# Assignment3_Group 1

```Presented by:
Group 2

Two-level PSK (BPSK)

Uses two phases to represent binary digits

binary 1
 A cos2f ct 
s t   
binary 0


A
cos
2

f
t



c

which equals

 A cos2f ct 


 A cos2f ct 
binary 1
binary 0
Where we can consider the above two functions to be multiplied by
+1 and -1 for a binary 1 and binary 0 respectively

Differential PSK (DPSK)

Phase shift with reference to previous bit
▪
▪
Binary 0 – signal burst of same phase as previous signal
burst
Binary 1 – signal burst of opposite phase to previous signal
burst

The term differential is used because the phase shift
is with reference to the previous bit

Doesn’t require an accurate receiver oscillator matched with
the transmitter for the phase information but obviously
depends to the preceding phase (information bit) being
 Four-level

Each element represents more than one bit


st   




A cos 2f c t  
4

3 

A cos 2f c t 

4 

3 

A cos 2f c t  
4 



A cos 2f c t  
4

11
01
00
10
I stream (in-phase)
 OQPSK
has phase transitions between every halfbit time that never exceeds 90 degrees (π/2 radians)
 Results in much less amplitude variation of the
bandwidth-limited carrier
 BER is the same as QPSK
 When amplified, QPSK results in significant
bandwidth expansion, whereas OQPSK has much
less bandwidth expansion especially if the channel
has non-linear components

Multilevel PSK

Using multiple phase angles with each angle having
more than one amplitude, multiple signals elements can
be achieved
R
R
D 
L log 2 M
D = modulation rate, baud
▪ R = data rate, bps
(note the difference in baud and bps)
▪ M = number of different signal elements = 2L
▪ L = number of bits per signal element
If L = 4 bits in each signal element using M = 16 combinations of
amplitude and phase, then if the data rate is 9600 bps,
the line signaling speed/modulation rate is 2400 baud
▪

 QAM

is a combination of ASK and PSK
Two different signals sent simultaneously on the same
carrier frequency
st   d1 t cos 2f ct  d 2 t sin 2f ct
BPSK transmission
-1.2
10
-1.3
10
-1.4
BER
10
-1.5
10
-1.6
10
-1.7
10
-1.8
10
0
5
10
Eb/N0 (dB)
15
BPSK transmission under one path fading
0
BER
10
-1
10
-2
10
0
5
10
Eb/N0 (dB)
15
 Program
BPSK performance
0
10
BPSK
BPSK
BPSK
BPSK
-1
BER
10
AWGN theory
AWGN
Rayleigh theory
Rayleigh
-2
10
-3
10
-4
10
0
5
10
15
20
E b/N0 [dB]
25
30
35
40
BPSK In Flat Fading AWGN Channel - BER Vs SNR
0
10
-1
10
-2
BER
10
-3
10
-4
10
-5
10
-10
-5
0
5
10
15
SNR
20
25
30
35
40
16 QAM Flat Fading Channel Performance Comparison
0
10
-1
10
-2
BER
10
-3
10
Amplitude Distortion Only
-4
10
Amplitude &amp; Phase Distortion
-5
10
-10
-5
0
5
10
15
SNR
20
25
30
35
40
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-1.5
-1
-0.5
0
In-Phase
0.5
1
Amplitude + Phase Distortion
1.5
-1.5
-1
-0.5
0
In-Phase
0.5
Amplitude Distortion
1
1.5
Input
 sr=256000.0; % Symbol rate
 ipoint=2^03;
% Number of oversamples
 ncc=1;
 %******************* Filter initialization
********************

 irfn=21;
% Number of filter taps
Nyquist Pulses
1.4
Beta=0.5
Beta=0
Beta=1
1.2
1
Pulse Shape
0.8
0.6
0.4
0.2
0
-0.2
-0.4
0
1
2
3
4
5
time
6
7
8
9
-5
x 10
Frequency Response
9
8
7
6
5
4
3
2
1
0
-1.5
-1
-0.5
0
0.5
1
1.5
6
x 10
1.2
Transmitter filter coefficients
1
0.8
0.6
0.4
0.2
0
-0.2
0
20
40
60
80
100
120
140
160
180
 Number
of symbols (nd) = 10
 data=rand(1,nd)&gt;0.5
 data1=data.*2-1
data1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1
2
3
4
5
6
7
8
9
10
 [data2]
= oversamp( data1, nd , IPOINT)
data2
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
0
10
20
30
40
50
60
70
80
 data3
= conv(data2,xh)
data3
1.5
1
0.5
0
-0.5
-1
-1.5
0
50
100
150
200
250
inoise=randn(1,length(data3)).*attn
5
4
3
2
1
0
-1
-2
-3
-4
-5
0
50
100
150
200
250
data4=data3+inoise
5
4
3
2
1
0
-1
-2
-3
-4
-5
0
50
100
150
200
250
data5=conv(data4,xh2)
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2.5
0
50
100
150
200
250
300
350
400
450
data6 = data5(sampl:8:8*nd+sampl-1)
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2.5
1
2
3
4
5
6
7
8
9
10
 demodata=data6
&gt;0
demodata=data6 &gt; 0
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1
2
3
4
5
6
7
8
9
10
Source: Wikipedia

Graphical eye pattern showing an example of
two power levels in an OOK modulation scheme.
