Chapter VII: Minimum Datalength for Integer Period Estimation
7.8 Minimal Non-Contiguous Sampling For Period Estimation
TTTTCT TTTTCT TTTTCT TTTTCT… (b)Period 6 sequence:
Figure 7.6: Demonstration of Theorem 7.2.1 using DNA repeats. (a) Period 5 repeats used in DNA fingerprinting applications (GenBank M86525). (b) A period 6 sequence constructed such that its first 9 samples (shown in black) are the same as in the sequence (a). See text for details.
Then, y(n)is also a periodic signal, with its period being a divisor of (M,P)P . ♦
Proof.
y
n+ P (M,P)
= x
Mn+ M P (M,P)
= x(Mn)
= y(n)
So (M,P)P is a repetition index of y(n). A repetition index of a signal must always be
a multiple of its period (see Lemma 3 in [9]). 5 5 5
We also need the following lemma.
Lemma 7.8.2. Letx(n)be a periodic signal with periodP,1. Let Mbe an integer coprime to P, and let y(n) = x(Mn). Then, y(n)is a periodic signal with period
,1. ♦
Proof. From Lemma 7.8.1, it follows that y(n) must be a periodic signal whose period is a divisor of P. So y(n)can have period 1 if and only if all the entries in the following set are equal:
Y= {y(0),y(1), . . . ,y(P−1)} (7.49) Using (7.48), we can re-writeYas:
Y= {x(0),x(M), . . . ,x(M P−M)} (7.50)
Sincex(n)is periodic with periodP, we can re-write this further as:
Y= {x(0 modP),x(M modP), . . . ,y(M P−M mod P)}
It is a well known result in Number Theory that when(M,P)=1, the following two sets are the same (permuted versions of each other):
{(0 mod P),(M modP), . . . ,(M P−M modP)}
= {0,1, . . . ,P−1}
Since x(n) has period P , 1, all the elements in the above set cannot be equal to each other. Hence, there are at least two unequal entries in (7.49), due to which, the
period ofy(n)cannot be 1. 5 5 5
We will now prove the sufficiency part of Theorem 7.8.1. We shall do this by considering two distinct cases as follows:
Sufficiency whenP1|P2
In this case, Theorem 7.8.1 claims that P2 samples are sufficient. To prove this, we can use the Nested Periodic Matrices proposed in Chapter 3. Let x(n) be a signal whose period is eitherP1orP2, and letxbe aP2×1 vector consisting ofP2 consecutive samples of x(n). Further, let Abe a P2×P2Nested Periodic Matrix.
Then, sinceAis a full rank matrix, the following system of equations has a unique solution fors:
x=As (7.52)
Moreover, it follows from Lemma 3.2.3 that the Least Common Multiple of the periods of those columns ofAthat are multipled by non-zero entries in s, is equal to the period of x(n)(either P1 orP2in this case). Hence, P2 samples of x(n)are sufficient to find its period.
Sufficiency whenP1 - P2
WhenP1- P2, P1samples are sufficient to find the period. We will first prove this for the case when P1 and P2 are coprime, and then extend the proof to the more general case.
WhenP1andP2are coprime, consider downsamplingx(n)byM = P2to yeildy(n). That is,y(n)= x(Mn). In this case, ifx(n)had periodP2, then, the period ofy(n)is 1 from Lemma 7.8.1. However, if x(n)had periodP1, then the output y(n)cannot have period 1 according to Lemma 7.8.2. To check whether y(n)has period 1, we need atmost P1 samples of y(n), since its period can be any divisor of P1. This proves thatP1samples are sufficient to find the period in this case.
Let us now consider the more general case when (P1,P2) = G (G , P1, since P1 - P2). We now propose to use the setup shown in Fig. 7.7. Let us first consider
P2
y0(n) P2
P2
P2
y1(n) y2(n)
yG-1(n) z -1
x (n)
z -1 z -1
z -1
Figure 7.7: Finding the period of x(n)when(P1,P2)=G, P1. See text for details.
the case when x(n) has period P2. Then, it is easy to see that all the outputs y0(n),y1(n), . . . ,yG−1(n)will have period 1.
