Digital to Analog Modulation

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Digital-to-analog modulation

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Types of digital-to-analog modulation

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Bit rate: The number of bits transmitted during 1 second.

Baud rate: It refers to the number of signal units per second that are required to represent those bits.

A signal unit is composed of 1 or more bits.

Baud rate is less than or equal to the bit rate.

Note:Note:

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Example of bit and baud rate In transportation, a baud is analogous to a

car, and a bit is analogous to a passenger. A car can carry one or more passenger.

Here the number of cars, not the number of passengers determines the traffic and, therefore, the need for wider highway.

Similarly, the number of bauds determines the required bandwidth, not the number of bits.

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There are 2 bits at each signal unit

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Example Example

An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate

SolutionSolution

Baud rate = number of signal units per secondBaud rate = number of signal units per second = 1000 bauds per second= 1000 bauds per second

Bit rate= baud rate * number of bits per signal unitBit rate= baud rate * number of bits per signal unit = 1000x4 = 4000 bps= 1000x4 = 4000 bps

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Alternative solution:

In this case, number of bit in each signal unit, r = 4, baud rate, S = 1000, and bit rate, N is unknown. We can find the value of N from

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SolutionSolution

Math:Math:An analog signal has a bit rate of 8000 bps and a baud An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do by each signal element? How many signal elements do we need?we need?

In this example, baud rate, S = 1000, bit rate. N = 8000, number of bit in each signal unit ,r and signal element, L are unknown.

We find first the value of r and then the value of L. S=N * (1/r) = > r = N/S = > r = (8000/1000)= > r = 8 bits/baudAgain, we know thatr= log2 L=> L = 2^r => L = 2^8 =256 r

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ASK The two binary values are represented by

two different amplitudes of the carrier signal. (Commonly one level is 0)

Frequency and phase are remaining constant.

The ASK signal is

00

1)2cos()(

binary

binarytfAts c

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Demodulation: only the presence or absence of a sinusoid in a given time interval needs to be determined

Advantage: simplicity

Disadvantage: ASK is very susceptible to noise.

Application: ASK is used to transmit digital data over optical fiber.

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Relation bit rate and baud rate in ASK

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In ASK, the minimum bandwidth required for transmission is equal to the baud rate.

Bandwidth requires for ASK are calculated using the formula Bandwidth = (1+d) x N baud

Bandwidth of ASK:

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Example Example

Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate?

SolutionSolution

In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps.

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Example Example

Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in the two directions.

SolutionSolution

For full-duplex ASK, the bandwidth for each direction isBW = 10000 / 2 = 5000 Hz

The carrier frequencies can be chosen at the middle of each band (see Fig. 5.5).

fc (forward) = 1000 + 5000/2 = 3500 Hzfc (backward) = 11000 – 5000/2 = 8500 Hz

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Solution

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Frequency Shift Keying (FSK)• The frequency of the modulated signal is

constant for the duration of one signal element, but changes for the next signal element if data element changes.

• In FSK, two binary values are represented by two different frequencies near the carrier frequency.

• Amplitude and phase are remaining constant.

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Demodulation: demodulator must be able to determine which of two possible frequencies is present at a given time.

Advantage: FSK is less susceptible to noise.

Disadvantage: FSK spectrum is 2x ASK spectrum

Application: over voice lines, in high-frequency radio transmission, paging, LAN over coaxial cable etc.

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Relation bit rate and baud rate in FSK

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bandwidth in FSK

Bandwidth = fc1 – fc0 + N baud

i.e. The bandwidth required for FSK transmission is equal to the baud rate of the signal plus the frequency shift ( difference between the two carrier frequencies).

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Example Example

Find the maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode.

SolutionSolution

Because the transmission is full duplex, only 6000 Hz is allocated for each direction. BW = baud rate + fc1 BW = baud rate + fc1 fc0 fc0 Baud rate = BW Baud rate = BW (fc1 (fc1 fc0 ) = 6000 fc0 ) = 6000 2000 = 4000 2000 = 4000But because the baud rate is the same as the bit rate, the bit rate is 4000 bps.

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Phase shift keying - PSK Phase of the carrier signal is varied to

represent binary 1 or 0

Peak amplitude and frequency remain constant during each bit interval Example: binary 1 is represented with a phase of 0º, while binary 0is represented with a phase of 180º = π rad

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Demodulation: demodulator must be able to determine the phase of received sinusoid with respect to some reference phase.

Advantage: PSK is less susceptible to noise than ASK and requires same bandwidth as ASK and less than FSK.

Disadvantage: more complex signal detection / recovery process, than in ASK and FSK.

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NOTE:

Bit contains at each variation = log2 x baud rate

or log2 X n Where n = number of variation

Bit rate = Baud rate x Bit contains at each variation

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Relation bit rate and baud rate in PSK

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Constellation diagram: It can help us defining the amplitude and

phase of a signal element, particularly when we are using two carrier ( one for amplitude and other for phase)

PSK or 2-PSK constellation

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The 4-PSK method

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The QPSK or 4- PSK method

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The 4-PSK characteristics

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The 8-PSK characteristics

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bandwidth in PSK

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ExampleExample

Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode.

SolutionSolution

For PSK the baud rate is the same as the bandwidth.Here,Bit contains at each variation , log n = log 4 = 2

Baud rate = bit rate / bit contains at each variation = ( 2000/2) = 1000 baudSo, bandwidth = 1000 bps

22

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Example Example

Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate?

