1. Vector Concept of Signals

2. Topics to be Covered Vector space concept
Representation of signals using vectors
Representation of noise using vectors
Representation of modulation schemes using vector space concept 2

3. Signal Space
Digitally modulated signal consists of selection / transmission of one out of a finite set of elements.
The signals space concept is similar to vector space.
Concept of vector space can be applied to represent digital modulated signals.
It allows convenient visualization, mathematical modeling and comparison of various modulation schemes
Energy

4. Orthogonal Space of Basis Functions
An N-dim orthogonal space consisting of linearly independent basis functions
Basis functions are denoted by
Basis functions must satisfy the following conditions:
When Kj are nonzero, the signal space is called Orthogonal; When Kj=1, the signal space is called Orthonormal.

5. Orthogonal Space of Basis Functions
Any finite set of waveforms can be expressed as a linear combination of N orthogonal waveforms
Also called Generalized Fourier Series

6. Energy of a Signal
For orthonormal functions

7. Topics to be Covered Matched filter Correlator receivers
Probability of error 7

8. OPTIMUM RECEIVER FOR BINARY DIGITAL MODULATION SCHEMES
The function of a receiver in a digital communication
system is to distinguish between all the M transmitted signals sm(t), in the presence of noise
The performance of the receiver is usually measured
in terms of the probability of error and the receiver
is said to be optimum if it yields the minimum
probability of error
m=1,2,…,M 4

9. Received Signal + Noise Vector
The shape of the waveforms depends on the type of the modulation used .The output of the modulator passes through a bandpass channel Hc(f) which
for purposes of analysis is assumed to be an ideal channel with adequate bandwidth so the signal passes through without suffering any distortion other then propagation delay .the channel nosie n(t) is assumed to be a zero mean stationary, Gaussian random process with a known power spectral .(density Gn(f 8

10. Demodulation of Digital Signals

11. Additive white Gaussian noise (AWGN) with zero mean and variance N0 /2
Matched Filter
Detection of signal in presence of additive noise
Receiver knows what pulse shape it is looking for
Channel memory ignored (assumed compensated by other means, e.g. channel equalizer in receiver)
s(t)
Transmitted signal
w(t)
x(t)
h(t)
y(t)
t = T
y(T)
Matched filter
T is the symbol period
Additive white Gaussian noise (AWGN) with zero mean and variance N0 /2

12. Matched Filter Derivation
Design of matched filter
Maximizes signal power i.e. power at t = T
Minimizes noise i.e. power of noise
Combined design criteria

13. Matched Filter Derivation
PSDs are related by
Noise power spectrum SW(f)
AWGN
Matched filter

14. Matched Filter Derivation
Find h(t) that maximizes pulse peak SNR

15. Matched Filter Derivation
Schwartz’s inequality
For functions: upper bound reached if

16. Matched Filter Derivation
For real s(t)

17. Matched Filter
Given transmitted signal s(t) of duration T, matched filter is given by hopt(t) = k s(T-t) for any k
Duration and shape of impulse response of the optimal filter is determined by pulse shape s(t)
hopt(t) is scaled, time-reversed, and shifted version of s(t)
Optimal filter maximizes peak pulse SNR
SNR Does not depend on pulse shape s(t)
Directly proportional to signal energy (energy per bit) E
Inversely proportional to power spectral density of noise

18. Matched Filter Receiver

19. Matched filter Example
Received SNR is maximized at time T
Matched Filter: optimal receive filter for maximized
example:
transmit filter
receive filter
(matched)

20. Correlation Filter
At t=T
At t=T, matched filter becomes a correlator

21. Correlator and Matched Filter Outputs for a Sine Wave Input

22. Matched Filter Receiver

23. Matched filter Example
Received SNR is maximized at time T
Matched Filter: optimal receive filter for maximized
example:
transmit filter
receive filter
(matched)

