ECE 2714 Exam 1 Review PDF

Exam 1 Policy  Grading Rubric 5 - Fully correct 4.5 - Single math error 4 - Most of the algebra is correct with only one algebra mistake 3.5 - Some manipulation of the equations is done with some substitutions but not all substitutions are done correctly 3 - Basic engineerin...

Exam 1 Policy  Grading Rubric 5 - Fully correct 4.5 - Single math error 4 - Most of the algebra is correct with only one algebra mistake 3.5 - Some manipulation of the equations is done with some substitutions but not all substitutions are done correctly 3 - Basic engineering concepts are there (e.g., the starting equations) but no manipulation of the equations is done in any way beyond the basic equations 2 - At least one equation relevant to the problem is present 1 - Work is present but no applicable equations 0 - No attempt  CT / DT Signals and Properties Important CT/DT functions: unit impulse, unit step, exponential (complex or real), sin / cos, Euler’s formula. Properties: even / odd / neither, periodic / aperiodic, power / energy / neither. If a signal is periodic, what is its fundamental period / fundamental frequency? Fundamental Period: the smallest positive real number T (CT) or positive integer N (DT) such that 𝑥𝑥 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 + 𝑇𝑇 (CT signals) or 𝑥𝑥[𝑛𝑛] = 𝑥𝑥[𝑛𝑛 + 𝑁𝑁] (DT signals). Notice the similarities as well as important difference between CT and DT signals when it comes to periodicity. If two signals (CT / DT) are periodic (i.e., 𝑥𝑥1 and 𝑥𝑥2 ), how about their sum / product (i.e., 𝑥𝑥1 + 𝑥𝑥2 , 𝑥𝑥1 𝑥𝑥2 )? How to calculate the energy / power of CT / DT signals?  CT / DT System Properties Stability How to determine whether a CT/ DT Invertibility system possess these properties: Memory Definition. Properties of impulse response Causality function. Time Invariance Construction of examples (i.e., an inverse system) or counter-examples. Linearity  Convolution ∞ 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝜏𝜏 ℎ 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝜏𝜏 How to evaluate any given −∞ convolution (𝑦𝑦 = 𝑥𝑥 ∗ ℎ) ∞ based on their definition: 𝑦𝑦 𝑛𝑛 = 𝑥𝑥 𝑘𝑘 ℎ 𝑛𝑛 − 𝑘𝑘 𝑘𝑘=−∞ Properties of (DT / CT) convolution: commutative, distributive, associative, time shift, as well as their combinations How to use convolution tables (if the problem allows you to do so). CT Convolution Table in Exam DT Convolution Table in Exam  Impulse Response Function (CT) 𝑑𝑑 𝑁𝑁 𝑦𝑦 𝑡𝑡 𝑑𝑑 𝑁𝑁−1 𝑦𝑦 𝑡𝑡 𝑑𝑑𝑦𝑦 𝑡𝑡 𝑎𝑎𝑁𝑁 + 𝑎𝑎𝑁𝑁−1 + ⋯ + 𝑎𝑎1 + 𝑎𝑎0 𝑦𝑦 𝑡𝑡 Always set 𝑎𝑎𝑁𝑁 = 1 𝑑𝑑𝑡𝑡 𝑁𝑁 𝑑𝑑𝑡𝑡 𝑁𝑁−1 𝑑𝑑𝑡𝑡 𝑀𝑀 𝑀𝑀−1 𝑑𝑑 𝑥𝑥 𝑡𝑡 𝑑𝑑 𝑥𝑥 𝑡𝑡 𝑑𝑑𝑥𝑥 𝑡𝑡 = 𝑏𝑏𝑀𝑀 𝑀𝑀 + 𝑏𝑏𝑀𝑀−1 𝑀𝑀−1 + ⋯ + 𝑏𝑏1 + 𝑏𝑏0 𝑥𝑥 𝑡𝑡 𝑄𝑄 𝐷𝐷 𝑦𝑦 = 𝑃𝑃 𝐷𝐷 𝑥𝑥 𝑑𝑑𝑡𝑡 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑 Only include this term if 𝑁𝑁 = 𝑀𝑀 Homogenous solution ℎ 𝑡𝑡 = 𝑏𝑏𝑁𝑁 𝛿𝛿 𝑡𝑡 + 𝑃𝑃 𝐷𝐷 𝑦𝑦ℎ 𝑡𝑡 𝑢𝑢(𝑡𝑡) 𝑦𝑦ℎ 𝑡𝑡 = 𝐶𝐶1 𝑦𝑦ℎ1 𝑡𝑡 + ⋯ + 𝐶𝐶𝑖𝑖 𝑦𝑦ℎ𝑖𝑖 𝑡𝑡 + ⋯ + 𝐶𝐶𝑁𝑁 𝑦𝑦ℎ𝑁𝑁 𝑡𝑡 Real roots / complex roots / repeated roots. The constants 𝐶𝐶𝑖𝑖 are determined by using auxiliary conditions: 𝑦𝑦ℎ 0+ = 0 ⋯ 𝐷𝐷 𝑖𝑖 𝑦𝑦ℎ 0+ = 0 ⋯ 𝐷𝐷 𝑁𝑁−1 𝑦𝑦ℎ 0+ = 1 Linear Constant Coefficient Difference Equation 𝑁𝑁 𝑀𝑀 Advance Form: 𝑎𝑎𝑘𝑘 𝑦𝑦[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] = 𝑏𝑏𝑘𝑘 𝑥𝑥[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] 𝑘𝑘=0 𝑘𝑘=0 𝑁𝑁 𝑀𝑀 Delay Form: 𝑎𝑎𝑘𝑘 𝑦𝑦[𝑛𝑛 − 𝑘𝑘] = 𝑏𝑏𝑘𝑘 𝑥𝑥[𝑛𝑛 − 𝑘𝑘] 𝑘𝑘=0 𝑘𝑘=0 𝑄𝑄 𝐸𝐸 𝑦𝑦 𝑛𝑛 = 𝑃𝑃 𝐸𝐸 𝑥𝑥 𝑛𝑛 𝑄𝑄 𝐸𝐸 = 𝑎𝑎0 𝐸𝐸 𝑁𝑁 + 𝑎𝑎1 𝐸𝐸 𝑁𝑁−1 + ⋯ + 𝑎𝑎𝑁𝑁−1 𝐸𝐸 + 𝑎𝑎𝑁𝑁 𝑃𝑃 𝐸𝐸 = 𝑏𝑏0 𝐸𝐸 𝑁𝑁 + 𝑏𝑏1 𝐸𝐸 𝑁𝑁−1 + ⋯ + 𝑏𝑏𝑁𝑁−1 𝐸𝐸 + 𝑏𝑏𝑁𝑁  Impulse Response Function (DT) 𝑄𝑄 𝐸𝐸 ℎ 𝑛𝑛 = 𝑃𝑃 𝐸𝐸 𝛿𝛿 𝑛𝑛 𝑏𝑏𝑁𝑁 ℎ 𝑛𝑛 = 𝛿𝛿 𝑛𝑛 + 𝑦𝑦ℎ 𝑛𝑛 𝑢𝑢[𝑛𝑛] 𝑎𝑎𝑁𝑁 Distinct roots: 𝑦𝑦ℎ 𝑛𝑛 = 𝑐𝑐1 𝜆𝜆1𝑛𝑛 + ⋯ + 𝑐𝑐𝑖𝑖 𝜆𝜆𝑛𝑛𝑖𝑖 + ⋯ + 𝑐𝑐𝑁𝑁 𝜆𝜆𝑛𝑛𝑁𝑁 Determine the values of 𝑐𝑐𝑖𝑖 by iteratively calculating the values of: ℎ0 ℎ1 ⋯ ℎ 𝑁𝑁 − 1 Under the conditions of: 𝑥𝑥 𝑛𝑛 = 𝛿𝛿[𝑛𝑛] ℎ 𝑛𝑛 = 0 For all n 8 𝑦𝑦 𝑡𝑡 = 3 6=9 𝜏𝜏 = 𝑡𝑡 − 3 2 𝜏𝜏 = 𝑡𝑡 𝜏𝜏 = 2 𝜏𝜏 𝜏𝜏 = 𝑡𝑡 − 6  Convolution: Graphic Understanding Signal 𝑥𝑥 𝑡𝑡 : Unit impulse response function ℎ 𝑡𝑡 : 𝑥𝑥 𝑡𝑡 = 𝑢𝑢 𝑡𝑡 − 2 ℎ 𝑡𝑡 = 12 − 2𝑡𝑡 𝑢𝑢 𝑡𝑡 − 3 − 𝑢𝑢 𝑡𝑡 − 6 6 𝑡𝑡 = 2 𝑡𝑡 𝑡𝑡 ∞ 𝑡𝑡 = 3 𝑡𝑡 = 6 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝜏𝜏 ℎ 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝜏𝜏 = 𝑥𝑥 𝑡𝑡 ∗ ℎ 𝑡𝑡 −∞ For 5 < 𝑡𝑡 < 8 𝑡𝑡−3 𝜏𝜏 = 𝑡𝑡 − 3 𝑦𝑦 𝑡𝑡 = 2 𝜏𝜏 − 𝑡𝑡 − 6 𝑑𝑑𝜏𝜏 𝜏𝜏 = 𝑡𝑡 2 𝑦𝑦 𝑡𝑡 = 16𝑡𝑡 − 𝑡𝑡 2 − 55 𝜏𝜏 𝜏𝜏 = 𝑡𝑡 − 6 𝜏𝜏 = 2  Using Convolution Table 𝑥𝑥 𝑡𝑡 = 𝑢𝑢 𝑡𝑡 − 2 ℎ 𝑡𝑡 = 12 − 2𝑡𝑡 𝑢𝑢 𝑡𝑡 − 3 − 𝑢𝑢 𝑡𝑡 − 6 ∞ 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝜏𝜏 ℎ 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝜏𝜏 = 𝑥𝑥 𝑡𝑡 ∗ ℎ 𝑡𝑡 −∞ 𝑦𝑦 𝑡𝑡 = 2𝑢𝑢 𝑡𝑡 − 2 ∗ 6 − 𝑡𝑡 𝑢𝑢 𝑡𝑡 − 3 − 2𝑢𝑢 𝑡𝑡 − 2 ∗ 6 − 𝑡𝑡 𝑢𝑢 𝑡𝑡 − 6 𝑥𝑥1 𝑡𝑡 − 𝑇𝑇1 ∗ 𝑥𝑥2 𝑡𝑡 − 𝑇𝑇2 = 𝑐𝑐 𝑡𝑡 − 𝑇𝑇1 − 𝑇𝑇2 1 2 𝑢𝑢 𝑡𝑡 ∗ 𝑡𝑡𝑢𝑢 𝑡𝑡 = 𝑡𝑡 𝑢𝑢 𝑡𝑡 2 2𝑢𝑢 𝑡𝑡 − 2 ∗ 6 − 𝑡𝑡 𝑢𝑢 𝑡𝑡 − 6 = − 𝑡𝑡 − 8 2 𝑢𝑢 𝑡𝑡 − 8 2𝑢𝑢 𝑡𝑡 − 2 ∗ 6 − 𝑡𝑡 𝑢𝑢 𝑡𝑡 − 3 = 6 𝑡𝑡 − 5 𝑢𝑢 𝑡𝑡 − 5 − 𝑡𝑡 − 5 2 𝑢𝑢 𝑡𝑡 − 5 𝑦𝑦 𝑡𝑡 = 16𝑡𝑡 − 𝑡𝑡 2 − 55 𝑢𝑢 𝑡𝑡 − 5 + −16𝑡𝑡 + 𝑡𝑡 2 + 64 𝑢𝑢 𝑡𝑡 − 8  Reading Textbook (O&W): Chapter 2.4. Notes (Prof. Wyatt): Chapter 5 (TLO 5) After reading, solve on your own, examples given in the textbook, during lectures, and in Prof. Wyatt’s note.  (CT) Differential and (DT) Difference Equations ∆𝑡𝑡 𝑑𝑑𝑑𝑑 𝑥𝑥 𝑛𝑛 − 𝑥𝑥 𝑛𝑛 − 1 𝑥𝑥 𝑡𝑡 𝑑𝑑𝑑𝑑 ∆𝑡𝑡 First-order differential equation: 𝑑𝑑𝑑𝑑 𝑡𝑡 = −2𝑥𝑥 𝑑𝑑𝑑𝑑 𝑛𝑛 − 1 ∆𝑡𝑡 𝑛𝑛∆𝑡𝑡 Replace differentiation First-order difference equation: with finite-difference 𝑥𝑥 𝑛𝑛 − 𝑥𝑥 𝑛𝑛 − 1 𝑥𝑥 𝑛𝑛 + 𝑥𝑥 𝑛𝑛 − 1 1 + ∆𝑡𝑡 𝑥𝑥 𝑛𝑛 + −1 + ∆𝑡𝑡 𝑥𝑥 𝑛𝑛 − 1 = 0 = −2 ∆𝑡𝑡 2  Linear Constant-Coefficient Difference Equation 𝑁𝑁 𝑀𝑀 𝑄𝑄[𝐷𝐷]𝑦𝑦 = 𝑃𝑃[𝐷𝐷]𝑥𝑥 Delay Form: 𝑎𝑎𝑘𝑘 𝑦𝑦[𝑛𝑛 − 𝑘𝑘] = 𝑏𝑏𝑘𝑘 𝑥𝑥[𝑛𝑛 − 𝑘𝑘] N: order of the system 𝑘𝑘=0 𝑘𝑘=0 Example: 2𝑦𝑦 𝑛𝑛 − 3𝑦𝑦 𝑛𝑛 − 1 + 7𝑦𝑦 𝑛𝑛 − 2 = 1.5𝑥𝑥 𝑛𝑛 − 2.5𝑥𝑥[𝑛𝑛 − 1] 𝑎𝑎2 = 7 𝑎𝑎1 = −3 𝑎𝑎0 = 2 𝑁𝑁 = 2 𝑦𝑦 𝑛𝑛 − 2 𝑦𝑦 𝑛𝑛 − 1 𝑦𝑦 𝑛𝑛 𝑛𝑛 − 3 𝑛𝑛 − 2 𝑛𝑛 − 1 𝑛𝑛 𝑥𝑥 𝑛𝑛 − 1 𝑥𝑥 𝑛𝑛 𝑏𝑏1 = −2.5 𝑏𝑏0 = 1.5 𝑁𝑁 𝑀𝑀 Advance Form (Replace n with 𝑎𝑎𝑘𝑘 𝑦𝑦[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] = 𝑏𝑏𝑘𝑘 𝑥𝑥[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] n+N): 𝑘𝑘=0 𝑘𝑘=0 𝑄𝑄[𝐸𝐸]𝑦𝑦 = 𝑃𝑃[𝐸𝐸]𝑥𝑥 2𝑦𝑦 𝑛𝑛 + 2 − 3𝑦𝑦 𝑛𝑛 + 1 + 7𝑦𝑦 𝑛𝑛 = 1.5𝑥𝑥 𝑛𝑛 + 2 − 2.5𝑥𝑥[𝑛𝑛 + 1]  Linear Constant-Coefficient Difference Equation (LCCDE) 𝑁𝑁 𝑀𝑀 Delay Form: 𝑎𝑎𝑘𝑘 𝑦𝑦[𝑛𝑛 − 𝑘𝑘] = 𝑏𝑏𝑘𝑘 𝑥𝑥[𝑛𝑛 − 𝑘𝑘] 𝑘𝑘=0 𝑘𝑘=0 𝑁𝑁 𝑀𝑀 Advance Form: 𝑎𝑎𝑘𝑘 𝑦𝑦[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] = 𝑏𝑏𝑘𝑘 𝑥𝑥[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] 𝑘𝑘=0 𝑘𝑘=0 Advance operator 𝐸𝐸: 𝐸𝐸𝐸𝐸 𝑛𝑛 = 𝑥𝑥 𝑛𝑛 + 1 𝐸𝐸 𝑚𝑚 𝑥𝑥 𝑛𝑛 = 𝑥𝑥 𝑛𝑛 + 𝑚𝑚 Delay operator D: 𝐷𝐷𝑥𝑥 𝑛𝑛 = 𝑥𝑥 𝑛𝑛 − 1 𝐷𝐷 𝑚𝑚 𝑥𝑥 𝑛𝑛 = 𝑥𝑥 𝑛𝑛 − 𝑚𝑚 Advance by 1 Delay by 1 1 1 1 0 𝑛𝑛 0 𝑛𝑛 0 𝑛𝑛 𝛿𝛿[𝑛𝑛] 𝐸𝐸𝐸𝐸 𝑛𝑛 = 𝛿𝛿 𝑛𝑛 + 1 𝐷𝐷𝐷𝐷 𝑛𝑛 = 𝛿𝛿 𝑛𝑛 − 1  Linear Constant-Coefficient Difference Equation (LCCDE) Delay Form: 2𝑦𝑦 𝑛𝑛 − 3𝑦𝑦 𝑛𝑛 − 1 + 7𝑦𝑦 𝑛𝑛 − 2 = 1.5𝑥𝑥 𝑛𝑛 − 2.5𝑥𝑥[𝑛𝑛 − 1] 2 − 3𝐷𝐷 + 7𝐷𝐷 2 𝑦𝑦 𝑛𝑛 = 1.5 − 2.5𝐷𝐷 𝑥𝑥 𝑛𝑛 𝑄𝑄[𝐷𝐷]𝑦𝑦 = 𝑃𝑃[𝐷𝐷]𝑥𝑥 𝑄𝑄[𝐷𝐷] = 2 − 3𝐷𝐷 + 7𝐷𝐷 2 𝑃𝑃[𝐷𝐷] = 1.5 − 2.5𝐷𝐷 Advance Form: 2𝑦𝑦 𝑛𝑛 + 2 − 3𝑦𝑦 𝑛𝑛 + 1 + 7𝑦𝑦 𝑛𝑛 = 1.5𝑥𝑥 𝑛𝑛 + 2 − 2.5𝑥𝑥[𝑛𝑛 + 1] 2𝐸𝐸 2 − 3𝐸𝐸 + 7 𝑦𝑦 𝑛𝑛 = 1.5𝐸𝐸 2 − 2.5𝐸𝐸 𝑥𝑥 𝑛𝑛 𝑄𝑄[𝐸𝐸]𝑦𝑦 = 𝑃𝑃[𝐸𝐸]𝑥𝑥 𝑄𝑄[𝐸𝐸] = 2𝐸𝐸 2 − 3𝐸𝐸 + 7 𝑃𝑃[𝐸𝐸] = 1.5𝐸𝐸 2 − 2.5E Notice that 𝑄𝑄[𝐷𝐷] and 𝑄𝑄[𝐸𝐸] (also 𝑃𝑃[𝐷𝐷] and 𝑃𝑃[𝐸𝐸]) are NOT the same polynomials.  