Lecture 7

# Lecture 7 - ECE 3101 Signals and Systems Reading...

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Lecture 7, 2/15/11 ECE 3101, Copyright P. B. Luh 1 Reading Assignment: Sections 2.4 Problem Set 2: Due today Quiz 1: Sorry for mixing up the context. Everyone gets 100 Exam. 1. Thursday. Four problems. To the end of Chap. 1. Four cheat sheets. Will go over sample problems today Makeup classes Next Monday, 2/21/11, 6:30-7:45 pm, MSB 411 Wednesday, 3/2/11, 6:30-7:45 pm, MSB 411 ECE 3101 Signals and Systems

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Lecture 7, 2/15/11 ECE 3101, Copyright P. B. Luh 2 Last Time: Time Domain Analysis of LTIC Systems y (N) + a 1 y (N-1) + . . + a N-1 y (1) + a N y=b 0 u (N) + b 1 u (N-1) + . . + b N u Q(D)y = P(D)u y(t) = y ZI (t) + y ZS (t) for linear systems Zero Input Response: Characteristic polynomial and characteristic modes Unit Impulse Response: h(t) = b 0 δ (t) + [P(D)y n (t)]U(t) y n (t): Unit impulse response to Q(D)w(t) = u(t), or zero input response to that system with y n (N-1) (0 + ) = 1; y n (0 + ) = y n (1) (0 + ) = . . = y n (N-2) (0 + ) = 0 y ZS (t): Convolution, building on linearity and time invariance τ τ τ = τ τ τ = t 0 t 0 ZS d ) t ( u ) ( h d ) t ( h ) ( u ) t ( y
Lecture 7, 2/15/11 ECE 3101, Copyright P. B. Luh 3 Convolution of Continuous-Time Signals Direction Integration Based on Definition Semi-graphical Method Laplace Transform (Later) Today: Sample problems for Exam. 1 Semi-graphical Method Properties of Convolution Examples Next Time: Sections 2.5, 3.6, 3.7

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Lecture 7, 2/15/11 ECE 3101, Copyright P. B. Luh 4 Sample Problems Determine whether the following systems are causal or noncausal; instantaneous or dynamic; are linear or nonlinear; are time invariant or time varying. Briefly justify your answers y 1 (t) = |u(t)|; y 3 [k+1] = -ky[k] + (u[k]) 2 τ τ = + 1 t 1 t 2 d ) ( u ) t ( y
Lecture 7, 2/15/11 ECE 3101, Copyright P. B. Luh 5 1. Evaluate the following integrals a. ( ) [] δ + δ + 4 4 2 dt ) 2 t ( 3 ) t ( 2 t () ( ) ∫∫ δ + + δ + = 4 4 4 4 2 2 dt ) 2 t ( 2 t 3 dt ) t ( 2 t ( ) δ + + δ + = 4 4 4 4 2 dt ) 2 t ( 2 2 3 dt ) t ( 2 0 = 2 + 3·6 = 20 b. δ + δ + + δ 4 4 2 dt ) 5 t ( ) t ( ) 2 t ( t δ + + δ = 4 4 2 4 4 2 dt ) t ( t dt ) 2 t ( t δ + + δ = 4 4 2 4 4 2 dt ) t ( 0 dt ) 2 t ( ) 2 ( = 4 τ τ δ = τ τ δ τ d ) ( ) 0 ( f d ) ( ) ( f τ τ δ = τ τ δ τ d ) t ( ) t ( f d ) t ( ) ( f 1 1 1

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6 2. Consider the circuit a. Find the I/O equation b. Find the state and output equation in matrix form R 1 R 2 C 1 C 2 - + - + y(t) - + f(t) + - R 3 ; 0 R ) t ( f dt ) t ( dv C 1 1 = + 0 R ) t ( v R ) t ( y dt ) t ( dy C 2 3 2 = + + v(t) ; 0 C R R ) t ( f dt ) t ( dy R 1 dt ) t ( y d C 1 2 1 3 2 2 2 = + 1 2 1 3 2 2 2 C R R ) t ( f dt ) t ( dy R 1 dt ) t ( y d C = + ; 0 R ) t ( f dt ) t ( dv C 1 1 1 = + ; 0 R ) t ( v R ) t ( v dt ) t ( dv C 2 1 3 2 2 2 = + + y = v 2 ; f 0 C R 1 v v C R 1 C R 1 0 0 dt dv dt dv 1 1 2 1 2 3 2 2 2 1 +
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## This note was uploaded on 03/29/2011 for the course ECE 3101 taught by Professor Luh during the Spring '11 term at UConn.

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Lecture 7 - ECE 3101 Signals and Systems Reading...

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