電路知識：Modulation
工具：Qucs
簡介：發射端將低頻訊號處理成高頻訊號以後，再傳送出去
主要的調變方法
- 振幅調變（amplitude modulation，AM）
- 頻率調變（Frequency modulation，FM）
- 相位調變（Phase Modulation，PM）
AM (振幅調變)
$$
\begin{align*}
c(t) &= A_c cos(2\pi f_ct+\phi _c)\\
m(t) &= M cos(2\pi f_mt+\phi )\\
AM(t) &= [1+m(t)]\cdot c(t)\\
&=\small{A_c cos(2\pi f_ct+\phi _c)+\frac{A_cM}{2} [cos(2\pi(f_c-f_m)t+\phi _c-\phi)+cos(2\pi(f_c+f_m)t+\phi _c+\phi)]}\\
M &< 1\\
\end{align*}
$$
$$
\begin{align*}
AM(t) &= [1+m(t)]\cdot c(t) \\
&=[1+M cos(2\pi f_mt+\phi )]\cdot A_c cos(2\pi f_ct+\phi _c)\\
&=A_c cos(2\pi f_ct+\phi _c)+M cos(2\pi f_mt+\phi )\cdot A_c cos(2\pi f_ct+\phi _c)\\
&=A_c cos(2\pi f_ct+\phi _c)+A_cM cos(2\pi f_ct+\phi _c) cos(2\pi f_mt+\phi )\\
&=A_c cos(2\pi f_ct+\phi _c)+\frac{A_cM}{2} [cos(2\pi f_ct+\phi _c-(2\pi f_mt+\phi) )+cos(2\pi f_ct+\phi _c+(2\pi f_mt+\phi))]\\
&=A_c cos(2\pi f_ct+\phi _c)+\frac{A_cM}{2} [cos(2\pi(f_c-f_m)t+\phi _c-\phi)+cos(2\pi(f_c+f_m)t+\phi _c+\phi)]\\
\end{align*}
$$
必要條件：\(1+m(t)>1\)，故 \(M < 1\)，否則會 overmodulation，訊號就不對了，因會使得輸出振幅為負的
頻寬 \(BW = 2f_m\)
電路驗證
角調變 定義
$$
\begin{align*}
c(t) &= A_c cos(2\pi f_ct+\phi _c)\\
XM(t) &= A_c cos(2\pi f_ct+\phi _c+\phi _m(t))\\
\\
\theta(t) &= 2\pi f_ct+\phi _c+\phi _m(t)\\
f(t)&=f_c+\frac{1}{2\pi}\frac{\mathrm{d} \phi _m(t)}{\mathrm{d} t}\\
\end{align*}
$$
相位偏差：\(\phi _m(t) \)
頻率偏差 \(\Delta f\)：\(\frac{1}{2\pi}\frac{\mathrm{d} \phi _m(t)}{\mathrm{d} t}\)
$$
\begin{align*}
\frac{\mathrm{d} \theta(t)}{\mathrm{d} t} &=w(t)\\
&=2\pi f(t)\\
&=2\pi f_c+\frac{\mathrm{d} \phi _m(t)}{\mathrm{d} t}\\
f(t)&=f_c+\frac{1}{2\pi}\frac{\mathrm{d} \phi _m(t)}{\mathrm{d} t}
\end{align*}
$$
FM (頻率調變)
$$
\begin{align*}
c(t) &= A_c cos(2\pi f_ct+\phi _c)\\
m(t) &= M cos(2\pi f_mt+\phi )\\
\\
\phi _m(t)&=k_f\int_{0}^{t}m(\tau) d\tau\\
FM(t) &=A_c cos\left (2\pi f_ct+\phi _c+\phi _m(t) \right)\\
&=A_c \sum_{k=-\infty }^{\infty}J_{k}(\beta)cos(2\pi (f_c+kf_m)t+\phi _c+k\phi )\\
\beta&=\frac{k_f M}{2\pi f_m}\\
\Delta f &= \beta f_m\\
\end{align*}
$$
\(k_f\) 為自定頻率偏差函數，單位為 Hz/Volt
