EE3054 Signals and Systems Continuous Time Convolution Yao Wang Polytechnic University Some slides included are extracted from lecture presentations prepared by McClellan and Schafer License Info for SPFirst Slides This work released under a Creative Commons License with the following terms: Attribution The licensor permits others to copy, distribute, display, and perform the work. In return, licensees must give the original authors credit. Non-Commercial The licensor permits others to copy, distribute, display, and perform the work. In return, licensees may not use the work for commercial purposes—unless they get the licensor's permission. Share Alike The licensor permits others to distribute derivative works only under a license identical to the one that governs the licensor's work. Full Text of the License This (hidden) page should be kept with the presentation 3/14/2008 © 2003, JH McClellan & RW Schafer 2 LECTURE OBJECTIVES Review of C-T LTI systems Evaluating convolutions Examples Impulses LTI Systems Stability and causality Cascade and parallel connections 3/14/2008 © 2003, JH McClellan & RW Schafer 3 Linear and Time-Invariant (LTI) Systems If a continuous-time system is both linear and time-invariant, then the output y(t) is related to the input x(t) by a convolution integral y (t ) = ∞ ∫ x(τ )h(t − τ )dτ = x(t ) ∗ h(t ) −∞ where h(t) is the impulse response of the system. 3/14/2008 © 2003, JH McClellan & RW Schafer 4 Evaluating a Convolution x (t ) = u(t − 1) y (t ) = h (t ) = e − t u (t ) ∞ h ( τ ) x ( t − τ ) d τ = h ( t ) ∗ x ( t ) ∫ −∞ 3/14/2008 © 2003, JH McClellan & RW Schafer 5 “Flipping and Shifting” x (τ ) “flipping” g(τ ) = x(−τ ) = u(− τ −1) “flipping and shifting” g(τ − t) = x(−(τ − t)) = x(t − τ ) 3/14/2008 t −1 © 2003, JH McClellan & RW Schafer t 6 Evaluating the Integral 3/14/2008 0 t −1 y (t ) = −τ e dτ ∫ 0 © 2003, JH McClellan & RW Schafer t −1 < 0 t −1 > 0 7 Solution y (t ) = = y (t ) = 0 3/14/2008 t −1 ∫e 0 −τ 1− e dτ = − e −τ t −1 −( t −1) 0 t ≥1 t<1 © 2003, JH McClellan & RW Schafer 8 Convolution GUI 3/14/2008 © 2003, JH McClellan & RW Schafer 9 Another Example x (t ) = e − at u(t ) y (t ) = h(t ) = e − bt u (t ), b ≠ a ∞ ∫ x(τ )h(t − τ )dτ = x(t ) ∗ h(t ) −∞ t ∞ −bt − aτ bτ e e e dτ −aτ −b ( t −τ ) ∫ u ( t − τ ) dτ = = ∫ e u(τ )e 0 −∞ 0 e −at − e −bt e −at − e −bt t>0 u (t ) = = −a+b b−a 0 t<0 3/14/2008 © 2003, JH McClellan & RW Schafer t>0 t<0 10 Special Case: u(t) x (t ) = e − at u(t ), a ≠ 0 y (t ) = h (t ) = u (t ) ∞ ∫ x(τ )h(t − τ )dτ = x(t ) ∗ h(t ) −∞ = 1 − at (1 − e )u(t ) a if a = 2 1 −2 t y (t ) = (1 − e )u(t ) 2 3/14/2008 © 2003, JH McClellan & RW Schafer 11 Convolve Unit Steps x (t ) = u (t ) y (t ) = h (t ) = u (t ) ∞ ∫ x(τ )h(t − τ )dτ = x(t ) ∗ h(t ) −∞ t ∞ ∫ 1 dτ t > 0 = ∫ u (τ )u(t − τ )dτ = 0 0 −∞ t<0 t t > 0 = = t u (t ) Unit Ramp 0 t < 0 3/14/2008 © 2003, JH McClellan & RW Schafer 12 “Flipping and Shifting” x(τ ) “flipping” “flipping and shifting” 3/14/2008 g(τ ) = x(−τ ) = u(− τ −1) g(τ − t) = x( −(τ − t)) = x(t − τ ) t −1 © 2003, JH McClellan & RW Schafer t 13 More examples Rectangular pulses Another Convolution Example −t h(t) = e u(t) ∞ y(t) = ∫ x(τ )h(t − τ )dτ = x(t) ∗ h(t) −∞ 3/14/2008 © 2003, JH McClellan & RW Schafer 15 Evaluating the Integral y(t) = t = ∫e 1 = 3/14/2008 t <1 0 −(t−τ ) 2 ∫e 1 dτ −(t−τ ) dτ 1≤ t ≤ 2 2≤t © 2003, JH McClellan & RW Schafer 16 Solution t y(t) = ∫e 2 −(t−τ ) 1 = ∫e −(t−τ ) 1 y(t) = 0 3/14/2008 dτ dτ −(t−τ ) t =e 1 −(t−τ ) 2 =e 1 =1- e −(t−1) 1≤ t ≤ 2 = e−(t−2) - e−(t−1) 2 ≤ t t <1 © 2003, JH McClellan & RW Schafer 17 Convolution GUI 3/14/2008 © 2003, JH McClellan & RW Schafer 18 Convolution with Impulses, etc. Convolution with impulses x(t ) * δ (t − t1 ) = x(t − t1 ) Convolution with step function = integrator 3/14/2008 © 2003, JH McClellan & RW Schafer 19 Convolution is Commutative h(t ) ∗ x(t ) = ∞ h x t d ( ) ( ) τ − τ τ ∫ −∞ let σ = t − τ and dσ = −dτ −∞ h(t ) ∗ x(t ) = − ∫ h(t − σ ) x(σ )dσ = ∞ ∞ ∫ h(t − σ ) x(σ )dσ = x(t ) ∗ h(t ) −∞ 3/14/2008 © 2003, JH McClellan & RW Schafer 20 Stability A system is stable if every bounded input produces a bounded output. A continuous-time LTI system is stable if and only if ∞ ∫ h(t) dt < ∞ −∞ 3/14/2008 © 2003, JH McClellan & RW Schafer 21 Integrator is unstable Causal Systems A system is causal if and only if y(t0) depends only on x(τ) for τ< t0 . An LTI system is causal if and only if h(t ) = 0 for t < 0 3/14/2008 © 2003, JH McClellan & RW Schafer 23 Convolution is Linear Substitute x(t)=ax1(t)+bx2(t) ∞ y (t ) = ∫ [ax1 (τ ) + bx2 (τ )]h(t − τ )dτ −∞ ∞ ∞ −∞ −∞ = a ∫ x1 (τ )h(t − τ )dτ + b ∫ x2 (τ )h(t − τ )dτ = ay1 (t ) + by2 (t ) Therefore, convolution is linear. 3/14/2008 © 2003, JH McClellan & RW Schafer 24 Convolution is Time-Invariant Substitute x(t-t0) ∞ w(t) = ∫ h(τ )x((t − τ ) − t )dτ o −∞ ∞ = ∫ h(τ )x((t − t o ) − τ )dτ −∞ = y(t − to ) 3/14/2008 © 2003, JH McClellan & RW Schafer 25 Cascade of LTI Systems δ (t) h1 (t) h1 (t) ∗ h2 (t) h(t) = h1 (t) ∗ h2 (t) = h2 (t) ∗ h1(t) δ (t) 3/14/2008 h2 (t) © 2003, JH McClellan & RW Schafer h2 (t) ∗ h1(t) 26 Parallel LTI Systems h1 (t) δ (t) h1 (t) + h2 (t) h2 (t) h(t) = h1 (t) + h2 (t)(t 3/14/2008 © 2003, JH McClellan & RW Schafer 27 Example: More complicated combinations READING ASSIGNMENTS This Lecture: Chapter 9, Sects. 9-6, 9-7, and 9-8 Other Reading: Ch. 9, all Next Lecture: Start reading Chapter 10
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