Linear Constant-Coefficient Difference Equation 𝑁𝑁 𝑀𝑀 𝑎𝑎𝑘𝑘 𝑦𝑦[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] = 𝑏𝑏𝑘𝑘 𝑥𝑥[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] 𝑄𝑄[𝐸𝐸]𝑦𝑦 = 𝑃𝑃[𝐸𝐸]𝑥𝑥 𝑘𝑘=0 𝑘𝑘=0 I will often use advance form in mathematical calculations, even though advance operator (E) is not causal, and can be difficult to implement in practice. 𝑥𝑥 𝑛𝑛 𝑦𝑦 𝑛𝑛 𝑦𝑦 𝑛𝑛 = 𝐸𝐸𝐸𝐸 𝑛𝑛 = 𝑥𝑥 𝑛𝑛 + 1 E Thus the present output (@ n) depends on the future input (@ n+1) 𝑎𝑎0 𝑦𝑦 𝑛𝑛 + 𝑁𝑁 + 𝑎𝑎1 𝑦𝑦 𝑛𝑛 + 𝑁𝑁 − 1 + ⋯ + 𝑎𝑎𝑁𝑁−1 𝑦𝑦 𝑛𝑛 + 1 + 𝑎𝑎𝑁𝑁 𝑦𝑦 𝑛𝑛 = 𝑏𝑏𝑁𝑁−𝑀𝑀 𝑥𝑥 𝑛𝑛 + 𝑀𝑀 + 𝑏𝑏𝑁𝑁−𝑀𝑀+1 𝑥𝑥 𝑛𝑛 + 𝑀𝑀 − 1 + ⋯ + 𝑏𝑏𝑁𝑁−1 𝑦𝑦 𝑛𝑛 + 1 + 𝑏𝑏𝑁𝑁 𝑥𝑥 𝑛𝑛 Typically, we assume: 1) 𝑀𝑀 ≤ 𝑁𝑁 (otherwise non-causal), and 2) 𝑎𝑎0 = 1 (for convenience). If 𝑀𝑀 < 𝑁𝑁, we can still “increase” M to N, while selecting the corresponding coefficients for b to be zero. For example, 𝑏𝑏𝑁𝑁−𝑀𝑀 = 𝑏𝑏0 = 0. Thus in general, we choose M=N.  Linear Constant-Coefficient Difference Equation The general Nth order advance-form difference equation becomes: 𝑎𝑎0 𝑦𝑦 𝑛𝑛 + 𝑁𝑁 + 𝑎𝑎1 𝑦𝑦 𝑛𝑛 + 𝑁𝑁 − 1 + ⋯ + 𝑎𝑎𝑁𝑁−1 𝑦𝑦 𝑛𝑛 + 1 + 𝑎𝑎𝑁𝑁 𝑦𝑦 𝑛𝑛 = 𝑏𝑏0 𝑥𝑥 𝑛𝑛 + 𝑁𝑁 + 𝑏𝑏1 𝑥𝑥 𝑛𝑛 + 𝑁𝑁 − 1 + ⋯ + 𝑏𝑏𝑁𝑁−1 𝑦𝑦 𝑛𝑛 + 1 + 𝑏𝑏𝑁𝑁 𝑥𝑥 𝑛𝑛 𝑄𝑄 𝐸𝐸 𝑦𝑦 𝑛𝑛 = 𝑃𝑃 𝐸𝐸 𝑥𝑥 𝑛𝑛 𝑄𝑄 𝐸𝐸 = 𝑎𝑎0 𝐸𝐸 𝑁𝑁 + 𝑎𝑎1 𝐸𝐸 𝑁𝑁−1 + ⋯ + 𝑎𝑎𝑁𝑁−1 𝐸𝐸 + 𝑎𝑎𝑁𝑁 𝑃𝑃 𝐸𝐸 = 𝑏𝑏0 𝐸𝐸 𝑁𝑁 + 𝑏𝑏1 𝐸𝐸 𝑁𝑁−1 + ⋯ + 𝑏𝑏𝑁𝑁−1 𝐸𝐸 + 𝑏𝑏𝑁𝑁  Iterative Solution Example: 𝑦𝑦 𝑛𝑛 + 1 + 𝑦𝑦 𝑛𝑛 = 𝑥𝑥 𝑛𝑛 + 1 𝑦𝑦 −1 = 1 𝑥𝑥 𝑛𝑛 = 𝑢𝑢 𝑛𝑛 𝑦𝑦 𝑛𝑛 = −𝑦𝑦 𝑛𝑛 − 1 + 𝑥𝑥 𝑛𝑛 Step 1 Step 2 Step 3 𝑦𝑦 −1 = 1 𝑦𝑦 0 = 0 𝑦𝑦 1 = 1 𝑦𝑦 2 = 0 − − − 𝑦𝑦 𝑛𝑛 −1 0 1 2 + + + 𝑥𝑥 𝑛𝑛 −1 0 1 2 𝑥𝑥 −1 = 0 𝑥𝑥 0 = 1 𝑥𝑥 1 = 1 𝑥𝑥 2 = 1  DT Impulse Response Function Input x Output y ℎ[𝑛𝑛] 1 𝛿𝛿[𝑛𝑛] 0 𝑛𝑛 0 𝑛𝑛 𝑁𝑁 𝑀𝑀 𝑎𝑎𝑘𝑘 𝑦𝑦[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] = 𝑏𝑏𝑘𝑘 𝑥𝑥[𝑛𝑛 + 𝑁𝑁 − 𝑘𝑘] 𝑄𝑄 𝐸𝐸 ℎ 𝑛𝑛 = 𝑃𝑃 𝐸𝐸 𝛿𝛿 𝑛𝑛 𝑘𝑘=0 𝑘𝑘=0 How to determine the unit impulse response of a DT system? For n>0, 𝑥𝑥 𝑛𝑛. Thus we have a homogenous equation: 𝑄𝑄 𝐸𝐸 𝑦𝑦 𝑛𝑛 = 0 We can assume a trial solution to the homogenous equation of the form: 𝑦𝑦ℎ 𝑛𝑛 = 𝜆𝜆𝑛𝑛 𝜆𝜆 is a complex constant  Characteristics Equation of the DT System 𝑄𝑄 𝐸𝐸 𝑦𝑦ℎ 𝑛𝑛 = 0 𝑄𝑄 𝐸𝐸 = 𝑎𝑎0 𝐸𝐸 𝑁𝑁 + 𝑎𝑎1 𝐸𝐸 𝑁𝑁−1 + ⋯ + 𝑎𝑎𝑁𝑁−1 𝐸𝐸 + 𝑎𝑎𝑁𝑁 Assume: 𝑦𝑦 𝑛𝑛 = 𝜆𝜆𝑛𝑛 𝐸𝐸𝐸𝐸 𝑛𝑛 = 𝜆𝜆𝑦𝑦 𝑛𝑛 𝐸𝐸𝐸𝐸 𝑛𝑛 = 𝑦𝑦 𝑛𝑛 + 1 𝑎𝑎0 𝐸𝐸 𝑁𝑁 + 𝑎𝑎1 𝐸𝐸 𝑁𝑁−1 + ⋯ + 𝑎𝑎𝑁𝑁−1 𝐸𝐸 + 𝑎𝑎𝑁𝑁 𝑦𝑦ℎ 𝑛𝑛 = 0 𝑎𝑎0 𝜆𝜆𝑁𝑁 + 𝑎𝑎1 𝜆𝜆𝑁𝑁−1 + ⋯ + 𝑎𝑎𝑁𝑁−1 𝜆𝜆 + 𝑎𝑎𝑁𝑁 𝑦𝑦ℎ 𝑛𝑛 = 0 𝑎𝑎0 𝜆𝜆𝑁𝑁 + 𝑎𝑎1 𝜆𝜆𝑁𝑁−1 + ⋯ + 𝑎𝑎𝑁𝑁−1 𝜆𝜆 + 𝑎𝑎𝑁𝑁 = 0 Such an Nth order polynomial equation has N roots. The N roots may be distinct, or some may repeat. The roots can be real of complex.  Characteristics Equation of the DT System 𝑄𝑄 𝐸𝐸 𝑦𝑦ℎ 𝑛𝑛 = 0 Assume: 𝑦𝑦ℎ 𝑛𝑛 = 𝜆𝜆𝑛𝑛 𝑄𝑄 𝜆𝜆 = 0 𝜆𝜆1 ≠ ⋯ ≠ 𝜆𝜆𝑖𝑖 ≠ ⋯ ≠ 𝜆𝜆𝑁𝑁 All roots are distinct: 𝑄𝑄 𝜆𝜆 = 𝜆𝜆 − 𝜆𝜆1 ⋯ 𝜆𝜆 − 𝜆𝜆𝑖𝑖 ⋯ 𝜆𝜆 − 𝜆𝜆𝑁𝑁 = 0 For n>0: 𝑦𝑦ℎ 𝑛𝑛 = 𝑐𝑐1 𝜆𝜆1𝑛𝑛 + ⋯ + 𝑐𝑐𝑖𝑖 𝜆𝜆𝑛𝑛𝑖𝑖 + ⋯ + 𝑐𝑐𝑁𝑁 𝜆𝜆𝑛𝑛𝑁𝑁 𝑐𝑐𝑖𝑖 , 𝑖𝑖 = 1,2, ⋯ , 𝑁𝑁 are N independent (real or complex) constants. Suppose the 1st root repeat r times. Some roots repeat: 𝑟𝑟 𝑄𝑄 𝜆𝜆 = 𝜆𝜆 − 𝜆𝜆1 𝜆𝜆 − 𝜆𝜆𝑟𝑟+1 ⋯ 𝜆𝜆 − 𝜆𝜆𝑁𝑁 = 0 For n>0: 𝑦𝑦ℎ 𝑛𝑛 = 𝑐𝑐1 + 𝑐𝑐2 𝑛𝑛 + ⋯ + 𝑐𝑐𝑟𝑟 𝑛𝑛𝑟𝑟−1 𝜆𝜆1𝑛𝑛 + 𝑐𝑐𝑟𝑟+1 𝜆𝜆𝑛𝑛𝑟𝑟+1 + ⋯ + 𝑐𝑐𝑁𝑁 𝜆𝜆𝑛𝑛𝑁𝑁 𝑐𝑐𝑖𝑖 , 𝑖𝑖 = 1,2, ⋯ , 𝑁𝑁 are N independent (real or complex) constants.  