extend([above[0] + below[i] for i in range(nelem)]) elif abs(d) < TOLERANCE: # plane which crosses the orig r1 = r0[zcos==d][0] maybe_c2x.append(r1) r2 = numpy.dot(rotation_mat(zaxis, numpy.pi*2/n), r1) if abs(r1+r2).sum() > TOLERANCE: maybe_c2x.append(r1+r2) ...
一、abs()函数——获取绝对值 功能:返回数字的绝对值 语法:abs(x) 例子: 循环输出列表中定义的数字的绝对值 tupleabs = [12.45,0,-19.69] for i in tupleabs: print(abs(i)) 输出结果: 12.45 0 19.69 2.定义函数,用于输出2个数字相乘后的绝对值 def aabs(value1,value2): return abs(value1*value2...
Y=20*log10(abs(H))abs(H))是取H的绝对值 然后这是一个计算式子，y等于20乘以log以10为底的abs(H).你这是什么语言？matlab？
一、abs()函数——获取绝对值功能:返回数字的绝对值 语法:abs(x) 例子:循环输出列表中定义的数字的绝对值tupleabs = [12.45,0,-19.69] for i in tupleabs: print(abs(i)) 输出结果: 12.45 0 19.692.定义函数,用于输出2个数字相乘后的绝对值def aabs(value1,value2): return abs(v ...
(1.0, pi ) ); v1.push_back( log10 ( vc6 ) ); cout << "The complex components log10 (vci), where abs (vci) = 1" << "\n& arg (vci) = i * pi / 6 of the vector v1 are:\n" ; for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ ) cout << *...
log10 * to calculate the natural logarithm and * the base-10 logarithm of 9,000. */#include<math.h>#include<stdio.h>intmain(void){doublex =9000.0;doubley; y =log( x );printf("log( %.2f ) = %f\n", x, y ); y =log10( x );printf("log10( %.2f ) = %f\n", x, y ...
请问这个Y=20*log10(abs(H))是什么意思啊?Rp=3;Rs=60;fs=2;f=[1/4,5/16];a=[1,0];>> dev=[(10^(Rp/20)-1)/(10^(Rp/20)+1),10^(-Rs/20)];>> [N,fo,ao,w]=firpmord(f,a,dev,fs);>> h=firpm(N,fo,ao,w);>> [H,w1]=freqz(h,1);Y=20*log10(abs(H));f1=w1...
- 选择一计算:gcd(abs(a - b), n) - 选择二计算:gcd(a + b > n ? (a + b) - n : a + b, n) 参数: - n: 待分解的大整数 - iterations: 最大迭代次数,默认值为 1000 返回: - (p, q): 如果成功,返回两个质因子 p 和 q - None: 如果失败,返回 None """ if FJP(n): return...
log10((abs(self.get_model_output(self.cfs[i]) - 0.000001))/(1 - abs(self.get_model_output(self.cfs[i]) - 0.000001))) criterion = torch.nn.ReLU() temp_loss = criterion(0.5 - (temp_logits*self.target_cf_class))[0] yloss += temp_loss return yloss/self.total_CFs ...
【请问这个Y=20*log10(abs(H))是什么意思啊?Rp=3;Rs=60;fs=2;f=[1/4,5/16];a=[1,0];>>dev=[(10^(Rp/20)-1)/(10^(Rp/20)+1),10^(-Rs/20)];>>[N,fo,ao,w]=firpmord(f,a,dev,fs);>>h=firpm(N,fo,ao,w);>>[H,w1]=freqz(h,1);Y=20*log10(abs(H));f1=w1/pi...