# Lesson 32 Flashcards — Line of Sight and Wireless Channels

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<summary><strong>1. Context only (not assessed): Name the four wireless propagation modes.</strong></summary>
<div class="card-answer"><p><strong>Surface wave</strong>, <strong>sky wave</strong>, <strong>forward scatter</strong>, and <strong>line of sight</strong>. Which you get is decided by frequency, not preference.</p></div>
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<summary><strong>2. Context only (not assessed): Describe surface-wave propagation.</strong></summary>
<div class="card-answer"><p>The wave hugs the Earth's surface by diffraction. Works at VLF, LF and MF (carrier <span class="math notranslate nohighlight">\(f_c \lesssim 3\)</span> MHz) and reaches beyond the horizon — used for AM broadcast and submarine communications.</p></div>
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<summary><strong>3. Context only (not assessed): Describe sky-wave propagation.</strong></summary>
<div class="card-answer"><p>The signal bounces off the <strong>ionosphere</strong>. Useful in HF (<span class="math notranslate nohighlight">\(f_c \lesssim 30\)</span> MHz); a larger angle of incidence — a more grazing ray — gives a better bounce and a longer hop. Used for long-haul radio and over-the-horizon radar, but it varies with time of day and solar activity.</p></div>
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<summary><strong>4. Context only (not assessed): Describe forward scatter.</strong></summary>
<div class="card-answer"><p>The signal strikes discontinuities in the <strong>troposphere</strong> and scatters in all directions; a small fraction reaches a distant receiver. Useful at VHF, UHF and SHF (from about 50 MHz up), very inefficient, but gives beyond-line-of-sight range without a satellite.</p></div>
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<summary><strong>5. What is special about line-of-sight propagation?</strong></summary>
<div class="card-answer"><p><strong>All</strong> electromagnetic signals propagate by LOS, in every band. Its geometric limit is the curvature of the Earth; the other limit is power.</p></div>
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<summary><strong>6. Give the LOS distance equation used in this course, with units.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(r = \sqrt{2h}\)</span> with <span class="math notranslate nohighlight">\(h\)</span> in <strong>feet</strong> and <span class="math notranslate nohighlight">\(r\)</span> in <strong>statute miles</strong>. The equation sheet writes it <span class="math notranslate nohighlight">\(R_{LOS,mi}\)</span> and <span class="math notranslate nohighlight">\(h_{ft}\)</span>: the subscripts are the units.</p></div>
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<summary><strong>7. Why is that equation 'improper', and why is it used anyway?</strong></summary>
<div class="card-answer"><p>Because the units do not balance — the square root of feet should not give miles. It works because the 4/3 Earth radius model happens to make 5280 miles coincide with 5280 feet per mile. Engineers accept it because it is correct and fast.</p></div>
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<summary><strong>8. Why is the Earth radius adjusted to 4/3 of its true value?</strong></summary>
<div class="card-answer"><p>Because decreasing atmospheric pressure with altitude bends radio waves slightly downward, extending range beyond the geometric horizon.</p></div>
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<summary><strong>9. How do you handle two elevated antennas?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R_{LOS,mi} = \sqrt{2h_{ft,1}} + \sqrt{2h_{ft,2}}\)</span> — each antenna reaches its own horizon and the ranges add. An end on the surface has a height of zero and adds nothing.</p></div>
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<summary><strong>10. Which conversions go with the LOS equation?</strong></summary>
<div class="card-answer"><p>The equation sheet's two: <strong>1 ft = 0.305 m</strong> and <strong>1 mi = 1.61 km</strong>. Turn a height in meters into feet <em>before</em> the square root, and the answer in statute miles into kilometers <em>after</em> it.</p></div>
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<summary><strong>11. A helicopter at 2000 ft searches for a ground radio. How close must it get?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R_{LOS,mi} = \sqrt{2(2000)} + 0 = \sqrt{4000} = 63.25\)</span> miles, which is <span class="math notranslate nohighlight">\(63.25 \times 1.61 = 101.8\)</span> km.</p></div>
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<summary><strong>12. Why does altitude have diminishing returns?</strong></summary>
<div class="card-answer"><p>Range grows only as <span class="math notranslate nohighlight">\(\sqrt{h}\)</span> — quadrupling altitude only doubles range. You cannot build a tower out of the problem, which is why airborne relays and satellites exist.</p></div>
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