根據角調變
$$
\begin{align*}
\frac{\mathrm{d} \phi _m(t)}{\mathrm{d} t}&=k_fm(t)\\
\phi _m(t)&=k_f\int_{0}^{t}m(\tau) d\tau\\
FM(t) &= A_c cos\left (2\pi f_ct+\phi _c+\phi _m(t) \right)\\
&= A_c cos \left (2\pi f_ct+\phi _c+k_f\int_{0}^{t}m(\tau) d\tau \right )\\
&= A_c cos\left (2\pi f_ct+\phi _c+k_f\int_{0}^{t}M cos(2\pi f_m\tau+\phi)\right ) \\
&= A_c cos\left (2\pi f_ct+\phi _c+k_f\frac{1}{2\pi f_m}(M sin(2\pi f_mt+\phi)) \right )\\
&= A_c cos\left (2\pi f_ct+\phi _c+\frac{k_f M}{2\pi f_m}sin(2\pi f_mt+\phi )\right ) \\
&= A_c cos\left (2\pi f_ct+\phi _c+\beta sin(2\pi f_mt+\phi )\right ) \\
&=A_c [cos(2\pi f_ct+\phi _c)cos(\beta sin(2\pi f_mt+\phi ))-sin(2\pi f_ct+\phi _c)sin(\beta sin(2\pi f_mt+\phi ))]\\
\end{align*}
$$
用上面倒數第二式求 \(\Delta f\)
$$
\begin{align*}
\frac{\mathrm{d} \theta }{\mathrm{d} t} &= \frac{\mathrm{d} (2\pi f_ct+\phi _c+\beta sin(2\pi f_mt+\phi ))}{\mathrm{d} t}\\
&= 2\pi f_c+2\pi f_m \beta cos(2\pi f_mt+\phi )\\
&= w(t)\\
&= 2 \pi f(t)\\
f(t) &= f_c+\beta f_m cos(2\pi f_mt+\phi )\\
&= f_c+\Delta f \cdot cos(2\pi f_mt+\phi )
\end{align*}
$$
借由 bessel function
$$
\begin{align*}
cos(zsin\theta ) & = J_0(z)+2\sum_{k=1}^{\infty }J_{2k}(z)cos(2k\theta )\\
sin(zsin\theta ) & = 2\sum_{k=0}^{\infty }J_{2k+1}(z)sin((2k+1)\theta )\\
J_{-n}(z)&=(-1)^nJ_n(z)
\end{align*}
$$
可得
$$
\begin{align*}
&cos(2\pi f_ct+\phi _c)cos(\beta sin(2\pi f_mt+\phi ))\\
=&cos(2\pi f_ct+\phi _c)\left [ J_0(\beta)+2\sum_{k=1}^{\infty }J_{2k}(\beta)cos(2k(2\pi f_mt+\phi ) ) \right ]\\
=&J_0(\beta)cos(2\pi f_ct+\phi _c)+ 2\sum_{k=1}^{\infty }J_{2k}(\beta)cos(2\pi f_ct+\phi _c)cos(2k(2\pi f_mt+\phi ) ) \\
=&J_0(\beta)cos(2\pi f_ct+\phi _c)+ \sum_{k=1}^{\infty }J_{2k}(\beta)[cos(2\pi (f_c+2kf_m)t+\phi _c+2k\phi )+cos(2\pi (f_c-2kf_m)t+\phi _c-2k\phi ) ] \\
=&J_0(\beta)cos(2\pi f_ct+\phi _c)+ \sum_{k=1}^{\infty }J_{2k}(\beta)cos(2\pi (f_c+2kf_m)t+\phi _c+2k\phi )+\sum_{k=1}^{\infty }J_{2k}(\beta)cos(2\pi (f_c-2kf_m)t+\phi _c-2k\phi ) \\
