Delta ModulationEE 442 – Spring Semester

Lecture 10

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Voice Input

Decoder Out

Encoder Out

(t)

m(t)

mq(t)

Read: Section 5.6, pp. 252-255 in Agbo & Sadiku

2

Key Attributes About Delta Modulation

1. Delta modulation (DM) is the simplest method for analog-to-digital conversion (ADC).

2. Delta modulation uses 1-bit per sampling period (TS) – it is a 1-bit ADC

3. Delta modulation requires a sampling rate much greater than theNyquist rate (commonly four or more times the Nyquist rate).

4. DM is closely related to differential PCM.

5. In DM we use a first-order predictor (one time delay TS is the predictor).

6. DM uses very simple hardware and is very low cost for that reason.

7. The transmitted output is a binary stream of pulses at fS. It givesa stepwise approximation mq(t) to the message signal m(t).

3

Delta ModulationTransmitter:

d(t)

Comparator

Samplerdq[k]m(t)

m(t)

mq(t)

time

TS

+ +

-

Integrator

mq(t) dq[k]

timedq[k]

m(t

) o

r m

q(t

)

-

(accumulator)

step size

To thetransmission

channel

SampledValues

4

Delta Modulation Waveforms

From: Leon W. Couch, II, Digital and Analog Communication Systems, 8th ed., Pearson Educational, Inc., 2013. Chapter 3, Section 3.8, page 200.

Start-up intervalSlope Overload Noise

Band-Limited Analog Signal m(t)

Integrator Output mq(t)

Granular Noise

dq(t)

-

time

timed

q[k] 0 0 0 1 0 1 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 0 1 0 . . .

TS

5

mpred(k)

Time DelayTS

m(k)+

+

+

Comparator (binary quantizer)

( )q k =

Accumulator

If ( ) ( ) is positive, then ( )

If ( ) ( ) is negative, then ( )

pred q

pred q

m k m k k

m k m k k

− = −

− = +

Another View of the Delta Modulation Process

We are generating

a predictionerror

Output isa binary

waveform

Sampling rate Rb = fS =(1/TS); BT > Rb/2

6

Delta Demodulation

Now that is really simple!

dq[k]

Integrator Low-pass filter

m(t)

(accumulator)

mq(t)

From thechannel

7

Special Case: Adaptive Delta Modulation

We can address the slope overload error problem with adaptive DM.Of course, it does add more complexity to its implementation.

m(t)

1 0 1 1 1 1 0 1 0 1 0 1 0 1 0 . . .

2

3

5

-3

Staircase with DM

Staircase with ADM

2

mq(t)

8

Comparing PCM, DPCM and DM

Parameters PCM DPCM DM

Number of bits/sample

4, 8, 16, and so on.

Typically 4 to 6 bits < PCM

1 bit

Bandwidth Highest required Lower than PCM Lowest

Step Size Fixed Fixed Fixed

Sampling Rate 8 kHz 8 kHz 64 to 128 kHz

Bit Rate 56 and 64 kbps 32 to 48 kbps 64 to 128 kbps

QuantizationError

Depends upon q Slope overload and granular noise

Slope overload and granular noise

9

Slope-Overload Distortion and Granular Noise

Slope

Delta modulation transmits the derivative of signal m(t). Suppose themessage signal is the sinusoidal signal Am cos(2fmt), then

where fm < B. Slope overload means mq(t) can’t follow m(t). We then have the condition, mAm < fS or mAmTS < .

max

( )2 2m m

dm tf A B

dt

=

Source: Figure 7.30 in Haykin & Moher, Communication Systems, 5th ed., John Wiley & Sons, Inc.,New York, 2009; page 269.

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Quantization or Granular Noise in Delta Modulation

TS

m(t) is a sinusoidal waveform

mq(t)

Error = m(t) - mq(t)

22

3qN

=

11

q

S

N

Step size

Granular noisedominates

Slope overloaddominates

opt

Granular Noise versus Slope Overload Tradeoff in DM

From: Leon W. Couch, II, Digital and Analog Communication Systems, 8th ed., Pearson Educational, Inc., 2013. Chapter 3, Section 3.8, page 201.

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Signal-to-Quantized-Noise Ratio (SQNR) for Delta Modulation

Assuming the modulation signal is sinusoidal,

where the overload condition is just met andB is the bandwidth of the receiver.

320

2 2

3 3( )

8

S S

q m m m

S f fm t

N A f f B

= =

Conclusion: The SQNR increases by 9 dB for each doubling in fS.

From: Leon W. Couch, II, Digital and Analog Communication Systems, 8th ed., Pearson Educational, Inc., 2013. Chapter 3, Section 3.8.

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Worked Example for DM

Problem:A Delta modulated system is designed to operate at five times the NyquistRate. The signal bandwidth B at its input port is 3 kHz and the quantizedstep is 250 millivolts (0.25 volt). For this problem we assume a 2 kHz sinusoidal input – Find the maximum amplitude Am of this 2 kHz tone that avoids slope overload..

Solution:We know that B = 3 kHz, fm = 2 kHz and = 250 mV.

The Nyquist rate is 3,000 2 = 6,000 Hz. So five time the Nyquist rate is30,000 Hz = fS.

Using the relationship given at the bottom of slide 8 allows us to write

( )0.25 30,0000.60 volt

2 2 (2,000)S

m

m

fA

f

= =

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Maximum Information Rate in Communications

Basic relationship in digital communications:

A maximum of 2B independent elements of information per second can be transmitted, error-free, over a noiseless channelof bandwidth B Hz.

Remember the sampling theorem states that a low-pass signal g(t) ofbandwidth B Hz can be fully recovered from uniform samples taken atthe rate of 2B samples per second.

The sampling theorem is important in signal analysis, digital signalprocessing and transmission because it allows us to replace an analogsignal with a discrete sequence of numbers (i.e., digital signal). Whensampling at a rate exceeding the Nyquist rate it is called “oversampling.”

Recall from prior lecture:

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Comparing PCM With DM

Problem:A one kilohertz (1 kHz) signal m(t) is sampled at 8 kHz with 12-bit encoding for PCM transmission.(a) How many bits are transmitted per second in in PCM? What is the

bandwidth required in this case? (b) Now switch to using DM with 8 kHz sampling. How many bits are

transmitted per second using DM? What is the bandwidth required in using DM?

Solution:The signal frequency is fm = 1 kHz and the sampling rate is 8 kHz.

(a) For PCM we have 8,000 samples per second and 12 bits per sample;which equals 96,000 bits/second. The bandwidth is one-half of thisgiving 48,000 Hz.

(b) Now for DM we have 1 bit per sample at 8,000 samples per second.Thus, we have 8,000 bits per second and a bandwidth of 4,000 Hz.

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http://www.eeeguide.com/delta-modulation/

Delta Modulation Transmission System

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Questions