The invention relates to data coding, and particularly to data coding for achieving bit and byte synchronization using a single data field.
Bit synchronization refers to the synchronization of a clock for receiving or reading incoming data with the data being received or read. Normally, bit synchronization is achieved when a field of, say, 1s (ones), is written as the data is stored (or transmitted) in a partial-response, maximum-likelihood sequence (“PRML”) channel, and it is sampled using an acquisition loop. In a PRML channel, byte synchronization refers to a field with a data pattern in the data that marks the first bit of a symbol. For byte synchronization, this field is followed by a pattern that, when recognized, determines the start of data. Thus, data can be sampled and the information retrieved in the usual form.
As pointed out in U.S. Pat. No. 6,089,749, the conventional byte synchronization approach has the disadvantage of a long synchronization pattern, with a significant possibility of synchronization failure. The '749 patent proposes a byte synchronization scheme using a byte synchronization pattern between 16 and 18 bits in length.
Our purpose is to unify both the bit synchronization and byte synchronization fields into a single field and achieve bit and byte synchronization simultaneously using the single field (“bit-byte synchronization”). The idea is that the new field is short, allowing for significant savings in magnetic disk real estate. Alternatively, the new field can be used in a hybrid way. In order to combat events like thermal asperity (TA) that wipes out a whole synchronization field, a dual synchronization architecture has been proposed: the bit-byte synchronization field is repeated twice, so if the first bit-byte synchronization field is wiped out, then the system relies on the second one to achieve synchronization. One of the problems associated with TA is loss of both bit and byte synchronization. For that reason, it would be useful to have a synchronization field that recovers bit and byte synchronization simultaneously. Such a capability is disclosed in the following specification.
This invention combines bit and byte synchronization into a single data field u. A data string x has pre-pended to it a short string of 1s (ones), followed by u to yield a string y= . . . 1111, u, x. The string y is normally precoded by convolving it with 1/(1⊕D2), where the symbol ⊕ denotes modulo 2 addition and the operator D denotes a delay of one, D2 a delay of 2, etc. Thus, for a string x0,x1,x2,x3 . . . then (1⊕D2)(x0,x1,x2,x3 . . . ) denotes the string:
x0,x1,x0⊕x2,x1⊕x3,x2⊕x4 . . .
The operation 1/(1⊕D2) denotes the inverse of 1⊕D2. PRML-sampling of y starts at an initial phase, and vectors are obtained from that string by sampling at pre-selected phases following the initial sampling point. The vectors of y are compared with vectors corresponding to PRML samples of an initial set of bits in u obtained at predetermined phases. The pair of y, u vectors exhibiting the minimum Euclidian distance yields a sampling correction value by which the initial sampling phase is corrected and a new initial sampling point preceding x is determined. Here, bit and byte synchronization have been achieved and sampling of x proceeds at the corrected phase, from the new initial sampling point.
Refer to
In the discussion that follows, a 20-bit bit-byte synchronization pattern is set forth to illustrate certain principles. However, it will be clear to those skilled in the art that the ideas and algorithms to be presented can be adapted to a bit-byte synchronization pattern of any size.
The bit-byte synchronization pattern is preceded by a sufficient number of is such that sampling starts before the pattern is encountered. The exemplary 20-bit bit-byte synchronization pattern is:
u=(10110111111011011111) (1)
The string . . . 1111, u, x is transmitted, where x is the data. Sampling of the string starts at any moment prior to u. The idea is that once u is read, both byte synchronization and very close bit-synchronization have been achieved.
The method of bit-byte synchronization proceeds as follows. Consider the string y=. . . 1111, u, x, with u and x as described above. The first step is to precode y by convolving it with 1/(1⊕D2), where ⊕ denotes modulo-2 addition. This precoded string is then sampled at an initial unknown point f using PRML. In a noiseless environment, when the first 16 bits of u, are sampled, i.e., (1011011111101101), at phases 0, 0.125, 0.25, 0.375, 0.5, 0.625, 0.75 and 0.875 respectively using PRML, the following 8 vectors are obtained (notice that precoding makes the samples of u0 below coincide in absolute value with the first 16 bits of u):
Consider m=min{d(ui, uj), 0≦i≦j≦7}, where d(ui, uj) denotes the Euclidean distance between ui and uj. The 16-bit vector (1011011111101101) was chosen in such a way that it maximizes m. Of course other choices are possible, but this vector is the one that gives the best results with the algorithms for bit-byte synchronization to be described next. However, this is only a preferred embodiment and several others are obvious to those skilled in the art.
The synchronizer operates by implementing one of the following algorithms.
Algorithm 2.1 Assume the string . . . 1111, u, x, where u is defined in (1) and x is random data. This string is precoded by convolution with 1/(1⊕D2) and PRML sampling of the precoded string starts at points f, f+1, f+2, . . . before the start of u, where the initial sampling point f is random and unknown. The sampled signal will be possibly subject to noise. Denote this sampled and possibly noisy signal by v0, v1, v2, . . . Then, the algorithm proceeds as follows:
An example of Algorithm 2.1 is as follows.
