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howto:hambasics:temp [2019/11/25 17:54] ve7hzfhowto:hambasics:temp [2019/11/25 18:31] – [Old] ve7hzf
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 {{ :howto:hambasics:travelingwaves.png }} {{ :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? 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?
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 Now, here's a related question: how long does it take for each wave to complete one cycle? Now, here's a related question: how long does it take for each wave to complete one cycle?
  
-  * For the blue wave, we know that it oscillates 150,000,000 times / second, so only one of those time would be 150,000,000<sup>th</sup> of a second.  That is: $\frac{1}{150,000,000}\text{s} = 6.67 \times 10^-9 \text{s} = 6.67 \text{ns}$ +  * 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}or 6.67 10<sup>-9</sup> or 6.67 ns.(("ns" means nanosecond. "Nano" means a billionth of ___)) 
-  *  +  * 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 time to complete one full cycle is called the //period (T)// and is the reciprocal of the frequency:
  
 +<latex> \qquad  $$f = \frac{1}{T}  \qquad  \text{or}   \qquad T = \frac{1}{f}$$</latex>
  
  
-Look at the following two waves.  How are they different? 
  
-{{wave1.png?400}}{{wave2.png?400}} 
  
-At first sight: 
-  - 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: 
-  - the //amplitude//: **vertical** height from the centre of the wave to its highest (or lowest) point. 
-  - the //period//: **horizontal** length of one complete cycle. 
- 
-{{  wave3.png  }} 
- 
-So the previous two waves have: 
-  - 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 $f=\frac{c}{\lambda}$