Optimizing SNR with Matched Filter and QPSK Modulation

Learn about matched filter and QPSK modulation for improved SNR in radar and image processing. Understand the impulse response of matched filter and differential detection. Optimize bit rate and probability of error with MATLAB implementation.

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1

QUESTION 1
Section A
A matched filter is an optimal linear filter used to maximize the SNR in the presence of
additional zero mean white Gaussian noise with power spectral density. Some of the common
applications are in radar where signals are sent out and the reflected signals are measured and
compared to the sent signal. They are also used in image processing as an improved SNR shows
an improved SNR for x-ray pictures.
The signal shows a rectangular pulse given as,
s ( t ) = { A , for 0 ≤ t ≤T
0 , otherwise
The power spectral density is given as,
N 0
2 watts /Hz
Part 1
The matched filter output is y(t), and the filter input is given as, x(t),
x (t )=s ( t ) +n ( t ) 0 ≤ t ≤T
The output is expressed as,
y ( t ) =s0 ( t ) +n (t)
2
S(t)
n(t)

The output has a combination of the signal and noise components of the input signal. To
determine the maximum peak pulse SNR,
η= |s0 ( t )|2
E [ n2 ( t ) ]
Part 2
The white noise is further described based on the spectral power density as,
H ( f ) =2 K
N0
S¿ ( f ) e− j w t 0
When the input noise is the zero mean white Gaussian noise, the impulse response of the
matched filter is given as,
h( t)=Cs (t 0−t )
h ( t ) =F−1 [ H ( f ) ] = 2 K
N0
∫
−∞
∞
S¿ ( f ) e j w t0
e− j wt df
¿ 2 K
N0 [∫
−∞
∞
S¿ ( f ) e j 2 πf ( t0−t ) df ]¿
¿ 2 K
N0
[ s ( t0 −t ) ] ¿
Part 3
Since the signal S(t) is a real life signal, the maximum signal to noise ratio is given as,
2 K
N0
=Cs ...impulse response
The impulse response signal shows a slight backward shift of the original input signal before the
noise input.
Part 4
3

To show that
y ( T ) =∫
0
T
x ( τ ) h(T −τ ) dτ
y ( T ) = [ x ( t ) , s ( t ) ]L2
Section B
4

Part 1
Transmitting the QPSK requires a probability of bit error, PB. The received signals are,
Sm ( t )= √ ( 2 Pr ) cos [ 2 π f 0 t +ϕ ( m ) ] 0 ≤ t< T
ϕ ( m ) ϵ { π
4 , 3 π
4 , 5 π
4 , 7 π
4 }
Pr −received average signal power
The channel output is given as,
r ( t )=sm ( t )+ n(t)
The maximum bit rate transmitted in bps. The PB requirement needs to be met such that
Pr
N 0
, Pr= 1
T ∫
0
T
sm
2 ( t ) dt
y=Q ( x )= 1
√2 π ∫
−∞
∞
e
−α 2
2 dα , x =Q−1 ( y )
The QPSK constellation of the carrier phase with the ϕ ( m ) as given is,
The striking result of the QPSK signal output is that the bit error probability of the QPSK is
identical to the binary phase shift keying, although the data sent through the channel has double
5

the bandwidth. QPSK provides twice the spectral efficiency as the same energy efficiency of the
BPSK.
SQPSK = {√ Es cos [ (i−1 ) π
2 ] ϕ1 ( t )− √ Es sin [ ( i−1 ) ] ϕ2 ( t ) }i=1,2,3,4 , …
It’s differentially encoded to allow non-coherent detection. The bit rate in either I or Q channel
are ½ input data rate.
Part 2
Matlab Implementation of the maximum bit rate of the QPSK signals being transmitted.
clear
N = 10^5; % number of symbols
Es_N0_dB = [-3:20]; % multiple Eb/N0 values
ipHat = zeros(1,N);
for ii = 1:length(Es_N0_dB)
ip = (2*(rand(1,N)>0.5)-1) + j*(2*(rand(1,N)>0.5)-1); %
s = (1/sqrt(2))*ip; % normalization of energy to 1
n = 1/sqrt(2)*[randn(1,N) + j*randn(1,N)]; % white guassian noise, 0dB
variance
y = s + 10^(-Es_N0_dB(ii)/20)*n; % additive white gaussian noise
figure
semilogy(Es_N0_dB,theorySer_QPSK,'b.-');
hold on
semilogy(Es_N0_dB,simSer_QPSK,'mx-');
axis([-3 15 10^-5 1])
grid on
legend('theory-QPSK', 'simulation-QPSK');
xlabel('Es/No, dB')
ylabel('Symbol Error Rate')
title('Symbol error probability curve for QPSK(4-QAM)')
6

