, the fundamental period of π£ π‘ is π0 = π1 π01 = 21 × • This period can be confirmed by examination of Figure 2.11(a). 2π = 12π 3.5 π . Test Signals for Aperiodicity Another signal, π₯4 π‘ = 3π ππ 5ππ‘ , is added to π£ π‘ to give: π€ π‘ = π₯1 π‘ + π₯2 π‘ + π₯3 π‘ + π₯4 π‘ . • This sum of four periodic signals is shown in Figure 2.11(b). The periodicity of this new sum of signals must be determined as was done previously for π£ π‘ . The addition of π₯4 π‘ gives the additional ratio of periods, 2π π01 3.5 = 5π . = 2π π04 3.5 5π • Since π is an irrational number, this Ratio is NOT Rational. Therefore, the signal π€(π‘) is NOT Periodic/Aperiodic. Individual Activity 2.2 1. Do the following problems at the end of chapter 2 in Phillips as preparation for the tutorial session: • Phillips et al: 4th edition 2·13(a)(b); 2·14 • Phillips et al: 5th edition 2·11(a)(b); 2·12 Questions? SS 2.3 – Basic Signals Study Outcomes: • Know and explain Euler’s relationship. • Write a sine or cosine function in terms of exponential functions (and deduce these formulae). • Present a sine or cosine function in phasor notation. • Know and describe in your own words the definition of harmonic-related complex exponential signals. • Know the definitions of time-constants and enveloping and practically work with these magnitudes. • Know and write down the definitions of the unit step function, rectangularshaped pulse function, unit impulse function and unit gradient function. • Present functions in terms of linear combinations of unit step functions, rectangular-shaped pulse functions and straight-line segments. • Know and be able to name the characteristics of the unit impulse function. • Plot any signal graphically. • Describe any signal with the aid of a mathematical function. Euler’s Relation Formula: π ππ = cos π + π sin π Further Expressions: π −ππ = cos π − π sin π π ππ + π −ππ cos π = 2 π ππ − π −ππ sin π = 2π Euler’s Relation in Polar Form Formula: π π ππ = π ∠π Determining the Magnitude: π ππ = 2 1 2 2 cos π + sin π =1 Determining the Angle: arg π ππ = tan −1 sin π =π cos π Euler’s Relation Special Cases Euler Case 1 π₯ π‘ = πΆπ ππ‘ = πΆπ −π‘ π , π > 0 and πΆ ∈ β and π ∈ β Examples Transient response of RL and RC circuits τ - Time Constant Number of Time Constants [π] Level of Decay [%] Level of Increase [%] 1 36.79 63.21 2 13.53 86.47 3 4.98 95.02 4 1.83 98.17 5 0.67 99.33 Euler’s Relation Special Cases Euler Case 2 π₯ π‘ = πΆπ ππ‘ , where πΆ = π΄π ππ = π΄∠π and π = ππ0 π > 0 and πΆ ∈ β and π ∈ π Therefore, π₯ π‘ = π΄π ππ π ππ0 π‘ = π΄π π(π0 π‘+π) = π΄ cos π0 π‘ + π + ππ΄ sin π0 π‘ + π π0 is the Frequency And 2π = π0 is the Period π0 Examples Transient response of LC circuits Harmonic Related Complex Exponentials A set of functions with frequencies related by integers π₯π π‘ = π΄π π πππ0 π‘ , π = ±1, ±2, … . Euler’s Relation Special Cases Euler Case 3 ππ‘ ππ π₯ π‘ = πΆπ , where πΆ = π΄π and π = π0 + ππ0 π > 0 and πΆ ∈ β and π ∈ β Therefore, Envelope: ± π΄π π0 π‘ Undamped Sinusoid (Case 2) If π0 = 0 π₯ π‘ = π΄π ππ π (π0 +ππ0 )π‘ = π΄π π0 π‘ π π(π0 π‘+π) = π΄π π0 π‘ cos π0 π‘ + π + ππ΄π π0 π‘ sin π0 π‘ + π = π₯π π‘ + ππ₯π π‘ Underdamped Sinusoid If π0 < 0 Examples Transient response of RLC