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howto:hambasics:wavemodulationmath [2019/12/12 18:46] – [FM] va7fi | howto:hambasics:sections:wavemodulationmath [2021/01/03 08:01] – [AM] va7fi |
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====== More Details: AM / FM ====== | ====== More Details: AM / FM ====== |
| Here are a few more details about the AM, SSB, and FM modulation schemes introduced on the [[wavemodulation |Wave Modulation]] page. |
Here are a few more details about the AM, SSB, and FM modulation schemes introduced on the [[wavemodulation |Wave Modulation]] page. I made the interactive animations for this section using [[https://www.geogebra.org/ | Geogebra 4.2]] (later versions can't export as html5). | |
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For both AM and FM examples, we'll Let: | For both AM and FM examples, we'll Let: |
* <latex> $c(t) = \cos(2 \pi f_c t)$ </latex> be the <fc #4682b4>radio carrier</fc> with frequency <latex>$f_c$</latex> | * \$c(t) = \cos(2 \pi f_c t)\$ be the <fc #4682b4>radio carrier</fc> with frequency \$f_c\$ |
* <latex> $s(t) = \cos(2 \pi f_s t)$ </latex> be the <fc #ff0000>baseband audio signal</fc> with frequency <latex>$f_s$</latex> | * \$s(t) = \cos(2 \pi f_s t)\$ be the <fc #ff0000>baseband audio signal</fc> with frequency \$f_s\$ |
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{{ am01.png }} | {{ am01.png }} |
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====== AM ====== | ====== AM ====== |
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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: | 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: |
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<WRAP centeralign> | \begin{align*} |
<latex> | |
\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) \\ | \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)\\ | &= \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} | &= \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} | \end{align*} |
</latex> | |
</WRAP> | |
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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: | 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: |
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<WRAP centeralign> | |
<latex> | |
\begin{align*} | \begin{align*} |
\cos(A+B) =& \cos(A)\cos(B) - \sin(A)\sin(B) \\ | \cos(A+B) =& \cos(A)\cos(B) - \sin(A)\sin(B) \\ |
\cos(A-B) =& \cos(A)\cos(B) + \sin(A)\sin(B) \\ | \cos(A-B) =& \cos(A)\cos(B) + \sin(A)\sin(B) \\ |
\end{align} | \end{align*} |
</latex> | |
</WRAP> | |
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Which gives: | Which gives: |
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<WRAP centeralign> | |
<latex> | |
\begin{align*} | \begin{align*} |
&\cos(A+B) + \cos(A-B) = 2 \cos(A)\cos(B) \\ | &\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) | \Rightarrow &\cos(A)\cos(B) = \frac{1}{2}\cos(A+B) + \frac{1}{2}\cos(A-B) |
\end{align} | \end{align*} |
</latex> | |
</WRAP> | |
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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. | 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. |
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Some things to try: | Some things to try: |
* Set <fc #ff0000>f<sub>s</sub> at 10</fc> and <fc #4682b4>f<sub>c</sub> at 200</fc> and check only the transmitted signal. You can easily imagine what the envelope (the baseband signal) should be that produced that signal. But... | * 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 (the baseband signal) should be that produced that signal. But... |
* Decrease <fc #4682b4>f<sub>c</sub></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. | * 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. |
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| ===== 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\$ |
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| \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*} |
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| 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\$ |
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| 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. |
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====== FM ====== | ====== 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: | 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: |
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<WRAP centeralign> | \$$2\pi f_c \quad \rightarrow \quad 2\pi f_c + 2\pi k s(t)\$$ |
<latex> $$2\pi f_c \quad \rightarrow \quad 2\pi f_c + 2\pi k s(t)$$ </latex>. | |
</WRAP> | |
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* Here, $f_c$ is the frequency of the carrier, which is a constant (this is important), | * 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. | * 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. |
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<WRAP round alert box center 80%> | <WRAP round alert box center 80%> |
Now, it might be tempting to simply substitute this sum in the wave like so: | Now, it might be tempting to simply substitute this sum in the wave like so: |
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<WRAP centeralign> | \$$ \cos(2\pi f_c t) \quad \rightarrow \quad \cos\Big(\big(2\pi f_c + 2\pi k s(t)\big) t\Big) \$$ |
<latex> $$ \cos(2\pi f_c t) \quad \rightarrow \quad \cos\Big(\big(2\pi f_c + 2\pi k s(t)\big) t\Big) $$ </latex> | |
</WRAP> | |
but that's not quite right because the frequency is derived from the change in angle. | but that's not quite right because the frequency is derived from the change in angle. |
</WRAP> | </WRAP> |
To solve this properly, we need some calculus and deduce the angle from our new frequency: | To solve this properly, we need some calculus and deduce the angle from our new frequency: |
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<WRAP centeralign> | \$$ \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 \$$ |
<latex> $$ \frac{d}{dt}\theta(t) = 2\pi f_c + 2\pi k s(t) \qquad \Rightarrow \qquad \theta(t) = 2\pi f_c t + 2\pi k \int_0^{t}s(\tau) d\tau $$ </latex> | |
</WRAP> | |
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The frequency modulated transmission is actually given by: | The frequency modulated transmission is actually given by: |
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<WRAP centeralign> | \$$ \cos\Big(2\pi f_c t + 2\pi k \int_0^{t}s(\tau) d\tau\Big) \$$ |
<latex> $$ \cos\Big(2\pi f_c t + 2\pi k \int_0^{t}s(\tau) d\tau\Big) $$ </latex> | |
</WRAP> | |
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In our particular example, with <latex> $s(t) = \cos(2 \pi f_s t)$ </latex>, the modulated radio signal becomes: | |
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<WRAP centeralign> | In our particular example, with \$s(t) = \cos(2 \pi f_s t)\$, the modulated radio signal becomes: |
<latex> | |
\begin{align} | \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 + 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) | &= \cos\Big(2 \pi f_c t + k \sin(2 \pi f_s t)\Big) |
\end{align} | \end{align*} |
</latex> | |
</WRAP> | |
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For more details about FM, see: [[http://www.ece.umd.edu/~tretter/commlab/c6713slides/ch8.pdf]] | 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. | 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. |
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Some things to try: | Some things to try: |
* Set <fc #ff0000>f<sub>s</sub> at 10</fc> and <fc #4682b4>f<sub>c</sub> 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... | * 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<sub>c</sub></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. | * 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". | * 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". |
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See [[https://electronicspost.com/narrow-band-fm-wide-band-fm/]] | See [[https://electronicspost.com/narrow-band-fm-wide-band-fm/]] |
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====== PM ====== | ====== 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. | //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 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 $\cos\Big(2 \pi f_c t + 2\pi k s(t)\Big)$ | * With PM, it leads to \$\cos\Big(2 \pi f_c t + 2\pi k s(t)\Big)\$ |
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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. | 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. |
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[[sections |{{/back.png }}]] [[mathbasics |{{ /next.png}}]] | [[..:sections|{{/back.png }}]] [[mathbasics |{{ /next.png}}]] |