-2 0 2 4 6 8 10 12 14
Es/No, dB
10 -5
10 -4
10 -3
10 -2
10 -1
10 0
Symbol Error Rate
Symbol error probability curve for QPSK(4-QAM)
theory-QPSK
simulation-QPSK
QUESTION 2
Binary communications system used to transmit bits.
π0 + π1=1
The channel output is a continuous random variable, R, with the conditional probability density
functions. The priori probabilities are modeled as equiprobables and are used as multistage
detection processes in which,
P0∧P1 =1−P0 be arbitrary
V →rv ( conditiona l probability density )
f v∨u { v|am }−finite∧non−zero ∀ v ∈ R
m∈ {0,1 }
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Part A
The modifications for zero densities are discrete and complex. These conditional densities are
likelihoods in the argon of the hypothesis testing. The marginal density is given by,
f v ( u ) =P0 f v∨u ( v|a0 ) + P1 f v∨u ( v|a1 )
The posteriori probability of U for m=0 or 1 is given by
pu∨v ( am|v )= ( Pm f v∨u ( v|am ) )
f v ( u )
For the MAP decision, the likelihood ratio is given as,
A ( v ) = f v∨u ( v|a0 )
f v∨u ( v|a1 ) , p1
p0
=η
This denotes the threshold and depends on the priori probabilities. Adjusting the threshold value
may relate the equation,
p0= p1 η=1
When the threshold is unity,
^U ( v ) =a0
f v∨u ( v |a0 ) ≥ f v∨u (v ∨a1 )
^U ( v ) =1 ,otherwise
8

It is the maximum likelihood rule or test. The priori probabilities are usually equal and the MAP
reduces to ML. the probability of error is derived for the MAP detection. The probability of error
is conditional on each hypothesis,
Pr {e|U =a1 }∧Pr { e|U =a0 }
The overall probability of the error is given as,
Pr {e }= p0 Pr {e|U =a0 }+ p1 {e|U =a1 }
Part B
For MAP detection with a threshold η,
Pr { e|U =a0 }=Pr { ^U=a1|U=a0 } =Pr { Λ ( v ) < η|U=a0 }
Pr {e|U =a1 }=Pr { ^U =a0|U=a1 }=Pr { Λ ( v ) ≥ η|U=a1 }
Using the values provided in the question,
The probability density functions are given as f R ( r|0 ) and f R ( r|1 )
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Therefore,
P ( error )=P ( 1−decided , 0−transmitted ) +P ( 0−decided ,1−transmitted )
P ( error )=P ( 1D ,OT ) + P ( OD , 1T )
P ( error )=P ( OT ) ∫
0
∞
f ( r|OT ) dr + P(1T ) ∫
−∞
0
f ( r|1T ) dr
Part C
Since for the a-priori probabilities P ( OT ) ∧P ( 1T )−0.5 , the equation is re-written as,
P ( error )= 1
2 ∫
0
∞
f ( r|OT ) dr + 1
2 ∫
−∞
0
f ( r|1T ) dr
For the continuous density functions,
S0 = √ Es
P(1D∨OT )=Q
( √ Es
N0
2 )
P(O∨1T )=Q
( √ Es
N0
2 )
10

Part D
Replacing in the previous equation,
P ( error ) =0.5 Q
( √ Es
N 0
2 ) +0.5Q
( √ Es
N 0
2 )
P ( error ) =Q ( √ 2 Es
N 0 ) =0.5 erfc ( √ Es
N0 )
QUESTION 3
Binary transmission of information
Part A
Signal set of two signals with apriori probabilities given as,
π0=Pr { transmits s0 ( t ) }= 1
2
π1=Pr {transmits s1 ( t ) }= 1
2
The signal transmission is given by,
H0 :r ( t ) =S0 (t ) +n ( t ) , 0 ≤ t <T
H1 : r ( t ) =S1 ( t ) + n ( t ) , 0 ≤t <T
n ( t )−zero mean whiteGaussian noise with p ower spectral density
Sn ( f )= N 0
2 ,−∞< f <∞
s0 ( t )= {A ,0 ≤ t ≤ 3
4 T
0 , otherwise
s1 ( t ) = {A , 1
4 T ≤ t ≤ T
0 , otherwise
11

Part B
The bit error is given as,
Eb =π0∫
0
T
s0
2 ( t ) dt+ π1∫
0
T
s1
2 ( t ) dt
Signals are measured across 1 ohm and the optimal minimum PB detection is used. It is based on
the Bayes theorem of conditional probability.
ωc= 2 π
T s
=N ( 2 π
T b )
ρ= A2 T b
2η
SN Re=10 log10 [ A2 Tb
2 η ]
Pexp= total number of erroneous bits
total number of transmitted bits
Part C
The probability of bit error as a function of Eb/N0 is given as,
Pb=erfc ( √ Eb
N0 ) [ 1− 1
2 erfc ( √ Eb
N 0 ) ]
QUESTION 4
Part A
12

