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Update 3.4 format
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AXYZdong committed May 7, 2024
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Expand Up @@ -22,7 +22,10 @@ $$h(t)= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega) e^{j \omega t}d\omega$

<font color=red>例:某LTI系统的 $H(j\omega)$ 和 $\theta(\omega)$ 如图,若 $f(t)=2+4\cos(5t)+4\cos(10t)$ ,求系统的响应。</font>

![在这里插入图片描述](https://img-blog.csdnimg.cn/20200403175356286.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQzMzI4MzEz,size_16,color_FFFFFF,t_70#pic_center =300x )<center><sup><sup>$H(j\omega)$ 和 $\theta(\omega)$ 图像</sup></sup>
<br>
![在这里插入图片描述](https://img-blog.csdnimg.cn/20200403175356286.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQzMzI4MzEz,size_16,color_FFFFFF,t_70#pic_center =300x )
</p><center><sup><code>$H(j\omega)$ 和 $\theta(\omega)$ 图像</code></sup></center><p></p>
<br>

解:
$f(t) 的基波角频率 \Omega = 5rad/s\\
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