Unit 2 Notes (1). .p df ppt presentation

Presentation

1. Unit II Digital Modulation-I (07 Hrs.)
Baseband Signal Receiver: Probability of Error, Optimal Receiver Design.
Digital Modulation: Generation, Reception, Signal Space Representation and
Probability of Error Calculation for Binary Phase Shift Keying (BPSK),
Binary Frequency Shift Keying (BFSK), Quadrature Phase Shift Keying
(QPSK), M-ary Phase Shift Keying (MPSK).
CO2: Understand and explain various digital modulation techniques used in
digital communication systems and analyze their performance in presence of
AWGN noise.

2. Learning Resources

Text Books:
1. Taub, Schilling and Saha, “Principles of Communication Systems”, McGraw-Hill, 4 th Edition,
2. B.P. Lathi, Zhi Ding , “Modern Analog and Digital Communication System”, Oxford University
Press, 4th Edition.

Reference Books:
1. Bernard Sklar, Prabitra Kumar Ray, “Digital Communications Fundamentals and Applications”,
Pearson Education, 2nd Edition
2. Wayne Tomasi, “Electronic Communications System”, Pearson Education, 5th Edition
3. A.B Carlson, P B Crully, J C Rutledge, “Communication Systems”, Tata McGraw Hill Publication, 5t
h Edition
4. Simon Haykin, “Communication Systems”, John Wiley & Sons, 4th Edition 5. Simon Haykin,
“Digital Communication Systems”, John Wiley & Sons, 4 th Edition.

MOOC / NPTEL Courses:
1. NPTEL Course on “Digital Communications” Link of the Course: https://nptel.ac.in/courses/108/102/108102096/

3. Objectives
• To understand base band receiver design and analysis of performance parameter
probability of error.
• To study generation and reception of digital modulation methods of BPSK, BFSK,
QPSK and M-ary PSK.
• To study signal space representation and its significance for various digital
modulation techniques.
• Derive mathematical expression for Probability of digital modulation techniques
BPSK, BFSK, QPSK and M-ary PSK.

4. Outcomes
[Contributing to Placement, Higher Education, Entrepreneurship]
Students Will
• Able to analyze baseband receiver for digital communication system.
• Compute probability of error for optimum filter
• Appraise the concept of generation and reception of different digital modulation
techniques such as BPSK, BFSK, QPSK, MPSK.
• Understand significance of signal space representation and power spectral density of
BPSK, BFSK, QPSK, MPSK.
• Compute probability of error for digital modulation techniques.

5. Base Band Receiver

6. 6
Detection of Binary Signal in Presence of Noise






• For any binary channel, the transmitted signal over a symbol interval (0,T) is:
s (t) =
 s
1 (t) 0  t  T for a binary 1
i 
s
(t) 0  t  T for a binary 0
 2
• The received signal r(t) degraded by noise n(t) and possibly degraded by the impulse response of the
channel h
c(t), is
r(t) = s
i (t) * h
c (t) + n(t) i = 1,2
Where n(t) is assumed to be zero mean AWGN process
• For ideal distortion less channel where h
c(t) is an impulse function and convolution with h
c(t) produces
no degradation, r(t) can be represented as:
r(t) = s
i (t) + n(t) i = 1,2 0  t  T

7. Detection of Binary Signal in Presence of Noise
• The recovery of signal at the receiver consist of two parts
• Filter
• Reduces the effect of noise (as well as Tx induced ISI)
• The output of the filter is sampled at t=T. This reduces the received signal to a single
variable z(T) called the test statistics

• Detector (or decision circuit)
• Compares the z(T) to some threshold level 0 , i.e.,
H
1

z(T )


 0
H
2

where H1 and H2 are the two possible binary hypothesis





7

8. Optimum Filter
• It is used for detecting baseband signals
• Requirements of signal receivers must have
• Improved SNR
• Checked at instant in bit period when SNR is Maximum
• Minimum Pe
• Decision Threshold in Optimum Filter
• Let us assume that received signal is binary waveform & is represented in Polar NRZ
signaling
• For Binary 1 : S1(t) = +A for T duration
• For Binary 0 : S2(t) = -A for T duration

9. Optimum Filter
• Thus the input signal can either be S1(t) or S2(t) depending upon the polarity of
NRZ Signaling






• In absence of Noise, receiver output is
• r(T) = S01(T) if s(t) = s1(t)
• r(T) = S02(T) if s(t) = s2(t)
• Thus in absence of noise, decisions are taken clearly.

