1. ASK, OOK, MASK
 The amplitude (or height) of the sine wave varies to transmit the
ones and zeros
 One amplitude encodes a 0 while another amplitude encodes a 1
(a form of amplitude modulation)
EE 541/451 Fall 2006

2. Binary amplitude shift keying, Bandwidth
 d ≥ 0  related to the condition of the line
B = (1+d) x S = (1+d) x N x 1/r
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3. Implementation of binary ASK
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4. OOK and MASK
 OOK (On-OFF Key)
– 0 silence.
– Sensor networks: battery life, simple implementation
 MASK: multiple amplitude levels
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5. Pro, Con and Applications
 Pro
– Simple implementation
 Con
– Major disadvantage is that telephone lines are very susceptible to
variations in transmission quality that can affect amplitude
– Susceptible to sudden gain changes
– Inefficient modulation technique for data
 Applications
– On voice-grade lines, used up to 1200 bps
– Used to transmit digital data over optical fiber
– Morse code
– Laser transmitters
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6. Example
 We have an available bandwidth of 100 kHz which spans from
200 to 300 kHz. What are the carrier frequency and the bit
rate if we modulated our data by using ASK with d = 1?
 Solution
– The middle of the bandwidth is located at 250 kHz. This
means that our carrier frequency can be at fc = 250 kHz.
We can use the formula for bandwidth to find the bit rate
(with d = 1 and r = 1).
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7. Frequency Shift Keying
 One frequency encodes a 0 while another frequency encodes a 1
(a form of frequency modulation)
 Represent each logical value with another frequency (like FM)
 A cos( 2πf t )
 binary 1
s (t ) =  1
 A cos( 2πf 2t )

binary 0
EE 541/451 Fall 2006

8. FSK Bandwidth
 Limiting factor: Physical capabilities of the carrier
 Not susceptible to noise as much as ASK
 Applications
– On voice-grade lines, used up to 1200bps
– Used for high-frequency (3 to 30 MHz) radio transmission
– used at higher frequencies on LANs that use coaxial cable
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9. Example
 We have an available bandwidth of 100 kHz which spans from
200 to 300 kHz. What should be the carrier frequency and the
bit rate if we modulated our data by using FSK with d = 1?
 Solution
– This problem is similar to Example 5.3, but we are modulating
by using FSK. The midpoint of the band is at 250 kHz. We
choose 2Δf to be 50 kHz; this means
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10. Multiple Frequency-Shift Keying (MFSK)
 More than two frequencies are used
 More bandwidth efficient but more susceptible to error
si (t ) = A cos 2π i t
f 1 ≤i ≤ M
x f i = f c + (2i – 1 – M)f d
x f c = the carrier frequency
x f d = the difference frequency
x M = number of different signal elements = 2 L
x L = number of bits per signal element
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11. Phase Shift Keying
 One phase change encodes a 0 while another phase change
encodes a 1 (a form of phase modulation)
A cos( 2πf t )
 binary 1
s (t ) =  c
A cos( 2πf c t + π ) binary 0

EE 541/451 Fall 2006

12. DBPSK, QPSK
 Differential BPSK
– 0 = same phase as last signal element
– 1 = 180º shift from last signal element
 Four Level: QPSK  π
A cos2π c t + 

