Chapter VII: Minimum Datalength for Integer Period Estimation
7.3 Mixtures of Periodic Signals
if we just use this minimum number. This is demonstrated in Fig. 7.3. In this plot, P= {1,2, . . . ,6}. Theorem 7.2.3 shows that at least 10 samples of the signalx(n)are needed for period estimation. However,x(n)now is contaminated by AWGN with 0 dB SNR. For each data-length shown, 10,000 independent realizations ofx(n)with period 3 were generated. A least squares modification of (7.15) was used to estimate the period, where, if zmax is the entry inzwith the maximum absolute value, then all the entries in z less than a threshold times |zmax| were set to zero. The error rate (the fraction of times the period was estimated incorrectly) is plotted for three different values of the threshold. As shown, the error is large when data length is close to the the theoretical minimum, and improves significantly as more datalength becomes available. Once again, there are many excellent works such as [16], [95]
for estimating the period under noise. Fig. 7.3 is just a simple demonstration that one may typically need more than the minimum data-length for reliable estimation under noise.
A special case of Theorem 7.2.1 is the following result:
Corollary 7.2.1. For the special case whenP = {1,2,3, . . . ,Pmax}, the minimum necessary and sufficient datalength in Theorem 7.2.1 becomes:
Lmin =2Pmax −2 (7.16)
♦
Proof. It is easy to see thatPmax andPmax −1, being co-prime, maximize (7.5) in
Theorem 7.2.1. 5 5 5
So when P = {1,2,3, . . . ,Pmax}, Lmin in (7.16) turns out to be close to the 2Pmax rule of thumb used in prior works such as [95]. The difference is that, we have now proved that necessity of (7.44) in fact is an algorithm-independent fundamental bound.
Figure 7.3: Error Rate vs Data-Length for a fixed SNR. “Th" refers to the threshold.
SNR = 0dB. See text for details.
need a precise mathematical definition of what we mean by the ‘hidden periods’
in a mixture. Intuitively, we may visualize hidden periods in the following way.
Given a signal, if we can decompose it as a sum of periodic signals with integer periods {P1,P2, . . . ,PN}, then we may want to call these integers as the hidden periods. However, there are some subtle issues with this, that have never been formally addressed in prior literature.
For example, let xP(n)be a periodPsignal. Let dbe any divisor ofP. ThenxP(n) can be trivially rewritten as
xP(n)= xd(n)+ x˜P(n) (7.17) where xd(n)is almost any arbitrary period d signal, and ˜xP(n) = xP(n) −xd(n)is still a period Psignal. Thus, for a periodP signal, it is meaningless to declare any of its divisorsdas another hidden period.
By insisting that the distinct hidden periods in a mixture should constitute an M-set as defined below, we can eliminate this redundancy:
Definition 7.3.1. M-set: A set of numbersFis defined to be an M-set, if the following holds:
fi6 | fj ∀ fi, fj ∈F, fi , fj (7.18) That is, no member of an M-set is a divisor of another member3. ♦ For example,{6,7,10}is an M-set, whereas{3,6,10}is not.
3The ‘M’ stands for Mutliple-free.
There is a second type of ambiguity that arises while defining hidden periods.
We can always associate different M-sets to any mixture of periodic signals. For example, consider adding period 3 and period 8 signals to give a period 24 signal.
Since{3,8}is an M-set, we may want to call them as the hidden periods. However, given the periodic components{3,8}, we can add and subtract any arbitrary period 5 signal from these two components, to get periods {15,40}. These new signals with periods 15 and 40 will also add to give the same period 24 signal. Moreover, {15,40} is also an M-set. So now, which among {15,40} and {3,8} should one declare as the hidden periods of the period 24 signal?
In the above example, it turns out that periods {3,8} represent the finest periodic structurein the mixture. That is, one cannot further decompose the period 3 signal into a sum of signals with periods are strictly smaller than 3. Similarly, one cannot further decompose the period 8 signal into a sum of signals with periods strictly smaller than 8. On the other hand, in this example, both the period 15 and the period 40 signals can be further decomposed into signals with periods strictly smaller than 15 and 40 respectively (namely, into periods 5 and 3, and periods 5 and 8 respectively). This can also be seen when above period 15 signal is decomposed as in (7.6), where it will not have any terms of the formej2πk15nwith(k,15)= 1. And the same holds for the period 40 signal.
We will now show that if we define the hidden periods of a signal as those repre- senting the finest periodic structure (in the above sense), then the hidden periods become unique for any mixture of periodic signals. That is, we define hidden periods formally in the following way:
Definition 7.3.2. Hidden Periods: A signal x(n) is said to have hidden periods PH = {P1,P2, . . . ,PN}if the following hold true:
1. PH is an M-set (see Notations in Sec. 7.1).
2. x(n)can be written as follows:
x(n)= xP1(n)+xP2(n)+. . .+ xPN(n) (7.19)
where each xPi(n) has period Pi, and cannot be further decomposed into a sum of periodic signals, all of whose periods are strictly smaller thanPi.
♦
PH x n 4 PC
the set of periods of all those complex exponentials that have non-zero coefficients in(7.6), thenPH ⊆ PC. More precisely:
PH = {Pi ∈PC : M Pi<PC ∀M > 1} (7.20)
♦
Remark 1: In (7.20), the R.H.S. is the set of all those numbers inPCthat do not have any multiples also present inPC. For example, ifPC = {1,3,5,10}, then the R.H.S.
of (7.20) is{3,10}. We will call this as theextracted M-setofPC.
