old quotient sequence stuff REVD
DOCX · 19.4 KB
Open DOCX file
Short revised Word note dated 3.26.05, in Phil's Scrambler folder under 'Not needed anymore'. It expresses the remainder R(z) of dividing the input I(z) by H(z) as a convolution of H with the tail of the output sequence. He finds the remainder coefficients depend on earlier outputs, not just the k register contents, so his conjecture fails.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
This is the Title PhL 3.26.05
Note that page numbering is turned on in this template.
Ir(z)/H(z) = Q(r)(z) + R(r)(z)/H(z) . (1.3.18)
Ir(z) = Q(r)(z)H(z) + R(r)(z)
in = Σj qn-jhj + rn
rn = in - Σj qn-jhj
******************************8
We know that the contents of the k registers are the next k output symbols, so
q0 = or+1
q1 = or+2
...
qk-1 = or+k
Meanwhile,
R(z) = H(z) zr [ Σi=r+1∞ oiz-i ] = H(z) { zr Otail(z) }
We have to group the zr one way or the other, so do it with the tail. Then
Otail(z) ↔ oi i > r
zr Otail(z) ≡F(z) ↔ or+i i > r . // by (1.1.4)
Now consider as a convolution
R(z) = [zr Otail(z)] H(z) = F(z) H(z) .
The time domain projection of the above equation is this:
rn = Σj=0k fn-jhj .
Now we know that
fi = or+i
so then
fn-j = or+n-j
and we then get
rn = Σj=0k or+n-j hj = or+n h0 + or+n-1 h1 + ...... or+n-k hk
So this gives the remainder coefficients in terms of these k+1 values
or+n
or+n-1
...
or+n-k
So in general,
rn = rn(or+n, or+n-1 .... or+n-k )
Here are special cases
r0 = r0(or, or-1 .... or-k )
r1 = r1(or+1, or .... or-k+1 )
But all these oj have already been output ! So my conjecture is wrong! the other extreme is
rk-1 = rk-1(or+k-1, or+k-2 .... or-1 )
Conclusion: the remainder for Fig 1.1 is NOT a function of the k constants in the registers!!! It is a function of the outputs oj and only some of these are in the registers when we stop.
Convolution then says
rn = Σj=0k (otail)n-jhj