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零中频采样及信号重构

分类: 文章 • 2025-04-24 20:14:28

1 零中频采样

零中频采样过程如图所示：

零中频采样及信号重构
假设信号为：

s=Asin(ω1t)+Bsin(ω2t)s=A\mathrm{sin}(\omega_1t)+B\mathrm{sin}(\omega_2t)

为了计算简便，所有的频率中省略2π2\pi以及相位。

对这个信号进行零中频正交采样，得到：

I=s⋅sin⁡(ω1+ω22t)=A2[cos⁡ω1−ω22t−cos⁡3ω1+ω22t]+B2[cos⁡ω1−ω22t−cos⁡ω1+3ω22t]\begin{aligned} I &=s \cdot \sin \left(\frac{\omega_{1}+\omega_{2}}{2} t\right) \\ &=\frac{A}{2}\left[\cos \frac{\omega_{1}-\omega_{2}}{2} t-\cos \frac{3 \omega_{1}+\omega_{2}}{2} t\right]+\frac{B}{2}\left[\cos \frac{\omega_{1}-\omega_{2}}{2} t-\cos \frac{\omega_{1}+3 \omega_{2}}{2} t\right] \end{aligned}

Q=s⋅cos⁡(ω1+ω22t)=A2[sin⁡3ω1+ω22t+sin⁡ω1−ω22t]+B2[sin⁡ω1+3ω22t−sin⁡ω1−ω22t]\begin{aligned} Q &=s \cdot \cos \left(\frac{\omega_{1}+\omega_{2}}{2} t\right) \\ &=\frac{A}{2}\left[\sin \frac{3 \omega_{1}+\omega_{2}}{2} t+\sin \frac{\omega_{1}-\omega_{2}}{2} t\right]+\frac{B}{2}\left[\sin \frac{\omega_{1}+3 \omega_{2}}{2} t-\sin \frac{\omega_{1}-\omega_{2}}{2} t\right] \end{aligned}

过低通滤波器之后，变为：

I=A2cos⁡ω1−ω22t+B2cos⁡ω1−ω22t=A+B2cos⁡ω1−ω22t\begin{aligned} I &=\frac{A}{2} \cos \frac{\omega_{1}-\omega_{2}}{2} t+\frac{B}{2} \cos \frac{\omega_{1}-\omega_{2}}{2} t \\ &=\frac{A+B}{2} \cos \frac{\omega_{1}-\omega_{2}}{2} t \end{aligned}

Q=A2sin⁡ω1−ω22t−B2sin⁡ω1−ω22t=A−B2sin⁡ω1−ω22t\begin{aligned} Q &=\frac{A}{2} \sin \frac{\omega_{1}-\omega_{2}}{2} t-\frac{B}{2} \sin \frac{\omega_{1}-\omega_{2}}{2} t \\ &=\frac{A-B}{2} \sin \frac{\omega_{1}-\omega_{2}}{2} t \end{aligned}

这样就将信号降至零中频，也就是基带。这时，信号带宽为射频带宽的一半，即B2\frac{B}{2}。因此，可以用fs=Bf_s=B来对信号进行采样，继而进行后续处理。

2 信号重构

零中频采样之后信号的重构过程如图所示：

零中频采样及信号重构
以上面的零中频采样之后的信号为例。分别乘以同相和正交的频率之后，得到：

Ir=I⋅sin⁡(ω1+ω22t)=A+B2cos⁡(ω1−ω22t)sin⁡(ω1+ω22t)=A+B4[sin⁡(ω1−ω2+ω1+ω22t)−sin⁡(ω1−ω2−ω1−ω22t)]=A+B4(sin⁡ω1t+sin⁡ω2t)\begin{aligned} I_r &=I\cdot\sin \left(\frac{\omega_{1}+\omega_{2}}{2} t\right) \\&=\frac{A+B}{2} \cos \left(\frac{\omega_{1}-\omega_{2}}{2} t\right) \sin \left(\frac{\omega_{1}+\omega_{2}}{2} t\right) \\ &=\frac{A+B}{4}\left[\sin \left(\frac{\omega_{1}-\omega_{2}+\omega_{1}+\omega_{2}}{2} t\right)-\sin \left(\frac{\omega_{1}-\omega_{2}-\omega_{1}-\omega_{2}}{2} t\right)\right] \\ &=\frac{A+B}{4}\left(\sin \omega_{1} t+\sin \omega_{2} t\right) \end{aligned}

Qr=Q⋅cos⁡(ω1+ω22t)=A−B2sin⁡(ω1−ω22t)cos⁡(ω1+ω22t)=A−B4[sin⁡(ω1−ω2+ω1+ω22t)+sin⁡(ω1−ω2−ω1−ω22t)]=A−B4(sin⁡ω1t−sin⁡ω2t)\begin{aligned} Q_r &=Q \cdot \cos \left(\frac{\omega_{1}+\omega_{2}}{2} t\right) \\&=\frac{A-B}{2} \sin \left(\frac{\omega_{1}-\omega_{2}}{2} t\right) \cos \left(\frac{\omega_{1}+\omega_{2}}{2} t\right) \\ &=\frac{A-B}{4}\left[\sin \left(\frac{\omega_{1}-\omega_{2}+\omega_{1}+\omega_{2}}{2} t\right)+\sin \left(\frac{\omega_{1}-\omega_{2}-\omega_{1}-\omega_{2}}{2} t\right)\right] \\ &=\frac{A-B}{4}\left(\sin \omega_{1} t-\sin \omega_{2} t\right) \end{aligned}

最后两路信号相加：

s=Ir+Qr=A+B4(sin⁡ω1t+sin⁡ω2t)+A−B4(sin⁡ω1t−sin⁡ω2t)=A2sin⁡ω1t+B2sin⁡ω2t\begin{aligned} s = I_{r}+Q_{r} &=\frac{A+B}{4}\left(\sin \omega_{1} t+\sin \omega_{2} t\right)+\frac{A-B}{4}\left(\sin \omega_{1} t-\sin \omega_{2} t\right) \\ &=\frac{A}{2} \sin \omega_{1} t+\frac{B}{2} \sin \omega_{2} t \end{aligned}

完全恢复了原始信号。

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