線性調頻信號的表達式為
\[s_0(t) = \operatorname{rect}(\frac{t}{T_p})\exp(j2\pi f_c t + j\pi k t^2 + j\phi_0) \]
其傅里葉變換為
\[\begin{align} S_0(f) &= \int{s(t)\exp(-j2\pi ft)dt} \\ &= \int_{-\frac{T_p}{2}}^{\frac{T_p}{2}}{ \exp(j2\pi f_c t + j\pi k t^2 + j\phi_0 - j2\pi ft)dt } \\ &= \exp[-j\frac{\pi}{k}(f_c - f)^2 + j\phi_0] \int_{\frac{f_c - f}{k}-\frac{T_p}{2}}^{\frac{f_c - f}{k}+\frac{T_p}{2}}{ \exp(j\pi k t^2)dt } \\ &= \frac{1}{\sqrt{k}}\exp[-j\frac{\pi}{k}(f_c - f)^2 + j\phi_0] \int_{\frac{f_c - f}{\sqrt{k}}-\sqrt{k}\frac{T_p}{2}} ^{\frac{f_c - f}{\sqrt{k}}+\sqrt{k}\frac{T_p}{2}}{ \exp(j\pi x^2)dx } \\ &= \{ [ \frac{1}{\sqrt{k}}C(\frac{f_c - f}{\sqrt{k}}+\sqrt{k}\frac{T_p}{2}) - \frac{1}{\sqrt{k}}C(\frac{f_c - f}{\sqrt{k}}-\sqrt{k}\frac{T_p}{2}) ] \\ & + j[ \frac{1}{\sqrt{k}}S(\frac{f_c - f}{\sqrt{k}}+\sqrt{k}\frac{T_p}{2}) -\frac{1}{\sqrt{k}}S(\frac{f_c - f}{\sqrt{k}}-\sqrt{k}\frac{T_p}{2}) ]\} \\ & \exp[-j\frac{\pi}{k}(f_c - f)^2 + j\phi_0] \end{align} \]
其中， \(C(x), S(x)\) 為菲涅尓積分, 定義為
\[\begin{align} C(x) &= \int_{0}^{x}{\cos(\pi t^2)dt} \\ S(x) &= \int_{0}^{x}{\sin(\pi t^2)dt} \end{align} \]
它們都是奇函數，其函數圖像如下圖所示。
於是, 信號的頻譜幅度為
\[\begin{align} |S_0(f)| = \sqrt{ \frac{1}{k}[C(\frac{1}{\sqrt{k}}(f - f_c + \frac{kT_p}{2})) - C(\frac{1}{\sqrt{k}}(f - f_c - \frac{kT_p}{2}))]^2 \\ +\frac{1}{k}[S(\frac{1}{\sqrt{k}}(f - f_c + \frac{kT_p}{2})) - S(\frac{1}{\sqrt{k}}(f - f_c - \frac{kT_p}{2}))]^2 } \end{align} \]
不難看出，當
\[-\frac{kT_p}{2}< f - f_c < \frac{kT_p}{2} \]
時，該積分的模顯著地不為零。
當 \(f=f_c\) 時，
\[\begin{align} S_0(f_c) &= \{ [ \frac{1}{\sqrt{k}}C(\sqrt{k}\frac{T_p}{2}) - \frac{1}{\sqrt{k}}C(-\sqrt{k}\frac{T_p}{2}) ] \\ & + j[ \frac{1}{\sqrt{k}}S(\sqrt{k}\frac{T_p}{2}) -\frac{1}{\sqrt{k}}S(-\sqrt{k}\frac{T_p}{2}) ]\} \exp(j\phi_0) \end{align} \]
當 \(\sqrt{k}T_p = \sqrt{kT_p^2} = \sqrt{BT_p}\) 足夠大即時寬帶寬積足夠大時，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，
\[\begin{align} S_0(f_c) \approx \frac{1}{\sqrt{k}}\sqrt{\frac{\pi}{2}}e^{j\frac{\pi}{4}} \exp(j\phi_0) \\ S_0(f) \approx \sqrt{\frac{\pi}{2k}}\operatorname{rect}(\frac{f - f_c}{kT_p})\exp(j\frac{\pi}{4} + j\phi_0) \end{align} \]