10. • In presence of noise
Optimum Filter
• Select S1(t) if r(T) is closer to S01(T) than S02(T)
• Select S2(t) if r(T) is closer to S02(T) than S01(T)
• Therefore the decision boundary will be the midway between S01(T) &
S02(T)
i.e. decision boundary = (S01(T) + S02(T))/2

11. Probability of Error for Optimum Filter
• Suppose that S2(t) was transmitted, but S01(T) is greater than S02(T).
• If noise no(T) is positive & larger in magnitude then the voltage
difference [(S01(T)+S02(T))/2] - S02(T) then incorrect decision will be
taken.
• Error will be generated if
• no(T) ≥ [(S01(T) + S02(T))/2]-S02(T)
• no(T) ≥ [(S01(T) - S02(T))/2] …………… 1

12. Probability of Error for Optimum Filter

• Similarly, Suppose that S1(t) was transmitted, but S02(T) is greater
than S01(T).
• Incorrect decision will be taken if
• no(T) ≤ [(S02(T) - S01(T))/2] …………….. 2


Probability of error for S1(t) & S2(t) are obtained on evaluating Probability of
equation 1 & 2 respectively.

13. Probability of Error for Optimum Filter







P[n0(t)]=(S01(T)-S02(T))/2















S
02 (T ) − S
01 (T )
2
S
01 (T ) − S
02 (T )
2

14. Probability of Error for Optimum Filter
Gaussian Noise: pdf of a Gaussian distributed function for random variable x
is given by,
1

1  x − m 
2


fX (x) =

0

exp − 
2 

2  
0

 
 


To calculate pdf of a white Gaussian noise we replace x = no(T) & zero
mean, i.e. m=0


1

1  no(T)

2


fX (no(T)) =

0

exp − 
2 

2  
0




 


15. 
Probability of Error for Optimum Filter

• P[no(T) ≥ [(S
01(T) - S
02(T))/2] ] =


fX (no(T))dno(T)

P
e =


e
− n
0
2
(T )/ 2
2

2
S
01 (T )− S
02 (T )
2

dn
0 (T )

n (T )
S
01 (T )− S
02 (T )
2
2
x =
0
2
• To solve the above integral Put
• So integrals will change to
x =
n
0 (T )
2
dx = d (
n
0 (T )
)

2
n0 (T ) =   x =  dx =
1
dn
0

2
(T )
n (T ) = S
01 (T ) − S
02 (T )
 x =
S
01 (T ) − S
02 (T )
0
2 2 2
dn0 (T ) = 2dx

16. 
1 2
Probability of Error for Optimum Filter


P
e =
2 



e
−

x
2
dx
S
01 (T )− S
02 (T )
2 2

We use the standard erfc function

erfc( x) =
2
e
−

x
2
dx
 x

Q( x) =
1
erfc(
x
)
Probability of Error for Optimum Filter is given by
2 2

P =
1
erfc(
S
01 (T ) −
S
02 (T )
)

e
2 2 2
Optimum Filter has to maximize the ratio in such a way that Pe becomes maximum

17. �
�
?????? − �
Matched Filter
• An optimum filter yields a maximum ratio �
2(�)/??????
2 is called a matched filter when the
0 0
input noise is white.
• In other words the matched filter is the optimal filter for maximizing the signal-to-noise
ratio (SNR) in the presence of additive white Noise.
• So the noise spectral density will be �
?????? = ??????0/2.
• Transfer function of matched filter is given by
• ?????? = �
??????
∗
(�)
�
−�2���
??????0/2
• The impulsive response of the filter can be calculated by taking inverse Fourier transform
of H(f).
• ℎ = ??????
−1
??????
=
2??????
׬
∞
X
∗

�
−�2���
�
�2���
��

??????0
−∞
• ℎ(�) =
2??????
׬
∞
X
∗

�
�2��(�−�)
�� X
∗
= X

??????0
−∞
• ℎ(�) =
2??????
׬
∞

�
�2��(�−�)
��

??????0
−∞
• Transfer Function of Matched filter �� � =
????????????
??????

???????????? 17
�
� � �
� �
�
�

18. Matched Filter

• The impulse response of a filter producing maximum output signal-to-
noise ratio is the mirror image of message signal s(t), delayed by symbol
time duration T.

• The filter designed is called a MATCHED FILTER

ℎ(�) = ቊ
� �(� − �) 0 ≤ � ≤ �
0 ���� �ℎ���











18

19. Realization of Matched filter
• A filter that is matched to the waveform s(t), has an impulse response

ℎ(�) = ቊ
� �(� − �) 0 ≤ � ≤ �
0 ���� �ℎ���
• h(t) is a delayed version of the mirror image of the original signal
waveform




Signal Waveform
Mirror image of signal
waveform
Impulse response of
matched filter



19

20. �
�
Correlator
• In correlator the received signal plus noise n(t) is multiplied by a locally generated
waveform s1(t)-s2(t).
• The output of the multiplier is passed through an integrator whose output is sampled at
t=T.
• This type of receiver is called a correlator since we are correlating the received signal and
noise with the waveform s1(t) –s2(t).
• The output signal and noise of the correlator shown
• � =
1
׬
�
[ � - � ]dt where Si(t) is either S1(t) ot S2(t)
0

?????? 0
� 1 2
• ?????? =
1
׬
�
[ � - � ]dt
0

?????? 0
1 2


� � �
� � �

21. Correlator

• The matched filter and correlator are not simply two distinct, independent
techniques which happen to yield the same result. In fact they are two techniques of
synthesizing the optimum filter h(t).