Remark 2: The extracted M-set of PC is unique, since PC itself is unique for any signal (Theorem 7.2.2). Hence, the set of hidden periods,PH, is also unique.
Remark 3: As an example, consider the following three signals:
1. ej2πn6 has period 6. There are no hidden periods (other than 6 itself).
2. ej2πn2 + ej2πn3 has period 6. The hidden periods are 2 and 3, but 6 is not a hidden period.
3. ej2πn2 +ej2πn3 +ej2πn6 has period 6. There are no hidden periods other than 6 itself, since this signal can never be written as a sum of signals, all of whose periods are strictly smaller than 6. Also, as discussed earlier, calling 2 and 3 as hidden periods in addition to 6 is redundant in this case, since any period 6 signal x6(n)can always be written as
x6(n)= x2(n)+ x3(n)+x˜6(n) (7.21) for some non-zero signals x2(n), x3(n) and ˜x6(n), with periods 2, 3 and 6 respectively. So Definition 7.3.2 simply gives 6 as the only hidden period.
Proof of Theorem 7.3.1: The proof involves three steps:
1. Showing thatPH ⊆ PC.
4Any mixture of periodic signals must also be a periodic signal.
2. Showing thatPC ⊆ D.S.(PH).
3. Using Lemma 7.11.2 from the Appendix to conclude thatPH is as in (7.20).
Proof of Step 1: Since x(n) has hidden periods PH, it will have a decomposition similar to (7.19). Let us now decompose the L.H.S. and each term in the R.H.S. of (7.19) in a manner similar to (7.6). While the decomposition of the L.H.S. is given in (7.6), each xPi(n)in the R.H.S. can be written as:
xPi(n)= Õ
1≤dPi≤Pi dPi|Pi
Õ
1≤k≤dPi (k,dPi)=1
αk,
dPiej
2πk dPi
n (7.22)
This decomposition ofxPi(n)must have at least one complex exponential with period Pi with a non-zero coefficient. Else, xPi(n)can be written as a sum of signals, all of whose periods are strictly smaller than Pi, which contradicts Condition 2 in Definition 7.3.2.
Further, if xPi(n) has a complex exponential with period Pi, then that complex exponential cannot be canceled in the R.H.S. of (7.19) by the complex exponentials in some other xPj(n). This is because, such a cancellation can occur only if xPj(n) has a period Pi complex exponential, meaning Pi|Pj. This is not possible because PH is an M-set. Hence, when the R.H.S. of (7.19) is expressed as a sum of complex exponentials, there has to be at least one complex exponential with period Pi, for everyPi ∈PH.
Since complex exponentials are linearly independent signals, when both the L.H.S.
and the R.H.S. of (7.19) are expressed in terms of complex exponentials, we must have the same complex exponentials on both sides. This implies that the decompo- sition of x(n)in (7.6) must also have at least one complex exponential with period Pi, with a non-zero coefficient. Hence,PH ⊆ PC.
Proof of Step 2: Letq ∈PC. From (7.6), the DTFT ofx(n)must have an impulse at the frequency 2πrq , for some(r,q)=1. But ifq <D.S.(PH), then none of thexPi(n) in (7.19) have an impulse in their DTFT at the frequency 2πrq (This is because, from Theorem 7.2.2, a periodPsignal can have non-zero DTFT only at frequencies that are of the form 2πmd , where(m,d) =1 andd|P). Hence, we obtain a contradiction.
Soq∈ D.S.(PH), and hencePC ⊆ D.S.(PH).
Finally, using Lemma 7.11.2 from the Appendix, we can conclude thatPH = {Pi ∈ PC : M P<PC ∀M >1}.
P =lcm(PH) (7.23) i.e., thelcm of the hidden periods of a signal is exactly equal to the period of the
signal. ♦
Proof. From Theorem 7.2.2, it follows thatP = lcm(PC). Notice thatPC and the R.H.S. of (7.20) have the samelcm. And so (7.23) follows. 5 5 5 Remark 1: Readers familiar with Ramanujan Subspaces and Nested Periodic Sub- spaces (NPSs) [18], [37] may recall the following: Suppose we decompose x(n) along a set of NPSs [37]. IfPN PSis the set of periods of all the non-zero components in the decomposition, then the extracted M-set ofPN PS was intuitively interpreted as the hidden periods in those works. It can be proved that this in fact turns out to be exactly equal to the set of hidden periods as defined in Definition 7.3.2. Moreover, Theorem 7.3.1 shows that although PN PS for a given signal may depend on which NPS is used, the extracted M-set would always be the same.
Remark 2: In many applications such as speech, the signal of interest may arise naturally as a sum of periodic signals. While the hidden periods in Definition 7.3.2 represent thefinest periodic structurein a mixture, how do they relate to the original set of component periods that generated that mixture? It can be shown that if x(n) is a mixture of randomly generated periodic signals, say by periodically extending Gaussian random vectors with periodsP= {P1,P2, . . . ,PK}, then the extracted M- set fromPis in fact exactly equal to the hidden periods as given by Definition 7.3.2 with probability 1. We will skip the proof in this regard. Once again, determining the divisors of the integers in the extracted M-set as also component periods in the mixture is mathematically meaningless.