circuits Singularity Functions Two Singularity Functions Will Be Discussed: 1. Unit Step Function 2. Unit Impulse Function Unit Step Function (Heaviside) – π π π’ π = 1, 0, π’(π‘ − π‘0 ) = π>0 π<0 1, 0, π‘ > π‘0 π‘ < π‘0 Used to switch functions (Multiply the unit step function with the function) Singularity Functions – Unit Step Function Example: π= π‘−5 π’(π‘ − 5) = 1, 0, π‘−5>0⇒π‘ >5 π‘−5<0⇒π‘ <5 Properties: π’(π‘ − π‘0 ) = 1, 0, π‘ > π‘0 π‘ < π‘0 π’ π‘ − π‘0 = π’(π‘ − π‘0 ) π π’ ππ‘ − π‘0 = π’ π‘ − π‘0 /π , π ≠ 0 π’(π‘0 − π‘) = 1, 0, π‘ < π‘0 π‘ > π‘0 Singularity Functions – Unit Step Function Extra: another useful switching function Rectangular Pulse Function rect π‘ π = π is the period 1, 0, −π 2 < π‘ < π 2 otherwise Singularity Functions – Unit Step Function Rectangular pulse function can be defined with unit step functions in 3 ways: π’ π‘+π 2 −π’ π‘−π 2 rect π‘ π = π’ π 2 − π‘ − π’ −π 2 − π‘ π’ π‘+π 2 ×π’ π 2−π‘ Singularity Functions – Unit Step Function Time-Shifted rectangular pulse function rect π‘ − π‘0 π = 1, 0, π‘0 − π 2 < π‘ < π‘0 + π 2 otherwise Singularity Functions – Unit Step Function Example: Use switching functions π₯ π‘ = cos π‘ × π’ π‘ − π’ π‘ − 2π = cos π‘ , 0, 0 < π‘ < 2π ππ‘βπππ€ππ π Or π₯ π‘ = cos π‘ × rect π‘ − π 2π Periodic signal representation for pulse functions: π£ π‘ = π£1 π‘ + π£1 π‘ − π0 + π£1 π‘ − 2π0 + β― ∞ = π£1 π‘ − ππ0 π=0 Singularity Functions – Unit Impulse Function Unit Impulse Function (Dirac Delta) – πΉ(π) Integrate Unit Step Function π‘ π π‘ = π’ π − π‘0 ππ = 0 π‘ π‘ ππ = π π‘0 π‘0 = π‘ − π‘0 π π − ππ Unit Ramp Function π₯ π‘ = π‘ − π‘0 π’ π‘ − π‘0 Singularity Functions – Unit Impulse Function What is the Unit Impulse Function? • Energy in a physical system cannot be transferred instantaneously. Therefore, when zoomed in, a more practical unit step function looks like π π‘ , with π → 0. • Deriving π(π‘), results in the rectangular pulse π(π‘) with width π and amplitude → ∞. The integral = 1 (from π(π‘)), and thus, the area under π π‘ = 1. • We call the limit of this function (When π → 0) The Unit Impulse Function. lim π(π‘) = πΏ π‘ − π‘0 π→0 Therefore, The unit impulse function can be described by πΏ π‘ − π‘0 = 0, π‘ ≠ π‘0 ; ∞ πΏ(π‘ − π‘0 ) ππ‘ = 1 −∞ Singularity Functions – Unit Impulse Function Shifting Properties of The Unit Impulse Function ∞ π(π‘)πΏ(π‘ − π‘0 ) ππ‘ = π(π‘0 ) −∞ And thus, π π‘ πΏ π‘ − π‘0 = π(π‘0 )πΏ(π‘ − π‘0 ) Singularity Functions – Unit Impulse Function Summary of the Unit Impulse Function’s Properties Mathematical Functions for Signals Example: Function to Graph Mathematical Functions for Signals Graph to Function Procedure When the slope of a signal changes, a ramp function is added at that point, with the slope of this ramp function equal to the new slope minus the previous slope (π1 − π0 ). At any point that a step occurs in the signal, a step function is added. An example using this procedure is now given. π₯ π‘ = π0 π‘ − π‘0 π’ π‘ − π‘0 + π1 − π0 π‘ − π‘1 π’ π‘ − π‘1 + β― Example: Graph to Function π₯ π‘ = 3 π‘ + 2 π’ π‘ + 2 − 6 π‘ + 1 π’ π‘ + 1 + 3 π‘ − 1 π’ π‘ − 1 + 3π’ π‘ − 3 Another method is discussed in the study guide, which takes much longer Mathematical Functions for Signals Example: Graph