22. Digital Modulation

23. Digital modulation and demodulation:
– Modulation maps the digital information into an analog waveform appropriate for
transmission over the channel. Examples: ASK, FSK, PSK
In CW modulation , PM and FM signals are difficult to distinguish. In PSK and
FSK both have constant envelope while ASK does not.

– Demodulation recover the baseband digital information from a bandpass analog
signal at a carrier frequency that is very high compared to the baseband frequency.

24. Digital Modulation Techniques
➢ Modulation involves operations on one or more of the three characteristics of a carrier
signal: Amplitude, Frequency and Phase.
➢ The three basic modulation methods are:
– Amplitude Shift Keying (ASK)
– Phase Shift Keying (PSK)
–Frequency Shift Keying (FSK)
➢ These could be applied to binary or M-ary signals.

25. Waveforms for the three basic forms of signaling binary information
(a) Amplitude-shift keying.
(b) Phase-shift keying.
(c) Frequency-shift keying with continuous phase.

26. Binary Phase-Shift Keying (BPSK)

• In BPSK the transmitted signal is a sinusoid of fixed amplitude.
• It has one fixed phase when the data is at one level and when the data is at other
level the phase is different by 180 degree.

27. : t
,
Binary Phase-Shift Keying (BPSK)
▪ BPSK equation can be written as

V
BPSK
= b(t )
2E
b

T
b

cos(2f
c
t +)
▪ Where b(t) will be either 1 or -1

“1”

“0”

▪ , bit duration
▪ : carrier frequency, chosen to be for some fixed
integer or fc  1 / Tb
▪ transmitted signal energy per bit, i.e.
▪ The pair of signals differ only in a relative phase shift of
180 degrees

28. BPSK Transmitter

29. BPSK Receiver

30. Spectral Density of BPSK

31. PSD of NRZ Data b(t) & Binary PSK






BW= 2fb

32. Signal Space Representation for BPSK

There is one basis function of unit energy



▪ Then

▪ A binary PSK system is therefore characterized by having a signal space that is one-
dimensional (i.e. N=1), and with two message points (i.e. M = 2)




s
2
0
s
1

33. Decision rule of BPSK

▪ Assume that the two signals are equally likely, i.e. P(s1) = P(s2) = 0.5. Then the
optimum decision boundary is the midpoint of the line joining these two message
points
Region R2 Region R1


s
2
0 s
1


▪ Decision rule:
▪ Signal s1(t) (or binary 1) was transmitted if the received signal point r falls in
region R1
▪ Signal s2(t) (or binary 0) was transmitted otherwise

34. Probability of Error for BPSK
▪ The conditional probability of the receiver deciding in favor of symbol s2(t) given that s1(t)
is transmitted is










0
r

35.  
Probability of Error for BPSK


BPSK signal for 1 and 0



Probability of Error function for Optimum Filter



 x (T ) − x (T ) 
2


X ( f )
2
Maximum Signal to noise ratio of Optimum filter.
01 02
= 
df

  max
−
S
ni ( f )


PSD of White noise input is S
ni(f)=N0/2

36. X(t)=s1(t)-s2(t)
Probability of Error for BPSK







Probability of Error for BPSK is given by

37. Quadrature Phase Shift Keying (QPSK)
▪ PSK that uses phase shifts of 90º=π/2 rad
▪ 4 different signals generated, each representing 2 bits





▪ Advantage: higher data rate than in PSK (2 bits per bit interval), while bandwidth
occupancy remains the same
▪ 4-PSK can easily be extended to 8-PSK, i.e. n-PSK
▪ However, higher rate PSK schemes are limited by the ability of equipment to
distinguish small differences in phase

38. Mathematical Representation of QPSK

39. QPSK Waveforms

40. QPSK Transmitter

41. QPSK Receiver

42. 2 2
Signal Space Representation




Four quadrature signals v (t) = 2P cos

t + (2m +1)

m = 0,1, 2, 3
m s

U sin g two ort ho normal signals


0
4




u
1
(t) =
T
cos
0
t and u
2
(t) =
T
sin
0
t
v (t) =

PT cos(2m +1)


2
cost −

PT sin(2m +1)


2
sint
m 

s
4

 T
0 

s
4

 T
0

If b
e
=
2 cos(2m +1)


4
and b
o
= −
2 sin(2m +1)


4
then v
m
(t) = E
b
b
e
(t)u
1
(t) − E
b
b
o
(t)u
2
(t)

43. Signal Space Representation






d = 2

44. p. 44

45. PSD of QPSK





???????????? =
??????????????????
????????????�?????? ??????
In QPSK M=4

???????????? =
??????????????????
????????????�?????? ??????