f 11
 4
3π 



A cos2π c t +
f 
s (t ) = 
01
 4 
 3π 
A cos2π c t −
f  00
  4 

 
A cos2πc t − 
f
π 10
 4
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13. QPSK Example
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14. Bandwidth
 Min. BW requirement: same as ASK!
 Self clocking (most cases)
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15. Concept of a constellation diagram
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16. MPSK
 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
– M = number of different signal elements = 2L
– L = number of bits per signal element
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17. QAM – Quadrature Amplitude Modulation
 Modulation technique used in the cable/video networking world
 Instead of a single signal change representing only 1 bps –
multiple bits can be represented buy a single signal change
 Combination of phase shifting and amplitude shifting (8 phases, 2
amplitudes)
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18. QAM
 QAM
– As an example of QAM, 12
different phases are combined
with two different amplitudes
– Since only 4 phase angles have 2
different amplitudes, there are a
total of 16 combinations
– With 16 signal combinations, each
baud equals 4 bits of information
(2 ^ 4 = 16)
– Combine ASK and PSK such that
each signal corresponds to
multiple bits
– More phases than amplitudes
– Minimum bandwidth requirement
same as ASK or PSK
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19. QAM and QPR
 QAM is a combination of ASK and PSK
– Two different signals sent simultaneously on the same carrier frequency
s ( t ) = d1 ( t ) cos 2πf c t + d 2 ( t ) sin 2πf c t
– M=4, 16, 32, 64, 128, 256
 Quadrature Partial Response (QPR)
– 3 levels (+1, 0, -1), so 9QPR, 49QPR
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20. Offset quadrature phase-shift keying (OQPSK)
 QPSK can have 180 degree jump, amplitude fluctuation
 By offsetting the timing of the odd and even bits by one bit-period, or half a
symbol-period, the in-phase and quadrature components will never change at
the same time.
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21. Continuous phase modulation (CPM)
 CPM the carrier phase is modulated in a continuous manner
 constant-envelope waveform
 yields excellent power efficiency
 high implementation complexity required for an optimal
receiver
 minimum shift keying (MSK)
– Similarly to OQPSK, MSK is encoded with bits alternating between
quarternary components, with the Q component delayed by half a bit
period. However, instead of square pulses as OQPSK uses, MSK
encodes each bit as a half sinusoid. This results in a constant-modulus
signal, which reduces problems caused by non-linear distortion.
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22. Gaussian minimum shift keying
 GMSK is similar to MSK except it incorporates a premodulation Gaussian
LPF
 Achieves smooth phase transitions between signal states which can
significantly reduce bandwidth requirements
 There are no well-defined phase transitions to detect for bit synchronization
at the receiving end.
 With smoother phase transitions, there is an increased chance in intersymbol
interference which increases the complexity of the receiver.
 Used extensively in 2nd generation digital cellular and cordless telephone
apps. such as GSM
EE 541/451 Fall 2006

23. Project 2
 Due 11/15/06 midnight. I will be 10pm on 14th and 12pm on 15th
 Design your own modulation and demodulation
 Show time signal, eye diagram, and constellation for no noise, SNR=0,
SNR=5dB and SNR=10dB. (1 point)
 Calculate BER for SNR=0. SNR=2.5dB and SNR=5dB, compared with
theoretic result. Change symb to sufficiently large. (4 point for under, 2 point
for grad)
 For QPSK and 16QAM, redo the above step (2 point for grad)
 Transmit images (3 point)
– Test small image first
– Alignment for both sampling and data
– Calculate PSNR for SNR=0dB, SNR=2.5dB, and SNR=5dB.
– Print images
 Timing: sampling at the wrong time. 2 point
– 1/16, 2/16, … for BER vs. SNR, PSNR vs. SNR
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24. ISI
2
1.5
1
0.5
0
-0.5
-1
-1.5
0 2 4 6 8 10 12 14 16
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25. Scatter Plot
Scatter plot Scatter plot
1
1.5
0.8
0.6 1
0.4
0.5
0.2
Quadrature
Quadrature
0 0
-0.2
-0.5
-0.4
-0.6
-1
-0.8
-1 -1.5
-1 -0.5 0 0.5 1
-1.5 -1 -0.5 0 0.5 1 1.5
In-Phase
In-Phase
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26. Eye Diagram
E y e Diagram for In-P has e S ignal
2
1
Eye Diagram
A m plitude
1.5 0
1 -1
0.5 -2
-0.5 0 0.5
plitude
Tim e
0
E y e Diagram for Q uadrature S ignal
Am
2
-0.5
1
-1
A m plitude
0
-1.5
-0.5 0 0.5
T e
im -1
-2
-0.5 0 0.5
Tim e
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27. BER and PSNR vs. SNR
 Error Floor for sampling errors
Performance of Baseband QPSK
0
10
Theoretical SER
Theoretical BER
Simulated SER
Simulated BER
-1
10
SER and BER
PSNR
-2
10
-3
10
0 1 2
EbNo (dB)
3 4 5
SNR
EE 541/451 Fall 2006

