HOMEWORK #10 Read first 5 sections of Notes on Linear Sequential Circuits, now posted on the class website. Problems: In all these problems, + means XOR 1. Solve the following set of linear equations for the Xs in terms of the Zs. Z1 = X1 + X3 Z2 = X1 + X2 Z3 = X1 + X2 + X3 Z=AX, where Z and X are column vectors and A is the matrix 101 110 111 Convert A to the identity matrix by a series of row operations, 101 r3=r3+r2 101 r1=r1+r3 100 r2=r2+r1 100 110 110 110 010 111 001 001 001 Apply the same sequence of row operations to the I matrix to obtain the inverse. 100 r3=r3+r2 101 r1=r1+r3 111 r2=r2+r1 111 010 010 010 101 001 011 011 011 So, from the final matrix we obtain: X1=Z1+Z2+Z3 X2=Z1+ Z3 X3= Z2+Z3 2. Find all the null sequences of each of the following polynomials (D3 means the 3rd power of D, represented also as D^3, etc.) (a) D3 + D + I Z=(I+D+D3)X=0 X=DX+D3X Start with any initial sequence of 3 bits, excluding 000. Then use the recursion formula X=DX+D3X to generate the continuation of the sequence bit by bit until the initial sequence is repeated (or, we could stop if the sequence length reaches 7) Starting with 001 we obtain: 0011101001, so the period is (0011101). This is an m-sequence so we need go no further. (b) D4 + D3 + D2 + D + I. The recursion formula here is X=DX+D2X+D3X+D4X. Starting with 0001 we obtain: 000110001, so we have one sequence (00011) Choose another starting sequence of length 4 not in the above sequence. 010100101... (01010) Choose another starting sequence not in either of the above sequences. 0111101111... (01111) Now we have used all 15 non-zero subsequences of length 4 (alternatively, we can say that the total length of all the sequences found is 5+5+5=15. So we have them all. This prime polynomial is NOT an m-polynomial and it has the 3 null sequences, (00011), (01010), (01111) 3. Find the polynomial for which the following is the m-sequence (maximal length null sequence). 1 1 0 0 0 1 0 0 1 1 0 1 0 1 1. Since the sequence length is 15, which is 2^4 -1, the polynomial must be of degree 4. So the form is: D4+a3D3+a2D2+a1D+I, and we need to find the values of the coefficients a3, a2, a1. At every point in the sequence, the following equation must be satisfied: D4X+a3D3X+a2D2X+a1DX+X. Find a length 5 subsequence of sequence beginning and ending with 0 and containing only one 1. This will identify a coefficient that must be a 0. One such subsequence is 00100. It gives us the equation: 0 + a3x0 + a2 + a1x0 + 0 = 0, which tells us that a2 must be 0. A second such equation can be derived from the subsequence 00010, which generates the equation 0 + a3x0 + 0 + a1 + 0 = 0, which tells us that a1=0. If we know that the sequence is an m-sequence, then we can assume that a3=1, else the polynomial would be D^4 + I, which cannot be an m-polynomial because it has and even number of terms and is thus divisible by D+I so that it is not prime (being prime is a necessary, but not sufficient condition for a polynomial to be an m-polynomial). If we are not sure that the polynomial is an m-polynomial, then we could assume a3=1 and then find the m-sequences of the resulting polynomial, which is D4+D3+I. Or, we could identify a3 directly from the subsequence 01001, which gives us the equation, 0+a3+0+0+1=0. From this it is clear that a3=1. In order to verify that the D4+D3+I is an m-polynomial, we must then either generate the null-sequences directly from it and then check to see if it has an m-sequence that is the same as the given sequence, or else check all consecutive subsequences of length 5 of the given m-sequence to see if it the polynomial equation yields a 0 for each. 4. Factor the following polynomials: (a) D6 + D4 + I All exponents of this polynomial are even numbers, so it is a "square", namely, (D3+D2+I)^2 (check this out by multiplication). D3+D2+I is prime, so we need go no further. (b) D5 + D4 + D2 + I This polynomial has an even number of terms, so it is divisible by D+I. Divide by D+I as below: D4 + D + I ------------------ D+I|D5 + D4 + D2 + I D5 + D4 ---------------- D2 + I D2 + D ------ D+I D+I --- 0 So we have D5 + D4 + D2 + I = (D+I)(D4 + D + I) We cannot factor any further because (D4 + D + I) is an m-polynomial, which is not hard to determine by simply generating the m-sequence. 5. Show that if S is any m-sequence, then S^R, (S written backwards) is also an m-sequence. Indicate how the corresponding polynomials are related. Consider the polynomial P= D3 + D + I. This is an m-polynomial. We can represent it by a vector indicating the coefficient values starting with that of the D3 term. This gives us 1011. The null sequence of P is S = (1110100). If we match 1011 against any length-4 subsequence of S, the number of matching 1's will always be even. For example, in 1110, the only matching 1's are the first and third. In 0011 the third and 4th 1's match. In 0100 there are no matching 1's. This is a necessary and sufficient condition for the sequence to be a null sequence of the polynomial. If we reverse both S and P (obtaining 1101 or D3 + D2 + I) the same applies. So if any sequence is an m-sequence corresponding to some polynomial P, the reverse of the sequence is an m-sequence of the polynomial that is the reverse of P.