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--- howto:hambasics:temp [2019/11/25 17:32] – ve7hzf
+++ howto:hambasics:temp [2019/11/25 18:31] – [Old] ve7hzf
@@ Line 19: / Line 19: @@
 {{ :howto:hambasics:travelingwaves.png }}
-  * the **amplitude** is the **vertical** height from the centre of the wave to its highest (or lowest) point.  <fc #0014a8>The blue wave has an amplitude of 2</fc> and the <fc #008000>green wave has an amplitude of 1</fc>.
+  * the //amplitude// is the vertical height from the centre of the wave to its highest (or lowest) point.  <fc #0014a8>The blue wave has an amplitude of 2</fc> and the <fc #008000>green wave has an amplitude of 1</fc>.
-  * the **wavelength** is the **horizontal** distance of one full cycle.  <fc #0014a8>The blue wave has a wavelength of 2m </fc> and the <fc #008000>green wave has a wavelength of 6m</fc>.
+  * the //wavelength// is the horizontal distance of one full cycle.  <fc #0014a8>The blue wave has a wavelength of 2m </fc> and the <fc #008000>green wave has a wavelength of 6m</fc>.
+Now imagine that the animation is in super slow motion and that the waves are actually travelling at the speed of light, which is roughly 300,000,000 metres per second: How many times does each dot go up and down in one second?
+Another way of asking that question is: how many full cycles can you fit in 300,000,000 metres (since radio waves travel 300,000,000 metres each second).
+  * Since the blue wave has a wavelength of 2m, it'll take 150,000,000 cycles to reach 300,000,000 metres.  That means that the blue dot oscillates at 150,000,000 cycles per second, or 150,000,000 Hz, or 150 Mhz
+  * Similarly, since the green wave has a wavelength of 6m, its frequency is 50 Mhz.
-All EM waves (radio, light, etc) in vacuum travel at the speed, which is roughly 300,000 metres per second.  Now, let's take a snapshot of the two waves
+So a quick way to relate the frequency $f$ (in MHz) and the wavelength $\lambda$ (in metres):
-Look at the following two waves.  How are they different?
+<latex> $$ \lambda = \frac{300}{f} \qquad \text{or} \qquad f = \frac{300}{\lambda}$$ </latex>
-{{wave1.png?400}}{{wave2.png?400}}
+Note that the reason we're using just 300, instead of 300,000,000 is that we've cancelled 6 of the zeros so that the frequency is in MHz instead of in Hz.
-At first sight:
+Now, here's a related question: how long does it take for each wave to complete one cycle?
-  - the first one is "taller" than the second one.  That is, it goes up and down higher and lower.
-  - the first one is also "longer" than the second one.  That is, it stretches sideways more.  It's not as "tight".
-These two observations can be quantified very precisely as:
+  * For the blue wave, we know that it oscillates 150,000,000 times / second, so only one of those time would take 150,000,000<sup>**th**</sup> of a second, or $\frac{1}{150,000,000}$ s or 6.67 x 10<sup>-9</sup> s or 6.67 ns.(("ns" means nanosecond. "Nano" means a billionth of ___))
-  - the //amplitude//: **vertical** height from the centre of the wave to its highest (or lowest) point.
+  * Similarly, the green wave oscillates at 50,000,000 cycles per second, so only one of those cycle would take $\frac{1}{50,000,000}$ s or 2 x 10<sup>-8</sup> s or 20 ns.
-  - the //period//: **horizontal** length of one complete cycle.
-{{  wave3.png  }}
+The time to complete one full cycle is called the //period (T)// and is the reciprocal of the frequency:
-So the previous two waves have:
+<latex> \qquad  $$f = \frac{1}{T}  \qquad  \text{or}   \qquad T = \frac{1}{f}$$</latex>
-  - Amplitude = 2, Period = 0.05 ms
-  - Amplitude = 1, Period = 0.02 ms
-{{wave1.png?400}}{{wave2.png?400}}
-The amplitude is normally related to the strength of the signal (like the volume for sound).
-Since the period (//T//) is the amount of time it takes to complete one cycle, and the frequency (//f//) is the number of cycles in one second, the period and the frequency are inverses of each other:
-<latex> \qquad  $$f = \frac{1}{T}  \qquad  \Leftrightarrow   \qquad T = \frac{1}{f}$$</latex>
-<box 80% blue>
-In this course, we'll see a few formulas and it'll be tempting to memorize them but let's instead understand what they really mean...
-</box>
-For example:
-  * if the period is half a second, we can fit 2 full cycles in one second.
-  * If the period is a quarter of a second, the frequency is 4.
-  * If the period is a tenth of a second, the frequency is 10.
-  * If the period is T seconds, the frequency is $\frac{1}{T}$ ( $\frac{1}{0.5} = 2,  \quad \frac{1}{0.25} = 4, \quad \frac{1}{0.1} = 10$ )
-Right?
-So for the previous two waves, the frequencies would be:
-  * <latex>$$f = \frac{1}{0.05 \text{ ms}} = \frac{1}{0.00005 \text{ s}}$$</latex> = 20,000 Hz = 20 kHz
-  * <latex>$$f = \frac{1}{0.02 \text{ ms}} = \frac{1}{0.00002 \text{ s}}$$</latex> = 50,000 Hz = 50 kHz
-Recall that //Hz// means "cycle per seconds".  That's why when we divide a number of cycles by time, we get Hertz.
-Let's now look at three different ways to encode a signal on a radio wave.
-FIXME: add $f=\frac{c}{\lambda}$