Assume the string
It can be seen that the minimum is achieved at d4,3. Therefore, the phase will be corrected in ⅜=0.375. Since the original (unknown) sampling point was 0.35, this sampling point is reset as 0.35−0.375=−0.025. Since i=4, sampling starts at points
f+i+20,f+i+21,f+i+22 . . . =23.975 ,24.975 , 25.975, . . .
Notice that for perfect bit-byte synchronization, it would be necessary to sample at points 24, 25, 26.
The bit-byte synchronization method described in Algorithm 2.1 provides useful results. However, even in a noiseless situation, an error of 0.06 (or 6%) is common. Although such an error is not catastrophic, the noise can easily make synchronization even more difficult. One way to make the sampling more precise is by increasing the number of comparison vectors ui. In the preferred embodiment described above these vectors are obtained by sampling at skips of 0.125. Smaller sample intervals may be taken. Of course doing so would increase the complexity of the system.
Another alternative is to perform an interpolation process. One way to do this is to take, instead of the smallest sampled value, a number of smallest ones (for example, the three smallest ones) and interpolate between them. This process will be described in the next algorithm.
Algorithm 2.2 Assume the string . . . 1111, u, x, where u was defined in (1) and x is random data. This string is precoded by convolution with 1/(1⊕D2). Then, PRML sampling of the precoded string starts at points f, f+1, f+2, . . . before the start of u, where the initial sampling point f is random and unknown. The sampled signal will possibly be subject to noise. Denote this sampled and possibly noisy signal by v0, vi, v2, . . . . Then, proceed as follows:
Now, consider l, 0≦l≦13 such that d1, d1+1, d1+2, are the smallest values in d (without loss of generality, it may be assumed that these three values are consecutive, although this assumption is not necessary to the operation of the algorithm). Consider the vector of length 16:
As an example of Algorithm 2.2, similarly to Example 2.1, assume a string:
Next, assuming that sampling of the precoded string starts at the unknown initial point 1.97, the following samples are obtained:
Notice that, according to Algorithm 2.2, di,j≦2 is achieved at i=2, so
Manifestly, l=7, thus, (al, al+1, al+2)=(0.875, 1, 1.125) and (dl, dl+1, dl+2)=(0.5, 0.14, 1.13). This gives,
Since the initial (unknown) sampling point was 1.97 and i=2, according to Algorithm 2.2, f is reset to f←f−(r−1)=1.97−(0.99 −1)=1.98 and sampling starts at points
f+i+20, f+i+21, f+i+22 . . . =23.98, 24.98, 25.98, . . .
Again, notice that for perfect bit-byte synchronization, one would need to sample at points 24, 25, 26, . . .
The next algorithm introduces a modification that makes it slightly more efficient than Algorithm 2.2.
Algorithm 2.3 Assume the string . . . 1111, u, x, where u was defined in (1) and x is random data. This string is precoded by convolution with 1/(1⊕D2) and PRML sampling of the precoded string starts at points f, f+1, f+2, . . . before the start of u, where the initial sampling point f is random and unknown. The sampled signal will be possibly subject to noise. Denote this sampled and possibly noisy signal by v0, v1, v2, . . . Then, proceed as follows:
Consider 1, 0≦l≦13 such that dl+1 is the smallest value in d and, without loss of generality, dl, dl+1, and dl+2 are the three smallest values in d. Consider the vector of length 16:
If dl+2≦2dl or dl≦2dl+2, let
Make f←f−(r−1).
Start sampling the data at sampling points f+i+20, f+i+21, f+i+22 . . . to obtain a sampled sequence w=w0, w1, w2, . . . Apply maximum likelihood decoding to w to obtain an estimate of x.
This example describes the same situation as Example 2.2, where 1=7, (al, al+1, al+2)=(0.875, 1, 1.125) and (dl, dl+1, dl+2)=(0.5, 0.14, 1.13). Notice that dl+2=1.3>2(0.5)=2dl. Thus, according to Algorithm 2.3:
Since the initial (unknown) sampling point was 1.97 and i=2, according to Algorithm 2.3, f is reset according to f←f−(r−1)=1.97−(0.97−1)=2 and sampling starts at points
f+i+20, f+i+21, f+i+22 . . . =24, 25, 26, . . .
Thus, perfect bit-byte synchronization occurs in this case.
Representative simulation results obtained using Algorithms 2.1, 2.2 and 2.3 are tabulated below. In all cases, white Gaussian noise with a variance of 0.06 was added. A number of tests were run for each algorithm and for each case to measure how far away the results were from perfect bit-synchronization (byte synchronization was correct in all cases). For instance, in Example 2.1 this number is 0.025, in Example 2.2 it is 0.02 and in Example 2.3 it is is 0. The average of all these numbers, the standard deviation, the worst case and the percentage of cases above 0.05 are shown in the following table.
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