eqalign figures in peakeq

Paul Batchelor committed 6 years ago

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

Unified · Split

4 files

 \overfullrule 0pt
 \noindent
 $$
-y(n) = \alpha x(n) + \beta(1 + \alpha)x(n - 1) + x(n - 2)
+\eqalign {
-- \beta(1 + \alpha)z^{-1} y(n - 1)  - \alpha y(n - 2)
+y(n) = &\alpha x(n) + \beta(1 + \alpha)x(n - 1) + x(n - 2)
+\cr
+- &\beta(1 + \alpha)y(n - 1)  - \alpha y(n - 2)
+}
 $$
 \bye

 \overfullrule 0pt
 \noindent
 $$
-v(n) = x(n) - \beta(1 + \alpha)v(n - 1) - \alpha v(n - 2)
+\eqalign{
-y(n) = \alpha v(n) + \beta(1 + \alpha) v(n - 1) + v(n - 2)
+v(n) &= x(n) - \beta(1 + \alpha)v(n - 1) - \alpha v(n - 2)
+\cr
+y(n) &= \alpha v(n) + \beta(1 + \alpha) v(n - 1) + v(n - 2)
+}
 $$
 \bye

 \overfullrule 0pt
 \noindent
 $$
-v(n) = x(n) - a_1 v(n - 1) - a_2 v(n - 2)
+\eqalign{
-y(n) = b_0 v(n) + b_1 v(n - 1) + b_2 v(n - 2)
+v(n) &= x(n) - a_1 v(n - 1) - a_2 v(n - 2) \cr
+y(n) &= b_0 v(n) + b_1 v(n - 1) + b_2 v(n - 2)
+}
 $$
 \bye

 @!
 (fig "peakeq_df1_difference"
 ``
-y(n) = \alpha x(n) + \beta(1 + \alpha)x(n - 1) + x(n - 2)
+\eqalign {
-- \beta(1 + \alpha)z^{-1} y(n - 1)  - \alpha y(n - 2)
+y(n) = &\alpha x(n) + \beta(1 + \alpha)x(n - 1) + x(n - 2)
+\cr
+- &\beta(1 + \alpha)y(n - 1)  - \alpha y(n - 2)
+}
 ``)
 !@
 @!(fig "peakeq_df2_structure"
 ``
-v(n) = x(n) - a_1 v(n - 1) - a_2 v(n - 2)
+\eqalign{
-y(n) = b_0 v(n) + b_1 v(n - 1) + b_2 v(n - 2)
+v(n) &= x(n) - a_1 v(n - 1) - a_2 v(n - 2) \cr
+y(n) &= b_0 v(n) + b_1 v(n - 1) + b_2 v(n - 2)
+}
 ``)!@
 Converting to direct form II is a matter of putting
 @!(fig "peakeq_df2_difference"
 ``
-v(n) = x(n) - \beta(1 + \alpha)v(n - 1) - \alpha v(n - 2)
+\eqalign{
-y(n) = \alpha v(n) + \beta(1 + \alpha) v(n - 1) + v(n - 2)
+v(n) &= x(n) - \beta(1 + \alpha)v(n - 1) - \alpha v(n - 2)
+\cr
+y(n) &= \alpha v(n) + \beta(1 + \alpha) v(n - 1) + v(n - 2)
+}
 ``
 )!@

Keyboard shortcuts