当满足下面解释的条件时，傅里叶变换相当于用虚数 s = iω 或 s = 2πiψ评估双边拉普拉斯变换，\...
其中利用欧拉公式 \( e^{i\phi} = \cos\phi + i\sin\phi \)。4. **提取实部**:实部为: (Re)(z) = e^(-θ) ⋅ cos(1/2ln29) 代入\( θ = \arctan\left(\frac{2}{5}\right) \),得最终结果: (Re)(z) = e^(-arctan(2/5)) ⋅ cos(1/2ln29) 反馈...
令z=r(\cos\phi+\mathbb i\sin\phi)，则\phi\in(0,\pi),r>0，且f(z)=\frac12\left[\lef...
【解析】 $$ z ^ { 4 } = \cos 4 \theta + i \sin 4 \theta $$. 记$$ \varphi = 4 \theta $$,得$$ ( \overline { z } ) ^ { 4 } = \overline { z ^ { 4 } } = \cos \phi - i \sin \phi $$ $$ \omega = \frac { 1 - \cos \phi + i \sin \phi } {...
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1.规定源平面变量: ξ=ρcosϕ,η=ρsinϕ. 2.规定参考平面变量: x=rcosφ,y=rsinφ. 3.菲涅耳衍射公式: Uo(x,y)=exp(jkz)jλz∬−∞∞Ui(ξ,η)exp{jk2z[(ξ−x)2+(η−y)2]}dξdη,Uo(r,φ)=exp(jkz)jλz∫02π∫0∞Ui(ρ,ϕ)exp{jk2z[ρ2+r2...
实心小圆柱体质心运动方程为f-mg\sin \theta =m(R-r)\dfrac{{{\text{d}}^{2}}\theta }{\text{d}{{t}^{2}}} (3)N-mg\cos \theta =m{{\left( \dfrac{\text{d}\theta }{\text{d}t} \right)}^{2}}(R-r) (4)实心小圆柱体绕其自身中心轴的转动方程为{{I}_{\text{CM}}}\...
令\lambda = re^{i\phi},展开后的相位响应为 \theta = -\omega -2\arctan\frac{r\sin(\omega - \phi)}{1-r\cos(\omega - \phi)} 对\omega求导得到 \frac{\mathrm{d}\theta(\omega)}{\mathrm{d}\omega} = -\frac{1-r^2}{|1+re^{-i(\omega -\phi)}|} < 0 ...
考虑关于θ的线性项\( 2a\cos\theta - 4a\sin\theta \)的最大值。该表达式可表示为\( R\cos(\theta + \phi) \),其最大值为: R = √((2a)^2 + (-4a)^2) = 2√5|a|. 因此原不等式转化为: 5a^2 + 2√5|a| ≤q 3. 设\( t = |a| \geq 0 \),则二次不等式为: ...
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