to Function π π£ π‘ = π‘, π 0<π‘<π Or π π π£ π‘ = π‘ π’ π‘ − π’(π‘ − π) = π‘ rect π π π π‘− 2 π Mathematical Functions for Signals Example: Periodic signal representation for pulse functions (Pulse Train) From the previous example, π£1 π‘ can be written As: π π π π£1 π‘ = π‘ π’ π‘ − π’(π‘ − π) = π‘ rect π‘ − π π π 2 If shifted one period to the right; π π£1 π‘ − π = π‘ − π π’ π‘ − π − π’(π‘ − 2π) π π 3π = π‘ − π rect π‘ − π π 2 Therefore, for ππ < π‘ < (π + 1)π or π£1 π‘ − ππ π π£1 π‘ − ππ = π‘ − ππ π’ π‘ − ππ − π’(π‘ − ππ − π) , π The Sawtooth wave can be represented as: ∞ π£ π‘ = π£1 π‘ − ππ π=−∞ π∈β€ Individual Activity 2.3 Do the following problems at the end of Chapter 2 in Phillips as preparation for the tutorial session: • Phillips et al: 4th edition: • 2·8 • 2·12(c)(e) • 2·7(a)(d) • 2·11 • 2·20(c) • 2·21(a)(c) • 2·19(a)(b)(c) • Phillips et al: 5th edition: • 2.6 • 2.1 (c)(e) • 2·5(a)(d) • 2.9 • 2.19 (c) • 2·20(a)(c) • 2·26(a)(b)(c) Questions? SS 2.4 – Block Diagrams Study Outcomes: • Formulate mathematical equations from system block diagrams. • Draw up block diagrams representing system equations. Block Diagrams System – A process for which cause-and-effect relations exist π¦ π‘ = π[π₯ π‘ ] π[π₯ π‘ ] is referred to as the mathematical model. This is usually differential equations for continuous-time systems. Block Diagram Components 1. System Component – Block 2. Junction Components – Circle • Summing Junction • Product Junction Block Diagram Components 3. System Connections • Parallel Connection π¦ π‘ = π¦1 π‘ + π¦2 π‘ = π1 π₯ π‘ + π2 π₯ π‘ =ππ₯ π‘ • Series or Cascade Connection π¦ π‘ = π2 π¦1 π‘ = π2 π1 π₯ π‘ = π[π₯ π‘ ] Block Diagrams Example Example: π¦1 π‘ = π1 π₯ π‘ π¦2 π‘ = π2 π₯ π‘ π¦3 π‘ = π1 π₯ π‘ π¦4 π‘ = π3 π¦3 π‘ + π2 π₯ π‘ = π3 π1 π₯ π‘ + π2 π₯ π‘ π¦5 π‘ = π4 π₯ π‘ π¦ π‘ = π¦4 π‘ × π¦5 π‘ = π3 π1 π₯ π‘ + π2 π₯ π‘ × π4 π₯ π‘ Control Theory – Feedback-Control System (Automatic Control) • The plant is the physical system to be controlled. • The controller is a physical system inserted by the design engineers to give the total system certain desired characteristics. • The sensor measures the signal to be controlled, and the input signal represents the desired output. • The Error Signal (π π‘ ) is a measure of the difference between the desired output (π₯ π‘ ) and the measurement of the output (π¦ π‘ ). π π‘ = π₯ π‘ − π3 π¦ π‘ π¦ π‘ = π2 π π‘ = π2 π1 π π‘ π¦ π‘ = π2 π1 π₯ π‘ − π3 π¦ π‘ Individual Activity 2.4 1. Do the following problems at the end of Chapter 2 in Phillips as preparation for the tutorial session: • Phillips et al: 4th edition: 2·23(a)(b); 2·24; 2·25 • Phillips et al: 5th edition: 2.30; 2.28; 2.29 Questions? SS 2.5 – Characteristics of Systems Study Outcomes: • know the definitions for the following characteristics of continuous-time systems: memory; reversibility; system-inverse; causality; stability; timeindependency and linearity. • Determine which of the above characteristics are displayed by an arbitrary system. Characteristics of Systems π π Produces π π π₯ π‘ →π¦ π‘ π¦ π‘ =π π₯ π‘ The Six Properties of Continuous-Time Systems 1. 2. 3. 4. 5. 