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8solution - ECEN 455 Assignment 8 Problems 1(CSE 8.3 In a...

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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 ± 0 . 1) T . Assuming that t = ( m - 0 . 1) T without loss of generality, then the sampled sequence is y m = summationdisplay n = -∞ a n x (( m - 0 . 1 - n ) T + ν (( m - 0 . 1) T ) . If the signal pulse is rectangular with amplitude A and duration T , then summationdisplay n = -∞ a n x (( m - 0 . 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 ( - 0 . 1 T ) + a m - 1 x ( T - 0 . 1 T ) + ν (( m - 0 . 1) T ) = 0 . 9 a m A 2 T + 0 . 1 a m - 1 A 2 T + ν (( m - 0 . 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 0 2 A 2 T 2 sinc 2 ( fT ) . Thus, the variance of the noise is σ n u 2 = integraldisplay -∞ N 0 2 A 2 T 2 sinc 2 ( fT ) df = N 0 2 A 2 T 2 1 T = N 0 2 A 2 T and the SNR is SNR = (0 . 9) 2 2( A 2 T ) 2 N 0 A 2 T = 0 . 81 2 A 2 T N 0 . As it is observed, there is a loss of 10log 10 0 . 81 = - 0 . 9151 dB due to the mistiming. 1
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(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 0 . 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 0 A 2 T integraldisplay - A 2 T -∞ e - ν 2 N 0 A 2 T + 1 2 πN 0 A 2 T integraldisplay - 0 . 8 A 2 T -∞ e - ν 2 N 0 A 2 T = 0 . 5 Q radicalBigg 2 A 2 T N 0 + 0 . 5 Q radicalBigg parenleftbigg 8 10 parenrightbigg 2 2 A 2 T N 0 . 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 0 . 5 Q radicalBigg parenleftbigg 8 10 parenrightbigg 2 2 A 2 T N 0 - 0 . 5 Q radicalBigg 2 A 2 T N 0 . 2. (CSE: 8.7) Consider a four-phase PSK signal that is represented by the equivalent lowpass signal v ( t ) = summationdisplay n a n g ( t - nT ) , where a n takes on one of the four possible values ± 1 ± j 2 with equal probability. The sequence of information symbols { a n } is statistically independent. (a) Determine and sketch the power-spectral density of v ( t ) when g ( t ) = braceleftbigg A, 0 t T 0 , otherwise The autocorrelation function of the information symbols { a n } is R a ( k ) = E [ a * n a n + k ] = | a n | 2 4 δ ( k ) = δ ( k ) Thus, the power spectral density of v ( t ) is S V ( f ) = 1 T S a ( f ) | G ( f ) | 2 = 1 T | G ( f ) | 2 where G ( f ) = F [ g ( t )] . If g ( t ) = A Π
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