=&J_0(\beta)cos(2\pi f_ct+\phi _c)+ \sum_{k=1}^{\infty }J_{2k}(\beta)cos(2\pi (f_c+2kf_m)t+\phi _c+2k\phi )+\sum_{k=-\infty }^{-1}(-1)^{2k}J_{2k}(\beta)cos(2\pi (f_c+2kf_m)t+\phi _c+2k\phi ) \\
=&\sum_{k=-\infty }^{\infty }J_{2k}(\beta)cos(2\pi (f_c+2kf_m)t+\phi _c+2k\phi )\\
\\
&sin(2\pi f_ct+\phi _c)sin(\beta sin(2\pi f_mt+\phi ))\\
=&sin(2\pi f_ct+\phi _c)\left [ 2\sum_{k=0}^{\infty }J_{2k+1}(\beta)sin((2k+1)(2\pi f_mt+\phi ) ) \right ]\\
=& 2\sum_{k=0}^{\infty }J_{2k+1}(\beta)sin(2\pi f_ct+\phi _c)sin((2k+1)(2\pi f_mt+\phi ) ) \\
=& \sum_{k=0}^{\infty }J_{2k+1}(\beta)[cos(2\pi (f_c-(2k+1)f_m)t+\phi _c-(2k+1)\phi)-cos(2\pi (f_c+(2k+1)f_m)t+\phi _c+(2k+1)\phi) ] \\
=& \sum_{k=0}^{\infty }J_{2k+1}(\beta)cos(2\pi (f_c-(2k+1)f_m)t+\phi _c-(2k+1)\phi)-\sum_{k=0}^{\infty }J_{2k+1}(\beta)(cos(2\pi (f_c+(2k+1)f_m)t+\phi _c+(2k+1)\phi) ] \\
=& \sum_{k=-\infty}^{-1 }(-1)^{2k+1}J_{2k+1}(\beta)cos(2\pi (f_c+(2k+1)f_m)t+\phi _c+(2k+1)\phi)-\sum_{k=0}^{\infty }J_{2k+1}(\beta)(cos(2\pi (f_c+(2k+1)f_m)t+\phi _c+(2k+1)\phi) ] \\
=& -\sum_{k=-\infty }^{\infty }J_{2k+1}(\beta)(cos(2\pi (f_c+(2k+1)f_m)t+\phi _c+(2k+1)\phi) \\
\end{align*}
$$
故可得
$$
\begin{align*}
FM(t) &=A_c [cos(2\pi f_ct+\phi _c)cos(\beta sin(2\pi f_mt+\phi ))-sin(2\pi f_ct+\phi _c)sin(\beta sin(2\pi f_mt+\phi ))]\\
&=A_c \left [\sum_{k=-\infty }^{\infty }J_{2k}(\beta)cos(2\pi (f_c+2kf_m)t+\phi _c+2k\phi )+\sum_{k=-\infty }^{\infty }J_{2k+1}(\beta)(cos(2\pi (f_c+(2k+1)f_m)t+\phi _c+(2k+1)\phi) \right ]\\
&=A_c \left [\sum_{k=even }J_{k}(\beta)cos(2\pi (f_c+kf_m)t+\phi _c+k\phi )+\sum_{k=odd }J_{k}(\beta)(cos(2\pi (f_c+kf_m)t+\phi _c+k\phi) \right ]\\
&=A_c \sum_{k=-\infty }^{\infty}J_{k}(\beta)cos(2\pi (f_c+kf_m)t+\phi _c+k\phi )\\
\end{align*}
$$
當 \(\beta \leq 0.2\) 且 \(k\geq 2\)，\(J_{k}(\beta)\cong 0\)，則頻寬 \(BW\cong 2f_m\)
電路驗證
PM (相位調變)
$$
\begin{align*}
c(t) &= A_c cos(2\pi f_ct+\phi _c)\\
m(t) &= M cos(2\pi f_mt+\phi )\\
\\
PM(t) &=A_c cos\left (2\pi f_ct+\phi _c+m(t) \right)\\
&=A_c \sum_{k=-\infty }^{\infty}J_{k}(M)cos(2\pi (f_c-kf_m)t+\phi _c-k\phi +\frac{k}{2}\pi)\\