6. Memory Invertibility Causality Stability Time Invariance Linearity Properties of Continuous-Time Systems - Memory Memory (Dynamic System) A system has memory if its output at time π‘0 (i.e. π¦ π‘0 ) depends on input values other than π₯ π‘0 . Otherwise, the system is memoryless. System that Has Memory: Dynamic System Examples: Integrating Amplifier, Voltage of Capacitor System that Does not Have Memory: Static System Examples: Ideal Amplifier, Resistance, Squaring Circuit Properties of Continuous-Time Systems - Invertibility Invertibility A system is said to be invertible if distinct inputs result in distinct outputs. (For an invertible system, the system input can be determined uniquely from its output.) System that Is Invertible: Invertible System Examples: Ideal Amplifier System that Is Not Invertible: Non-Invertible System Examples: Squaring Circuit Special Case: Identity System – Output is equal to its input Examples: Ideal Amplifier with gain of unity Inverse of a System The inverse of a system (Denoted by π) is a second system (Denoted by ππ ) that, when cascaded with the system π, yields the identity system. π¦ π‘ =π π₯ π‘ ⇒ π₯ π‘ = ππ π¦ π‘ Transducers must have invertible systems Properties of Continuous-Time Systems – Causality Causality A system is causal if the output at any time π‘0 is dependent on the input only for π‘ ≤ π‘0 . System that is Causal: Non-Anticipatory System Examples: Any Physical System, Ideal Time Delay System that is Not Causal: Non-Real Time System Examples: Ideal Time Advance Properties of Continuous-Time Systems – Stability Bounded-Input-Bounded-Output (BIBO) Stability A system is stable if the output remains bounded for any bounded input. (i.e., for all time, the input signal (π₯ π‘ )) and output signal (π¦ π‘ ) are both bounded. System that is Stable: Stable System Examples: Ideal Amplifier, Squaring System System that is Not Stable: Unstable System Examples: Ramp Function (e.g., Integrating Unit Step) Special Case: Marginally Stable – Undefined Bounds Examples: Exponential Functions Properties of Continuous-Time Systems – Time Invariance Time Invariance A system is said to be time invariant if a time shift in the input signal results only in the same time shift in the output signal. π¦ π‘ − π‘0 = π π₯ π‘ − π‘0 π¦ π‘ − π‘0 = π¦π π‘ System that is Time-Invariant: Fixed System Examples: Recording System that is Time Varying: Examples: Time Reversal Operation Properties of Continuous-Time Systems – Linearity Linearity A System is linear if it meets the following two criteria: 1. Additivity: 2. Homogeneity: If π₯1 π‘ → π¦1 π‘ and π₯2 π‘ → π¦2 π‘ π₯1 π‘ + π₯2 π‘ → π¦1 π‘ + π¦2 π‘ . If π₯1 π‘ → π¦1 π‘ , then ππ₯1 π‘ → ππ¦1 π‘ ; ∀ π₯1 π‘ , π₯2 π‘ and π‘. I.E, Principle of Superposition Holds: π1 π₯1 π‘ + π2 π₯2 π‘ → π1 π¦1 π‘ + π2 π¦2 π‘ System that Has Linearity: Linear System Examples: Ideal Amplifier System that Does Not Have Linearity: Non-Linear System Examples: Squaring Circuit, Systems with Non-Zero Initial Conditions Examples Examples Individual Activity 2.5 1. Do the Following Problems at the End of Chapter 2 in Phillips as Preparation for the Tutorial Session: 2·31(a)(d) • Phillips et al: 4th edition: 2·31(a)(d) • Phillips et al: 5th edition: 2·35(a)(d) Questions?
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