# Since q x ex 2 2 the error probability can be upper

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Unformatted text preview: 0 = 0 dB) Density Data=-1 Data=+1 P e , +1 P e , −1 √ −E EECS 455 (Univ. of Michigan) √ Fall 2012 E Filter Output September 6, 2012 93 / 130 Lecture Notes 5 BPSK Modulation: Revisited Filter Output with Noise (E /N0 = 3 dB) Density Data=-1 Data=+1 P e , +1 P e , −1 √ −E EECS 455 (Univ. of Michigan) √ Fall 2012 E Filter Output September 6, 2012 94 / 130 Lecture Notes 5 BPSK Modulation: Revisited Bit Error Probability of BPSK Pe ,b = Q 2E N0 where ∞ Q (x ) = x EECS 455 (Univ. of Michigan) =Q 2Eb N0 1 −u 2 /2 e du 2π Fall 2012 September 6, 2012 95 / 130 Lecture Notes 5 BPSK Modulation: Revisited Bit Error Probability of BPSK For binary signals this is the smallest bit error probability, i.e. BPSK are optimal signals and the receiver shown above is optimum (in additive white Gaussian noise). For binary signals the energy transmitted per information bit Eb is equal to the energy per signal E . For Pe,b = 10−5 we need a bit-energy, Eb to noise density N0 ratio of Eb /N0 = 9.6dB. Note: Q (x ) is a decreasing function which is 1/2 at x = 0. There are efﬁcient algorithms (based on Taylor series 2 expansions) to calculate Q (x ). Since Q (x ) ≤ e{−x /2} /2 the error probability can be upper bounded by Pe ,b ≤ 1 {−Eb /N0 } e 2 which decreases exponentially with signal-to-noise ratio. EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 96 / 130 Lecture Notes 5 BPSK Modulation: Revisited Bit Error Probability of BPSK Error Probability of BPSK P e,b 1 10 -1 10 -2 10 -3 10 -4 10 -5 10 -6 10 -7 10 -8 10 -9 10 -10 0 2 4 6 8 10 12 14 16 Eb/N 0 (dB) Figure: Error Probability of BPSK. EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 97 / 130 Lecture Notes 5 BPSK Modulation: Revisited Bandwidth of BPSK The power spectral density is a measure of the distribution of power with respect to frequency. The power spectral density for BPSK has the form S (f ) = PT sinc2 ((f − fc )T ) + sinc2 ((f + fc )T ) 2 where sinc(x ) = Notice that ∞ sin(π x ) . πx S (f )df = P . −∞ EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 98 / 130 Lecture Notes 5 BPSK Modulation: Revisited Bandwidth of BPSK The power spectrum has zeros or nulls at f − fc = i /T except for i = 0; that is there is a null at f − fc = ±1/T called the ﬁrst null; a null at f − fc = ±2/T called the second null; etc. The bandwidth between the ﬁrst nulls is called the null-to-null bandwidth. For BPSK the null-to-null bandwidth is 2/T . Notice that the spectrum falls off as (f − fc )2 as f moves away from fc . (The spectrum of MSK falls off as the fourth power, versus the second power for BPSK). EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 99 / 130 Lecture Notes 5 BPSK Modulation: Revisited Bandwidth of BPSK It is possible to reduce the bandwidth of a BPSK signal by ﬁltering. If the ﬁltering is done properly the (absolute) bandwidth of the signal can be reduced to 1/T without causing any intersymbol interference; that is all the power is concentrated in the frequency range −1/(2T ) ≤ |f − fc | ≤ 1/(2T ). The drawbacks are that the signal loses its constant envelope property (useful for nonlinear ampliﬁers) and the sensitivity to timing errors is greatly increased. The timing sensitivity problem can be greatly alleviated by ﬁltering to a slightly larger bandwidth −(1 + α)/(2T ) ≤ |f − fc | ≤ (1 + α)/(2T ). EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 100 / 130 Lecture Notes 5 BPSK Modulation: Revisited Bandwidth of BPSK S(f) 0.50 0.40 0.30 0.20 0.10 0.00 -4 -3 -2 -1 0 1 2 3 4 (f-fc)T Figure: Spectrum of BPSK EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 101 / 130 Lecture Notes 5 BPSK Modulation: Revisited S(f) (dB) Bandwidth of BPSK 0 -20 -40 -60 -80 -100 -5 -4 -3 -2 -1 0 1 2 3 4 5 (f-f c )T Figure: Spectrum of BPSK EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 102 / 130 Lecture Notes 5 BPSK Modulation: Revisited Bandwidth of BPSK S(f) (dB) 0 -20 -40 -60 -80 -100 -10 -8 -6 -4 -2 0 2 4 6 8 10 (f-fc)T Figure: Spectrum of BPSK EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 103 / 130 Lecture Notes 5 BPSK Modulation: Revisited Bandwidth of BPSK Example Given: Noise power spectral density of N0 /2 = −150 dBm/Hz =10−18 Watts/Hz. Pr = 3 × 10−10 Watts Desired Pe = 10−7 . Find: The data rate that can be used and the bandwidth that is needed. Solution: Need Q ( 2Eb /N0 ) = 10−7 or Eb /N0 = 11.3dB or Eb /N0 = 13.52. But Eb /N0 = Pr T /N0 = 13.52. Thus the data bit must be at least T = 9.0 × 10−8 seconds long, i.e. the data rate 1/T must be less than 11 Mbits/second. Clearly we also need a (null-to-null) bandwidth of 22 MHz. EECS 455 (Univ. of Michigan) Fall 2012 September 6,...
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## This note was uploaded on 02/12/2014 for the course EECS 455 taught by Professor Stark during the Fall '08 term at University of Michigan.

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