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1.2 Continuous-Time Signal EECE 301 Signals & Systems Prof. Mark Fowler Note Set #2 • What are Continuous-Time Signals??? • Reading Assignment: Section 1.1 of Kamen and Heck 1/22 Course Flow Diagram The arrows here show conceptual flow between ideas. Note the parallel structure between the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis). Ch. 3: CT Fourier Ch. 5: CT Fourier Ch. 6 & 8: Laplace Signal Models System Models Models for CT New Signal Signals & Systems Models Fourier Series Frequency Response Transfer Function Periodic Signals Based on Fourier Transform Fourier Transform (CTFT) Ch. 1 Intro Non-Periodic Signals New System Model New System Model C-T Signal Model Functions on Real Line Ch. 2 Diff Eqs C-T System Model System Properties Differential Equations Ch. 2 Convolution LTI D-T Signal Model Causal Difference Equations New System Model C-T System Model Etc Zero-State Response Convolution Integral D-T Signal Model Zero-Input Response D-T System Model Functions on Integers Characteristic Eq. Convolution Sum Ch. 7: Z Trans. Ch. 4: DT Fourier Ch. 5: DT Fourier Models for DT New Signal Signal Models System Models Signals & Systems Model DTFT Freq. Response for DT Transfer Function (for “Hand” Analysis) Based on DTFT DFT & FFT Powerful New System (for Computer Analysis) New System Model Analysis Tool Model2/22 1.1 Continuous-Time Signal Our first math model for a signal will be a “function of time” Continuous Time (C-T) Signal: A C-T signal is defined on the continuum of time values. That is: f(t) for t ∈ℜ Real line f(t) t 3/22 Step & Ramp Functions These are common textbook signals but are also common test signals, especially in control systems. Unit Step Function u(t) u(t) 1⎧ ,t ≥ 0 u() t= ⎨ 1 . t 0⎩ ,t < 0 Note: A step of height A can be made from Au(t) 4/22 The unit step signal can model the act of switching on a DC source… t = 0 R + C Vs – R + C Vsu(t) – R + Vsu(t) C – 5/22 Unit Ramp Function r(t) Unit slope r(t) ⎧t, t≥ 0 1 . t r() t = ⎨ 0 ,t < 0 ⎩ 1 Note: ope A ramp with slm can be made from: mr(t) ⎧mt,≥ t 0 mr()= t ⎨ 0⎩ ,t < 0 6/22 Relationship between u(t) & r(t) t What is u () λ d λ ? −∫ ∞ Depends on t value ⇒function of t: f(t) t ⇒f()() t= uλ λd What is f(t)? −∫ ∞ -Write unit step as a function of λ -Integrate up to λ = t i.e., Find Area -How does area change as t changes? u(λ) t f⇒( t )= u (λ ) d= λ ⋅ 1 t = t = r ( t) 1 −∫ ∞ ⇒ t λ r()() t= uλ λ d λ = t −∫ ∞ Area = f(t) “Running Integral of step = ramp” 7/22 ⎧t, t≥ 0 Also note: For r () t = ⎨ 0 ,t < 0 we have: ⎩ dr() t 1⎧ ,t > 0 = ⎨ Not defined at t = 0! dt 0⎩ ,t < 0 dr() t Overlooking this, we can roughly say u() t = dt r(t) u(t) 1 . t 1 . 1 8/22 Time Shifting Signals Time shifting is an operation on a signal that shows up in many areas of signals and systems: • Time delays due to propagation of signals ─ acoustic signals propagate at the speed of sound ─ radio signals propagate at the speed of light • Time delays can be used to “build” complicated signals ─ We’ll see this later Time Shift: If you know x(t), what does x(t–t0) look like? For example… If t0 = 2: x(0 – 2) = x(–2) x(1 – 2) = x(–1) At t = 0, x(t – 2) takes the At t = 1, x(t – 2) takes the value of x(t) at t = –2 value of x(t) at t = –1 9/22 Example of Time Shift of the Unit Step u(t): u(t) 1 . -2 -1 1 2 3 4 t u(t-2) 1 . -2 -1 1 2 3 4 t u(t+0.5) General View: 1 . x(t ± t0) for t0 > 0 t -2 -1 1 2 3 4 “+t0”gives Left shift (Advance) “–t0”gives Right shift (Delay) 10/22 Other Names: Delta Function, The Impulse Function Dirac Delta Function One of the most important functions for understanding systems!! Ironically…it does not exist in practice!! ⇒ It is a theoretical tool used to understand what is important to know about systems! But… it leads to ideas that are used all the time in practice!! There are three views we’ll take of the delta function: Infinite height Rough View: a pulse with: