通信分析中常见的积分计算

1.符号间隔内的指数积分:

\[\begin{aligned} \langle s_{ml}(t),s_{nl}(t)\rangle&=\frac{2\epsilon}{T}\int_{0}^{T}e^{j2\pi(m-n)\Delta ft}dt\\ &=\frac{2\epsilon}{T}\frac{1}{j2\pi(m-n)\Delta f}[e^{j2\pi T(m-n)\Delta f}-1]\\ &=\frac{2\epsilon}{T}\frac{1}{j2\pi(m-n)\Delta f}[\cos(2\pi T(m-n)\Delta f)+j\sin(2\pi T(m-n)\Delta f)-1]\\ &=\frac{2\epsilon}{j2\pi T(m-n)\Delta f}[1-2\sin^{2}(\pi T(m-n)\Delta f)+j\sin(2\pi T(m-n)\Delta f)-1]\\ &=\frac{2\epsilon}{j2\pi T(m-n)\Delta f}[-2\sin^{2}(\pi T(m-n)\Delta f)+j2\sin(\pi T(m-n)\Delta f)\cos(\pi T(m-n)\Delta f)]\\ &=\frac{2\epsilon\sin(\pi T(m-n)\Delta f)}{j\pi T(m-n)\Delta f}[-\sin(\pi T(m-n)\Delta f)+j\cos(\pi T(m-n)\Delta f)]\\ &=\frac{2\epsilon\sin(\pi T(m-n)\Delta f)}{-1\pi T(m-n)\Delta f}[-j\sin(\pi T(m-n)\Delta f)-\cos(\pi T(m-n)\Delta f)]\\ &=\frac{2\epsilon\sin(\pi T(m-n)\Delta f)}{\pi T(m-n)\Delta f}[j\sin(\pi T(m-n)\Delta f)+\cos(\pi T(m-n)\Delta f)]\\ &=\frac{2\epsilon\sin(\pi T(m-n)\Delta f)}{\pi T(m-n)\Delta f}e^{j\pi T(m-n)\Delta f}\\ &=2\epsilon\ \text{sinc}(T(m-n)\Delta f)e^{j\pi T(m-n)\Delta f} \end{aligned}\]

另外

\[\begin{aligned} \langle s_{m}(t), s_{n}(t) \rangle &= \frac{1}{2}\bold{\text{Re}}[\langle s_{ml}(t),s_{nl}(t)\rangle]\\ &=\frac{\epsilon\sin(\pi T(m-n)\Delta f)}{\pi T(m-n)\Delta f}\cos(\pi T(m-n)\Delta f)\\ &=\epsilon\frac{\sin(2\pi T(m-n)\Delta f)}{2\pi T(m-n)\Delta f}\\ &=\epsilon\ \text{sinc}(2T(m-n)\Delta f) \end{aligned}\]

其中用到的公式有:

\[\cos2x=\cos^{2}x-\sin^{2}x=\cos^{2}x+\sin^{2}x-2\sin^{2}x=1-2sin^{2}x \]

\[\sin2x=2\sin x\cos x \]

\[e^{jx}=\cos x+j\sin x \]

\[\text{sinc}(x)=\frac{\sin(\pi x)}{\pi x} \]

posted @ 2023-06-25 11:30  Vinson88  阅读(94)  评论(0编辑  收藏  举报