By B. P. Lathi
Incorporating new difficulties and examples, the second one version of Linear platforms and Signals positive factors MATLAB® fabric in every one bankruptcy and in the back of the ebook. It provides transparent descriptions of linear structures and makes use of arithmetic not just to turn out axiomatic conception, but in addition to reinforce actual and intuitive understanding.
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Extra resources for Linear Systems and Signals (2nd Edition)
9: Example of time reversal. The instants −1 and −5 in x(t) are mapped into instants 1 and 5 in x(−t). Because x(t) = e t/2 , we have x(-t) = e 1/2 . The signal x(−t) is depicted in Fig. 9b. 2-4 Combined Operations Certain complex operations require simultaneous use of more than one of the operations just described. The most general operation involving all the three operations is x(at − b), which is realized in two possible sequences of operation: 1. Time-shift x(t) by b to obtain x(t − b). , replace t with at) to obtain x(at − b).
7). Thus Let x a (t) represent the function x(t) advanced (left-shifted) by 1 second as depicted in Fig. 5c. This function is x(t + 1); its mathematical description can be obtained from x(t) by replacing t with t + 1 in Eq. 7). 4 Write a mathematical description of the signal x 3 (t) in Fig. 3c. This signal is delayed by 2 seconds. Sketch the delayed signal. Show that this delayed signal x d (t) can be described mathematically as x d (t) = 2(t − 2) for 2 ≤ t ≤ 3, and equal to 0 otherwise. Now repeat the procedure with the signal advanced (left-shifted) by 1 second.
K = 0:2:11 k = 0 2 4 6 8 10 In this case, the termination value does not appear as an element of the vector. Negative and noninteger step sizes are also permissible. 0000 If a step size is not specified, a value of one is assumed. >> k = 0:11 k = 0 1 2 3 4 5 6 7 8 9 10 11 Vector notation provides the basis for solving a wide variety of problems. For example, consider finding the three cube roots of minus one, w 3 = −1 = e j(π + 2πk) for integer k. Taking the cube root of each side yields w = e j(π/3+2πk/3) To find the three unique solutions, use any three consecutive integer values of k and MATLAB's exp function.