24. Correlation Filter
At t=T
At t=T, matched filter becomes a correlator

25. Correlator and Matched Filter Outputs for a Sine Wave Input

26. Channel Model AWGN channel model: Signal vector is deterministic.
Elements of noise vector are i.i.d (independent and identically distributed) Gaussian random variables with zero-mean and variance The noise vector pdf is
The elements of observed vector are independent Gaussian random variables. Its pdf is

27. Correlation Receiver for BPSK/BASK
One basis function, so one branch
The received signal is input to a correlator
The correlator output is compared with a threshold to decide which symbol was transmitted.
Decision
Circuit
Compare r
with threshold.
Sample at T

28. Probability of Error BPSK/BASK
Output of the correlator:
a1(T) is the correlator output when s1(t) is sent
a2(T) is the correlator output when s2(t) is sent
n0(T) is the correlator output when AWGN is present at the input
Hypotheses:
H1: s1 is assumed to have been transmitted
H2: s2 is assumed to have been transmitted 29

29. Probability of Error: BPSK/BASK30

30. Probability of Error: BPSK/BASK31

31. Probability of Error BPSK/BASK32

32. Probability of Error BPSK/BASK
Where Q(x) is the Complementary Error Function
Now
And for BPSK 33

33. Coherent detection of BFSK
For correct detection, the carriers must be orthogonal.

34. Error Probability BPSK and BFSK
BPSK and BFSK with coherent detection:
“0”
“1”
“0”
“1”
BPSK
BFSK

35. Assignment#4 Due Before start of next class on Saturday 23 April, 2011
Q 1: Show that the probability of error for coherently detected BFSK is given by
Q 2: Derive the value of decision threshold if the binary transmitted symbols in BPSK are not equiprobable.
Hint: See Sec. 3.2 and App B.2.

36. Quiz#4
Q : (a) A BASK signal transmits 2000 bps. If the energy per bit is 4 Joules and the noise PSD is 1 W/Hz, how many erroneous bits will be received in 1 second?
(b) Also sketch the PDFs of the received signal for both the possible transmitted signals and indicate the area that corresponds to the probability of correct decision.

37. Coherent Detection of MPSK
Information resides in the phase, which needs to be detected
“110”
“000”
“001”
“011”
“010”
“101”
“111”
“100”
8-PSK
Compute
Choose
smallest
Decision variable

38. Coherent Detection of M-QAM (Amplitude Phase Shift Keying)
16-QAM

39. Signal space dimension
How many basis functions does it take to express a signal? It depends on the dimensionality of the signal
Some need just 1 some need an infinite number.
The number of dimensions is N and is always less than the number of signals in the set
N<=M
©2000 Bijan Mobasseri

40. Example: Fourier series
Remember Fouirer series? A signal was expanded as a linear sum of sines and cosines of different frequencies. Sounds familiar?
Sines and cosines are the basis functions and are in fact orthogonal to each other
©2000 Bijan Mobasseri

41. Interpretation
s1(t) is now condensed into just two numbers. We can “reconstruct” s1(t) like this
s1(t)=(1)1(t)+(-0.5)2(t)
Another way of looking at it is this 2 1 1
-0.5
©2000 Bijan Mobasseri

42. Signal constellation
Finding individual components of each signal along the two dimensions gets us the constellation 2 s4 s2 1
-0.5
0.5
-0.5 s1 s3
©2000 Bijan Mobasseri

43. Energy in simple language
What we just saw says that the energy of a signal is simply the square of the length of its corresponding constellation vector 2
E=9+4=13 3
©2000 Bijan Mobasseri

44. Constrained energy signals
Let’s say you are under peak energy Ep constraint in your application. Just make sure all your signals are inside a circle of radius sqrt(Ep )
©2000 Bijan Mobasseri

45. Correlation of two signals
A very desirable situation in is to have signals that are mutually orthogonal. How do we test this? Find the angle between them
transpose s1 s2 
©2000 Bijan Mobasseri

