1. Dr. Clincy Professor of CS
Chapter 4 Handout #4
Dr. Clincy
Professor of CS
Dr. Clincy
Lecture 2

2. Four Data-to-Signal Cases
DATA SIGNAL
D D A
A D
Already covered
Will cover today
Dr. Clincy
Lecture 2

3. ANALOG-TO-DIGITAL CONVERSION
We have seen in Chapter 3 that a digital signal is superior to an analog signal. The tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulation and delta modulation.

4. Components of PCM encoder
PCM – Pulse Code Modulation
1st: analog signal is sampled
2nd: sampled signal is quantized
3rd: quantized values are encoded as bit streams (or codes)
Analog signal is sampled every Ts seconds
Sample rate is fs = 1/Ts

5. Three different sampling methods for PCM
High-speed switch used – able to retain the shape of the signal
Ideal but complex
Sample-and-hold method that creates flat-top samples by using a circuit
Sampling process also called pulse amplitude modulation (PAM)

6. Recovery of a sampled sine wave for different sampling rates
Nyquist theorem states that the sampling rate must be at least 2 times the highest frequency of the signal
Catches the essence of the signal
Doesn’t improve the case
Doesn’t capture the essence of the signal

7. Quantization and encoding of a sampled signal
actual-amplitude/D
actual amplitude
Quantization steps:
Determine Vmin and Vmax
Divide range into L zones, each of height D
D = [Vmin - Vmax]/L
3. Assign quantized values of 0 to L-1 to midpoint of each zone
4. Map the sample value to a quantized value
Norm. Actual
Error between actual and nornalized
Quant. value for code
Code that represents the voltage level
Assume sample amplitudes between -20V and +20V
Let L = 8 (levels) – therefore, D = [ ]/8 = 5
Quantization error can contribute to Shannon’s SNR: SNRdB = 6.02nb where nb is bits per sample
Bit rate = sampling rate x # of bits per sample = fs x nb

8. Components of a PCM decoder
PCM decoder recovers the original signal
Smooths out the staircase signal
What is the minimum bandwidth of the filter the digitized signal will need ?
Bmin = c x nb x 2 x Banalog x 1/r (nb = # bits per sample)
If 1/r=1 and c=1/2, Bmin=nb x Banalog
If the data rate and number of signal levels are fixed, minimum bandwidth is
Bmin = N / [2 x log2 L]

9. The process of delta modulation
PCM is more complex than Delta modulation
PCM finds the amplitude of the signal; delta modulation simply finds the change in the signal from the previous sample
Delta modulation doesn’t use codes – bits are sent one after another
Positive changes are encoded as 1; negative changes are encoded as 0

10. TRANSMISSION MODES
The transmission of binary data across a link can be accomplished in either parallel or serial mode.
In parallel mode, multiple bits are sent with each clock tick.
In serial mode, 1 bit is sent with each clock tick.
While there is only one way to send parallel data, there are three subclasses of serial transmission: asynch, syn and iso approaches (asynchronous, synchronous, and isochronous.)

11. Parallel transmission

12. Serial transmission

13. Asynchronous transmission
In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each byte. There may be a gap between each byte.
Asynchronous here means “asynchronous at the byte level,” but the bits are still synchronized; their durations are the same.

14. Synchronous transmission
In synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the responsibility of the receiver to group the bits.

15. Isochronous Transmission
For realtime audio and video, uneven delays between frames is not acceptable – so synchronous transmission doesn’t work well
The entire stream of bits must be synchronized – this is isochronous transmission
Isochronous transmission guarantees data at a fixed rate

16. Dr. Clincy Professor of CS
Chapter 5 Handout #4
Dr. Clincy
Professor of CS
Dr. Clincy
Lecture 2

17. 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

18. 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

19. 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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Lecture

20. 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

21. 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

22. 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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23. 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

24. 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

25. 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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26. Binary phase shift keying
changing the original phase
Explain this
not changing the original phase
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Lecture

27. 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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28. 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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29. 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 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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Binary Phase Shift Keying – uses 2 different phases and the same amplitude

30. 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