In this chapter, we propose an efficient algorithm for finding the minimum sam-pling frequency for a signal consists of two bandpass signals. First we start up the analysis for a signal consists of two bandpass signals, which leads to four constraints of fs for causing no aliasing. This will be discussed in section 4.1.
In section 4.2 we introduce an efficient algorithm finding the minimum sampling frequency. Section 4.3 demonstrates the complexity analysis.
4.1 Conditions for Alias-free Sampling of Two Bandpass Signals
Conditions for alias-free sampling can be stated in various ways in terms of the band edges and bandwidths of the member bandpass signals. The conditions that are employed affect the complexity of ensuing algorithms. In this section, we derive a new set of conditions for alias-free sampling that will lead to an efficient algorithm in the next section.
First we consider the case of two bandpass signals for simplicity. Suppose we are to sample a signal X(f ) that consists of two bandpass signals X1(f ) and X2(f ) as shown in Fig. 4.1. Assume Xi(f ) 6= 0 for f`i < |f | < fhi, i = 1, 2, where f`i and fhi are band edges, and Wi = fhi− f`i are one-sided bandwidths as indicated in the figure. Let Xi+(f ), and Xi−(f ) denote respectively the positive frequency part and negative frequency part of Xi(f ). There are four signal bands,
X ( )f
Figure 4.1: An example of spectrum that consists of two bandpass signals.
including X1+(f ), X1−(f ), X2+(f ), and X2−(f ). Since the replicas of any two bands may overlap and result in aliasing after sampling, there are a total of C24 = 6 cases.
Note that X1+(f ) and X1−(f ) are symmetric with respect to 0, and so are X2+(f ) and X2−(f ). If X1+(f ) and X2+(f ) are not aliasing after sampling, then X1−(f ) and X2−(f ) will not be aliasing by symmetry. Similarly, if X1−(f ) and X2+(f ) are not aliasing after sampling, then X1+(f ) and X2−(f ) will not be aliasing. Thus, we need to consider only 4 cases:
(a) {X1+(f ), X1−(f )} this is the same as the case of one bandpass signal. For convenience, we will derive a condition in terms of the band edge fh1 and one-sided bandwidth W1. Upon sampling with frequency fs, replicas of X1+(f ) and X1−(f ) appear every fs, resulting in a periodic spectrum; we can simply consider the period [0, fs).
Since X1+(f ) and X1−(f ) are symmetric with respect zero, the replicas of X1+(f ) and X1−(f ) are symmetric with respect to f2s in the interval [0, fs) (Fig. 4.2(b)).
X ( )f spectrum for the interval [0, fs).
Observe that if 0 or fs is not contained inside the band of replicas of X1+(f ) and X1−(f ), there will not be aliasing. One necessary and sufficient condition for alias-free sampling is thus fh1 (mod f2s) = 0, or fh1 (mod f2s) ≥ W1. Equivalently, we (Fig. 4.3(b)). Observe that if 0 or fs is not contained inside the band of repli-cas of X2+(f ) and X2−(f ), there will not be aliasing. One necessary and sufficient condition for alias-free sampling is thus fh2 (mod f2s) = 0, or fh2 (mod f2s) ≥ W2. Equivalently, we have there will be no aliasing if and only if
2fh2 (mod fs) = 0,
or 2fh2 (mod fs) ≥ 2W2 (4.3)
0 spectrum for the interval [0, fs).
Case (c). Consider Fig. 4.4(a) where we have shown only the pair {X1+(f ), X2+(f )}.
First observe that there is no aliasing due to this pair if and only if there is no aliasing when we sample a shifted version of the pair {X1+(f + f0), X2+(f + f0)}
where f0 is the shift. For convenience we will consider the condition for alias-free sampling of the pair with a shift. Suppose we choose f0 as the midpoint of f`1 and fh2, i.e.,
f0 = (f`1 + fh2)/2.
Then the shifted pair is as shown in Fig. 4.4(b), where a = fh2 − f`1
2 ,
b = f`2 − (f`1 + fh2)/2, c = fh1 − (f`1 + fh2)/2.
If we consider the folded spectrum in the [0, fs) interval, the band edges a (mod fs) and (−a) (mod fs) are equal-distanced from fs/2. We now discuss two possible scenarios (i) a (mod fs) ≥ (−a) (mod fs) and (ii) a (mod fs) <
(−a) (mod fs). Examples of these two possible cases are shown respectively in Fig. 4.4(c) and (d).