\Delta f &= M f_m\\
\end{align*}
$$
$$
\begin{align*}
PM(t) &= A_c cos\left (2\pi f_ct+\phi _c+m(t) \right)\\
&= A_c cos\left (2\pi f_ct+\phi _c+M cos(2\pi f_m t+\phi)\right ) \\
&=A_c [cos(2\pi f_ct+\phi _c)cos(M cos(2\pi f_m t+\phi))-sin(2\pi f_ct+\phi _c)sin(M cos(2\pi f_m t+\phi))]\\
\end{align*}
$$
用上面倒數第二式求 \(\Delta f\)
$$
\begin{align*}
\frac{\mathrm{d} \theta }{\mathrm{d} t} &= \frac{\mathrm{d} (2\pi f_ct+\phi _c+M cos(2\pi f_m t+\phi))}{\mathrm{d} t}\\
&= 2\pi f_c-2\pi f_m M sin(2\pi f_m t+\phi)\\
&= w(t)\\
&= 2 \pi f(t)\\
f(t) &= f_c-M f_m sin(2\pi f_m t+\phi)\\
&= f_c+\Delta f \cdot sin(2\pi f_m t+\phi)
\end{align*}
$$
借由 bessel function
$$
\begin{align*}
cos(zsin\theta ) & = J_0(z)+2\sum_{k=1}^{\infty }J_{2k}(z)cos(2k\theta )\\
sin(zsin\theta ) & = 2\sum_{k=0}^{\infty }J_{2k+1}(z)sin((2k+1)\theta )\\
J_{-n}(z)&=(-1)^nJ_n(z)
\end{align*}
$$
可得
$$
\begin{align*}
&cos(2\pi f_ct+\phi _c)cos(M cos(2\pi f_m t+\phi))\\
=&cos(2\pi f_ct+\phi _c)cos(M sin(\frac{\pi}{2}-2\pi f_m t-\phi))\\
=&cos(2\pi f_ct+\phi _c)\left [ J_0(M)+2\sum_{k=1}^{\infty }J_{2k}(M)cos(2k(\frac{\pi}{2}-2\pi f_m t-\phi) ) \right ]\\
=&J_0(M)cos(2\pi f_ct+\phi _c)+ 2\sum_{k=1}^{\infty }J_{2k}(M)cos(2\pi f_ct+\phi _c)cos(2k(\frac{\pi}{2}-2\pi f_m t-\phi) ) \\
=&J_0(M)cos(2\pi f_ct+\phi _c)+ \sum_{k=1}^{\infty }J_{2k}(M)[cos(2\pi (f_c-2kf_m)t+\phi _c-2k\phi+k\pi )+cos(2\pi (f_c+2kf_m)t+\phi _c+2k\phi-k\pi ) ] \\
=&J_0(M)cos(2\pi f_ct+\phi _c)+\sum_{k=1}^{\infty }J_{2k}(M)cos(2\pi (f_c-2kf_m)t+\phi _c-2k\phi +k\pi) + \sum_{k=1}^{\infty }J_{2k}(M)cos(2\pi (f_c+2kf_m)t+\phi _c+2k\phi -k\pi) \\
=&J_0(M)cos(2\pi f_ct+\phi _c)+\sum_{k=1}^{\infty }J_{2k}(M)cos(2\pi (f_c-2kf_m)t+\phi _c-2k\phi +k\pi)+ \sum_{k=-\infty }^{-1}(-1)^{2k}J_{2k}(M)cos(2\pi (f_c-2kf_m)t+\phi _c-2k\phi +k\pi) \\
=&\sum_{k=-\infty }^{\infty }J_{2k}(M)cos(2\pi (f_c-2kf_m)t+\phi _c-2k\phi +k\pi)\\
\\
&sin(2\pi f_ct+\phi _c)sin(M cos(2\pi f_m t+\phi))\\