Zero width Unit area “A really narrow, really tall pulse that has unit area” 11/22 Slightly Less-Rough View: 1 (δt )= lim pε() t ε →0 ε 1 p() t 1 ε ε Beware of Fig 1.4 in the ε book… t does not show i 1 the real δ(t)… Here we define p ε() t as: ε −ε ε t So its vertical axis should NOT be labeled with δ(t) 2 2 Pulse having… height of 1/ε and width of ε … hich therefore has…w area of 1 (1 = ε×1/ε) So as ε gets smaller the pulse gets higher and narrower ays wlt abu has area of 1… In the limit it “becomes” the delta function 12/22 Precise Idea: δ(t) is not an ordinary function… It is defined in terms of its behavior inside an integral: The delta function δ(t) is defined as something that satisfies the following two conditions:( ) 0δ ,t = for t ≠any 0 ε ( )∫δt 1 , dt= for ε any> 0 −ε We show δ(t) on a plot using an arrow… (conveys infinite height and zero width) δ(t) Caution… this is NOT the vertical axis… it is the delta function!!! t 0 13/22 The Sifting Property is the most important property of δ(t): t0 +ε f( t )− ( tδ t ) = dt ( f ) t ∀ε > 0 ∫ 0 0 t0 −ε f(t) Integrating the product of f(t) f(t0) f(t0) and δ(t – to) returns a single t0 t number… the value of f(t) at δ(t- t0) the “location” of the shifted delta function t0 t As long as the integral’s limits surround the “location” of the delta… otherwise it returns zero 14/22 Steps for applying sifting property: Step 1: Find variable of integration t +ε 0 Step 2: Find the argument of δ(•) Step 3: Find the value of the f()()() t tδ − t0 dt = 0 f t ∫ variable of integration that causes t0 −ε the argument of δ(•) to go to zero. Step 4: If value in Step 3 lies inside limits of integration… Take everything that is multiplying δ(•) Example #1: and evaluate it at the value found in 7 step 3; Otherwise… “return” zero sin(t ) t ( dt 1 ) ? ∫ π δ− = Step 1: t Step 2: t –1 −4 Step 3: t 0 = –1⇒ t = 1 Step 4: t es in [–4,7] so = 1 li (δ t − 1 ) e…evaluat sin(π×1) = sin(π) = 0 sin(πt ) 7 3 12 t sin(∫ tπ ) t δ− ( 1dt = ) 0 −4 15/22 2 Example #2: sin(∫ tπ ) ( t δ− 2 . dt 5 = ) ? 0 Step 1: Find variable of integration: t Step 2: Find the argument of δ(•): t5 . –2 Step 3: Find the value of the variable of integration that causes the argument of δ(•) to go to zero: t – 2.5 = 0 ⇒ t = 2.5 Step 4: If value in Step 3 lies inside limits of integration… No! Otherwise… “return” zero… (δ t 2− . 5 ) 2 sin(πt ) sin(∫ tπ ) ( t δ− 2 . dt 5 = ) 0 0 3 12 t Range of Integration Does NOT “include delta function” 16/22 7 Example #3:sin(∫ tω )( t− 32 ) δ t ( + 3 dt =4 ) ? −4 Because of this… handle slightly differently! Step 0es: let : Change variablτ = 3t Î dτ = 3dt Î limits: τL = 3(-4) τL = 3(7) 21 1 sin(∫ ωτ / 3 )( τ− / 32 3δ + ) τd ( = τ 4 ) ? −12 3 Step 1: Find variable of integration: τ Step 2: Find the argument of δ(•): τ + 4 Step 3: Find the value of the variable of integration that causes the argument of δ(•) to go to zero: τ = –4 Step 4: If value in Step 3 lies inside limits of integration… Yes! Take everything that is multiplying δ(•): (1/3)sin(ωτ/3)(τ/3 – 3)2 e it at the value found in step 3:…and evaluat (1/3)sin(–4/3ω)(–4/3 – 3)2 = 6.26sin(–4/3ω) 7 sin( )(∫ 3tω )− t ( 32 δ t + 4 ) dt = 6 .() 26 −ω sin 4 / 3 −4 17/22 One Relationship Between δ(t) & u(t) t δ∫()() λd= λ u t − ∞ For t < 0: the integrand = 0 ⇒ integral = 0 for t < 0 δ(λ) λ = t < 0 0 λ Defines the unit tegrationRange of In step function For t > 0: we “integrate over” the delta ⇒ integral = 1 for t > 0 δ(λ) 0 λ = t > 0 λ tegrationRange of In 18/22 d Another Relationship Between δ(t) & u(t) ()()δ t = u t dt Derivative = 0 u(t) Derivative = 0 1 . t Derivative = “∞” (“Engineer Thinking”) Our view of the delta function having infinite height but zero width matches this interpretation of the values of the derivative of the unit step function!! 19/22 Periodic Signals Periodic signals are important because many human-made signals are periodic.
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