SolutionSolution

For PSK the baud rate is the same as the bandwidth.Here,Bit contains at each variation , log2 n = log2^8 = 3

bit rate = Baud rate * bit contains at each variation = ( 5000 * 3) = 15000 bps

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It is combination of ASK and PSK.

QAM is a logical expansion of Q-PSK

We could have x variations in phase and y variation in amplitude

QAM takes advantage of fact that it is possible to send two different signals simultaneously on the same carrier frequency, by using two copies of the carrier frequency, one is shift by 90 degree with respect to other.

At the receiver, the two signals are demodulated and results combined to produce the original binary input.

Quadrature Amplitude Modulation (QAM)Quadrature Amplitude Modulation (QAM)

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The minimum bandwidth required for QAM transmission is the same as that required for ASK and PSK.

Analogue QAM is used in NTSC, PAL and SECAM television systems, C-QUAM is used in AM stereo radio

Digital QAM, 64-QAM and 256-QAM are often used in Digital cable television.

QAM is popular for ADSL(Asymmetric digital subscriber)

Quadrature Amplitude Modulation (QAM)Quadrature Amplitude Modulation (QAM)

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Time domain for an 8-QAM signal

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The 4-QAM and 8-QAM constellations

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16-QAM constellations

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Bit and baud

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Table 5.1 Bit and baud rate comparison

ModulationModulation UnitsUnits Bits/BaudBits/Baud Baud rateBaud rate Bit Rate

ASK, FSK, 2-PSKASK, FSK, 2-PSK Bit 1 N N

4-PSK, 4-QAM4-PSK, 4-QAM Dibit 2 N 2N

8-PSK, 8-QAM8-PSK, 8-QAM Tribit 3 N 3N

16-QAM16-QAM Quadbit 4 N 4N

32-QAM32-QAM Pentabit 5 N 5N

64-QAM64-QAM Hexabit 6 N 6N

128-QAM128-QAM Septabit 7 N 7N

256-QAM256-QAM Octabit 8 N 8N

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Example Example

A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate?

SolutionSolution

The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is 4800 / 3 = 1600 baud

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Example Example

Compute the bit rate for a 1000-baud 16-QAM signal.

SolutionSolution

A 16-QAM signal has 4 bits per signal unit since log216 = 4. Thus,

(1000)(4) = 4000 bps

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Example Example

Compute the baud rate for a 72,000-bps 64-QAM signal.

SolutionSolution

A 64-QAM signal has 6 bits per signal unit since log2 64 = 6. Thus, 72000 / 6 = 12,000 baud

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Analog-to-analog modulation

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Types of analog-to-analog modulation

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Why Analog-to-Analog Modulation?

There are two principal reasons for combining an analog signal with a carrier at frequency fc

(1) Higher frequency may be needed for effective transmission, especially in unguided media. (2) Modulation permits FDM (frequency-division multiplexing)

Note:Note:

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Amplitude of carrier signal varies with changing amplitude of input/modulating signal; frequency and phase remain unchanged

Amplitude Modulation (AM)

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Bandwidth of Amplitude modulation

Bandwidth of an AM signal = 2x bandwidth of modulating signal, and covers a range centered on carrier frequency BWtotal = 2*BW modulating-signal

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AM band allocation

AM radioThe bandwidth of an audio signal (speech + music) is 5 kHzeach AM radio station needs a min bandwidth of 10 kHz

AM stations are allowed carrier frequencies anywhere between 530 - 1700 kHz; each station’s carrier frequency must be separated from those on either side by at least 10 kHz, to avoid interference

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Example Example

We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM? Ignore FCC (Federal Communication Commission) regulations.

SolutionSolution

An AM signal requires twice the bandwidth of the original signal:

BW = 2 x 4 KHz = 8 KHz

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Frequency of carrier signal varies with changing frequency of input/modulating signal; amplitude and phase remain unchanged

Frequency Modulation (FM)

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FM bandwidth

Bandwidth of an FM signal = 10xbandwidth of modulating signal,and covers a range centered on carrier frequency BWtotal = 10*BW modulating-signal

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FM band allocation

FM radio• The bandwidth of an audio signal (speech and music) in stereo is almost 15 kHz • Each FM radio station needs a minimum bandwidth of 150 kHz• FM stations are allowed carrier frequencies anywhere between 88 and 108 MHz; stations must be separated by at least 200 kHz to keep their bandwidths from overlapping

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Example Example

We have an audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM? Ignore FCC regulations.

SolutionSolution

An FM signal requires 10 times the bandwidth of the original signal:

BW = 10 x 4 MHz = 40 MHz

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AM vs. FMAM Advantages AM signals can be reflected from the ionospheric

layerback to earth, so that the signals can reach unintended places that are thousands of miles away.

FM advantages The effects of amplitude noise are minimized, since

the recovered audio is dependent only on the frequency, and not the strength.

The FM bandwidth can easily cover entire musical range and that is why FM radio sounds better than AM radio

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AM vs. FMAM Disadvantages Most natural and man made radio noise is AM in

nature, and AM receivers have no means of rejecting that noise.

Also, weak signals have lower amplitude than strong ones, which requires the receiver to have circuits to compensate for the signal level differences.

FM Disadvantages At the high(er)-frequency FM signals pass unreflected

through the ionosphere.

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AM radio has wider coverage than FM radio; FM radio has better sound quality than AM radio

AM signal propagates in sky mode and FM signal propagates in ground-wave mode.

Note:Note:

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Then the modulated signal,

This is because it tends to require more complex receiving hardware and there can be ambiguity problems with determining whether, for example, the signal has 0° phase or 180° phase

Phase Modulation (PM)