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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 bitenergy, 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 signaltonoise 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 nulltonull bandwidth. For BPSK the nulltonull
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 (ffc)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 (ff 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 (ffc)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 (nulltonull)
bandwidth of 22 MHz.
EECS 455 (Univ. of Michigan) Fall 2012 September 6,...
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