質問編集履歴
3
コード記載
test
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test
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@@ -158,12 +158,12 @@
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SNR = 0
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Ps = 1.0
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sigma = math.sqrt( Ps / (2 * math.pow(10.0, SNR/10)))
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Ps = 1.0
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-
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Pn = 2*sigma*sigma
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r0 = Ps/Pn
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2
コード記載
test
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test
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@@ -2,9 +2,9 @@
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2
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3
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4
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-
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s = [s1, s2, s3]
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-
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6
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-
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t = [t1, t2, t3]
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@@ -14,12 +14,16 @@
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(条件)
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・ pk = max{
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・ pk = max{ tk(sk - h), 0 } k = 1,2,3
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・ h は p1 + p2 + p3 = 1 を満たす為の定数
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この条件下で出来るだけ p を 0 にしないようなコードを書きたいと考えております。
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(0 を許可するコードなら既に書けております)
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どういったコードが考えられますでしょうか。
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@@ -32,7 +36,17 @@
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-
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返信が遅れまして大変申し訳ございません。
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またご回答くださり誠に有難うございます。
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以下、コードです。
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```python
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def calculate_send_power(bits, A, r0, a, b):
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@@ -91,3 +105,127 @@
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return p
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if __name__ == '__main__':
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for bits in bit_pattern:
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num_stream = len(bits)
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p = calculate_send_power(bits, A, r0, a, b)
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if len(p) != num_stream:
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for i in range(num_stream - len(p)):
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p.append(0)
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```
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かなり分かりにくいかと思いますが、 calculate_send_power() における over が s を表し、under が t を表し、 xi が h を表しているとお考えください。
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以下にそれぞれの変数を述べておきます。
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```ここに言語を入力
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a = [0,0,1/2,0,3/8,0,7/24,0,15/64]
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b = [0,0,2,0,10,0,42,0,170]
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bit_pattern = [[8], [6, 2], [4, 4], [4, 2, 2], [2, 2, 2, 2]]
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H = (4,4)の複素行列 実部、虚部ともに正規分布に従う乱数
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gram_H = np.conjugate(H.T) @ H
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U, A, Uh = svd(gram_H)
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SNR = 0
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sigma = math.sqrt( Ps / (2 * math.pow(10.0, SNR/10)))
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Ps = 1.0
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Pn = 2*sigma*sigma
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r0 = Ps/Pn
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```
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このコードは bit_pattern = [[8], [6, 2], [4, 4], [4, 2, 2], [2, 2, 2, 2]] のそれぞれに割り当てる電力(電力総和 = 1)を決定するもので以下のようになります。
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|bits|p(ower)|sum|
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|:--|:--:|--:|
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|[8]|[1.000]|1.0|
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|[6, 2]|[0.453, 0.546]|1.0|
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以下に実行結果を述べます。
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```ここに言語を入力
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for bits in bit_pattern:
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num_stream = len(bits)
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tmp = 0
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p = calculate_send_power(bits, A, r0, a, b)
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if len(p) != num_stream:
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for i in range(num_stream - len(p)):
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p.append(0)
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print("bits = {} , power = {}, sum = {}".format(bits, p, sum(p)))
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>>>
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bits = [8] , power = [1.00], sum = 1.0
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bits = [6, 2] , power = [0.39036578941718647, 0.6096342105828132], sum = 1.0
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bits = [4, 4] , power = [0.6905457297074978, 0.30945427029250216], sum = 1.0
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bits = [4, 2, 2] , power = [0.4747411414947003, 0.35540013052184316, 0.1698587279834565], sum = 1.0
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bits = [2, 2, 2, 2] , power = [0.3157712110211322, 0.40062777763337526, 0.2836010113454926, 0], sum = 1.0
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```
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と出来ているのですが、 bits = [2, 2, 2, 2] の4項目に割り当てる電力が 0 になってしまいます。この条件下で出来るだけ p を 0 にしないようなコードを書きたいと考えており、考え方でもご教授頂けると幸いです。よろしくお願い申し上げます。
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1
a
test
CHANGED
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File without changes
|
test
CHANGED
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@@ -27,3 +27,67 @@
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27
27
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考え方でもご教授頂けると幸いです。
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28
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29
29
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よろしくお願い申し上げます。
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def calculate_send_power(bits, A, r0, a, b):
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num_stream = len(bits)
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over = 0
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under = 0
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for k in range(num_stream):
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rk = A[k] * r0
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ak = a[bits[k]]
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bk = b[bits[k]]
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mk = bits[k]
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over += bk/rk * math.log(ak * mk * rk / bk)
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under += bk/rk
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xi = (over - 1) / under
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p = []
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for k in range(num_stream):
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rk = A[k] * r0
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ak = a[bits[k]]
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bk = b[bits[k]]
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mk = bits[k]
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if bk/rk * (math.log(ak * mk * rk / bk) - xi) > 0:
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p.append(bk/rk * (math.log(ak * mk * rk / bk) - xi))
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if len(p) != num_stream:
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p = calculate_send_power(bits[:-1], A, r0, a, b)
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91
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return p
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