cos(wt) 的欧拉变换是一个非常有趣且实用的数学变换。具体来说,cos(wt) 的欧拉变换可以表示为: [ \cos(wt) = \frac{e^{jwt} + e^{-jwt}}{2} ] 这里,我们使用了欧拉公式(Euler's Formula): [ e^{j\theta} = \cos(\theta) + j\sin(\theta) ] 其中,(e) 是自然对数的底数,(j) 是虚数...
它就不会变了。相当于根据theta随机抽了一个三角函数。只要知道几个时刻的样本值，就可以确定其它样本的...
c1coswt+c2coswt是否一定可以表示成Asin(wt+ψ)?试证明 答案 一定的当C1,C2都不是0时abs(C1)看成直角三角形一边,abs(C2)另外一边,夹角theta sintheta=C1/sqrt(C1*C1+C2*C2) costheta=C2/sqrt(C1*C1+C2*C2) A*sinthetacoswt+Acosthetasinwt =Asin(theta+wt) 相关...
一定的当C1,C2都不是0时abs(C1)看成直角三角形一边，abs(C2)另外一边，夹角theta sintheta=C1/sqrt(C1*C1+C2*C2)costheta=C2/sqrt(C1*C1+C2*C2)A*sinthetacoswt+Acosthetasinwt =Asin(theta+wt)
e^{-jwt}=cos(wt)-jsin(wt) 其实部对应cos虚部对应sin,与傅里叶级数的余弦展开(参考式2.14)+变-号,那么求其相位应该是: \varphi = \arctan\left( \frac{b_{n}}{a_{n}} \right) 同时,依据式3.5,其幅度是 c_{n}\ \ = \frac{2}{T}\sqrt{{a_{n}}^{2} + {b_{n}}^{2}} 离...
1=(sinwt)^2+(coswt)^2=x^2/4b^2+y^2/b^2 这是典型的半长轴为2b半短轴为b的椭圆 按照定义 v(x)=dx/dt=-2w*b*sin(wt)v(y)=dy/dt=w*bcos(wt)v=sqrt[v(x)^2+v(y)^2]=w*b*sqrt[3sin^2(wt)+1]v与x轴成夹角theta=arctan(v(y)/v(x))=arctan[0.5cot(wt)]...
ey=dy+64.05531*sin(120.*pi/180+a+r1+r2-pi); plot(ex,ey); 之后我想再把函数中的数据有序导出成表格可是不太会希望知道的同学帮帮忙 分享回复赞 广州89中吧 overthereis 球谐函数图像 l=0, m=0 Y=\frac{1}{\sqrt{4pi}} l=1, m=0 Y=\sqrt{\frac{3}{4pi}}\cos{\theta} l=1, m=±...
A wave has the form y(x, t) = A sin (kx- wt- \theta), where A = 10.0 cm, k = 0.0141 cm^{-1} w = 0.564 s^{-1} and \theta = 1.00 rad. (a) What is the wavelength of this wave? (b) What is the period of this wave? (c) What is ...
1=(sinwt)^2+(coswt)^2=x^2/4b^2+y^2/b^2 这是典型的半长轴为2b半短轴为b的椭圆 按照定义 v(x)=dx/dt=-2w*b*sin(wt)v(y)=dy/dt=w*bcos(wt)v=sqrt[v(x)^2+v(y)^2]=w*b*sqrt[3sin^2(wt)+1]v与x轴成夹角theta=arctan(v(y)/v(x))=arctan[0.5cot(wt)]...
Solve the differential equation \frac{dr}{d\theta} + r\sec \theta = \cos \theta Solve the differential equation: y"" + 2y" + y = cosx Determine the form of a particular solution to the differential equation. y'' + 5y' + 6y = \sin x - \cos 2x Solve the differential equation....