28. Digital Carrier System Baseband analysis
∞
Signal in baseband: xTp (t ) = T ∑ ( d ′(l ) + jd ′′(l ) ) gTx (t − lT )
l =−∞
∞
∫
2
mean symbol energy: ES = T 2 D 2
gTx (t )dt
−∞
signal in carrier band:
{
xBp (t ) = 2 Re xTp (t )e j 2π f0t }
 ∞ ∞

= 2T cos(2π f 0 ) ∑ d ′(l ) gTx (t − lT ) − sin(2π f 0 ) ∑ d ′′(l ) gTx (t − lT ) 
 l =−∞ l =−∞ 
 D′ 2 D′′ 2  ∞
mean symbol energy: EX = T 2 ⋅ 2 ⋅  +  ⋅ gTx (t )dt =ES
∫
2
 2 2  −∞

1442443 
D2
Conclusion: analysis of carrier band = base band. Fc=0 in project
EE 541/451 Fall 2006

29. Baud Rate, Bit Rate, Bandwidth Efficiency
 Remember channel capacity C=Wlog2 (1+ SNR)> fb
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30. Power Spectrum, ASK
 Baseband
 Sy(W)=Sx(W) P(W)
 ASK: Sy(t)=b Acoswct, Square wave convolute with sinusoid.
EE 541/451 Fall 2006

31. FSK Spectrum
 FSK: two sinc added together
 A cos( 2πf t )
 binary 1
s(t ) =  1
 A cos( 2πf 2t )

binary 0
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32. BPSK Spectrum
 BPSK: Sx(W): NRZ. P(t): raised cosine function. Sy(W)= P(W)
 Rb
baud rate
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33. QPSK Spectrum
 Same Rb
Narrow BW
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34. Pulse Shaped M-PSK
 Different α
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35. Bandwidth vs. Power Efficiency
 Bandwidth efficiency high, required SNR is high and low power efficiency
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36. QAM efficiencies
 For l =1  PSD for BPSK
 For l =2  PSD for QPSK, OQPSK …
 PSD for complex envelope of the bandpass multilevel signal is
same as the PSD of baseband multilevel signals
 Same baud rate, higher bit rate.
 Same bit rate, less bandwidth. But higher power
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37. Minimum Shift Keying spectra
 Continuous phase and constant envelop. So narrow spectrum
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38. GMSK spectral shaping
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39. Gray coding
 It is very unlikely that switches will change states exactly in synchrony. So
there might be misunderstanding. E.g. 011->100
 In a digital modulation scheme such as QAM where data is typically
transmitted in symbols of 4 bits or more, the signal's constellation diagram
is arranged so that the bit patterns conveyed by adjacent constellation points
differ by only one bit. By combining this with forward error correction
capable of correcting single-bit errors,
it is possible for a receiver to correct
any transmission errors that cause a
constellation point to deviate into the
area of an adjacent point. This makes
the transmission system less susceptible
to noise.
 Graduate student for 16-QAM
EE 541/451 Fall 2006

40. Coherent Reception
 An estimate of the channel phase and attenuation is recovered. It
is then possible to reproduce the transmitted signal, and
demodulate. It is necessary to have an accurate version of the
carrier, otherwise errors are introduced. Carrier recovery
methods include:
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41. Coherent BER
 PSK
– BPSK QPSK
– MPSK
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42. Coherent BER performance
 1 Eb 
 ASK Pb = 2(1 − L )Q
1
 L −1 N 