???????????? =
??????????????????
??????
???????????? = ????????????







p. 45

46. Probability of Error for QPSK
Error Probability of QPSK is given by
Pe
QPSK
= 2 x Pe
BPSK

P
eQPSK
= 2 x
1
erfc
2
E
2N
0



P
eQPSK = erfc
E
2N
0


In QPSK each symbol is of 2 bit duration long so E=2Eb
Therefore probability of error for QPSK is


P
eQPSK
= erfc
E
b

N
0

47. M ary PSK (MPSK)

• In M-ary data transmission, it sends one of M possible signals during
each signaling interval T.
• M=2^n and T=nTb, where n is an integer.
• Each of the M signals is called as symbol.
• In MPSK the phase of the carrier takes on M possible values.



m =
2(m
M
− 1)
, m

= 1,2...M

48. M ary PSK (MPSK)

The mathematical equation for MPSK is given by

s
m (t ) =
2E
s

T
cos[2f
c t +
2(m
M
− 1)
]

49. MPSK Transmitter

50. MPSK Receiver

51. Signal Space Representation

52. Signal Space Representation

53. Power Spectral Density of MPSK













BW =
2 fb
log
2 M

54. Probability of Error for MPSK

• Probability of MPSK is given by

55. Binary Frequency Shift Keying

• The output of a FSK modulated wave is high in frequency for a binary High
input and is low in frequency for a binary Low input.
• The binary 1s and 0s are called Mark and Space frequencies.


0 1 1 0 1 0 0 1

56. Binary Frequency Shift Keying

57. Binary Frequency Shift Keying

• Two balanced modulators are used one with carrier W
H and one with
carrier W
L.
• The voltage values of P
H(t) and P
L(t) are related to the voltage values
of d(t) in the following manner


d(t) P
H(t) P
L(t)
+1 V +1V 0V
-1V 0V +1V

58. Binary FSK Transmitter

59. Non Coherent BFSK Receiver

60. Coherent Binary FSK Receiver

61. Power Spectral Density of BFSK

62. Power Spectral Density of BFSK

63. PSD of BFSK







BW= 4fb

64. Signal Space for BFSK
▪ Unlike BPSK, here two orthogon normal basis functions are required to represent
s1(t) and s2(t).






▪ Signal space representation

65. ▪ Signal space diagram for binary FSK


Message point


Message point

66. Decision Regions of Binary FSK





Message
point
R
2
Decision boundary

R1

Message point



▪ The receiver decides in favor of s1 if the received signal point represented by the
observation vector r falls inside region R1. This occurs when r1 > r2
▪ When r1 < r2 , r falls inside region R2 and the receiver decides in favor of s2

67. Probability of Error for Binary FSK
• The binary FSK signal is represented as
• For binary 1
• For binary 0
S
H (t ) =
S
L (t ) =
2P
s
2P
s

cos(2f
H t )
cos(2f
Lt )
Probability of Error function for Optimum Filter
 S
01 (T ) − S
01 (T ) 
=
0.6Eb

 2 2

 No
Error Probability of BPSK is given by
Non-orthogonal FSK Orthogonal FSK


P
e =
1
erfc(
2
0.6E
b

N
0

P
e =
1
erfc(
2
E
b

2N
0

68. Probability of Error and the Distance Between Signals

Parameters BPSK QPSK BFSK
d
12 2E
b
2E
b
2E
b

P
e
½ erfc(E
b /N
o ) erfc(E
b /N
o ) ½ erfc(0.6E
b /N
o )
Bandwidth requirement 2f
b
f
b
4f
b

Modulation Linear Linear Non linear

69. Applications
• QPSK is used in various cellular wireless standards such as GSM, CDMA, LTE,
802.11 WLAN, 802.16 fixed and mobile WiMAX, Satellite and CABLE TV
applications.
• BPSK modulation is a very basic technique used in various wireless standards
such as CDMA, WiMAX (16d, 16e), WLAN 11a, 11b, 11g, 11n, Satellite, DVB,
Cable modem etc.
• Telephone line modem use FSK to transmit 300 bits/sec at two frequencies
1070Hz & 1270 Hz

70. Thank You….. !!