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howto:hambasics:wavemodulationmath [2020/08/10 15:19] – va7fi | howto:hambasics:sections:wavemodulationmath [2021/01/03 08:04] – [AM] va7fi |
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~~NOTOC~~ | ~~NOTOC~~ |
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| [[home |Ham Basics]] | [[test |About The Test]] | [[Reference |References]] ^ [[sections |Study Sections]] | | |
====== More Details: AM / FM ====== | |
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| ====== 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. |
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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*} |
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\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*} |
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</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> | \begin{align*} |
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\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*} |
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</WRAP> | |
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Which gives: | Which gives: |
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<WRAP centeralign> | \begin{align*} |
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\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*} |
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</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: |
* DC components (vertical shift) \$c_1\$ and \$c_2\$ | * DC components (vertical shift) \$c_1\$ and \$c_2\$ |
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<WRAP centeralign> | \begin{align*} |
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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) \\ \\ | \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 ) \\ | = & A_1 A_2 \cos(2 \pi f_1 t + \phi_1) \cos(2 \pi f_2 t + \phi_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)) \\ | = & \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 | & + 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} | \end{align*} |
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</WRAP> | |
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The last line looks like a real mess, but all it says is that the result is: | The last line looks like a real mess, but all it says is that the result is: |
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)\$$ |
\$$2\pi f_c \quad \rightarrow \quad 2\pi f_c + 2\pi k s(t)\$$. | |
</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), |
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) \$$ | \$$ \cos(2\pi f_c t) \quad \rightarrow \quad \cos\Big(\big(2\pi f_c + 2\pi k s(t)\big) t\Big) \$$ |
</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 \$$ | \$$ \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 \$$ |
</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) \$$ | \$$ \cos\Big(2\pi f_c t + 2\pi k \int_0^{t}s(\tau) d\tau\Big) \$$ |
</WRAP> | |
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In our particular example, with \$s(t) = \cos(2 \pi f_s t)\$, the modulated radio signal becomes: | In our particular example, with \$s(t) = \cos(2 \pi f_s t)\$, the modulated radio signal becomes: |
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<WRAP centeralign> | \begin{align*} |
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\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*} |
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</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: |
* 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. | * 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. |
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[[sections |{{/back.png }}]] [[mathbasics |{{ /next.png}}]] | [[..:sections|{{/back.png }}]] [[mathbasics |{{ /next.png}}]] |