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软件无线电的一个目标就是,尽量使用一个通用化的硬件平台,通过加载不同的软件,以满足不同通信体制的需求。传统信号调制方式,如AM调制、bpsk、2fsk、2ask等等,都可以用正交变换的方式进行解调。
以AM调制为例进行说明。

正交变换的最基本结构就是接收信号与两正交本地信号分别相乘,得到IQ支路信号,然后两路信号进行分别处理…
那么这IQ两正交信号不一定非得是cos和sin。比如可以是cos和-sin,这里是要看你后面的处理,以及期待的信号结果的。
假设接收到的AM信号为 s ( t ) = A ( 1 + cos ⁡ ( w n t ) ) cos ⁡ ( w c t + φ ) <\rm>t) = A(1 + \cos (t))\cos (t + \varphi ) s ( t ) = A ( 1 + cos ( w n ​ t ) ) cos ( w c ​ t + φ ) .
与本地同相正交信号分别相乘:
I = A ( 1 + cos ⁡ ( w n t ) ) IQ Option 缺点 cos ⁡ ( w c t + φ ) sin ⁡ ( ( w c + Δ w ) t ) = a ( t ) cos ⁡ ( w c t + φ ) sin ⁡ ( ( w c + Δ w ) t ) = a ( t ) [ cos ⁡ w c t cos ⁡ IQ Option 缺点 φ − sin ⁡ w c t sin ⁡ φ ] [ sin ⁡ w c t cos ⁡ Δ w t + cos ⁡ w c t sin ⁡ Δ w t ] = a ( t ) ( c o s w c t cos IQ Option 缺点 IQ Option 缺点 ⁡ φ sin ⁡ w c t cos ⁡ Δ w t − sin ⁡ w c t sin ⁡ φ sin ⁡ w c t cos ⁡ Δ w t + cos ⁡ w c t cos ⁡ φ cos ⁡ w c t sin ⁡ Δ w t − sin ⁡ w c t sin ⁡ φ cos ⁡ w c t sin ⁡ Δ w t ) \begin I = A(1 + \cos (t))\cos (t + \varphi )\sin (( + \Delta w)t)\\ = a(t)\cos (t + \varphi )\sin (( + \Delta w)t)\\ = a(t)\left[ <\cos t\cos \varphi - \sin t\sin \varphi > \right]\left[ <\sin t\cos \Delta wt + \cos t\sin \Delta wt> \right]\\ = a(t)(>ost\cos \varphi \sin t\cos \Delta wt - \sin t\sin \varphi \sin t\cos \Delta wt + \\ \cos t\cos \varphi \cos t\sin \Delta wt - \sin t\sin \varphi \cos t\sin \Delta wt) \end I = A ( 1 + cos ( w n ​ t ) ) cos ( w IQ Option 缺点 IQ Option 缺点 IQ Option 缺点 c ​ t + φ ) sin ( ( w c ​ + Δ w ) t ) = a ( t ) cos ( w c ​ t + φ ) sin ( ( w c ​ + Δ w ) t ) = a ( t ) [ cos w c ​ t cos φ − sin w IQ Option 缺点 c ​ t sin φ ] [ sin w c ​ t cos Δ w t + cos w c ​ t sin Δ w t ] = a ( t ) ( c o s w c ​ t cos φ sin w c ​ t IQ Option 缺点 IQ Option 缺点 cos Δ w t − sin w c ​ t sin φ sin w c ​ t cos Δ w t + IQ Option 缺点 cos w c ​ t cos φ cos w c ​ t sin Δ w t − sin w c ​ t sin φ cos w c ​ t sin Δ w t ) ​
上式最后一步,经低通滤波之后第一项、第四项直接没了,
剩下:
I = a ( t ) ( − sin ⁡ w IQ Option 缺点 c t sin ⁡ φ sin ⁡ w c t cos ⁡ Δ w t + cos ⁡ w c t cos ⁡ IQ Option 缺点 φ cos ⁡ w c t sin ⁡ Δ w t ) ≫ − 0.5 sin ⁡ φ cos ⁡ Δ w IQ Option 缺点 t + 0.5 cos ⁡ φ sin ⁡ Δ w t = 0.5 a ( t ) sin ⁡ ( Δ w t − φ ) \begin I = a(t)( - \sin t\sin \varphi \sin t\cos \Delta wt + \cos t\cos \varphi \cos t\sin \Delta IQ Option 缺点 wt)\\ \gg - 0.5\sin \varphi \cos \Delta wt + 0.5\cos \varphi \sin \Delta wt\\ = 0.5a(t)\sin (\Delta wt - \varphi ) \end I = a ( t ) ( − sin w c ​ t sin φ sin w c ​ t IQ Option 缺点 IQ Option 缺点 cos Δ w t + cos w c ​ t cos φ cos w c ​ t sin Δ w t ) ≫ − 0 . 5 sin φ cos Δ w t + 0 . 5 cos φ sin Δ w t = 0 . 5 a ( t ) IQ Option 缺点 sin ( Δ w t − φ ) ​
同理
Q = 0.5 a ( t ) cos ⁡ ( Δ w t − φ ) Q IQ Option 缺点 = 0.5a(t)\cos (\Delta wt - \varphi ) Q = 0 . 5 a ( t ) cos ( Δ w t − φ )

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AM信号的数字正交解调

者乎之类的 于 2020-06-06 08:34:45 发布 2674 收藏 31

软件无线电的一个目标就是,尽量使用一个通用化的硬件平台,通过加载不同的软件,以满足不同通信体制的需求。传统信号调制方式,如AM调制、bpsk、2fsk、2ask等等,都可以用正交变换的方式进行解调。
以AM调制为例进行说明。

正交变换的最基本结构就是接收信号与两正交本地信号分别相乘,得到IQ支路信号,然后两路信号进行分别处理…
那么这IQ两正交信号不一定非得是cos和sin。比如可以是cos和-sin,这里是要看你后面的处理,以及期待的信号结果的。
假设接收到的AM信号为 s ( t ) = A ( 1 + cos ⁡ ( w n t ) ) cos ⁡ ( w c t + φ ) <\rm>t) = A(1 + \cos (t))\cos (t + \varphi ) s ( t ) = A ( 1 + cos ( w n ​ t ) ) cos ( w c ​ t + φ ) .
与本地同相正交信号分别相乘:
I = A ( 1 + cos ⁡ ( w n t ) ) cos ⁡ ( w c t + φ ) sin ⁡ ( ( w c + Δ w ) t ) = a ( t ) cos ⁡ ( w c t + φ ) sin ⁡ ( ( w c + Δ w ) t ) = a ( t ) [ cos ⁡ w c t cos ⁡ IQ Option 缺点 φ − sin ⁡ w c t sin ⁡ φ ] [ sin ⁡ w c t cos ⁡ Δ w t IQ Option 缺点 + cos ⁡ w c t sin ⁡ Δ w t ] = a ( t ) ( c o s w c t cos IQ Option 缺点 ⁡ φ sin ⁡ w c t cos ⁡ Δ w t − sin ⁡ w c t sin ⁡ φ sin ⁡ IQ Option 缺点 w c t cos ⁡ Δ w t + cos ⁡ w c t cos ⁡ φ cos ⁡ w c t IQ Option 缺点 IQ Option 缺点 sin ⁡ Δ w t − sin ⁡ w c t sin ⁡ φ cos ⁡ w c t sin ⁡ Δ IQ Option 缺点 w t ) \begin I = A(1 + \cos (t))\cos (t + \varphi )\sin (( + \Delta w)t)\\ = a(t)\cos (t + \varphi )\sin (( + \Delta w)t)\\ = a(t)\left[ <\cos t\cos \varphi - \sin t\sin \varphi > \right]\left[ <\sin t\cos \Delta wt + \cos t\sin \Delta wt> \right]\\ = a(t)(>ost\cos \varphi \sin t\cos \Delta wt - \sin t\sin \varphi \sin t\cos \Delta wt + \\ \cos t\cos \varphi \cos t\sin \Delta wt - \sin t\sin \varphi \cos t\sin \Delta wt) \end I = A ( 1 + cos ( w n ​ t ) ) cos ( w c ​ t + φ ) sin ( ( w c ​ + Δ w ) t ) = a ( t ) cos ( w c ​ t + φ ) sin ( ( w c ​ + Δ w ) t ) = a ( t ) [ cos w c ​ t cos φ − sin w IQ Option 缺点 c ​ t sin φ ] [ sin w c ​ t cos Δ w t + cos w c ​ t IQ Option 缺点 sin Δ w t ] = a ( t ) ( c o s w c ​ t cos φ sin w c ​ t cos Δ w t − sin w c ​ t sin φ sin w c ​ t cos Δ w t + IQ Option 缺点 IQ Option 缺点 cos w c ​ t cos φ cos w c ​ t sin Δ w t − sin w c ​ t sin φ cos w c ​ t sin Δ w t ) ​
上式最后一步,经低通滤波之后第一项、第四项直接没了,
剩下:
I = a ( t ) ( − sin ⁡ w c t sin ⁡ φ sin ⁡ w c t cos ⁡ Δ w t + cos ⁡ w c t cos ⁡ φ cos ⁡ w c t sin ⁡ Δ w t ) ≫ − 0.5 sin ⁡ φ cos ⁡ Δ w IQ Option 缺点 t + 0.5 cos ⁡ φ sin ⁡ Δ w t = 0.5 a ( t ) sin ⁡ ( Δ w t − φ ) \begin I = a(t)( - \sin t\sin \varphi \sin t\cos \Delta wt + \cos t\cos \varphi \cos t\sin \Delta wt)\\ \gg - 0.5\sin \varphi \cos \Delta wt + 0.5\cos \varphi \sin \Delta wt\\ = 0.5a(t)\sin (IQ Option 缺点 \Delta wt - \varphi ) \end I = a ( t ) ( − sin w c ​ t sin φ sin w c ​ t IQ Option 缺点 cos Δ w t + cos w c ​ t cos φ cos w c ​ t sin Δ w t ) ≫ IQ Option 缺点 IQ Option 缺点 − 0 . 5 sin φ cos Δ w t + 0 . 5 cos φ sin Δ w t = 0 . 5 a ( t ) sin ( Δ w t − φ ) ​
同理
Q = 0.5 a ( t ) cos ⁡ ( Δ w t − φ ) Q = 0.5a(t)\cos (\Delta wt - \varphi ) Q = 0 . 5 a ( t ) cos ( Δ w t − φ )

