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--- hambasics:sections:wavemodulationmath [2026/07/10 10:29] – va7fi
+++ hambasics:sections:wavemodulationmath [2026/07/10 10:31] (current) – [FM] va7fi
@@ Line 1: / Line 1: @@
+====== More Optional Details ======
+Here are a few more details about the AM, SSB, and FM modulation schemes introduced on the [[wavemodulation |Wave Modulation]] page.
+For both AM and FM examples, we'll let:
+  * \$c(t) = \cos(2 \pi f_c t)\$ be the <fc #4682b4>radio carrier</fc> with frequency \$f_c\$
+  * \$s(t) = \cos(2 \pi f_s t)\$ be the <fc #ff0000>baseband audio signal</fc> with frequency \$f_s\$
+{{  am01.png  }}
+With the radio carrier frequency several times greater than the baseband audio signal.
+====== AM ======
+The resulting //Amplitude Modulated// radio wave is the **product** of the vertically shifted baseband signal and the radio carrier, which is also equal to the **sum** of the carrier and the two side bands:
+\begin{align*}
+\Big(s(t)+1\Big) \times c(t) &= \Big(\cos(2 \pi f_s t) + 1\Big) \times \cos(2 \pi f_c t) \\
+                 &=  \cos(2 \pi f_s t)\cos(2 \pi f_c t) + \cos(2 \pi f_c t)\\
+                 &= \underbrace{\frac{1}{2} \cos\Big(2 \pi (f_c + f_s) t\Big)}_\text{USB} + \underbrace{\frac{1}{2} \cos\Big(2 \pi (f_c - f_s) t\Big)}_\text{LSB} + \underbrace{\cos(2 \pi f_c t)}_\text{Carrier}
+\end{align*}
+In line 1, I distributed the bracket, which, in line 2, gave us the carrier (last term) and a product (first term).  To expend this product into the sum of the two side bands (line 3),  I added these two trig identities together:
+\begin{align*}
+\cos(A+B)  =& \cos(A)\cos(B) - \sin(A)\sin(B) \\
+\cos(A-B)  =& \cos(A)\cos(B) + \sin(A)\sin(B)
+\end{align*}
+Which gives:
+\begin{align*}
+&\cos(A+B) + \cos(A-B) = 2 \cos(A)\cos(B) \\
+\Rightarrow &\cos(A)\cos(B) = \frac{1}{2}\cos(A+B) + \frac{1}{2}\cos(A-B)
+\end{align*}
+Use this animation to see what happens when you vary the individual frequencies.  You can use the check boxes to show or hide different waves.
+FIXME: animation in wrong place
+{{ggb>/hambasics/sections/am.ggb 800,405}}
+Some things to try:
+  * Set <fc #ff0000>\$f_s\$ at 10</fc>  and <fc #4682b4>\$f_c\$ at 200</fc> and check only the transmitted signal.  You can easily imagine what the envelope should be, which is what the baseband signal is.  But...
+  * Decrease <fc #4682b4>\$f_c\$</fc> slowly.  At some point (around 20 or 30) the baseband signal becomes unrecoverable.  This illustrates the point that to transmit a high frequency baseband, a higher frequency carrier is needed (at least 3 to 4 times the frequency of the baseband signal.  This is why with digital signals, the higher the transfer speed, the higher the carrier frequency needs to be.
+===== Mixer =====
+Later on, we'll see that a //mixer// is an electronic component that multiplies two waves together, resulting in four different frequencies: \$f_1, f_2, f_1+f_2, \text{ and } f_1 - f_2\$.  Although it's not modulation, the math is very similar to the way AM is created, so this is a good place to have a look at it.  So let's multiply two very general waves with the following properties:
+  * Amplitudes \$A_1\$ and \$A_2\$
+  * Frequencies \$f_1\$ and \$f_2\$
+  * Phases (horizontal shift) \$\phi_1\$ and \$\phi_2\$
+  * DC components (vertical shift) \$c_1\$ and \$c_2\$
+\begin{align*}
+\Big( A_1 \cos(2 \pi f_1 t + \phi_1) + c_1 \Big) \times & \Big( A_2 \cos(2 \pi f_2 t + \phi_2) + c_2 \Big) \\ \\
+= & A_1 A_2 \cos(2 \pi f_1 t + \phi_1) \cos(2 \pi f_2 t + \phi_2 ) \\
+& + c_2 A_1 \cos(2 \pi f_1 t + \phi_1 ) + c_1 A_2 \cos(2 \pi f_2 t + \phi_2) +  c_1 c_2   \\ \\
+= & \frac{A_1 A_2}{2} \cos(2 \pi (f_1+f_2) t + (\phi_1+\phi_2)) + \frac{A_1 A_2}{2} \cos(2 \pi (f_1-f_2) t + (\phi_1-\phi_2))  \\
+& + c_2 A_1 \cos(2 \pi f_1 t + \phi_1 ) + c_1 A_2 \cos(2 \pi f_2 t + \phi_2) +  c_1 c_2
+\end{align*}
+The last line looks like a real mess, but all it says is that the result is:
+  * A wave with amplitude \$\frac{A_1 A_2}{2}\$, frequency \$f_1 + f_2\$, and phase \$\phi_1 + \phi_2\$
+  * A wave with amplitude \$\frac{A_1 A_2}{2}\$, frequency \$f_1 - f_2\$, and phase \$\phi_1 - \phi_2\$
+  * A wave with amplitude \$c_2 A_1\$, frequency \$f_1\$, and phase \$\phi_1\$
+  * A wave with amplitude \$c_1 A_2\$, frequency \$f_1\$, and phase \$\phi_2\$
+  * A DC component of \$c_1 c_2\$
+A mixer is useful to raise or lower the frequency of a signal.  For example, if a signal at 13 Mhz is mixed with a local oscillator signal at 14 MHz, two new signals will be produced (in addition to the original two): one at 1 Mhz and the other at 27 Mhz.  If we want the higher one, we can put the result through a high pass filter, which will discard the unwanted signals.
