Lesson_12 - EEL 3135 Signals and Systems Dr Fred J Taylor...

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EEL 3135: Signals and Systems Dr. Fred J. Taylor, Professor Lesson Title: Impulse Invariant FIR Lesson Number: 12 (Section 5-6 to 5-8) Background: In Chapter 5, elementary FIR filters are discussed. The Chapter ends with a discussion of some of the basic properties of FIRs and demonstrations. Recall that an FIR implements the convolution sum given by: [ ] [ ] = - = M k k k n x h n y 0 1. which defines the process by which input samples x[k] and converted into output samples y[k]. The convolution process describe in Equation 1 has a short hand notation and it is: y=h x ` 2 Example: Convolve the N = 4 point time series x={x[0], x[1], x[2], x[3]} = [1, 1, 1, 1] with the L = 4 tap FIR having an impulse response h={h[0], h[1], h[2], h[3]} = [ 1, 1, 1, 1]. The convolution sum, y=h x, is of length (N+L-1) = 7 and is given by: y[0]=h[0]x[0] = 1 y[1]=h[0]x[1] + h[1]x[0] = 2 y[2]=h[0]x[2] + h[1]x[1] + h[2]x[0] = 3 y[3]=h[0]x[3] + h[1]x[2] + h[2]x[1] + h[2]x[0] = 4 h[0] h[1] h[2] h[3] x[3] x[2] x[1] x[0] h[0] h[1] h[2] h[3] x[3] x[2] x[1] x[0] h[0] h[1] h[2] h[3] x[3] x[2] x[1] x[0] h[0] h[1] h[2] h[3] x[3] x[2] x[1] x[0] 1
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EEL 3135: Signals and Systems Dr. Fred J. Taylor, Professor y[4]=h[0]0 + h[1]x[3] + h[2]x[2] + h[2]x[1] = 3 y[5]=h[0]0 + h[1]0 + h[2]x[3] + h[2]x[2] = 2 y[6]=h[0]0 + h[1]0 + h[2]0 + h[2]x[3] = 1 The outcome is graphically interpreted in Figure 1. Figure 1: Convolution of two pulses. ------------------- end of example The example problem motivates an observation that the convolution sum values can be defined in terms of an inner product (dot product), y=h T x k = h i x k i , where h is a vector of impulse response coefficients and x k is a reversed in time listing of sample values, namely: ( 29 1 [ ],. .., 1 [ ], [ - - - = L k x k x k x x K k 3. Time Invariance An important class of filters is called time-invariant FIR filters. A time-invariant has a predictable response when signal delays are introduced. Assume a signal s(t) is presented to system S and shown in Figure 2. If a time-invariant system is presented with an input x[k], producing an output y[k], then when presented with an input x[k-k 0 ], the output is y[k-k 0 ], The study illustrated in Figure 3 can be used as a test for time-invariance. Time-invariant systems have constant coefficient and behave in the manner shown in Figure 2 and 3. If the systems coefficients are time varying, then the behavior shown in Figure 2 cannot be guaranteed in general.
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This note was uploaded on 08/21/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

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Lesson_12 - EEL 3135 Signals and Systems Dr Fred J Taylor...

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