DT Impulse Response Function For n>0, we have: ℎ[𝑛𝑛] = 𝑦𝑦ℎ 𝑛𝑛 𝑄𝑄 𝐸𝐸 𝑦𝑦ℎ 𝑛𝑛 = 0 For causal system and n 1 0 1 𝑡𝑡 Impulse Response ℎ(𝑡𝑡) 1 0 < 𝑡𝑡 < 2 ℎ 𝑡𝑡 = 0 𝑡𝑡 < 0; 𝑡𝑡 > 2 1 ∞ 𝑡𝑡 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝜏𝜏 ℎ 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝜏𝜏 0 2 −∞  Example Input 𝑥𝑥 𝑡𝑡 ∞ 1 𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝜏𝜏 ℎ 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝜏𝜏 −∞ 𝑡𝑡 0 1 𝑡𝑡 0 < 𝑡𝑡 < 1 𝑦𝑦 𝑡𝑡 = 𝑑𝑑𝜏𝜏 = 𝑡𝑡 0 1 1 < 𝑡𝑡 < 2 𝑦𝑦 𝑡𝑡 = 𝑑𝑑𝜏𝜏 = 1 Impulse Response ℎ(𝑡𝑡) 0 1 2 < 𝑡𝑡 < 3 𝑦𝑦 𝑡𝑡 = 𝑑𝑑𝜏𝜏 = 3 − 𝑡𝑡 𝑡𝑡−2 1 𝑡𝑡 𝑡𝑡 1 0 2 0 1 2 3  Reading Textbook (O&W): Chapter 1.1, 1.2, 1.3, and 1.4 Notes (Prof. Wyatt): Chapter 2 (TLO 2), Chapter 3 (TLO 3). Notes (Prof. Wyatt): Work on Chapter 2.7 (solved problems), problem 1, 2, and 3  Step Function (CT) In circuits, we use unit step-function to describe the action of opening or closing a switch. 1, 𝑡𝑡 ≥ 0 𝑢𝑢 𝑡𝑡 = 0, 𝑡𝑡 < 0  Unit Impulse Function (CT) Intuitively, it is often helpful to think 𝛿𝛿 𝑡𝑡 as: 𝛿𝛿 𝑡𝑡 = 0, if 𝑡𝑡 ≠ 0 0, 𝑡𝑡 ≠ 0 +∞ 𝛿𝛿 𝑡𝑡 = 𝛿𝛿 𝑡𝑡 𝑑𝑑𝑑𝑑 = 1 +∞, 𝑡𝑡 = 0 −∞ 𝛿𝛿 𝑡𝑡 1 𝜀𝜀 𝑡𝑡 When 𝜀𝜀 → 0, the “short” 𝜀𝜀 𝜀𝜀 𝑡𝑡 pulse becomes the unit − 2 2 impulse function Unit impulse function A short pulse with time duration 𝜀𝜀  Unit Impulse Function Properties F(t) is an arbitrary function of t. 𝐹𝐹 𝑡𝑡 𝛿𝛿 𝑡𝑡 − 𝑇𝑇 = 𝐹𝐹 𝑇𝑇 𝛿𝛿 𝑡𝑡 − 𝑇𝑇 𝐹𝐹 𝑡𝑡 𝛿𝛿 𝑡𝑡 = 𝐹𝐹 0 𝛿𝛿 𝑡𝑡 T is any real constant. ∞ ∞ 𝐹𝐹 𝑡𝑡 𝛿𝛿 𝑡𝑡 𝑑𝑑𝑑𝑑 = 𝐹𝐹 0 𝐹𝐹 𝑡𝑡 𝛿𝛿 𝑡𝑡 − 𝑇𝑇 𝑑𝑑𝑑𝑑 = 𝐹𝐹 𝑇𝑇 −∞ −∞ 𝑏𝑏 If we have 𝑎𝑎 < 0 & 𝑏𝑏 > 0 𝐹𝐹 𝑡𝑡 𝛿𝛿 𝑡𝑡 𝑑𝑑𝑑𝑑 = 𝐹𝐹 0 𝑎𝑎 𝑏𝑏 If we don’t have a < 0 < 𝑏𝑏 𝐹𝐹 𝑡𝑡 𝛿𝛿 𝑡𝑡 𝑑𝑑𝑑𝑑 = 0 𝑎𝑎 Unit Impulse Function and Unit Step Function 𝑢𝑢 𝑡𝑡 𝛿𝛿 𝑡𝑡 1 𝑡𝑡 𝑡𝑡 𝑡𝑡 We can show that the left-hand side 𝑢𝑢 𝑡𝑡 = 𝛿𝛿 𝜏𝜏 𝑑𝑑𝑑𝑑 and the right-hand side take the same −∞ value for any t. 𝑑𝑑𝑑𝑑 𝑡𝑡 𝛿𝛿 𝑡𝑡 = 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑡𝑡 − 𝑇𝑇 𝛿𝛿 𝑡𝑡 − 𝑇𝑇 = 𝑑𝑑𝑑𝑑  Exponential Functions (CT) 𝑎𝑎𝑎𝑎 In the most general case, both coefficients C 𝑥𝑥 𝑡𝑡 = 𝐶𝐶𝑒𝑒 and can be complex numbers. Special case 1: a is a real number. Examples: 𝑥𝑥 𝑡𝑡 = 3𝑒𝑒 2𝑡𝑡 𝑥𝑥 𝑡𝑡 = 4𝑒𝑒 −3𝑡𝑡 Special case 2: a is purely imaginary: 𝑎𝑎 = 𝑗𝑗𝜔𝜔 Examples: 𝑥𝑥 𝑡𝑡 = 3𝑒𝑒 𝑗𝑗𝜔𝜔𝑡𝑡 1 𝑗𝑗𝜔𝜔𝑡𝑡 cos 𝜔𝜔𝑡𝑡 = 𝑒𝑒 + 𝑒𝑒 −𝑗𝑗𝜔𝜔𝑡𝑡 2 Useful relations: 1 𝑗𝑗𝜔𝜔𝑡𝑡 sin 𝜔𝜔𝑡𝑡 = 𝑒𝑒 − 𝑒𝑒 −𝑗𝑗𝜔𝜔𝑡𝑡 2𝑗𝑗  Magnitude Scaling In the most general case, a can be 𝑥𝑥 𝑡𝑡 𝑎𝑎𝑎𝑎 𝑡𝑡 any complex constant. 𝑢𝑢 𝑡𝑡 1 𝑢𝑢 𝑡𝑡 𝑡𝑡 Example: 3 3𝑢𝑢 𝑡𝑡 3𝑢𝑢 𝑡𝑡 𝑡𝑡  Time Shifting 𝜏𝜏 > 0 delay 𝑥𝑥 𝑡𝑡 𝑥𝑥 𝑡𝑡 + 𝜏𝜏 𝜏𝜏 < 0 advance 𝑢𝑢 𝑡𝑡 𝑢𝑢 𝑡𝑡 1 Example: 𝑡𝑡 1 𝑢𝑢 𝑡𝑡 − 2 𝑢𝑢 𝑡𝑡 − 2 2 𝑡𝑡  Time Scaling 𝑡𝑡 𝜏𝜏 is typically a real number and 𝑥𝑥 𝑡𝑡 𝑥𝑥 𝜏𝜏 can be either positive or negative Example: 𝜏𝜏 = 2 𝑥𝑥 𝑡𝑡 𝑥𝑥 𝑡𝑡 2 1 1 𝑡𝑡 𝑡𝑡 −2 −1 𝑥𝑥 −𝑡𝑡 𝑡𝑡 1  Examples of CT Signals We can form more complex CT signals based on linear superpositions and transformations (e.g., scaling, multiplication) of basic CT signals such as unit step function, polynomials, exponential functions, etc. 6 𝑥𝑥 𝑡𝑡 −1 2 𝑡𝑡 𝑥𝑥 𝑡𝑡 = 6 𝑢𝑢 𝑡𝑡 + 1 − 𝑢𝑢 𝑡𝑡 − 2  Examples of CT Signals 6 𝑥𝑥 𝑡𝑡 −1 2 𝑡𝑡 𝑥𝑥 𝑡𝑡 = 𝑓𝑓 𝑡𝑡 𝑢𝑢 𝑡𝑡 + 1 − 𝑢𝑢 𝑡𝑡 − 2 (-1,6) 𝑓𝑓 𝑡𝑡 = −2𝑡𝑡 + 4 (2,0) 𝑥𝑥 𝑡𝑡 = −2𝑡𝑡 + 4 𝑢𝑢 𝑡𝑡 + 1 − 𝑢𝑢 𝑡𝑡 − 2  Signal Classifications Even signals: 𝑥𝑥 𝑡𝑡 = 𝑥𝑥 −𝑡𝑡 Odd signal: 𝑥𝑥 𝑡𝑡 = −𝑥𝑥 −𝑡𝑡 Note: unit impulse function 𝛿𝛿 𝑡𝑡 is an even function: 𝛿𝛿 𝑡𝑡 = 𝛿𝛿 −𝑡𝑡 Periodic signals: 𝑥𝑥 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 + 𝑇𝑇 Period: T Aperiodic signals: a signal that is not periodic  Periodic Signals (CT) Fundamental Period 𝑇𝑇0 : the smallest positive value of T such that 𝑥𝑥 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 + 𝑇𝑇 2𝜋𝜋 Fundamental (angular) frequency 𝜔𝜔0 = 𝑇𝑇 Example 𝑥𝑥 𝑡𝑡 = cos 2𝜋𝜋𝜋𝜋 + 3 𝑥𝑥 𝑡𝑡 = 𝑅𝑅𝑅𝑅 𝑒𝑒 𝑗𝑗 2𝜋𝜋𝜋𝜋+3 𝑒𝑒 𝑗𝑗 2𝜋𝜋𝜋𝜋+3 = 𝑒𝑒 𝑗𝑗 2𝜋𝜋𝜋𝜋+2𝜋𝜋𝜋𝜋+3 1 = 𝑒𝑒 𝑗𝑗2𝜋𝜋𝜋𝜋 2𝜋𝜋𝜋𝜋 = 0, 2𝜋𝜋, 4𝜋𝜋, ⋯ Period T: 𝑇𝑇 = 1, 2, 3, ⋯ Fundamental Period: 𝑇𝑇0 = 1 Fundamental frequency: 𝜔𝜔0 = 2𝜋𝜋  Periodic Signals (CT) What if two signals have different periods, and we add / multiply them together? If 𝑥𝑥1 𝑡𝑡 is periodic with period 𝑇𝑇1 and 𝑥𝑥2 𝑡𝑡 is periodic with period 𝑇𝑇2 , and if there exists positive integers m and n such that: 𝑚𝑚𝑇𝑇1 = 𝑛𝑛𝑇𝑇2 = 𝑃𝑃 Then 𝑥𝑥1 𝑡𝑡 + 𝑥𝑥2 𝑡𝑡 and 𝑥𝑥1 𝑡𝑡 𝑥𝑥2 𝑡𝑡 are periodic signals with period P.  