1. Dr. Clincy Professor of CS
Chapter 5 Handout #4
Dr. Clincy
Professor of CS
Per the syllabus, Exam 1 is scheduled for this Thursday covering lectures 1-8
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Lecture

2. Digital-to-analog conversion
Based on the digital data, the
Modulator changes characteristics of the “controllable” analog signal (bandpass analog signal) on the transmitter side to represent the digital data
Demodulator interprets the analog signal in re-creating the digital data on the receiver side
Terminology: “modulating digital data into an analog signal”
The analog signal we can control ? Sine Wave, Carrier Signal, Periodic Signal
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Lecture

3. Types of digital-to-analog conversion
Change amplitude to represent a bit
Change frequency to represent a bit
Change phase to represent a bit
Combination of changing both amplitude and phase to represent a set of bits
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Lecture

4. Recall
For digital transmission, bit rate (data rate) and signal rate (baud rate) relationship was
S = N x 1/r where r = # of data elements per signal element and N is the data rate in bps (and S is the signaling or baud rate)
For analog, r = log2L where L is the type of signal (versus level)
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5. Example
An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate.
Solution
In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from
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Lecture

6. Example
An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need?
Solution
In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L.
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Lecture

7. Binary amplitude shift keying
changing the original amplitude
Explain this
not changing the original amplitude
Bandwidth (B) is proportional to the signal rate (S) and depending on the modulation and filtering process, the required bandwidth can range between S to 2S (where middle bandwidth is fc). The value of d relates to the modulation and filtering process
B = (1 + d) x S
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8. 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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Lecture
S = N * 1/r

9. Binary frequency shift keying
changing the original frequency
Use two different carrier frequencies, f1 and f2, for 0 and 1
Explain this
not changing the original frequency
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Lecture

10. 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
The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means
The difference (delta) between the two frequencies
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11. Binary phase shift keying
changing the original phase
Explain this
not changing the original phase
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12. QPSK and its implementation
QPSK – Quadrature Phase Shift Keying
Use 2 bits in each signal element – decreases baud rate and bandwidth
Uses 4 possible phases (versus 2)
2 composite signals are created
Because the 2 signals are using the same bandwidth – each signal has ½ bandwidth
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Lecture

13. Example
Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.
Solution
For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz.
B = (1 + d) x S
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Lecture

14. Concept of a constellation diagram
Helps define the amplitude and phase of a signal element
the amplitude of the 2nd carrier
Peak Amplitude
Phase
This is the amplitude given using one carrier
Only use 1 carrier and phase is static and 2 amplitude levels
Only use 1 carrier and 1 amplitude and 2 phases (0o and 180o)
Uses 2 carriers and 1 amplitude and 4 phases (45o, 135o, -45o, -135o)
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Lecture

15. Constellation diagrams for some QAMs
QAM – Quadrature Amplitude Modulation
For QPSK, we only changed the phase
For QAM, we change both the phase and amplitude
Has a 0 amplitude and a positive amplitude (with 2 carriers)
Has a negative amplitude and a positive amplitude (with 2 carriers)
Has 2 positive amplitudes (with 2 carriers)
Has 4 negative levels and 4 positive levels (with 2 carriers)
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Lecture

16. ANALOG TO ANALOG
Analog-to-analog conversion is the representation of analog information by an analog signal. One may ask why we need to modulate an analog signal; it is already analog. Modulation is needed if the medium is bandpass in nature or if only a bandpass channel is available to us.
Bandpass – signal being shifted to a particular range
Lowpass – signal that IS NOT shifted to a particular range
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Lecture

17. Types of analog-to-analog modulation
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18. Amplitude modulation
Vary the amplitude of the carrier signal to mimic the changing voltage levels (amplitude) of the modulating signal
result
The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B.
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Lecture

19. Frequency modulation
Vary the frequency of the carrier signal to mimic the changes in voltage level (amplitude) of the modulating signal
result
The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + β)B.
Would be given
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20. Phase modulation
Vary the phase of the carrier signal to mimic the changes in voltage level (amplitude) of the modulating signal
This illustrates the signal starting at different phases
The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal: BPM = 2(1 + β)B.
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Lecture

21. Dr. Clincy Professor of CS
Chapter 5 and 6 Handout #5
Dr. Clincy
Professor of CS
Test on February 12th
Your test will cover lectures 1 – 8 and will be 75 minutes
You should use a calculator
You can view the handouts via your laptop or you can print them - you shouldn’t use the browser
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Lecture