The coherently demodulated M-ary PSK modulation.
The set of signals in the M-ary PSK are
ϕi ( t ) =Acos ( ωc t+θi ) 0<t ≤T s
The M phase angles are,
θi =0 , 2 π
M , … , 2 ( M −1 ) π
M
The equiprobables ones and zeros for the PSD are such that the M-ary PSK is,
sϕ ( ω ) = A2 T s S a2
[ ( ω−ωc ) T s
2 ]
The information or bit rate is given as,
T b= T s
log2 M
While the potential bandwidth efficiency is obtained as
f b
B =log2 M bps / Hz
For all the M-ary PSK signal constellations greater than 2, the probability of making a symbol
error is given by,
Pϵ ≈2 erfc √ 2 Es
η sin2 π
M m>2
When the SNR is large, the probability of error is given as,
Pϵ ≈2 erfc √ 2 Es
η sin2 π
√ 2 M
Implementing the gray code, the bit error is approximated to,
Pbe ≈ Pϵ
log2 M
The differential detection that is set to increase the power requirements by certain factors is
given as,
13

Γ =
sin2 π
M
sin2 π
√2 M
Part B
For a coherently demodulated M-ary FSK modulation
The MATLAB implementation is given as,
% for different rolloff values
a=[0 0.15 0.25 0.35 0.45 1.00]
tb=theorySer_QPSK*a(1);
tc=theorySer_QPSK*a(2);
td=theorySer_QPSK*a(3);
te=theorySer_QPSK*a(4);
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tf=theorySer_QPSK*a(5);
figure(2)
plot(Es_N0_dB,tb,'black-*');
hold on
plot(Es_N0_dB,tc,'cyan-*');
hold on
plot(Es_N0_dB,tb,'g-*');
hold on
plot(Es_N0_dB,te,'b-*');
hold on
plot(Es_N0_dB,tf,'magenta-*');
hold on
legend('aplha=0','aplha=0.15','aplha=0.25','aplha=0.35','aplha=0.45','aplha=1.
00');
xlabel('Es/No, dB')
ylabel('Symbol Error Rate')
title('Symbol error probability curve for QPSK')
grid on
QUESTION 5
Part A
Satellite earth station uses a 6.8 meter parabolic reflect with an efficiency of 0.55.
Satellite uplink parameters
1. Transmit power 10 Watts
2. Uplink frequency 30GHz
3. Satellite –earth distance 36000 km
4. Gr/Ts -3dB/K
Satellite downlink parameters
1. Downlink frequency 20GHz
2. Diameter parabolic dish, efficiency 0.55
3. Antenna temperature 50 Kelvin
4. Feedline 0.5dB attenuation
5. Noise figure 2.5dB
Transmit earth station EIRP (dBW)
EIR Ptransmit =PT GT
EIR Ptransmit =10 log ( Ptransmitter ) −Transmitte rfe eder loss
+ Transmitte r Antenna gain
− Antenn apointin gloss
( dBW )
EIR Ptransmit =10 log 10−0.5 dB+45.21−0.70
EIR Ptransmit =54.01 dB
15

The receive earth station Gr
T s
[dB/ K ]
Gr
T s [ dB
K ]=−4 dB/ K
The uplink and downlink free space path loss [dB]
Using the Friis formula,
Pr =Pt (1−|Γt |2
. gt . ( λ
4 πd )2
. 1
αm . Aeff (1−|Γr|2
) )
Aeff = λ2
4 π gr … .effective area of reception
Free space basic transmission loss,
Lbf =20 log10 ( 4 πd
λ )
For the uplink,
Lb f uplink
=20 log10
( 4 π∗36000
c
f )=20 log10 ( 144e6 π
0.01 )
Lb f uplink
=213.11dB
For the downlink,
16

Lb f downlink
=20 log10
( 4 π∗36000
c
f )=20 log10 ( 144e6 π
0.015 )
Lb f downlink
=209.5884 dB
QPSK implementation, data rate of satellite link in bps
The relation between the power of the modulated carrier and the noise power is given as,
C
N 0
=
EIRP
Lb
gr
k . t = EIRP
Lb ( gr
t ) 1
k
The signal to noise ratio is given as,
Eb
N o
=SNR . b
Rb
Rb −signal datarate
Part B
8-ary Phase shift keying modulation with a (94, 64) error correction code which is guaranteed to
correct 6 errors in the 94 bit code-word. The resulting constellation is as plotted below,
17

The error probability of each channel is given by,
P1=Pr {1 bit error |s1 (t)}
¿ Pr {s2 ( t )|s1 ( t ) }+ Pr {s4 ( t )|s1 ( t ) }+ Pr { s8 ( t )|s1 ( t ) }
18

19

20