95下信號與系çµ&plusmn

95下信號與系çµ&plusmn

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Unformatted text preview: Signals and Systems, Midterm Exam Solutions (Draft) Spring 2007, Edited by bypeng 1. [10] Consider designing a discrete-time inverse system to eliminate the distortion associated with multipath propagation in a data communication problem, where the propagation channel consists of a direct path and a number of reflected paths. For simplicity, let us consider the discrete-time model of a two-path communication channel depicted as follows: a) [2] Write down the difference equation describing the two-path communication channel system. b) [6] Find the impulse response of a causal inverse system that will recover [ ] x n from [ ] y n . c) [2] Check if the inverse system is stable and explain the physical meaning of the condition you have derived. Solution: a) [ ] [ ] [ 1] y n x n ax n = +- b) Since the inverse system is causal, the impulse response [ ] g n can be written [ ] [ ] l l g n g n l δ ∞ = =- ∑ We also observe that the system in LTI, and then by the definition of the inverse system we consider the convolution of [ ] y n and [ ] g n which is given by 1 1 [ ]* [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] ( [ ] [ 1]) ( [ ] [ 1]) ( [ 1] [ 2]) ... ( [ ] [ 1]) ... [ ] ( l l k k l k l k k k k k y n g n y n k g k y n k g k l y n k g k l g y n k g x n k ax n k g x n ax n g x n ax n g x n k ax n k g x n g a g δ δ ∞ ∞ ∞ ∞ ∞ =-∞ =-∞ = =-∞ = ∞ ∞ =-∞ =-∞ =- =-- =-- =- =- +- - = +- +- +- + +- +- - + = + + ∑ ∑ ∑ ∑ ∑ ∑ ∑ 1 2 1 ) [ 1] ( ) [ 2] ... ( ) [ ] ... [ ] k k x n g a g x n g a g x n k x n +- + +- + + +- + = Now we find that 1 g = , 1 g a = - , 2 2 g a = , …, and in general ( ) k k g a = - for each integer k . So [ ] ( ) [ ] ( ) [ ] l n l g n a n l a u n δ ∞ = =-- = - ∑ . c) The inverse system is stable if 1 a < . This means if the reflected path needs to be weaker or the signals cannot be recovered in practical approaches. 2. [12] Consider the signal 7 9 [ ] cos cos 16 16 x n n n π π = + a) [4] Compute the discrete-time Fourier transform (DTFT) of the signal. b) [8] Now compute the DTFT of only a portion of the signal by multiplying [ ] x n with a windowing function [ ] w n , 1, [ ] 0, n M w n n M ≤ = > Plot the DTFT of the truncated signal with 8 M = and 40 M = to evaluate the effect of truncating a signal on the DTFT. Solution: a) We have 7 9 7 9 ( ) cos cos cos cos 16 16 16 16 7 7 9 9 2 2 2 2 16 16 16 16 9 7 2 16 16 j l X e n n n n l l l l l ω π π π π π π π π π δ ω π δ ω π δ ω π δ ω π π π π δ ω π δ ω ∞ =-∞ = + = + =-- + +- +-- + +- = +- + +- ∑ F F F 7 9 2 2 2 16 16 l l l l π π π δ ω π δ ω π ∞...
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95下信號與系çµ&plusmn

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