Delta Modulation - web.sonoma.edu · Delta modulation requires a sampling rate much greater than the ... Adaptive Delta Modulation We can address the slope overload error problem

Delta ModulationEE 442 – Spring Semester

Lecture 10

1

Voice Input

Decoder Out

Encoder Out

(t)

m(t)

mq(t)

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 five times the Nyquist rate).

4. DM is closely related to DPCM.

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

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

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

)

SlopeOverload

-

(accumulator)

step size

To thetransmission

channel

4

Delta Modulation Waveforms

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 . . .

5

Delta Demodulation

Now that is really simple!

dq[k]

IntegratorLow-pass

filter

m(t)

(accumulator)

mq(t)

From thechannel

6

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)

7

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

8

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

9

Quantization or Granular Noise in DM

TS

m(t) is a sinusoidal waveform

mq(t)

Error = m(t) - mq(t)

22

3qN

10

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

11

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.

It is related to the sampling theorem:

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).

Recall from prior lecture:

12

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:We know that the signal frequency is fm = 1 kHz and the sampling rate is8 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.