22. Chapter 6: Bandwidth Utilization: Multiplexing and Spreading
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23. Multiplexing & Spreading (Physical Layer Issues)
Up to this point, you have learning about translating “data” into a “signal” – so that the “signal” can travel across the transport
It would be very efficient use of the transport’s bandwidth if multiple signals could travel on the transport at the same time ?
Also, it would be great if we could protect against eavesdropping
That efficiency can be achieved by multiplexing; privacy and anti-jamming can be achieved by spreading.
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Lecture

24. SPREAD SPECTRUM
In spread spectrum (SS), we combine signals from different sources to fit into a larger bandwidth, but our goals are to prevent eavesdropping and jamming. To achieve these goals, spread spectrum techniques add redundancy.
Typically used for wireless applications – privacy outweighs efficiency in this case
Frequency Hopping Spread Spectrum (FHSS) Direct Sequence Spread Spectrum Synchronous (DSSS)
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Lecture

25. Frequency selection in FHSS
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26. DSSS – Direct Sequence Spread Spectrum
Each bit sent by the Tx is replaced with a set of bits called a “chip code”
For the time it takes to send the original single bit, it now will take more time to send the chip code
Therefore, the data rate must be N times the original data rate, where N is the # of bits of the chip code
Also, the bandwidth for the chip code should N times greater than the original bit stream’s BW
Example of original bits being transmitted as 6-bit chip codes
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27. DSSS using polar NRZ encoding
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28. Multiplexing
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29. Dividing a link into channels – Multiplexing in general
Explain this
Categories of multiplexing
Will also cover Statistical Time-Division Multiplexing
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Lecture

30. Frequency-division multiplexing
Divide the link’s bandwidth into separate channels (guardband separating each channel)
Recall from the Modulation Lectures that – being able to modulate around different “carrier frequencies” was important to being able to adjust the modulated signal into a particular “band” (bandpass signal)
On the MULTIPLEXING SIDE
Resultant modulated signals are combined into a single composite signal
Signals modulate different carrier frequencies (based on amplitude in this case)
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Lecture

31. FDM demultiplexing example
On the DEMULTIPLEXING SIDE
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32. Example
Assume that a voice channel occupies a bandwidth of 4 kHz. We need to combine three voice channels into a link with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard bands.
Solution
We shift (modulate) each of the three voice channels to different bandwidth. We use the 20- to 24-kHz bandwidth for the first channel, the 24- to 28-kHz bandwidth for the second channel, and the 28- to 32-kHz bandwidth for the third one. Then we combine them into a single composite signal.
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Lecture

33. Wavelength-division multiplexing
Same as FDM but instead of electrical type signals – muxing optical signals (light signals)
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Lecture

34. Time Division Multiplexing (TDM)
All networking devices work off clock ticks (explain)
Do “tap” analogy
Explain this
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Lecture

35. Synchronous time-division multiplexing
Given n connections needing to be muxed, each frame is divided into n parts (for each slot)
Also notice that the time duration before muxing is 1/3 of the time duration after muxing
In this case, each frame is divided into 3 time slots
For synchronous TDM, the Tx and Rx must be in synch for the Rx to “pull out” of the frame the correct set of data (called interleaving)
For synchronous TDM, the data rate of the output link must be n times the data rate of the connection to guarantee the flow of data
In keeping the mux and demux in synch, synch bits (framing bits) are added at the beginning of each frame
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Lecture

36. Suppose the input data rates are different ?
Multilevel multiplexing
When input data rates are multiple of others – can be combined to make equal – for example, the two 20 kbps links could be muxed together as a 40 kbps link
Multi-slot multiplexing
Allocate more than 1 time slot in a frame to a single input – for example, the 50 kbps line gets 2 slots, while the 25 kbps lines get 1 slot each
Pulse Stuffing
Make the highest input data rate the dominate rate and then add dummy bits (stuffing) to the other input lines
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Lecture

37. Statistical TDM
For STATISTICAL TDM - Time slots are dynamically allocated based on previous history
Slots are reserved – could be wasted slots
Slots are allocated to Input Lines with data only – no wasted slots – because of this, the address of the Rx has to be carried with the data
The address needs to be n bits to define N output lines – with n = log2N (ie. need 5-bit address for 32 output lines)
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Lecture