Periodic Signals (CT) Exist positive integers m and n such that: 𝑚𝑚𝑇𝑇1 = 𝑛𝑛𝑇𝑇2 = 𝑃𝑃 𝑇𝑇1 = 2 𝑡𝑡 𝑇𝑇2 = 3 We can choose m=3 and n=2, such that: P=6 𝑡𝑡 𝑥𝑥1 𝑡𝑡 = cos 2𝜋𝜋 The period of signals such as 2 𝑥𝑥1 𝑡𝑡 + 𝑥𝑥2 𝑡𝑡 and 𝑥𝑥1 𝑡𝑡 𝑥𝑥2 𝑡𝑡 𝑡𝑡 should be P=6 𝑥𝑥2 𝑡𝑡 = sin 2𝜋𝜋 3 However, cos 2𝜋𝜋𝜋𝜋 + sin 5𝑡𝑡 is NOT periodic.  Signal Energy and Power The energy of a CT signal 𝑥𝑥 𝑡𝑡 is: Alternative notations: 𝑇𝑇 𝐸𝐸∞ and 𝑃𝑃∞ 𝐸𝐸𝑥𝑥 = lim 𝑥𝑥 𝑡𝑡 2 𝑑𝑑𝑑𝑑 𝑇𝑇→∞ −𝑇𝑇 Example: proportional to the total energy delivered to a load resistor whose voltage signal (with limited duration) is 𝑥𝑥 𝑡𝑡. The power of a CT signal 𝑥𝑥 𝑡𝑡 is: 1 𝑇𝑇 2 𝑑𝑑𝑑𝑑 𝑃𝑃𝑥𝑥 = lim 𝑥𝑥 𝑡𝑡 𝑇𝑇→∞ 2𝑇𝑇 −𝑇𝑇 Example: the time-averaged power consumed by a load resistor with AC voltage signal of the form 𝑥𝑥 𝑡𝑡 = 2 cos 𝜔𝜔𝜔𝜔. Read textbook section 1.1.2.  Signal Energy and Power 𝑇𝑇 2 𝑑𝑑𝑑𝑑 1 𝑇𝑇 2 𝑑𝑑𝑑𝑑 𝐸𝐸𝑥𝑥 = lim 𝑥𝑥 𝑡𝑡 𝑃𝑃𝑥𝑥 = lim 𝑥𝑥 𝑡𝑡 𝑇𝑇→∞ −𝑇𝑇 𝑇𝑇→∞ 2𝑇𝑇 −𝑇𝑇 Energy signal: A signal with finite 𝐸𝐸𝑥𝑥. As a result, 𝑃𝑃𝑥𝑥 → 0. Power signal: A signal with finite 𝑃𝑃𝑥𝑥. As a result, 𝐸𝐸𝑥𝑥 → ∞. Neither: Both 𝐸𝐸𝑥𝑥 and 𝑃𝑃𝑥𝑥 are infinite. Example 𝑥𝑥 𝑡𝑡 = 𝑡𝑡.  Discrete-Time (DT) Unit Step Signal We can generate DT signals by “sampling” continuous-time (CT) signals at regular time steps 𝑛𝑛Δ𝑡𝑡 𝑢𝑢 𝑡𝑡 1 𝑡𝑡 𝑢𝑢[𝑛𝑛] 1 1, 𝑛𝑛 ≥ 0 𝑢𝑢 𝑛𝑛 = 0, 𝑛𝑛 < 0 1 𝑛𝑛 0  DT Unit Impulse Signal 𝛿𝛿 𝑡𝑡 0, 𝑡𝑡 ≠ 0 𝛿𝛿 𝑡𝑡 = +∞, 𝑡𝑡 = 0 0+ 𝑡𝑡 𝛿𝛿 𝑡𝑡 𝑑𝑑𝑑𝑑 = 1 0− 𝛿𝛿[𝑛𝑛] 1 1, 𝑛𝑛 = 0 𝛿𝛿 𝑛𝑛 = 0, 𝑛𝑛 ≠ 0 1 𝑛𝑛 0 We can also refer to 𝛿𝛿 𝑛𝑛 as unit sample.  DT Unit Step vs Unit Impulse 𝑢𝑢[𝑛𝑛] 𝑢𝑢[𝑛𝑛 − 1] 1 minus 1 1 𝑛𝑛 1 𝑛𝑛 0 0 First-order difference 𝛿𝛿 𝑛𝑛 = 𝑢𝑢 𝑛𝑛 − 𝑢𝑢[𝑛𝑛 − 1] 𝑛𝑛 Running sum: 𝑢𝑢 𝑛𝑛 = 𝛿𝛿[𝑚𝑚] 𝛿𝛿[𝑛𝑛] 𝑚𝑚=−∞ 1 𝑢𝑢 𝑛𝑛 = ⋯ + 𝛿𝛿 −3 + 𝛿𝛿 −2 = 0 𝑛𝑛=−2 1 𝑛𝑛 𝑢𝑢 𝑛𝑛 = ⋯ + 𝛿𝛿 −1 + 𝛿𝛿 0 + 𝛿𝛿 1 = 1 𝑛𝑛=1 0  Differentiation (CT) vs Difference (DT) ∆𝑡𝑡 𝑑𝑑𝑑𝑑 The value of first- order derivative in the time-interval 𝑥𝑥 𝑡𝑡 𝑑𝑑𝑑𝑑 𝑛𝑛 − 1 ∆𝑡𝑡 ≤ 𝑡𝑡 ≤ 𝑛𝑛∆𝑡𝑡 𝑡𝑡 𝑥𝑥 𝑛𝑛 − 𝑥𝑥 𝑛𝑛 − 1 𝑛𝑛 − 1 ∆𝑡𝑡 𝑛𝑛∆𝑡𝑡 ∆𝑡𝑡 In many situations, first-order differentiation of CT signals corresponds to first-order finite-difference of DT signals  Integration vs Running Sum Integration of CT signals ∆𝑡𝑡 𝑛𝑛∆𝑡𝑡 𝑥𝑥 𝑡𝑡 𝑥𝑥 𝑡𝑡 𝑑𝑑𝑑𝑑 −∞ 𝑡𝑡 𝑛𝑛 𝑛𝑛∆𝑡𝑡 𝑥𝑥[𝑚𝑚]∆𝑡𝑡 𝑚𝑚 − 1 ∆𝑡𝑡 𝑛𝑛 − 1 ∆𝑡𝑡 𝑚𝑚=−∞ 𝑚𝑚∆𝑡𝑡 Running sum of DT signals 𝑛𝑛 𝑥𝑥[𝑚𝑚]∆𝑡𝑡 = ⋯ + 𝑥𝑥 𝑚𝑚∆𝑡𝑡 ∆𝑡𝑡 + ⋯ + 𝑥𝑥 𝑛𝑛∆𝑡𝑡 ∆𝑡𝑡 𝑚𝑚=−∞  DT Signal Energy and Power 𝑇𝑇 𝐸𝐸𝑥𝑥 = lim 𝑥𝑥 𝑡𝑡 2 𝑑𝑑𝑑𝑑 𝑇𝑇→∞ −𝑇𝑇 1 𝑇𝑇 CT Signals 𝑃𝑃𝑥𝑥 = lim 𝑥𝑥 𝑡𝑡 2 𝑑𝑑𝑑𝑑 𝑇𝑇→∞ 2𝑇𝑇 −𝑇𝑇 For DT Signals, we have: 𝑁𝑁 𝐸𝐸𝑥𝑥 = lim 𝑥𝑥 𝑛𝑛 2 Sum n from 𝑛𝑛 = −𝑁𝑁, to 𝑛𝑛 = 𝑁𝑁. 𝑁𝑁→∞ −𝑁𝑁 𝑁𝑁 1 2 𝑃𝑃𝑥𝑥 = lim 𝑥𝑥 𝑛𝑛 𝑁𝑁→∞ 2𝑁𝑁 + 1 −𝑁𝑁  Exponential Functions (DT) 𝛽𝛽𝛽𝛽 In the most general case, both coefficients C 𝑥𝑥 𝑛𝑛 = 𝐶𝐶𝑒𝑒 and 𝛽𝛽 can be complex numbers. Sometimes, we can express it alternatively as: 𝛽𝛽 𝑁𝑁 with 𝛼𝛼 = 𝑒𝑒 𝛽𝛽 𝑥𝑥 𝑛𝑛 = 𝐶𝐶 𝑒𝑒 = 𝐶𝐶𝛼𝛼 𝑁𝑁 An important example: 𝑥𝑥 𝑛𝑛 = 𝑒𝑒 𝑗𝑗𝜔𝜔0 𝑛𝑛 𝐶𝐶 = 1 𝛽𝛽 = 𝑗𝑗𝜔𝜔0 We notice that 𝑥𝑥 𝑛𝑛 = 𝑒𝑒 𝑗𝑗𝜔𝜔0 𝑛𝑛 and 𝑥𝑥 𝑛𝑛 = 𝑒𝑒 𝑗𝑗 𝜔𝜔0 +2𝜋𝜋 𝑛𝑛 are identical DT signals.  