(i) When a (mod fs) ≥ (−a) (mod fs) there will be no aliasing if and only if (−a) (mod fs) = a (mod fs) or if the interval ((−a) (mod fs), a (mod fs)) is large enough to accommodate the two replicas. That is,
a (mod fs) − ((−a) (mod fs)) = 0,
or a (mod fs) − ((−a) (mod fs)) ≥ W1+ W2. The equivalent conditions are
2a (mod fs) = 0,
or 2a (mod fs) ≥ W1+ W2 (4.4)
(ii) when a (mod fs) < (−a) (mod fs) as shown in Fig. 4.4(d), there is some space between the two replicas and the space is of length ((−a) (mod fs)−a (mod fs)). There will be no aliasing if and only if the remaining part of the [0, fs) interval is large enough to take in the two replicas. That is,
fs− ((−a) (mod fs) − a (mod fs)) ≥ W1+ W2. Or equivalently
2a (mod fs) ≥ W1 + W2 This is the same as the second condition in (4.4).
Substituting a = (fh2 − f`1)/2 to (4.4), we obtain one necessary and sufficient condition for alias-free sampling
(fh2 − f`1) (mod fs) = 0,
or (fh2 − f`1) (mod fs) ≥ W1 + W2 (4.5)
Case (d). Similarly, for the pair {X1−(f ), X2+(f )} as shown in Fig. 4.5(a), we can use the technique in case (c) to consider the condition for alias-free sampling of the pair with a shift where we choose f0 as the midpoint of −fh1 and fh2, i.e.,
f0 = (fh2 − fh1)/2.
X (f+f )
Then the shifted pair is as shown in Fig. 4.5(b), where a = fh1 + fh2
2 ,
b = f`2 − (fh2 − fh1)/2, c = −f`1 − (fh2 − fh1)/2.
If we consider the folded spectrum in the [0, fs) interval, the band edges a (mod fs) and (−a) (mod fs) are equal-distanced from fs/2. We can discuss two possible scenarios (i) a (mod fs) ≥ (−a) (mod fs) and (ii) a (mod fs) < (−a) (mod fs) as case (c) similarly and examples of these two possible cases are shown respectively in Fig. 4.5(c) and (d). Substituting a = (fh1 + fh2)/2 to (4.4), we obtain one necessary and sufficient condition for alias-free sampling
(fh1 + fh2) (mod fs) = 0,
or (fh1 + fh2) (mod fs) ≥ W1+ W2 (4.6) Summarizing, for a given sampling frequency fs, there will not be aliasing if the following four conditions are satisfied.
1. 2fh1 (mod fs) = 0 or 2fh1 (mod fs) ≥ 2W1 2. 2fh2 (mod fs) = 0 or 2fh2 (mod fs) ≥ 2W2
3. (fh2 − f`1) (mod fs) = 0 or (fh2 − f`1) (mod fs) ≥ W1+ W2
4. (fh1 + fh2) (mod fs) = 0 or (fh1 + fh2) (mod fs) ≥ W1+ W2
4.2 Proposed Algorithm for finding the Mini-mum Sampling Frequency of Two Bandpass Signals
In this section we propose an efficient algorithm for finding the minimum sampling frequency. For simplicity, first consider the case of two bandpass signals, which we have derive four alias-free conditions in section 4.1. For each of the four
X (f+f )
cases, we derive the minimum increment in sampling frequency such that the corresponding condition for alias-free sampling can be satisfied.
Case (a). Suppose the condition in (4.2) is not satisfied for a given sampling frequency fs. Consider the folded spectrum for the interval [0, fs). We discuss the two cases (i) 0 < fh1 (mod fs) < fs/2 and (ii) fs/2 < fh1 (mod fs) < fs separately.
(i) 0 < fh1 (mod fs) < fs/2: When we gradually increase the sampling fre-quency the band edge fh1 (mod fs) of replica X1+(f ) moves towards 0 while the band edge −fh1 (mod fs) of replica X1−(f ) moves towards fs. When the sampling frequency is increased such that fh1 (mod fs) decreases to 0, then the condition in (4.2) becomes satisfied.
(ii) fs/2 < fh1 (mod fs) < fs: Similarly the condition in (4.2) becomes satisfied when fh1 (mod fs) decreases to fs/2.
