Delta Modulation EE 442 – Spring 2017

Lecture 10

1

Voice Input

Decoder Out

Encoder Out

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 the Nyquist 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 gives a stepwise approximation m(t) to m(t).

3

Delta Modulation Transmitter:

d(t)

Comparator

Sampler dq[k] m(t)

m(t)

m(t)

time

TS

+ +

-

Integrator

m(t) Dq[k]

time dq[k]

m(t

) o

r m

(t)

Slope Overload

-

(accumulator)

step size

To the channel

4

Delta Modulation Waveforms

Start-up interval Slope Overload Noise

Band-Limited Analog Signal m(t)

Integrator Output mq(t)

Granular Noise

dq(t)

-

time

time dq[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]

Integrator

Low-pass filter

m(t)

(accumulator)

mq(t)

From the channel

6

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

Quantization Error

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 the message 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, 2fmAm < fS or B Am < fS .

max

( )2 2m m

dm tf A B

dt

9

Quantization Noise in DM

TS

m(t) is a sinusoidal waveform

m(t)

= m(t) - m(t)

22

3qN

10

Worked Example for DM

Problem: A Delta modulated system is designed to operate at five times the Nyquist Rate. The signal bandwidth B at its input port is 3 kHz and the quantized step 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. 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 is 30,000 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 channel of bandwidth B Hz.

It is related to the sampling theorem:

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

The sampling theorem is important in signal analysis, digital signal processing and transmission because it allows us to replace an analog signal 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 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 this giving 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.