Periodic Signals (DT) A DT signal 𝑥𝑥 𝑛𝑛 is periodic with period N, where N is a positive integer, if: 𝑥𝑥 𝑛𝑛 = 𝑥𝑥 𝑛𝑛 + 𝑁𝑁 for any n values. Fundamental period 𝑁𝑁0 : the smallest period. Fundamental frequency: 2𝜋𝜋⁄𝑁𝑁0. Example: 𝑥𝑥 𝑛𝑛 = 3𝑒𝑒 𝑗𝑗3𝜋𝜋 𝑛𝑛+0.5 ⁄5 3𝜋𝜋 𝑗𝑗 𝑁𝑁 3𝑒𝑒 𝑗𝑗3𝜋𝜋 𝑛𝑛+0.5 ⁄5 = 3𝑒𝑒 𝑗𝑗3𝜋𝜋 𝑛𝑛+𝑁𝑁+0.5 ⁄5 1 = 𝑒𝑒 5 3𝜋𝜋 10 N = 0, 2𝜋𝜋, 4𝜋𝜋, ⋯ , 2𝑀𝑀𝑀𝑀, ⋯ N= 𝑀𝑀 5 3 N = 10, 20, 30, ⋯ 𝑁𝑁0 = 10  Periodic Signals (DT) Let us consider a DT exponential signal x 𝑛𝑛 = 𝑒𝑒 𝑗𝑗𝜔𝜔0𝑛𝑛. We want to answer the following questions: under what conditions is x 𝑛𝑛 periodic / aperiodic? 𝑥𝑥 𝑛𝑛 = 𝑥𝑥 𝑛𝑛 + 𝑁𝑁 𝑒𝑒 𝑗𝑗𝜔𝜔0𝑛𝑛 = 𝑒𝑒 𝑗𝑗𝜔𝜔0 𝑛𝑛+𝑁𝑁 1 = 𝑒𝑒 𝑗𝑗𝜔𝜔0𝑁𝑁 2𝜋𝜋 𝑁𝑁 = 𝑚𝑚 𝜔𝜔0 𝑁𝑁 = 2𝜋𝜋𝜋𝜋 𝑚𝑚 = 1,2,3, ⋯ 𝜔𝜔0 Once we have determined the value of 𝜔𝜔0 : If there is a pair of integers If 𝜔𝜔0 is an irrational number 𝜔𝜔 𝑚𝑚 2𝜋𝜋 such that 0 = (i.e., rational) 2𝜋𝜋 𝑁𝑁 x 𝑛𝑛 is periodic x 𝑛𝑛 is aperiodic  Examples 2𝜋𝜋 𝑗𝑗 𝑛𝑛 𝑥𝑥 𝑛𝑛 = 𝑒𝑒 3 Periodic signal with fundamental period 𝑁𝑁0 = 3 3𝜋𝜋 𝑥𝑥 𝑛𝑛 = 𝑒𝑒 𝑗𝑗 4 𝑛𝑛 Periodic signal with fundamental period 𝑁𝑁0 = 8 𝑥𝑥 𝑛𝑛 = 𝑒𝑒 𝑗𝑗3𝑛𝑛 Aperiodic signal 2𝜋𝜋 3𝜋𝜋 𝑥𝑥 𝑛𝑛 = 𝑒𝑒 𝑗𝑗 3 𝑛𝑛 + 𝑒𝑒 𝑗𝑗 4 𝑛𝑛 Periodic signal with fundamental period 𝑁𝑁0 = 24  Review of Complex Algebra a = ar + jai b = br + jbi Suppose we have two arbitrary complex a + b = [ar + br ] + j[ai + bi ] numbers a and b. a − b = [ar − br ] + j[ai − bi ] 𝑎𝑎∗ = 𝑎𝑎𝑟𝑟 − 𝑗𝑗𝑎𝑎𝑖𝑖 a ⋅ b = [ar br − ai bi ] + j[ai br + ar bi ] Complex conjugate a a ⋅ b* (ar + jai ) ⋅ (br − jbi ) [ar br + ai bi ] + j[ai br − ar bi ] = = = b b⋅b * (br + jbi ) ⋅ (br − jbi ) br2 + bi2 Please review and refresh your knowledge of complex algebra! Two complex numbers equal to 𝑎𝑎𝑟𝑟 = 𝑏𝑏𝑟𝑟 𝑅𝑅𝑅𝑅(𝑎𝑎) = 𝑅𝑅𝑅𝑅(𝑏𝑏) each other means that real part 𝑎𝑎 = 𝑏𝑏 𝑎𝑎𝑖𝑖 = 𝑏𝑏𝑖𝑖 equals to real part, imaginary 𝐼𝐼𝐼𝐼(𝑎𝑎) = 𝐼𝐼𝐼𝐼(𝑏𝑏) part equals to imaginary part.  Euler’s Formula For any real number 𝜑𝜑 (angle e jϕ = cos ϕ + j sin ϕ Im in radians) a = ar + jai = a e jϕ a ai = a sin ϕ Amplitude / a = ar2 + ai2 magnitude: ϕ ar = a cos ϕ Re −1  ai  Phase: ϕ = tan    ar  How to represent a complex number on a complex plane. You must have an intuitive understanding of this one-to-one correspondence. Also, notice that for real number c and d, we have 𝑒𝑒 𝑐𝑐+𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑐𝑐 𝑒𝑒 𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑐𝑐 (cos 𝑑𝑑 + 𝑗𝑗 sin 𝑑𝑑)  Complex Numbers and Phasors In circuit theory, we use phasor to represent complex numbers that denote the amplitude and phase of an AC voltage / current signal. Mathematically, you can simply treat any phasor as a complex number in polar form. Rectangular form: 𝑧𝑧 = 𝑥𝑥 + 𝑗𝑗𝑗𝑗 Polar form: 𝑧𝑧 = 𝑟𝑟∠𝜙𝜙 Exponential form: 𝑧𝑧 = 𝑟𝑟𝑒𝑒 𝑗𝑗𝜙𝜙 All three forms are equivalent.  Phasors and AC Signals Let us consider an AC voltage signal of the form: 𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝜔𝜔 + 𝜙𝜙 𝑉𝑉 = 𝑉𝑉𝑚𝑚 ∠𝜙𝜙 Time domain signal Phasor domain signal You should be used to the equivalence between the time-domain signal and the phasor domain signal. It happens in many sub-disciplines of ECE Generally speaking, all time-varying EM fields can be denoted in phasor form.  Addition of AC Signals Let us consider what will happen if we want to add two AC signals of the same frequency together. 