Applications of digital signal processing - Continuous and Discrete Time Signals and Systems - page 769

The block diagrams of DPCM encoding and decoding systems are shown in
Fig. 17.16. Example 17.7 illustrates various steps of the DPCM coding.

Example 17.7
Assume that the first four samples of a digital audio sequence are given by [70,
75, 80, 82]. The audio samples are encoded using DPCM with the first-order
predictor defined in Eq. (17.18). The error samples obtained by subtracting the
predicted sample values from the actual audio sample values are divided by a
quantization factor of 2 and then rounded to the nearest integer. Determine the
values of the reconstructed signal.

Solution
In DPCM, the first sample value is encoded independent of other samples in
the sequence. In this example, we assume that the first audio sample, at k = 0,
with a value of 70 is encoded without any quantization error. In other words,
e [0] =

′

[0] = 70.
At k = 1, the predicted sample, the associated error, and the quantized error
are given by

e [0] = 0 and the reconstructed sample value s

predicted value s [1] = 0 . 97 70 = 67 . 9;

e [1] = 75 − 67 . 9 = 7 . 1;

error

e [1] = round(7 . 1 / 2) = 4 .

quantized error

The reconstructed value of the sample at k = 1 is therefore given by

′

s [1] = 0 . 97 70 + 4 2 = 75 . 9 .

At k = 2, the predicted sample, the associated error, and the quantized error
are given by

predicted value s [2] = 0 . 97 75 . 9 = 73 . 623;

error e [2] = 80 − 73 . 623 = 6 . 377;

quantized error e [2] = round(6 . 377 / 2) = 3 .

The reconstructed value of the sample at k = 2 is therefore given by

′

[2] = 0 . 97 75 . 9 + 3 2 = 79 . 623 . s

At k = 3, the predicted sample, the associated error, and the quantized error
are given by

predicted value s [3] = 0 . 97 79 . 623 = 77 . 2343;

error e [3] = 82 − 77 . 2343 = 4 . 7657;

quantized error e [3] = round(4 . 7657 / 2) = 2 .