X t d produces an impulse response x ht

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Unformatted text preview: ulse response x(τ )h(t − τ )dτ . The output is the integral of all of these responses ￿∞ y (t) = x(τ )h(t − τ ) dτ −∞ Another perspective is just to look at the integral. • h(t − τ ) is the impulse response delayed to time τ • If we consider h(t − τ ) to be a function of τ , then h(t − τ ) is delayed to time t, and reversed. h(t − τ ) τ h(t − τ ) t t τ EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 9 • This is multiplied point by point with the input, x(τ ) t h(t − τ ) τ x(τ )h(t − τ ) t τ • Then integrate over τ to find y (t) for this t. Graphically, to find y (t): • flip impulse response h(τ ) backwards in time (yields h(−τ )) • drag to the right over t (yields h(t − τ )) • multiply pointwise by x (yields x(τ )h(t − τ )) ￿∞ • integrate over τ to get y (t) = x(τ )h(t − τ ) dτ −∞ EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 10 Simple Example 2 x(!) 1 2 3! 1 2 3! 1 0 -1 2 3! 2 3! 1 2 h(!) 1 0 -1 2 h(−!) 1 0 -1 2 1 0 -1 h(t − !) 1 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly h(t − !) -1 2 x(!) 0 1 1 -1 1 x(!) h(t − !) 0 1 0 x(!) 2 h(t − !) -1 x(!) 1 0 0<t <1 1 1 -1 0 3! EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly x(!) h(t − !) 1 2<t <3 3! 2 y(t ) = (x ∗ h)(t ) 2 t >3 3! 2 2 1<t <2 3 2 h(t − !) 1 3! 2 2 -1 t <0 1 11 1 -1 0 1 2 3! 12 Communication channel, e.g., twisted pair cable x(t ) y(t ) ∗h(t ) Impulse response: 1.5 h(t ) h 1 0.5 0 0 2 4 tt 6 8 10 This is a delay ≈ 1, plus smoothing. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 13 Simple signaling at 0.5 bit/sec; Boolean signal 0, 1, 0, 1, 1, . . . 1 x(t ) u 0.5 0 0 2 4 tt 6 8 y 1 10 y(t ) 0.5 0 0 2 4 tt 6 8 10 Output is delayed, smoothed version of input. 1’s & 0’s easily distinguished in y EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 14 Simple signalling at 4 bit/sec; same Boolean signal x(t ) 1 u 0.5 0 0 2 4 6 t 8 y(t ) 1 y 10 0.5 0 0 2 4 6 tt 8 10 Smoothing makes 1’s & 0’s very hard to distinguish in y . EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 15 Examples: Try these: x(t) h(t) 1 1 0 1 2 1 1 0 1 1 2 1 1 0 2 1 1 1 2 2 0 1 2 0 2 δ (t − 1) 0 1 1 2 0 1 2 1 2 1 0 1 1 1 0 0 2 1 0 (x ∗ h)(t) 1 0 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 1 2 16 Properties of Convolution For any two functions f and g the convolution is ￿ (f ∗ g )(t) = ∞ −∞ f (τ )g (t − τ ) dτ If we make the substitution τ1 = t − τ , then τ = t − τ1, and dτ = −dτ1. (f ∗ g )(t) = = ￿ −∞ ∞ ￿∞ −∞ f (t − τ1)g (τ1) (−dτ1) g (τ )f (t − τ1) dτ1 = (g ∗ f )(t) This means that convolution is commutative. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 17 Practically, If we have two signals to convolve, we can choose either to be the signal we hold constant, and which...
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This note was uploaded on 09/28/2013 for the course EE 108b taught by Professor Mukamel during the Fall '13 term at Singapore Stanford Partnership.

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