取同相正交分量平方和再开根号便可以得到a(IQ Option 缺点 t)信号,即 A ( 1 + cos ⁡ ( w n t ) ) A(1 + \cos (t)) A ( 1 + cos ( w n ​ IQ Option 缺点 IQ Option 缺点 t ) ) ,然后去掉直流分量便得到原始信号了。到此解调完成。所以可以看到,本地sin信号前面不取负号也是可以的(这是目前的理解,可能还有实际问题没考虑到)。
ps:不用加负号也可,见另一篇博客实现了的AM调制解调
所以AM正交解调不要求严格同频同相。
传统通信专业其实真的挺难的,学了好几年,如果让我不照抄其他人设计,感觉最简单的一个通信收发模型也实现不出来,可能都设计不出来。例如上例中,一个AM解调,其实就已经很复杂了,这还没考虑采样、滤波、射频处理模块,还有如果有噪声干扰怎么办,人家提出的种种指标问题如何满足等等,但这已经是通信调制解调中最简单的方式了。就每每学一点东西,书本上可能一句话就讲完了,啥也不懂的时候就感觉懂了,然后如果一细想如何实现,就发现内容太多了,太复杂了。感觉现在追求创新的导向,对于绝大多数人来说,是有问题的,可能会引导大家不去静下心来研究这些最基本的东西。前几天听了个博士答辩,有位老教授提出的问题都是原理性、最基本的问题,但是答辩人显然对这些问题没有过思考,只是人家这样做我也就这样做了,只去追求后面算法的创新先进性。这样子,虽然可能也能解决实际问题,但是总归是不牢靠的。所以还是很佩服那些老教授的,整个系统的细枝末节原理性问题都能非常清楚,本着一个严谨踏实的态度对待科研。就我个人而言,已经在有意追求这方面了,但是因为能力不足,环境导向问题,以及由于环境导向导致的我想关注的问题网上解释资源很少的原因,还是很难搞懂这些问题。就只能边学边回顾边反思边总结了。见得多了,很多问题也就明白了。

cosAsin(A+B)=0.5sinB,其中A是高频wt,B是低频 φ \varphi φ (接收相位与本地相位的差值)
cosAcos(A+B)=0.5cosB.
若设接收信号为 cos ⁡ ( Δ w IQ Option 缺点 IQ Option 缺点 t + φ ) \cos (\Delta wt +\varphi ) cos ( Δ w t + φ )
cos(A+B)sinA=-0.5sinB
cos(A+B)cosA=0.5cosB
上图就是这种模型,所以本地信号前就加了个负号,使后面符合人们习惯。 IQ Option 缺点 IQ Option 缺点
补充:传统模拟的AM解调方案使非相干二极管检波。

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