Digital Signal Processing 2 Part 2 - 4

1

1 L1 L1

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This can happen if and only if W M = W M or the set M terms W M

coincides with the set M of terms W M with 0 ≤ lM - 1.

-In case 1 = M we have:




c) Represent the beat transformation in the frequency domain

Evaluating X(z), Y↑↓ M/L (z) and Y↓↑ M/L (z) on the unit circle in the z-plane we have

Have:



Example 3.2.4.3

Given a string signal with spectrum as shown in Figure 3.2.4.4 below:




Prize

Figure 3.2.4.4


Solve using the graph in figure 3.2.4.4 below:

From Figure 3.2.4.4 we see that Y↓↑ 2/3 (e j ) = Y↑↓ 2/3 (e j )


3.3. SAMPLING TRANSFORM FILTER

3.3.1. DIVISION FILTER

a) Overview

In the above section, we have studied the division and the divider. The results show that the signal x(n) when passing through the divider ↓M, in the frequency domain will create M-1 nominal components (alíaing), these nominal components will cause the phenomenon of spectral overlap. But if x(n)

has a frequency band in the range

MM

that is stopband edge frequency

L

will cause overlapping phenomenon




To do this we can precede the ↓M divider with a low-pass filter

has c M


This low pass filter is responsible for removing frequency components, which will avoid spectral overlap.

M


so

The general diagram of the split filter is given in Figure 3.3.1.1


h(n): impulse response of the filter

Figure 3.3.1.1

To be concise we can use the following operator representation:


b) Represent the partition filter in the domain of variable n

The operations in a partition filter occur as follows in the domain of the variable n:


One thing to note is that division does not have the distributive property on convolution, that is:





c) Representation of the partition filter in the z-domain

In the z-domain the partitioning filter is described as follows:


As in the domain of variable n, we note that division does not have the distributive property to multiplication in the domain z, that is:



Note: Consider the following split filter:


Now consider the following partition filter:


We see that Y H↓M (z) = Y ↓MH (z) so these two filters are equivalent. Therefore, the implementation diagram of these two filters is identical, see figure 3.3.1.2


a) b)

Figure 3.3.1.2

- The identity of the two diagrams (a) and (b) in Figure 3.3.1.2 is very valuable in

Practical applications for implementing filters and filter banks.

- For brevity, we denote the filter operation in Figure 3.3.1.2 (a) and (b) as |H(z M ) ↓ M| and

|↓MH(z)|. So we can write:


(3.3.1.10)

H(z M )↓M ≡ ↓MH(z)

d) Representation of the division filter in the frequency domain

Evaluating X(z), H(z) and Y H↓M (z) on the unit circle in the z-plane, we have the following representation of the division filter in the frequency domain:



If Y H (e j ) is the frequency response of an ideal low-pass filter with

c M

then the

The nominal component will not cause information corruption, that is, there is no overlapping phenomenon.

Then we can extract the first component (I = 0 ) without distorting its form:

and if H(e j ) is an ideal low-pass filter, then the passband |H(e j )|=1 stopband

|H(e j )|=0 then we have the first component (I = 0) as follows:


In the above example at the splitter ↓M = 2 there is no spectral overlap, because

The frequency ω s of the low-pass filter is equal to

, it has cut off the high frequency component

2

.

2

So if our information is contained only in the spectral component

2

then the filter will not

loss of information. What if our information is contained in both components?

universal

2

then cutting off the high frequency components has lost their information

e) Properties of split filtering

If


Then

H(z) = H 1 (z) + H 2 (z)


Y H (z) = Y(z)[ H 1 (z) + H 2 (z)]

= X(z). H 1 (z) + X(z). H 2 (z)



= ↓ M[X(z).H 1 (z)]+ ↓ MX(z).H 2 (z)] (3.3.1.13)

So division is distributive to addition.

3.3.2. INTERPOLATION FILTER

a) Overview

We have studied interpolation and interpolators, the interpolation results have been inserted

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