However, if x(n) has period P1, then atleast one of y0(n),y1(n), . . . ,yG−1(n) will have period> 1. To prove this, we re-draw Fig. 7.7 as Fig. 7.8. If x(n)had period P1, and if P1 , G, then at least one of u0(n),u1(n), . . . ,uG−1(n)must have period
> 1. This is because, if all of them are period 1 signals, then x(n)must satisfy:
x(n+G)= x(n)∀n∈Z (7.53) which then necessitates thatP1|G(since the period must always divide any repetition index). This is possible only if G = P1, which contradicts our assumption that P1- P2.
Let us assume thatui(n) has period > 1. Because of Lemma 7.8.1, the period of ui(n) must be a divisor of PG1. Further, since PG1 and PG2 are always co-prime, any divisor of PG1 is also coprime to PG2. So using Lemma 7.8.2, we can conclude that the period ofyi(n)must be> 1.
Notice that, using Lemma 7.8.1,u0(n),u1(n), . . . ,uG−1(n)can have their periods as any divisors ofPG1, and so the outputsy0(n),y1(n), . . . ,yG−1(n)can have their periods as any divisors of PG1. So PG1 samples of each output are sufficient to check if they have period 1. Since there areG such outputs, PG1 × G = P1 samples of x(n) are sufficient to determine whether the period isP1or P2. This completes the proof of the sufficiency side of Theorem 7.8.1.
We will now prove the necessity part of Theorem 7.8.1.
G y1(n) y2(n)
yG-1(n) z -1
z -1
z -1
1
u2(n)
uG-1(n) 𝑃2/𝐺
𝑃2/𝐺
𝑃2/𝐺 G
G
Figure 7.8: Re-drawing Fig. 7.7 for analysis. See text for details.
Proving Necessity
We will first show that at least P1 samples are necessary to find the period from the set{P1,P2}, irrespective of whetherP1|P2orP16 |P2. Later, we will show that whenP1|P2,P2samples are necessary.
Theorem 7.8.2. Given any set of L < P1time indicesNT = {n1,n2, . . . ,nL} ⊂ Z, there exist periodic signals xP1(n)and xP2(n) with periods P1 and P2respectively such that
xP1(n)= xP2(n)∀n∈NT (7.54)
♦
Proof. Since L < P1, there exists at least one integer in the set {0,1, . . . ,P1−1}
that does not belong to the set{(n1mod P1),(n2modP1), . . . ,(nL modP1)}. Letm be such an integer. We define xP1(n)as follows:
xP1(n)=
0 ifnmodP1=m 1 otherwise
(7.55) It is easy to see thatxP1(n)has periodP1(a zero occurs only once everyP1samples).
Notice that xP1(n) = 1 ∀ n ∈ NT. In the same way, we can construct a periodP2
signalxP2(n)that satisfiesxP2(n)=1∀n ∈NT. Clearly, for thesexP1(n)andxP2(n),
(7.54) is satisfied. 5 5 5
We will now prove that whenP1|P2, one needs at least P2samples to estimate the period.
Theorem 7.8.3. LetP1|P2. Given any set ofL < P2time indicesNT = {n1,n2, . . . ,nL} ⊂ Z, and any periodP1signalxP1(n), there exists a periodP2signalxP2(n)such that
xP1(n)= xP2(n)∀n∈NT (7.56)
♦
Proof. LetxP2(n)be defined to be equal toxP1(n)for alln∈NT. Doing so will not violate the following condition:
xP2(n+P2)= xP2(n)∀n∈NT (7.57) since,n+P2 =n+k P1for some integer k.
Further, since L < P2, there exists at least one integer in the set {0,1, . . . ,P2−1}
that does not belong to the set{(n1mod P2),(n2modP2), . . . ,(nL modP2)}. Letm be such an integer. Moreover, letuandvbe integers such thatu> maxnxP1(n)and v < maxnxP1(n). We definexP2(n)as follows:
xP2(n)=
xP1(n) ifnmodP2∈NT
u nmodP2= m v otherwise
(7.58)
It is easy to see that xP2(n)has period P2(u occurs only once every P2 samples).
Hence, we have constructed a period P2 signal xP2(n)satisfying the conditions of
the theorem. 5 5 5
This completes the proof of Theorem 7.8.1.