Delta EE 442 – Spring Semester Lecture 10

m(t)

Voice Input (t) Encoder Out

mq(t) Decoder Out

1 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 (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 mq(t) to m(t).

2 Delta Modulation Transmitter: Sampler m(t) d(t) + + dq[k] To the  - transmission Comparator channel

mq(t) dq[k] Integrator (accumulator)

 time dq[k] -

step size

) t

(  mq(t) q Slope

m Overload

) or ) or

t (

m m(t)

TS time 3 Delta Modulation Waveforms

Start-up interval Slope Overload Noise

 Band-Limited Analog Signal m(t)

Integrator Output mq(t)

Granular Noise time

dq(t) 

- 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 . . .

4 Delta Demodulation

mq(t)  dq[k] m(t) From the Low-pass Integrator channel filter (accumulator)

Now that is really simple!

5 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.

-3 Staircase with ADM  2 m(t) 5

3 Staircase with DM

mq(t) 2 

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

6 Comparing PCM, DPCM and DM

Parameters PCM DPCM DM Number of 4, 8, 16, Typically 4 to 6 bits < 1 bit bits/sample and so on. PCM Bandwidth Highest required Lower than PCM Lowest Step Size Fixed Fixed Fixed Sampling Rate 8 kHz 8 kHz 64 to 128 kHz 56 and 64 kbps 32 to 48 kbps 64 to 128 kbps Quantization Depends upon q Slope overload and Slope overload and Error granular noise granular noise

7 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(2fmt), then dm() t  22fmm A B dt max where fm < B. Slope overload means mq(t) can’t follow m(t). We then have the condition, mAm < fS or mAmTS <  .

8 Quantization or Granular Noise in DM

mq(t)  m(t) is a sinusoidal waveform

2 2  N   q 3

Error = m(t) - mq(t)

 TS

9 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 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 is

30,000 Hz = fS.

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

 fS 0.25 30,000 Am    0.60 volt 2fm 2 (2,000)

10 Maximum Rate in Communications Recall from prior lecture: 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).

11 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.

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