Therefore we can conclude that the alias-free condition (4.2) can be satisfied by increasing the sampling frequency such that fh1 becomes an integer multiple of fs/2. The smallest new sampling fs,new for this to happen can be computed as follows. Let
fh1 = nh1fs/2 + rh1,
where rh1 = fh1 (mod fs/2) and nh1 = bfh1/(fs/2)c. Then we have fh1 = nh1fs,new/2, or equivalently
fs,new = 2fh1
nh1 = 2fh1
bfh1/(fs/2)c = 2fh1
b2fh1/fsc, (4.7) where we have used the fact that nh1 can also be computed using nh1 = b2fh1/fsc.
Case (b). Similar to case (a), if the condition in (4.3) is not satisfied, we can increase sampling frequency to
fs,new = 2fh2
b2fh2/fsc, (4.8)
then (4.3) will become satisfied.
Case (c). Suppose the condition in (4.5) is not satisfied. Consider again the shifted spectrum in Fig. 4.4(b). Using the steps in case (a), we can verify that there will be not aliasing if we increase the sampling frequency so that a (mod fs) to be equal to 0 or f2s. Moreover the new sampling frequency can be obtained by
fs,new = 2a
ba/f2sc = fh2 − f`1
b(fh2 − f`1)/fsc (4.9)
Case (d). Like case (c), if the condition in (4.6) is not satisfied, we can increase the sampling frequency to
fs,new = fh1 + fh2
b(fh1 + fh2)/fsc (4.10) then (4.6) will be satisfied.
Proposed iterative algorithm
Using the conditions for alias-free sampling in section 4.1 and the methods for computing new sampling frequency for each case, we have the following iterative algorithm for finding the minimum sampling frequency. To start off, let fs = 2(W1+ W2), which is the lowest possible sampling frequency for no aliasing.
1. Examine if the condition for case (a) in (4.2) is satisfied. If it is, go to the next step. If it is not satisfied, compute the new sampling frequency using (4.7) and go to the next step.
2. If the condition (4.3) for case (b) is satisfied, go to the next step. If it is not satisfied, compute the new sampling frequency using (4.8) and go to step 1.
3. If the condition (4.5) for case (c) is satisfied, go to the next step. If it is not, compute the new sampling frequency using (4.9) and go to step 1.
4. If the condition (4.6) for case (d) is not satisfied, compute the new sampling frequency using (4.10) and go to step 1. If it is satisfied then we have found the minimum sampling frequency.
Usually not all four steps are performed in one iteration.
4.3 Complexity
In this section we will analyze the computation of the algorithm. In the algo-rithm for two bandpass signals, the main computations are in the inspection of conditions in (4.2), (4.3), (4.5) and (4.6), and the computation of new sampling frequency in (4.7)-(4.10). Few computations are required for these equations as we can borrow results from earlier evaluations. For example in step 1 we compute 2fh1 (mod fs) in (4.2). In the process we can also obtain the integer nh1 which is used in computing the new sampling frequency (4.7). Similar conclusions can be drawn for steps 2. In step 3, we need to evaluate fh2 − f`1 (mod fs) which can be written as
(fh2 − f`1) (mod fs)
= (fh2 (mod fs) − f`1 (mod fs))
| {z }
call this x
(mod fs)
=
½ x , x ≥ 0, x + fs , otherwise.
(4.11)
When we are in step 3, the conditions in step 1 are already satisfied, we can obtain f`1 (mod fs) using
f`1 (mod fs) = fh1 (mod fs) − W1.
if fh1 (mod fs) 6= 0. When fh1 (mod fs) = 0, f`1 (mod fs) = fs− W1. fh1 and 2fh1 can be expressed as follows.
fh1 = mfs+ fh1 (mod fs), where m =
¹fh1 fs
º
2fh1 = nfs+ (2fh1) (mod fs), where n =
¹2fh1 fs
º
Comparing the above two equations, fh1 (mod fs) = (2fh1) (mod fs)/2 if n is even and fh1 (mod fs) = ((2fh1) (mod fs) + fs)/2 if n is odd. Thus both fh1
(mod fs) and fh2 (mod fs) can be obtained from steps 1 and 2. The evaluation
for step 3 requires at most 5 additions. Similarly in step 4, we need to evaluate fh1 + fh2 (mod fs) which can be written as
(fh1 + fh2) (mod fs)
= (fh1 (mod fs) + fh2 (mod fs))
| {z }
call this y
(mod fs)
=
½ y , y < fs, y + fs , otherwise.
(4.12)
fh1 (mod fs) and fh2 (mod fs) are already obtained from step 3. The evaluation requires at most 2 additions.