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Convolution2Lecture

# Convolution2Lecture - Graphical Convolution(contd Linear...

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Graphical Convolution (contd) Linear Systems and Signals – Lecture 8 Dr. J. K. Aggarwal The University of Texas at Austin

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Example 1 ± 2 () () ? t t xt e ut ht yt xt ht = = = ∗= 2( ) 0 2 0 2 2 (1 ) 0 t t t t tt e e d ee d for t ττ τ −− = = =− = −≥ 2 ,( ) 0 0 , () ( ) () Moreover y t so e e ut = <
Example 2 ± 2 2 1,2 44 0 (2 ) 0 2 repeated λλ λ ++ = += ∴= Char. Eqn: 2 0 () t yt t e u t = ( ) 2 01 2 22 2 2 2 t tt c c t e c c t e c e + =− + + ± 02 (0) 0 0 (0) 1 1 yc = ⇒= = ± Initial conditions: Zero-input response: 2 ( 4 4) ( ) 9) ( ) D Dy t Dx t = + (Notice the special initial conditions)

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Example 2 …contd ( ) ( ) () 0 2 0 22 2 2 2 9 () () 2 9 09 2 2 ( 2 ) 25 ( ) n t tt t t ht b t D y t ut bt D t e u t te e t e te ut δ −− =+ + + + + Impulse Response: ± In this case, the initial conditions for finding y n (t) are the same as for y 0 (t) – a bit unusual
Example 3 ± Problem (4 ) ( 3 ) 0 λ + += 1,2 4, 3 Char. Eqn: =− − ( ) 34 0 () tt y tee u t −− =− 12 3 4 o o yt c e c e c e c e ∴= + ± 00 (0) 0, (0) 1 yy = = ± Zero-input response: Initial conditions: ( ) 2 0 (7 1 2 ) ( ) 2 ( ) DD y tD x t ++ = + 1 2 03 4 1 cc c c ⇒+ = = 1; 1 ==

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Example 3 ...contd Impulse Response: (we use the same function for because of initial cond.)
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Convolution2Lecture - Graphical Convolution(contd Linear...

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