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Unformatted text preview: Chapter 3 LTIC Systems 1. CT Linear, Time Invariant (LTIC) Systems 2. Impulse Response of LTIC Systems 3. Convolution Integral and Properties 4. Output of LTIC Systems using Convolution Sections: 3.1 – 3.7 ELEC242: CT Signals and Systems Series RLC Circuit Activity I: 1. Derive the differential equation representation of the above series RLC circuit between x(t) and w(t)? 2. Solve the differential equation for L = 1/12H, R = 7/12Ω, and C = 1F when input x(t) = 2exp(−t)u(t). Assume that the capacitor has an initial charge of 5V. ELEC242: CT Signals and Systems System Response 6 5 4 3 2 w(t ) 1 wzs (t ) 0 wzi (t ) t −1 0 0.5 1 1.5 2 2.5 3 3.5 4 Solution to Activity I: [ w (t ) = [8e w(t ) = [− 7e wzi (t ) = 20e −3t − 15e − 4t u (t ) zs − 4t − 12e −3t + 4e −t u (t ) − 4t + 8e −3t + 4e −t u (t ) ELEC242: CT Signals and Systems 4.5 5 Approximation of CT Signals x(t) δΔ(t) 1/Δ t 0 Δ (a) −5Δ −3Δ −Δ 0 Δ 3Δ 5Δ (b) 7Δ Activity II: 1. Express x(t) as a function of δΔ(t)? 2. Determine the impulse response of the systems represented by following differential equations 1. y (t ) = x(t − 1) + 2 x(t − 3) 2. dy + 4 y (t ) = 2 x(t ) dt Solution: 1. h(t ) = δ(t − 1) + 2δ(t − 3) 2. h(t ) = 2e −4t u (t ) ELEC242: CT Signals and Systems t Impulse Response The impulse response of an LTIC system is given by 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 −2 −1 0 1 2 3 4 5 6 7 8 9 t Sketch the output for the input x(t) = δ(t – 1) + 3δ(t – 2) + 2δ(t – 6)? 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 −2 −1 0 1 2 3 4 5 6 ELEC242: CT Signals and Systems 7 8 9 t Output Response of a system Show that the output of a LTI with impulse response h(t) is given by ∞ y (t ) = ∫ x(τ)h(t − τ)dτ −∞ x(t ) = ∞ ∫ x(τ)δ(t − τ)dτ LTIC System h (t ) y (t ) = x(t ) ∗ h(t ) = ∞ ∫ x(τ)h(t − τ)dτ −∞ −∞ ELEC242: CT Signals and Systems Example: Output of an LTI system Activity IV: Determine the output response of an LTIC Systems for Input: Impulse Response: x(t) = exp(−t) u(t) h(t) = exp(−2t)u(t). Solution: y(t) = (e−t – e−2t) u(t). 0.4 0.3 0.2 0.1 t 0 −2 −1 0 1 2 3 4 5 6 7 ELEC242: CT Signals and Systems 8 9 Convolution Activity: IV x(τ) Step-1 1 e−τ u(τ) τ 0 h(τ) Step-2 1 e−2τu(τ) τ 0 h(−τ) Step-3 1 e2τu(−τ) τ 0 h(t−τ) Step-4 1 e2(t−τ)u(t−τ) t 0 ELEC242: CT Signals and Systems τ Convolution Activity: IV x(τ) h(t−τ) Case 1: t < 0 1 Step-5a t 0 Step-5b Case 2: t > 0 τ x(τ) h(t−τ) 1 0 t τ y(t) Step-6 0.25 0 0.693 ELEC242: CT Signals and Systems t Example II: Output of an LTI system Activity V: Determine the output response of an LTIC Systems for Input: Impulse Response: ⎧⎪1.5 x(t ) = ⎨ ⎪⎩ 0 −2≤t ≤3 ⎧⎪2 h(t ) = ⎨ ⎪⎩ 0 −1 ≤ t ≤ 2 otherwise otherwise. ELEC242: CT Signals and Systems Convolution: Activity V x(τ) Step-1 1.5 0 −2 3 τ h(τ) 2.0 Step-2 −1 0 2 τ h(−τ) 2.0 Step-3 −2 Step-4 0 t+1 τ h(t − τ) 2.0 t−2 1 0 ELEC242: CT Signals and Systems τ Convolution: Activity V Step-5a Case 1: t < −3 x(τ) h(t−τ) 2.0 t−2 t + 1 −2 1.5 0 τ 3 Step-5b Case 2: −3 ≤ t < 0 x(τ) h(t−τ) 2.0 t−2 Step-5c −2 t + 1 0τ 1.5 τ 3 x(τ) h(t−τ) Case 3: 0 ≤ t < 1 2.0 1.5 −2 t − 2 0 x(τ) h(t−τ) Case 4: 2 ≤ t < 5 1.5 t + 13 2.0 Step-5d −2 0 t−2 3 ELEC242: CT Signals and Systems t+1 τ Convolution: Activity V Step-5e x(τ) h(t−τ) Case 5: t ≥ 5 1.5 2.0 0 −2 3 t−2 τ t+1 10 Step-6 8 6 4 τ 2 0 −6 t −4 −2 0 2 4 ELEC242: CT Signals and Systems 6 8 Properties of Convolution 1. Commutative property: x1 (t)∗ x2 (t) = x2 (t)∗ x1 (t) 1. Distributive Property: x1 (t)∗ [ x2 (t) + x3 (t)] = x1 (t)∗ x2 (t) + x1 (t)∗ x3 (t) 2. Associative Property: x1 (t)∗ [ x2 (t)∗ x3 (t)] = [ x1 (t)∗ x2 (t)] ∗ x3 (t) 3. Shift Property: x1 (t − T1 )∗ x2 (t − T2 ) = g(t − T1 − T2 ) 4. Convolution with Impulse Response: x(t)∗ δ (t − T0 ) = x(t − T0 ) 5. Convolution with Unit Step Function: t x(t)∗ u(t) = ∫ x(α ) dα −∞ 6. Duration of Convolution: : Let the non-zero durations (or widths) of the convolution operands x1(t) and x2(t) be denoted by T1 and T2 time units, respectively. It can be shown that the non-zero duration (or width) of the convolution x1(t) x2(t) is T1 + T2 time units. ELEC242: CT Signals and Systems Properties of LTIC Systems 1. Memoryless System: An LTIC system will be memoryless if and only if its impulse response h(t) = 0 for t ≠ 0. 1. Causal System: An LTIC system will be causal if and only if its impulse response h(t) = 0 for t < 0. 2. Stable System: ∞ ∫ h(α ) dα < ∞ −∞ 3. Invertible System: h(t)∗ hi (t) = δ (t) Activity: Determine if systems with the impulse responses i) h(t) = δ(t) – δ(t – 2). ii) h(t) = 2rect(t / 2). iii) h(t) = 2exp(−4t) u(t). iv) h(t) = [1 − exp(−4t)] u(t). are memoryless, causal, and stable. ELEC242: CT Signals and Systems ...
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