 
 1 Eb 
Pb = 2(1 − L )Q
1
 L −1 N 

 
 1.217 Eb 
Pb = Q



 FSK  N 
 MSK: less bandwidth but the same BER
 MQAM
EE 541/451 Fall 2006

43. Non-coherent detection
 Non-coherent detection
– does not require carrier phase recovery (uses differentially encoded mod.
or energy detectors) and hence, has less complexity at the price of higher
error rate.
 No need in a reference in phase with the received carrier
 Differentially coherent detection
– Differential PSK (DPSK)
x The information bits and previous symbol, determine the phase of the
current symbol.
 Energy detection
– Non-coherent detection for orthogonal signals (e.g. M-FSK)
x Carrier-phase offset causes partial correlation between I and Q
braches for each candidate signal.
x The received energy corresponding to each candidate signal is used
for detection.
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44. Differential Reception
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45. Differential Coherent
 DBPSK
 3dB loss
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46. Non-coherent detection of BFSK
2 / T cos(ω t )
1
T z11
( )2
∫
0 2 2
z11 + z12
2 / T sin(ω t )
1
T z12
r (t )
∫
0
( )2 + z (T )
Decision stage:
ˆ
m
2 / T cos(ω2t ) if z (T ) > 0, m = 1
ˆ
z 21 if z (T ) < 0, m = 0
ˆ
T
( )2 -
∫
0
2 2
2 / T sin(ω2t ) z21 + z 22
T z 22
∫
0
( )2
EE 541/451 Fall 2006

47. Non-coherent detection BER
 Non-coherent detection of BFSK
1 1
PB = Pr( z1 > z 2 | s 2 ) + Pr( z 2 > z1 | s1 )
2 2
= Pr( z1 > z 2 | s 2 ) = E [ Pr( z1 > z 2 | s 2 , z 2 )]
∞
= ∫ Pr( z1 > z 2 | s 2 , z 2 ) p ( z 2 | s 2 )dz 2 = ∫
∞
 ∞ p ( z | s )dz  p ( z | s )dz
0 0  ∫z2
 1 2 1
 2 2 2
1  Eb 
PB = exp −
 2N 
Rayleigh pdf Rician pdf
2  0 
 Similarly, non-coherent detection of DBPSK
1  E 
PB = exp − b 
 N 
2  0 
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48. BER Example
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49. Example of samples of matched filter output
for some bandpass modulation schemes
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50. Comparison of Digital Modulation
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51. Comparison of Digital Modulation
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52. Spectral Efficiencies in practical radios
 GSM- Digital Cellular
– Data Rate = 270kb/s, bandwidth = 200kHz
– Bandwidth Efficiency = 270/200 =1.35bits/sec/Hz
– Modulation: Gaussian Minimum Shift Keying (FSK with
orthogonal frequencies).
– “Gaussian” refers to filter response.
 IS-54 North American Digital Cellular
– Data Rate = 48kb/s, bandwidth = 30kHz
– Bandwidth Efficiency = 48/30 =1.6bits/sec/Hz
– Modulation: pi/4 DPSK
EE 541/451 Fall 2006

53. Modulation Summary
 Phase Shift Keying is often used, as it provides a highly
bandwidth efficient modulation scheme.
 QPSK, modulation is very robust, but requires some form of
linear amplification. OQPSK and p/4-QPSK can be
implemented, and reduce the envelope variations of the signal.
 High level M-ary schemes (such as 64-QAM) are very
bandwidth efficient, but more susceptible to noise and require
linear amplification.
 Constant envelope schemes (such as GMSK) can be employed
since an efficient, non-linear amplifier can be used.
 Coherent reception provides better performance than
differential, but requires a more complex receiver.
EE 541/451 Fall 2006