46. Find the angle between s1 and s2
Given that s1=(1,2)T and s2=(2,1)T, what is the angle between the two?
©2000 Bijan Mobasseri

47. Distance between two signals
The closer signals are together the more chances of detection error. Here is how we can find their separation 2 1
©2000 Bijan Mobasseri

48. Constellation building using correlator banks
We can decompose the signal into its components as follows s1 1 s2
N components
s(t) 2 sN N
©2000 Bijan Mobasseri

49. Detection in the constellation space
Received signal is put through the filter bank below and mapped to a point s1 1 s2
s(t) 2
components
mapped to a single point sN N
©2000 Bijan Mobasseri

50. Constellation recovery in noise
Assume signal is contaminated with noise. All N components will also be affected. The original position of si(t) will be disturbed
©2000 Bijan Mobasseri

51. Actual example
Here is a 16-level constellation which is reconstructed in the presence of noise
©2000 Bijan Mobasseri

52. Detection in signal space
One of the M allowable signals is transmitted, processed through the bank of correlators and mapped onto constellation question is based on what we see , what was the transmitted signal?
received signal
which of the four did it
come from
©2000 Bijan Mobasseri

53. Minimum distance decision rule
It can be shown that the optimum decision, in the sense of lowest BER, is to pick the signal that is closest to the received vector. This is called maximum likelihood decision making
this is the most likely
transmitted signal
received
©2000 Bijan Mobasseri

54. Defining decision regions
An easy detection method, is to compute “decision regions” offline. Here are a few examples
decide s2
decide s1
decide s1 s2 s1
decide s1
measurement s2 s1 s1 s4 s3
decide s2
decide s3
decide s4
©2000 Bijan Mobasseri

55. More formally...
Partition the decision space into M decision regions Zi, i=1,…,M. Let X be the measurement vector extracted from the received signal. Then
if XZi
si was transmitted
©2000 Bijan Mobasseri

56. P{error|si}=P{X does not lie in Zi|si was transmitted}
Error probability
we can write an expression for error like this
P{error|si}=P{X does not lie in Zi|si was transmitted}
Generally
©2000 Bijan Mobasseri

57. Example: BPSK (Binary Phase Shift Keying)
BPSK is a well known digital modulation obtained by carrier modulating a polar NRZ signal. The rule is
1: s1=Acos(2πfct)
0:s2= - Acos(2πfct)
1’s and 0’s are identified by 180 degree phase reversal at bit transitions
©2000 Bijan Mobasseri

58. Signal space for BPSK
Look at s1 and s2. What is the basis function for them? Both signals can be uniquely written as a scalar multiple of a cosine. So a single cosine is the sole basis function. We have a 1-D constellation
cos(2pifct) -A A
©2000 Bijan Mobasseri

59. Eb= A2Tb/2 --->A=sqrt(2Eb/Tb)
Bringing in Eb
We want each bit to have an energy Eb. Bits in BPSK are RF pulses of amplitude A and duration Tb. Their energy is A2Tb/2 . Therefore
Eb= A2Tb/2 --->A=sqrt(2Eb/Tb)
We can write the two bits as follows
©2000 Bijan Mobasseri

60. BPSK basis function
As a 1-D signal, there is one basis function. We also know that basis functions must have unit energy. Using a normalization factor
E=1
©2000 Bijan Mobasseri

61. Formulating BER BPSK constellation looks like this -√Eb √Eb received
if noise is negative enough, it will push
X to the left of the boundary, deciding 0
instead
X|1=[√Eb+n,n]
noise
-√Eb
√Eb
noise
transmitted
©2000 Bijan Mobasseri

62. Finding BER Let’s rewrite BER
But n is gaussian with mean 0 and variance No/2
-sqrt(Eb)
©2000 Bijan Mobasseri

63. BER for BPSK
Using the trick to find the area under a gaussian density(after normalization with respect to variance)
BER=Q[(2Eb/No)0.5]
©2000 Bijan Mobasseri