Constant binary 1 and 0 levels are shown, as well
as transistions from 0 to 1, 1 to 0, 0 to 1 to 0,
and 1 to 0 to 1
Q-Channel (qch4)
I-Channel (ich4)
2
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
-2
-2.5
-2.5
0
50
100
150
200
250
300
350
400
450
0
50
100
150
200
250
300
350
400
450
 MSK
is a continuous phase FSK (CPFSK) where
the frequency changes occur at the carrier
zero crossings.
 MSK is unique due to the relationship
between the frequency of a logic 0 and 1.


The difference between the frequencies is
always &frac12; the data rate.
This is the minimum frequency spacing that
allows 2 FSK signals to be coherently orthogonal.
 The
baseband modulation starts with a bitstream of
0’s and 1’s and a bit-clock.
 The baseband signal is generated by first
transforming the 0/1 encoded bits into -1/1 using
an NRZ filter.
 This signal is then frequency modulated to produce
the complete MSK signal.
 The amount of overlap that occurs between bits
will contribute to the inter-symbol interference
(ISI).
 1200
bits/sec baseband MSK data signal
 Frequency spacing = 600Hz
a) NRZ data
b) MSK signal
 Since
the MSK signals are orthogonal and minimal
distance, the spectrum can be more compact.
 The detection scheme can take advantage of the
orthogonal characteristics.
 Low ISI (compared to GMSK)
 The
fundamental problem with MSK is that the
spectrum has side-lobes extending well above
the data rate (See figure on next slide).
 For wireless systems which require more
efficient use of RF channel BW, it is necessary
to reduce the energy of the upper side-lobes.
 Solution – use a pre-modulation filter:


Straight-forward Approach: Low-Pass Filter
More Efficient/Realistic Approach: Gaussian Filter

Gaussian Filter


Impulse response defined by a Gaussian
Distribution – no overshoot or ringing
(see lower figure)
BT – refers to the filter’s -3dB BW and
data rate by:
BT 





f 3dB
BitRate
Notice that a bit is spread over more
than 1 bit period. This gives rise to ISI.
interfere with each other more than for
BT=0.5
GMSK with BT=∞ is equivalent to MSK.
suppression (top and bottom figures)
The higher the ISI, the more difficult the
detection will become.
 An
important application of GMSK is GSM,
which is a time-division multiple-access
system.
 For this application, the BT is standardized
at 0.3, which provides the best compromise
between increased bandwidth occupancy and
resistance to ISI.
 Ninety-nine percent of the RF power of GMSK
signals is specified to confine to 250kHz (+/25kHz margin from the signal), which means
that the sidelobes need to be virtually zero
outside this frequency band and the ISI
should be negligible.
 The
program bpsk.m prints the BER in each
simulation loop, and this causes the program
to run slowly, therefore, I stopped printing
those results. Instead, I plotted the BER vs.
EbN0 with a counter that displays the current
value of EbN0.
I
tried to plot the eye diagram for QPSK, but
I didn’t succeed in that.
Wikepedia.com
 Haykin, S. 2001: “Communication Systems”. 4th
ed. New York, NY. John Wiley &amp; Sons.
 Introduction to GMSK
www.eecs.tufts.edu/~gcolan01
 GMSK: Practical GMSK Data Transmission
http://www.eetchina.com/ARTICLES/2003AUG/PD
F/2003AUG29_NTEK_AN.PDF
 Minimum Shift Keying: A Spectrally Effiecient
Modulation