History

eqalign figures in peakeq

Paul Batchelor committed 6 years ago

Commit 574e7f01 parent 52b273d4 files · +28 −12Browse files

Unified · Split

4 files
  1. _fig/peakeq_df1_difference.tex +5 −2
  2. _fig/peakeq_df2_difference.tex +5 −2
  3. _fig/peakeq_df2_structure.tex +4 −2
  4. peakeq.org +14 −6
_fig/peakeq_df1_difference.tex+5−2
@@ -4,7 +4,10 @@ · 3 unmodified lines
44 \overfullrule 0pt
55 \noindent
66 $$
7−y(n) = \alpha x(n) + \beta(1 + \alpha)x(n - 1) + x(n - 2)
8−- \beta(1 + \alpha)z^{-1} y(n - 1) - \alpha y(n - 2)
7+\eqalign {
8+y(n) = &\alpha x(n) + \beta(1 + \alpha)x(n - 1) + x(n - 2)
9+\cr
10+- &\beta(1 + \alpha)y(n - 1) - \alpha y(n - 2)
11+}
912 $$
1013 \bye
_fig/peakeq_df2_difference.tex+5−2
@@ -4,7 +4,10 @@ · 3 unmodified lines
44 \overfullrule 0pt
55 \noindent
66 $$
7−v(n) = x(n) - \beta(1 + \alpha)v(n - 1) - \alpha v(n - 2)
8−y(n) = \alpha v(n) + \beta(1 + \alpha) v(n - 1) + v(n - 2)
7+\eqalign{
8+v(n) &= x(n) - \beta(1 + \alpha)v(n - 1) - \alpha v(n - 2)
9+\cr
10+y(n) &= \alpha v(n) + \beta(1 + \alpha) v(n - 1) + v(n - 2)
11+}
912 $$
1013 \bye
_fig/peakeq_df2_structure.tex+4−2
@@ -4,7 +4,9 @@ · 3 unmodified lines
44 \overfullrule 0pt
55 \noindent
66 $$
7−v(n) = x(n) - a_1 v(n - 1) - a_2 v(n - 2)
8−y(n) = b_0 v(n) + b_1 v(n - 1) + b_2 v(n - 2)
7+\eqalign{
8+v(n) &= x(n) - a_1 v(n - 1) - a_2 v(n - 2) \cr
9+y(n) &= b_0 v(n) + b_1 v(n - 1) + b_2 v(n - 2)
10+}
911 $$
1012 \bye
peakeq.org+14−6
@@ -34,8 +34,11 @@ This has the following difference equation: · 33 unmodified lines
3434 @!
3535 (fig "peakeq_df1_difference"
3636 ``
37−y(n) = \alpha x(n) + \beta(1 + \alpha)x(n - 1) + x(n - 2)
38−- \beta(1 + \alpha)z^{-1} y(n - 1) - \alpha y(n - 2)
37+\eqalign {
38+y(n) = &\alpha x(n) + \beta(1 + \alpha)x(n - 1) + x(n - 2)
39+\cr
40+- &\beta(1 + \alpha)y(n - 1) - \alpha y(n - 2)
41+}
3942 ``)
4043 !@
4144
@@ -45,8 +48,10 @@ represented in this generalized difference equation: · 3 unmodified lines
4548
4649 @!(fig "peakeq_df2_structure"
4750 ``
48−v(n) = x(n) - a_1 v(n - 1) - a_2 v(n - 2)
49−y(n) = b_0 v(n) + b_1 v(n - 1) + b_2 v(n - 2)
51+\eqalign{
52+v(n) &= x(n) - a_1 v(n - 1) - a_2 v(n - 2) \cr
53+y(n) &= b_0 v(n) + b_1 v(n - 1) + b_2 v(n - 2)
54+}
5055 ``)!@
5156
5257 Converting to direct form II is a matter of putting
@@ -56,8 +61,11 @@ following difference equation: · 3 unmodified lines
5661
5762 @!(fig "peakeq_df2_difference"
5863 ``
59−v(n) = x(n) - \beta(1 + \alpha)v(n - 1) - \alpha v(n - 2)
60−y(n) = \alpha v(n) + \beta(1 + \alpha) v(n - 1) + v(n - 2)
64+\eqalign{
65+v(n) &= x(n) - \beta(1 + \alpha)v(n - 1) - \alpha v(n - 2)
66+\cr
67+y(n) &= \alpha v(n) + \beta(1 + \alpha) v(n - 1) + v(n - 2)
68+}
6169 ``
6270 )!@
6371

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