64. BPSK Example
Data is transmitted at Rb=106 b/s. Noise PSD is 10-6 and pulses are rectangular with amplitude 0.2 volt. What is the BER?
First we need energy per bit, Eb. 1’s and 0’s are sent by
©2000 Bijan Mobasseri

65. Solving for Eb Since bit rate is 106, bit length must be 1/Rb=10-6
Therefore,
Eb=20x10-6=20 w-sec
©2000 Bijan Mobasseri

66. Solving for BER Noise PSD is No/2 =10-6. We know for BPSK
BER=0.5erfc[(Eb/No)0.5]
What we have is then
Finish this using erf tables
©2000 Bijan Mobasseri

67. M-ary signaling
Binary communications sends one of only 2 levels; 0 or 1
There is another way: combine several bits into symbols
Combining two bits at a time gives rise to 4 symbols; a 4-ary signaling
©2000 Bijan Mobasseri

68. QPSK constellation 00 01 45o √E 10 11 Basis functions
S=[0.7 √E,- 0.7 √E]
©2000 Bijan Mobasseri

69. QPSK decision regions 00 01 10 11 Decision regions re color-coded
©2000 Bijan Mobasseri

70. QPSK error rate Symbol error rate for QPSK is given by
This brings up the distinction between symbol error and bit error. They are not the same!
©2000 Bijan Mobasseri

71. Symbol error
Symbol error occurs when received vector is assigned to the wrong partition in the constellation
When s1 is mistaken for s2, 00 is mistaken for 11 s2 s1 11 00
©2000 Bijan Mobasseri

72. Symbol error vs. bit error
When a symbol error occurs, we might suffer more than one bit error such as mistaking 00 for 11.
It is however unlikely to have more than one bit error when a symbol error occurs 00 10 10 11 10
Sym.error=1/10
Bit error=1/20 00 11 10 11 10
10 symbols = 20 bits
©2000 Bijan Mobasseri

73. QPSK vs. BPSK Let’s compare the two based on BER and bandwidth
BER Bandwidth
BPSK QPSK BPSK QPSK
Rb Rb/2
EQUAL
©2000 Bijan Mobasseri

74. M-phase PSK (MPSK)
If you combine 3 bits into one symbol, we have to realize 23=8 states. We can accomplish this with a single RF pulse taking 8 different phases 45o apart
©2000 Bijan Mobasseri

75. 8-PSK constellation
Distribute 8 phasors uniformly around a circle of radius √E
45o
Decision region
©2000 Bijan Mobasseri

76. Quadrature Amplitude Modulation (QAM)
MPSK was a phase modulation scheme. All amplitudes are the same
QAM is described by a constellation consisting of combination of phase and amplitudes
The rule governing bits-to-symbols are the same, i.e. n bits are mapped to M=2n symbols
©2000 Bijan Mobasseri

77. 16-QAM constellation using Gray coding
16-QAM has the following constellation
Note gray coding where adjacent symbols
differ by only 1 bit
0000
0001
0011
0010
1000
1001
1011
1010
1100
1101
1111
1110
0100
0101
0111
0110
©2000 Bijan Mobasseri

78. Vector representation of 16-QAM
There are 16 vectors, each defined by a pair of coordinates. The following 4x4 matrix describes the 16-QAM constellation
©2000 Bijan Mobasseri

79. What is energy per symbol in QAM?
We had no trouble defining energy per symbol E for MPSK. For QAM, there is no single symbol energy. There are many
We therefore need to define average symbol energy Eavg
©2000 Bijan Mobasseri

80. Eavg for 16-QAM
Using the [ai,bi] matrix and using E=ai^2+bi^2 we get one energy per signal
Eavg=10
©2000 Bijan Mobasseri

81. Relation between Eb/N0 & S/N
With R=Rb
In M-ary schemes
In most cases e.g. FSK

82. Time Division Multiplexing