𝑣𝑣1 𝑡𝑡 = 𝑉𝑉1 cos 𝜔𝜔𝜔𝜔 + 𝜙𝜙1 𝑉𝑉 1 = 𝑉𝑉1 ∠𝜙𝜙1 = 𝑉𝑉1 𝑒𝑒 𝑗𝑗𝜙𝜙1 𝑒𝑒 𝑗𝑗𝜔𝜔𝑡𝑡 𝑣𝑣2 𝑡𝑡 = 𝑉𝑉2 cos 𝜔𝜔𝜔𝜔 + 𝜙𝜙2 𝑉𝑉 2 = 𝑉𝑉2 ∠𝜙𝜙2 = 𝑉𝑉1 𝑒𝑒 𝑗𝑗𝜙𝜙2 𝑒𝑒 𝑗𝑗𝑗𝑗𝑗𝑗 𝑣𝑣1 𝑡𝑡 + 𝑣𝑣2 𝑡𝑡 = 𝑅𝑅𝑅𝑅 𝑉𝑉1 𝑒𝑒 𝑗𝑗𝜙𝜙1 𝑒𝑒 𝑗𝑗𝑗𝑗𝑗𝑗 + 𝑉𝑉2 𝑒𝑒 𝑗𝑗𝜙𝜙2 𝑒𝑒 𝑗𝑗𝑗𝑗𝑗𝑗 𝑣𝑣1 𝑡𝑡 + 𝑣𝑣2 𝑡𝑡 = 𝑅𝑅𝑅𝑅 𝑉𝑉1 𝑒𝑒 𝑗𝑗𝜙𝜙1 + 𝑉𝑉2 𝑒𝑒 𝑗𝑗𝜙𝜙2 𝑒𝑒 𝑗𝑗𝑗𝑗𝑗𝑗 = 𝑅𝑅𝑅𝑅 𝑉𝑉𝑠𝑠 𝑒𝑒 𝑗𝑗𝜙𝜙𝑠𝑠 𝑒𝑒 𝑗𝑗𝑗𝑗𝑗𝑗 𝑉𝑉1 ∠𝜙𝜙1 + 𝑉𝑉2 ∠𝜙𝜙2 = 𝑉𝑉𝑠𝑠 ∠𝜙𝜙𝑠𝑠 𝑉𝑉𝑠𝑠 ∠𝜙𝜙𝑠𝑠 = 𝑉𝑉𝑆𝑆 cos 𝜔𝜔𝜔𝜔 + 𝜙𝜙𝑠𝑠  Phasor Operations 𝑧𝑧1 = 𝑟𝑟1 ∠𝜙𝜙1 = 𝑟𝑟1 𝑒𝑒 𝑗𝑗𝜙𝜙1 Multiplication: 𝑧𝑧2 = 𝑟𝑟2 ∠𝜙𝜙2 = 𝑟𝑟2 𝑒𝑒 𝑗𝑗𝜙𝜙2 𝑧𝑧1 𝑧𝑧2 = 𝑟𝑟1 𝑟𝑟2 ∠ 𝜙𝜙1 + 𝜙𝜙2 𝑧𝑧1 𝑧𝑧2 = 𝑟𝑟1 𝑟𝑟2 𝑒𝑒 𝑗𝑗 𝜙𝜙1 +𝜙𝜙2 Division: 𝑧𝑧1 𝑟𝑟1 𝑗𝑗 𝜙𝜙 −𝜙𝜙 Reciprocal: = 𝑒𝑒 1 2 1 1 −𝑗𝑗𝜙𝜙 𝑧𝑧2 𝑟𝑟2 = 𝑒𝑒 1 𝑧𝑧1 𝑟𝑟1 𝑧𝑧1 𝑟𝑟1 = ∠ 𝜙𝜙1 − 𝜙𝜙2 1 1 𝑧𝑧2 𝑟𝑟2 = ∠ −𝜙𝜙1 𝑧𝑧1 𝑟𝑟1  Phasor Operations 𝑧𝑧1 = 𝑟𝑟1 ∠𝜙𝜙1 = 𝑟𝑟1 𝑒𝑒 𝑗𝑗𝜙𝜙1 = 𝑥𝑥1 + 𝑗𝑗𝑦𝑦1 Complex Conjugation: 𝑧𝑧2 = 𝑟𝑟2 ∠𝜙𝜙2 = 𝑟𝑟2 𝑒𝑒 𝑗𝑗𝜙𝜙2 = 𝑥𝑥2 + 𝑗𝑗𝑦𝑦2 𝑧𝑧1 = 𝑥𝑥1 + 𝑗𝑗𝑦𝑦1 𝑧𝑧1 ∗ = 𝑥𝑥1 − 𝑗𝑗𝑦𝑦1 Square Root: 𝑧𝑧1 ∗ = 𝑟𝑟1 ∠−𝜙𝜙1 = 𝑟𝑟1 𝑒𝑒 −𝑗𝑗𝜙𝜙1 𝑗𝑗 𝜙𝜙1 ⁄2 𝑧𝑧1 = 𝑟𝑟1 𝑒𝑒 Addition and Subtraction: 𝑧𝑧1 + 𝑧𝑧2 = 𝑥𝑥1 + 𝑥𝑥2 + 𝑗𝑗 𝑦𝑦1 + 𝑦𝑦2 𝑧𝑧1 = 𝑟𝑟1 ∠ 𝜙𝜙1 ⁄2 𝑧𝑧1 − 𝑧𝑧2 = 𝑥𝑥1 − 𝑥𝑥2 + 𝑗𝑗 𝑦𝑦1 − 𝑦𝑦2  From Rectangular to Polar or Exponential Form Imaginary 𝑧𝑧 = 𝑟𝑟∠𝜙𝜙 𝑧𝑧 = 𝑥𝑥 + 𝑗𝑗𝑗𝑗 𝑗𝑗𝜙𝜙 II Quadrant I Quadrant 𝑧𝑧 = 𝑟𝑟𝑒𝑒 How to convert a phasor from the rectangular 𝑟𝑟 form to the polar or exponential form? 𝜙𝜙 𝑟𝑟 = 𝑥𝑥 2 + 𝑦𝑦 2 Real If the phasor is in the first or the fourth quadrant (use your judgement based on the rectangular form): III Quadrant IV Quadrant −1 ⁄ 𝜙𝜙 = tan 𝑦𝑦 𝑥𝑥 If the phasor is in the second or the third quadrant (use your judgement based on the rectangular form): 𝜙𝜙 = 𝜋𝜋 + tan−1 𝑦𝑦⁄𝑥𝑥 Notice that the tan−1 𝑢𝑢 function can only generate a phase angle from −𝜋𝜋 to 𝜋𝜋, which does not cover the full range of possible phase angles.  Examples 𝑧𝑧 = 1 + 𝑗𝑗 3 2 2 𝑟𝑟 = 12 + 3 = 2 3 60° 𝜙𝜙 = tan−1 3⁄1 = 1.047 𝑟𝑟𝑟𝑟𝑟𝑟 = 60° 1 𝑧𝑧 = 2 ∠1.047 = 2∠60° 𝑧𝑧 = 2 𝑒𝑒 𝑗𝑗1.047 = 2𝑒𝑒 𝑗𝑗𝑗𝑗𝑗  Examples 𝑧𝑧 = −1 + 𝑗𝑗 3 2 3 2 120° 𝑟𝑟 = 12 + 3 = 2 −1 tan−1 − 3⁄1 = −1.047 In this class, I will generally limit 𝜙𝜙 = −1.047 + π = 2.094 = 120° phase angle within the range of − 180° to 180°. Depending on how you calculate 𝑧𝑧 = 2 ∠2.094 = 2∠120° your phase angles, you may need to add or subtract 𝜋𝜋.  Additional Reading and Exercises Textbook (O & W): page 71 Textbook basic problems: page 57, P1.1 and 1.2.  Reading Textbook (O&W): Chapter 1 Notes (Prof. Wyatt): Chapter 2 and 3 (TLO2 and 3) Input-Output Description of System Models 𝑥𝑥 𝑡𝑡 𝑦𝑦 𝑡𝑡 Input Output System The input and output signals can be a variety of parameters such as voltage, current, mechanical force / torque, optical intensity, electric field intensity, etc. The system can contain a variety of components such as capacitors, inductors, diodes, transistors, mechanical beams, DC / AC motors, optical devices, cells (biological).  Circuits 𝑥𝑥 𝑡𝑡 System 𝑦𝑦 𝑡𝑡 Input Output Input 𝑖𝑖 Output + 𝑥𝑥 𝑡𝑡 + 𝑑𝑑𝑑𝑑 𝑣𝑣 𝑦𝑦 𝑡𝑡 − 𝑖𝑖 = 𝐶𝐶 𝑑𝑑𝑑𝑑 −  Communication Network 𝑥𝑥 𝑡𝑡 𝑦𝑦 𝑡𝑡 Input Output Communication System Alice Bob A communication system where Alice send Bob data (electrical, optical, etc) through a variety of communications networks (fiber optical network, wireless, etc).  Optical Systems 𝑥𝑥 𝑡𝑡 Optical systems 𝑦𝑦 𝑡𝑡 Input Output Optical and electrical Incident Light components Transmitted Light An optical fiber that allows optical signals in specific spectral range (e.g., near infra-red light) to pass through  Electromechanical Systems 𝑥𝑥 𝑡𝑡 𝑦𝑦 𝑡𝑡 Input Output Electromechanical Systems DC / AC Voltage Mechanical Torque The electromechanical system can contain components such as circuit components and AC / DC motors.  Linear vs Nonlinear Systems 𝑥𝑥 𝑡𝑡 𝑦𝑦 𝑡𝑡 Input Output System In linear system, its output 𝑦𝑦 𝑡𝑡 is proportional to its input x 𝑡𝑡. The linear system response is additive: 𝑥𝑥1 𝑡𝑡 𝑦𝑦1 𝑡𝑡 𝑥𝑥1 𝑡𝑡 + 𝑥𝑥2 𝑡𝑡 𝑦𝑦1 𝑡𝑡 + 𝑦𝑦2 𝑡𝑡 𝑥𝑥2 𝑡𝑡 𝑦𝑦2 𝑡𝑡 Linear scaling: 𝑥𝑥1 𝑡𝑡 𝑦𝑦1 𝑡𝑡 𝑐𝑐𝑐𝑐1 𝑡𝑡 𝑐𝑐𝑐𝑐1 𝑡𝑡 𝑐𝑐: any real or complex number (constant)  A Nonlinear System 𝑖𝑖𝐷𝐷 + − 𝑅𝑅 + 𝑣𝑣𝐷𝐷 𝑦𝑦 𝑡𝑡 𝑥𝑥 𝑡𝑡 − Any circuit containing diode(s) is in general nonlinear.  