给出IQ两路求幅角，请教！
时间：10-02
整理：3721RD
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偶想用xinlinx ise 中的 cordic 实现求幅角 ，输入为I，Q两路，但是看不懂它的输入数据的格式，这是怎么转换的呢?我的数据是2的补码形式的，怎么样才能使用这个ipcore?另外不用这个方法的话，还有什么招可以求幅角的?请指点
The input vector (Xin, Yin) is expressed as a pair of signed 1QN numbers. The output angle, Pout radians, isexpressed as a signed 2QN number. In this example, the input/output width is set to 10 bits.
Xin : .0010100000. => 00.10100000 => 0.625
Yin : .0010000000. => 00.10000000 => 0.500
Pout : .0001010110. => 000.1010110=> 0.672
1QN Format Data
SB D8 D7 D6 D5 D4 D3 D2 D1
+1 0 1 0 0 0 0 0 0 0
-1 1 1 0 0 0 0 0 0 0
+Pi/4 0 0 1 1 0 0 1 0 0
-Pi/4 1 1 0 0 1 1 0 1 1
^ <---Binary Point
2QN Format Phase
SB D8 D7 D6 D5 D4 D3 D2 D1
+1 0 0 1 0 0 0 0 0 0
-1 1 1 1 0 0 0 0 0 0
+Pi 0 1 1 0 0 1 0 0 1
-Pi 1 0 0 1 1 0 1 1 1
^ <---Binary Point
In 2Q10 format values, +Pi and -Pi, are represented:
"01100100100" => 011.00100100 => +3.14
"10011011100" => 100.11011100 => - 3.14
The input vector (Xin, Yin) is expressed as a pair of signed 1QN numbers. The output angle, Pout radians, isexpressed as a signed 2QN number. In this example, the input/output width is set to 10 bits.
Xin : .0010100000. => 00.10100000 => 0.625
Yin : .0010000000. => 00.10000000 => 0.500
Pout : .0001010110. => 000.1010110=> 0.672
1QN Format Data
SB D8 D7 D6 D5 D4 D3 D2 D1
+1 0 1 0 0 0 0 0 0 0
-1 1 1 0 0 0 0 0 0 0
+Pi/4 0 0 1 1 0 0 1 0 0
-Pi/4 1 1 0 0 1 1 0 1 1
^ <---Binary Point
2QN Format Phase
SB D8 D7 D6 D5 D4 D3 D2 D1
+1 0 0 1 0 0 0 0 0 0
-1 1 1 1 0 0 0 0 0 0
+Pi 0 1 1 0 0 1 0 0 1
-Pi 1 0 0 1 1 0 1 1 1
^ <---Binary Point
In 2Q10 format values, +Pi and -Pi, are represented:
"01100100100" => 011.00100100 => +3.14
"10011011100" => 100.11011100 => - 3.14
[求助]给出IQ两路求幅角，请教！
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