线性调频信号的傅里叶变换
线性调频信号的表达式为
\[s_0(t) = \operatorname{rect}(\frac{t}{T_p})\exp(j2\pi f_c t + j\pi k t^2 + j\phi_0)
\]
其傅里叶变换为
\[\begin{align}
S_0(f) &= \int{s(t)\exp(-j2\pi ft)dt} \\
&= \int_{-\frac{T_p}{2}}^{\frac{T_p}{2}}{
\exp(j2\pi f_c t + j\pi k t^2 + j\phi_0 - j2\pi ft)dt
}
\\
&= \exp[-j\frac{\pi}{k}(f_c - f)^2 + j\phi_0]
\int_{\frac{f_c - f}{k}-\frac{T_p}{2}}^{\frac{f_c - f}{k}+\frac{T_p}{2}}{
\exp(j\pi k t^2)dt
}
\\
&= \frac{1}{\sqrt{k}}\exp[-j\frac{\pi}{k}(f_c - f)^2 + j\phi_0]
\int_{\frac{f_c - f}{\sqrt{k}}-\sqrt{k}\frac{T_p}{2}}
^{\frac{f_c - f}{\sqrt{k}}+\sqrt{k}\frac{T_p}{2}}{
\exp(j\pi x^2)dx
}
\\
&= \{ [
\frac{1}{\sqrt{k}}C(\frac{f_c - f}{\sqrt{k}}+\sqrt{k}\frac{T_p}{2})
- \frac{1}{\sqrt{k}}C(\frac{f_c - f}{\sqrt{k}}-\sqrt{k}\frac{T_p}{2})
] \\
& + j[ \frac{1}{\sqrt{k}}S(\frac{f_c - f}{\sqrt{k}}+\sqrt{k}\frac{T_p}{2})
-\frac{1}{\sqrt{k}}S(\frac{f_c - f}{\sqrt{k}}-\sqrt{k}\frac{T_p}{2})
]\} \\
& \exp[-j\frac{\pi}{k}(f_c - f)^2 + j\phi_0]
\end{align}
\]
其中， \(C(x), S(x)\) 为菲涅尓积分, 定义为
\[\begin{align}
C(x) &= \int_{0}^{x}{\cos(\pi t^2)dt} \\
S(x) &= \int_{0}^{x}{\sin(\pi t^2)dt}
\end{align}
\]
它们都是奇函数，其函数图像如下图所示。
于是, 信号的频谱幅度为
\[\begin{align}
|S_0(f)| = \sqrt{
\frac{1}{k}[C(\frac{1}{\sqrt{k}}(f - f_c + \frac{kT_p}{2})) - C(\frac{1}{\sqrt{k}}(f - f_c - \frac{kT_p}{2}))]^2 \\
+\frac{1}{k}[S(\frac{1}{\sqrt{k}}(f - f_c + \frac{kT_p}{2})) - S(\frac{1}{\sqrt{k}}(f - f_c - \frac{kT_p}{2}))]^2
}
\end{align}
\]
不难看出，当
\[-\frac{kT_p}{2}< f - f_c < \frac{kT_p}{2}
\]
时，该积分的模显著地不为零。
当 \(f=f_c\) 时，
\[\begin{align}
S_0(f_c)
&= \{ [
\frac{1}{\sqrt{k}}C(\sqrt{k}\frac{T_p}{2})
- \frac{1}{\sqrt{k}}C(-\sqrt{k}\frac{T_p}{2})
] \\
& + j[ \frac{1}{\sqrt{k}}S(\sqrt{k}\frac{T_p}{2})
-\frac{1}{\sqrt{k}}S(-\sqrt{k}\frac{T_p}{2})
]\}
\exp(j\phi_0)
\end{align}
\]
当 \(\sqrt{k}T_p = \sqrt{kT_p^2} = \sqrt{BT_p}\) 足够大即时宽带宽积足够大时，有
\[\begin{align}
C(\sqrt{k}\frac{T_p}{2}) &= - C(-\sqrt{k}\frac{T_p}{2})
&\approx \sqrt{\frac{\pi}{8}} \\
S(\sqrt{k}\frac{T_p}{2}) &= - S(-\sqrt{k}\frac{T_p}{2})
&\approx \sqrt{\frac{\pi}{8}}
\end{align}
\]
因此，
\[\begin{align}
S_0(f_c) \approx \frac{1}{\sqrt{k}}\sqrt{\frac{\pi}{2}}e^{j\frac{\pi}{4}} \exp(j\phi_0) \\
S_0(f) \approx \sqrt{\frac{\pi}{2k}}\operatorname{rect}(\frac{f - f_c}{kT_p})\exp(j\frac{\pi}{4} + j\phi_0)
\end{align}
\]
