Notes Lecture 06 DC

Lecture # 06 & 7Course Instructor:Engr. Sana Ziafat

We use three levels, positive, negative and zero

The voltage level for one data element is at zero, while the voltage level for the other element alternates between positive and negative

In bipolar encoding, we use three levels: positive, zero, and negative.

Note

AMI: Alternate Mark Inversiono It uses three voltage levelsoUnlike RZ 0 level is used to represent a 0oBinary 1’s are represented by alternating positive and negative.

Pseudoternaryo variation of AMI encoding in which 1 bit is encoded as zero voltage level and 0 bit is encoded as alternating positive and negative voltages

It is solution for the synchronization problem associated with Bipolar Schemes.

Two common scrambling techniques are:1.B8ZS (Eight consecutive zero levels are replaced by

sequence 000VB0VB)2.HDB3 (Four consecutive zero levels are replaced by

000V 0r B00V)

Figure Two cases of B8ZS scrambling technique

B8ZS substitutes eight consecutive zeros with 000VB0VB.

Note

Figure Different situations in HDB3 scrambling technique

HDB3 substitutes four consecutive zeros with 000V or B00V depending

on the number of nonzero pulses after the last substitution.

Note

Digital-to-analog conversion is the process of changing one of the characteristics of an analog signal based on the information in digital data.

High frequency signal that acts as a base for the information signal.

Process of modifying one of the characteristics of carrier signal based on information contained in digital data is known as “Modulation”

Bit rate is the number of bits per second. Baud rate is the number of signal

elements per second.

In the analog transmission of digital data, the baud rate is less than

or equal to the bit rate.

Note

An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate.

SolutionIn this case, r = 4, S = 1000, and N is unknown. We can find the value of N from

Example

Amplitude of carrier signal is varied while keeping the frequency and phase constant.

In ASK, the two binary values are represented by to different amplitudes of the carrier frequency

The resulting modulated signal for one bit time is

Susceptible to noise Inefficient modulation technique used for

up to 1200bps on voice grade lines very high speeds over optical fiber

0,01),2cos(

)(binarybinarytfA

ts c

Figure Binary amplitude shift keying

Binary Amplitude Shift Keying

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?SolutionThe 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).

Example

In data communications, we normally use full-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure 5.5. The figure shows the positions of two carrier frequencies and the bandwidths. The available bandwidth for each direction is now 50 kHz, which leaves us with a data rate of 25 kbps in each direction.

Figure Bandwidth of full-duplex ASK used in Example 5.4

Binary Frequency Shift Keying

QPSK ( Quadrature Phase Shift Keying) is a phase modulation algorithm.

5.28

Quadrature amplitude modulation is a combination of ASK and PSK.

Note

5.30

Figure Types of analog-to-analog modulation

5.31

A carrier signal is modulated only in amplitude value

The modulating signal is the envelope of the carrier

The required bandwidth is 2B, where B is the bandwidth of the modulating signal

Since on both sides of the carrier freq. fc, the spectrum is identical, we can discard one half, thus requiring a smaller bandwidth for transmission.

5.32

Figure 5.16 Amplitude modulation

5.33

The total bandwidth required for AM can be determined

from the bandwidth of the audio signal: BAM = 2B.

Note

5.34

The modulating signal changes the freq. fc of the carrier signal

The bandwidth for FM is high It is approx. 10x the signal frequency

5.35

The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + β)B. Where is usually 4.

Note

5.36

Figure 5.18 Frequency modulation

5.37

Figure 5.19 FM band allocation

5.38

The modulating signal only changes the phase of the carrier signal.

The phase change manifests itself as a frequency change but the instantaneous frequency change is proportional to the derivative of the amplitude.

The bandwidth is higher than for AM.

5.39

Figure 5.20 Phase modulation

5.40

The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal:

BPM = 2(1 + β)B.Where = 2 most often.

Note

Chapter 5 (B. A Forouzan)◦ Section 5.1◦ Section 5.2