Time-Invariant vs Time-Varying Systems A system whose parameters (e.g., capacitor’s capacitance) remain constant is time-invariant. In time-invariant system, if input is 𝑥𝑥 𝑡𝑡 → 𝑦𝑦 𝑡𝑡 delayed by T, then system response is also delayed by T. 𝑥𝑥 𝑡𝑡 − 𝑇𝑇 → 𝑦𝑦 𝑡𝑡 − 𝑇𝑇 Input 𝑖𝑖 Output Linear time- + invariant system 𝑥𝑥 𝑡𝑡 + 𝑑𝑑𝑑𝑑 𝑣𝑣 𝑦𝑦 𝑡𝑡 is often denoted − 𝑖𝑖 = 𝐶𝐶 as LTI system. 𝑑𝑑𝑑𝑑 −  Instantaneous vs Dynamic Systems A system whose output at time t only depends on input at time t is called instantaneous or memoryless. Any circuits containing only resistors are instantaneous. A system is dynamic if its output response at time t depends on its past is called dynamic (or system with memory). Any circuits that contain components such as capacitors or inductors are dynamic.  Continuous Time (CT) Signals Mathematically speaking, we can simply define signals as a function y that maps elements from set A (domain) to elements in set B (co-domain). 𝑦𝑦: 𝐴𝐴 → 𝐵𝐵 Example 1 A is the set of real number that represent t (−∞ < 𝑡𝑡 < ∞) in unit of seconds. B is the set of real numbers that represent 𝑣𝑣 (−∞ < 𝑡𝑡 < ∞), in unit of volts. 𝑦𝑦 𝑡𝑡 = 3 cos 10𝑡𝑡 𝑉𝑉 Example 2 A represent time and B is the set of complex 20𝑡𝑡+𝜋𝜋⁄3 𝑦𝑦 𝑡𝑡 = 5𝑒𝑒 𝑗𝑗 𝐴𝐴 numbers representing phasor currents: Example 3 A corresponds spatial position z (unit m), and B corresponds to voltage / electric potential (unit V) at any given position. 𝑦𝑦 𝑧𝑧 = 2𝑧𝑧  Discrete Time (DT) Signals For discrete-time signals, you can think of them as 𝑦𝑦: 𝐴𝐴 → 𝐵𝐵 produced by “sampling” continuous time signals at specific time points. In this case, A is typically a set of integers, and B is the set of real or complex numbers. From Previous Example 1 𝑦𝑦 𝑡𝑡 = 3 cos 10𝑡𝑡 𝑉𝑉 Now imagine that we sample the voltage signal every 0.5 second (i.e., time interval is Δ𝑡𝑡 = 0.5 𝑠𝑠. We can then produce the following DT signal 𝑦𝑦[𝑛𝑛]: 𝑦𝑦 𝑛𝑛 = 3 cos 10𝑛𝑛Δ𝑡𝑡 𝑉𝑉 n ⋯ −𝟏𝟏 𝟎𝟎 𝟏𝟏 𝟐𝟐 𝟑𝟑 𝟒𝟒 ⋯ y[n] (V) ⋯ 0.85 3 0.85 -2.52 -2.28 1.22 ⋯  CT Signal Example 𝑦𝑦 𝑡𝑡 = 3 cos 10𝑡𝑡 𝑉𝑉  Matlab Script % an example of CT signal clear; N_array = 1000; t1 = -1.5; t2 = 2.5; t_array = linspace(t1,t2,N_array); y_array = 3*cos(10*t_array); plot(t_array,y_array,'LineWidth',2); grid on; xlabel('t (s)','FontSize',14); ylabel('y(t) (V)','FontSize',14); print('CT_signal_1','-dpng');  DT Signal Example 𝑦𝑦 𝑛𝑛 = 3 cos 10𝑛𝑛Δ𝑡𝑡 𝑉𝑉 Δ𝑡𝑡 = 0.5 𝑠𝑠 n ⋯ −𝟏𝟏 𝟎𝟎 𝟏𝟏 𝟐𝟐 𝟑𝟑 𝟒𝟒 ⋯ y[n] (V) ⋯ 0.85 3 0.85 -2.52 -2.28 1.22 ⋯  Matlab Script % an example of DT signal clear; delta_t = 0.5; n_array = -1:4; y_array = 3*cos(10*n_array*delta_t); stem(n_array,y_array,'LineWidth',2); grid on; xlabel('n','FontSize',14); ylabel('y(n) (V)','FontSize',14); print('DT_signal_1','-dpng'); Neural Network as a Nonlinear System This example illustrates how to use a neural network to correctly recognize an digital image that contains “3”. “How Deep Learning Works” IEEE Spectrum, page 32, Oct. 2021.  A Single Neuron 𝑦𝑦 𝑤𝑤0 𝑤𝑤1 𝑤𝑤𝐼𝐼 𝑥𝑥0 = 1 ⋯ 𝑥𝑥𝑖𝑖 IEEE Spectrum: How Deep Learning Works, page 32, 𝑥𝑥1 𝑥𝑥𝐼𝐼 Oct. 2021 Architecture of a single neuron: A number (𝐼𝐼) of inputs (𝑥𝑥𝑖𝑖 ) and one output (y). Each input (𝑥𝑥𝑖𝑖 ) is associated with a weight (𝑤𝑤𝑖𝑖 ). There may also be a bias (𝑤𝑤0 ) whose input (𝑥𝑥0 ) is always 1. The signal neuron is a feedforward device (input -> output). Neural Network and Attention Mechanisms B. Ghojogh and A. Ghodsi GPT (Generative Pre- trained Transformer)