8solution - ECEN 455: Assignment 8 Problems: 1. (CSE: 8.3)...

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Unformatted text preview: ECEN 455: Assignment 8 Problems: 1. (CSE: 8.3) In a binary PAM system, the clock that specifies the sampling of the correlator output is offset from the optimum sampling time by 10%. (a) If the signal pulse used is rectangular, determine the loss in SNR due to mistiming. If the transmitted signal is r ( t ) = ∞ summationdisplay n =-∞ a n h ( t- nT ) + n ( t ) , then the output of the receiving filter is y ( t ) = ∞ summationdisplay n =-∞ a n x ( t- nT ) + ν ( t ) where x ( t ) = h ( t ) ⋆ h ( t ) and ν ( t ) = n ( t ) ⋆ h ( t ) . If the sampling time is off by 10%, then the samples at the output of the correlator are taken at t = ( m ± . 1) T . Assuming that t = ( m- . 1) T without loss of generality, then the sampled sequence is y m = ∞ summationdisplay n =-∞ a n x (( m- . 1- n ) T + ν (( m- . 1) T ) . If the signal pulse is rectangular with amplitude A and duration T , then ∞ summationdisplay n =-∞ a n x (( m- . 1- n ) T is nonzero only for n = m and n = m- 1 . The sampled sequence is therefore given by y m = a m x (- . 1 T ) + a m- 1 x ( T- . 1 T ) + ν (( m- . 1) T ) = 0 . 9 a m A 2 T + 0 . 1 a m- 1 A 2 T + ν (( m- . 1) T ) . The power spectral density of the noise at the output of the correlator is S ν ( f ) = S n ( f ) | H ( f ) | 2 = N 2 A 2 T 2 sinc 2 ( fT ) . Thus, the variance of the noise is σ n u 2 = integraldisplay ∞-∞ N 2 A 2 T 2 sinc 2 ( fT ) df = N 2 A 2 T 2 1 T = N 2 A 2 T and the SNR is SNR = (0 . 9) 2 2( A 2 T ) 2 N A 2 T = 0 . 81 2 A 2 T N . As it is observed, there is a loss of 10 log 10 . 81 =- . 9151 dB due to the mistiming. 1 (b) Determine the amount of intersymbol interference introduced by the mistiming and determine its effect on performance. Recall from the previous part that the sampled sequence is y m = 0 . 9 a m A 2 T + 0 . 1 a m- 1 A 2 T + ν m . The term . 1 a m- 1 A 2 T expresses the ISI introduced to the system. If a m = 1 is transmitted, then the probability of error is P ( e | a m = 1) = 0 . 5 P ( e | a m = 1 ,a m- 1 = 1) + 0 . 5 P ( e | a m = 1 ,a m- 1 =- 1) = 1 2 √ πN A 2 T integraldisplay- A 2 T-∞ e- ν 2 N A 2 T dν + 1 2 √ πN A 2 T integraldisplay- . 8 A 2 T-∞ e- ν 2 N A 2 T dν = 0 . 5 Q radicalBigg 2 A 2 T N + 0 . 5 Q radicalBigg parenleftbigg 8 10 parenrightbigg 2 2 A 2 T N . Since the symbols of the binary PAM system are equiprobable the previous derived expression is the probability of error when a symbol by symbol detector is employed. Comparing this with the probability of error of a system with no ISI, we observe that there is an increase of the probability of error by . 5 Q radicalBigg parenleftbigg 8 10 parenrightbigg 2 2 A 2 T N - . 5 Q radicalBigg 2 A 2 T N ....
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This note was uploaded on 11/13/2011 for the course ECEN 455 taught by Professor Staff during the Spring '08 term at Texas A&M.

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8solution - ECEN 455: Assignment 8 Problems: 1. (CSE: 8.3)...

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