八點快速富利葉轉換(FFT)
程式編寫日期: 2006年7月11日
程式需要在 REG Quad 模式下執行，因此在輸入程式前請先按 Mode Mode 2 → 3 進入REG Quad模式。
注意: 藍色的英文字為統計模式中的變數(Σx3 按 Shift 1 → → 1，Σy 按 Shift 1 → 2，Σxy Shift 1 →3)，而3√是按shift x3 ， πr 是按 Shift EXP Shift Ans 2
程式 (153 bytes)
Stat clear: ?→M: 3√(M - 1 DT: ?→M: 1 , M DT: ?→M:
0 , M - Σy DT: ?→A: ?→B: ?→C: ?→D: ?→X: 0→M:
Lbl 0: πr M÷4→Y: Σx3 + Σxy cos Y + Σy cos2Y +
A cos3Y + B cos4Y + C cos5Y + D cos6Y + X cos7Y◢
- Σxy sinY - Σy sin2Y - A sin3Y - B sin4Y - C sin5Y -
D sin6Y - X sin7Y◢ M + 1→M: M - 8 => Goto 0: Norm 1
Example: Perform a 8-sample FFT for the input sequence x[n]={ 1, 2, 0, 1, 2, 1, 0, 0}
Press Prog 1 1 EXE 2 EXE 0 EXE 1 EXE 2 EXE 1 EXE 0 EXE 0
EXE (read-out: 7) EXE (read-out: 0)
EXE (read-out: - 1) EXE (read-out:-1.4142)
EXE (read-out: 3) EXE (read-out: - 2)
EXE (read-out: - 1) EXE (read-out: -1.4142)
EXE (read-out: - 1) EXE (read-out: 0)
EXE (read-out: - 1) EXE (read-out: 1.4142)
EXE (read-out: 3) EXE (read-out: 2)
EXE (read-out: - 1) EXE (read-out: 1.4142)
Therefore, the result is:
X[0] = 7
X[1] = - 1 - 1.4142 j
X[2] = 3 - 2j
X[3] = - 1 - 1.4142j
X[4] = - 1
X[5] = - 1 + 1.4142j
X[6] = 3 + 2j
X[7] = - 1 + 1.4142j