=&sin(2\pi f_ct+\phi _c)sin(M sin(\frac{\pi}{2}-2\pi f_mt-\phi ))\\
=&sin(2\pi f_ct+\phi _c)\left [ 2\sum_{k=0}^{\infty }J_{2k+1}(M)sin((2k+1)(\frac{\pi}{2}-2\pi f_mt-\phi ) ) \right ]\\
=& 2\sum_{k=0}^{\infty }J_{2k+1}(M)sin(2\pi f_ct+\phi _c)sin((2k+1)(\frac{\pi}{2}-2\pi f_mt-\phi ) ) \\
=& \sum_{k=0}^{\infty }J_{2k+1}(M)[cos(2\pi (f_c+(2k+1)f_m)t+\phi _c+(2k+1)\phi-(2k+1)\frac{\pi}{2})-cos(2\pi (f_c-(2k+1)f_m)t+\phi _c-(2k+1)\phi+(2k+1)\frac{\pi}{2}) ] \\
=& \sum_{k=0}^{\infty }J_{2k+1}(M)cos(2\pi (f_c+(2k+1)f_m)t+\phi _c+(2k+1)\phi-(2k+1)\frac{\pi}{2})-\sum_{k=0}^{\infty }J_{2k+1}(M)(cos(2\pi (f_c-(2k+1)f_m)t+\phi _c-(2k+1)\phi+(2k+1)\frac{\pi}{2}) \\
=& \sum_{k=-\infty}^{-1 }(-1)^{2k+1}J_{2k+1}(M)cos(2\pi (f_c-(2k+1)f_m)t+\phi _c-(2k+1)\phi+(2k+1)\frac{\pi}{2})-\sum_{k=0}^{\infty }J_{2k+1}(M)(cos(2\pi (f_c-(2k+1)f_m)t+\phi _c-(2k+1)\phi+(2k+1)\frac{\pi}{2}) \\
=& -\sum_{k=-\infty }^{\infty }J_{2k+1}(M)(cos(2\pi (f_c-(2k+1)f_m)t+\phi _c-(2k+1)\phi+(2k+1)\frac{\pi}{2}) \\
\end{align*}
$$
故可得
$$
\begin{align*}
PM(t) &=A_c [cos(2\pi f_ct+\phi _c)cos(M cos(2\pi f_m t+\phi))-sin(2\pi f_ct+\phi _c)sin(M cos(2\pi f_m t+\phi))]\\
&=A_c \left [\sum_{k=-\infty }^{\infty }J_{2k}(M)cos(2\pi (f_c-2kf_m)t+\phi _c-2k\phi +k\pi)+\sum_{k=-\infty }^{\infty }J_{2k+1}(M)(cos(2\pi (f_c-(2k+1)f_m)t+\phi _c-(2k+1)\phi+(2k+1)\frac{\pi}{2}) \right ]\\
&=A_c \left [\sum_{k=even }J_{k}(M)cos(2\pi (f_c-kf_m)t+\phi _c-k\phi +\frac{k}{2}\pi)+\sum_{k=odd }J_{k}(M)(cos(2\pi (f_c-kf_m)t+\phi _c-k\phi+\frac{k}{2}\pi) \right ]\\
&=A_c \sum_{k=-\infty }^{\infty}J_{k}(M)cos(2\pi (f_c-kf_m)t+\phi _c-k\phi +\frac{k}{2}\pi)\\
\end{align*}
$$
當 \(M \leq 0.2\) 且 \(k\geq 2\)，\(J_{k}(\beta)\cong 0\)，則頻寬 \(BW\cong 2f_m\)
電路驗證
參考
類比與數位調變
何謂 I/Q 資料 (I/Q Data)？
瞭解無線的基礎科技
電視RF訊號, 為何影像要用AM調變 / 聲音要用FM調變
頻率調製
振幅調變
Why is NTSC color carrier frequency 3.57954545 MHz and not some other number that can be remembered easily?
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Phase modulation
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