+====== FM ======
+Mathematically, FM is less intuitive and more complicated than AM to understand.  The first step is to modulate the frequency by adding a scaled baseband function to it:
+\$$2\pi f_c \quad \rightarrow \quad 2\pi f_c + 2\pi k s(t)\$$
+  * Here, \$f_c\$ is the frequency of the carrier, which is a constant (this is important),
+  * and \$k\$ is a scaling factor we can use to decide how much of a variation we allow the baseband signal to impart on the carrier frequency.  When \$k = 0\$, there is no modulation, and the greater \$k\$ becomes, the bigger the effect is.
+<WRAP round alert box center 80%>
+It might be tempting to simply substitute this sum in the wave like so:
+\$$ \cos(2\pi f_c t) \quad \rightarrow \quad \cos\Big(\big(2\pi f_c + 2\pi k s(t)\big) t\Big) \$$
+but that's not quite right because the frequency is derived from the change in angle.
+</WRAP>
+To solve this properly, we need some calculus to deduce the angle from our new frequency:
+\$$ \frac{d}{dt}\theta(t) = 2\pi f_c + 2\pi k s(t) \qquad  \Rightarrow \qquad \theta(t) = 2\pi f_c t + 2\pi k \int_0^{t}s(\tau) d\tau \$$
+The frequency modulated transmission is actually given by:
+\$$ \cos\Big(2\pi f_c t + 2\pi k \int_0^{t}s(\tau) d\tau\Big) \$$
+In our particular example, with \$s(t) = \cos(2 \pi f_s t)\$, the modulated radio signal becomes:
+\begin{align*}
+\cos\Big(2 \pi f_c t + 2\pi k \int_0^{t}s(\tau)d\tau\Big) &= \cos\Big(2 \pi f_c t + 2\pi k \int_0^{t}\cos(2 \pi f_s \tau)d\tau\Big) \\
+                 &=  \cos\Big(2 \pi f_c t + k \sin(2 \pi f_s t)\Big)
+\end{align*}
+For more details about FM, see: [[http://www.ece.umd.edu/~tretter/commlab/c6713slides/ch8.pdf]]
+Use this animation to see what happens when you vary the individual frequencies. You can use the check boxes to show or hide different waves.
+FIXME: animation in wrong place
+{{ggb>/hambasics/sections/fm.ggb 800,350}}
+Some things to try:
+  * Set <fc #ff0000>\$f_s\$ at 10</fc> and <fc #4682b4>\$f_c\$ at 200</fc> and check only the transmitted signal.  Notice how when the <fc #ff0000>baseband</fc> is high, the **transmitted wave** is "tight" (ie, its frequency is high), and vise-versa.  But...
+  * Decrease <fc #4682b4>\$f_c\$</fc> slowly.  At some point (around 20 or 30) that pattern becomes unnoticeable.  Again, this illustrates the point that to transmit a high frequency baseband, a higher frequency carrier is needed (at least 3 to 4 times the frequency of the baseband signal).  This is why with digital signals, the higher the transfer speed, the higher the carrier frequency needs to be.
+  * Increase and decrease **k** to see the effect it has on the transmitted wave.  The greater **k**, the more bandwidth the resulting signal uses.  This dictates the difference between "Narrow Band FM" and "Wide Band FM".
+See [[https://electronicspost.com/narrow-band-fm-wide-band-fm/]]
+====== PM ======
+//Phase Modulation// is not usually discussed in ham radio courses, but after understanding FM, we pretty much get PM for free...  Recall that for the wave \$\cos(2\pi f + \phi)\$, \$f\$ is the frequency and \$\phi\$ is the phase shift.  For a pure tone, both of these are constant.
+  * With FM, we saw that modulating the frequency led to  \$\cos\Big(2 \pi f_c t + 2\pi k \int_0^{t}s(\tau)d\tau\Big)\$.
+  * With PM, it leads to \$\cos\Big(2 \pi f_c t + 2\pi k s(t)\Big)\$
+Essentially, with PM, we simply let \$\phi\$ vary with the baseband \$s(t)\$.  But the thing to notice is that PM looks a lot like FM.  In fact, an FM signal modulated by \$s(t)\$ is the same as a PM signal modulated by \$\int_0^{t}s(\tau)d\tau\$.  In other words, the receiver needs to know if the signal was modulated in FM or PM since both wave forms look similar.
+[[..:sections|{{/back.png }}]] [[mathbasics |{{  /next.png}}]]