# Lesson 33 Flashcards — The Friis Equation

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<summary><strong>1. What is a link budget?</strong></summary>
<div class="card-answer"><p>Bookkeeping on power from transmitter to receiver: everything in between either adds gain or subtracts loss. If what arrives meets or exceeds the receiver's sensitivity <span class="math notranslate nohighlight">\(P_{R,\min}\)</span>, the link closes.</p></div>
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<summary><strong>2. Write the Friis equation.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(P_R = P_T G_T G_R \lambda^2/(4\pi R)^2\)</span>, with every gain a <strong>linear ratio</strong> (no units).</p></div>
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<summary><strong>3. Where does the 1/R² in Friis come from?</strong></summary>
<div class="card-answer"><p>The surface area of an expanding sphere. Transmitted power spreads over <span class="math notranslate nohighlight">\(4\pi R^2\)</span> — that is the one-way spreading loss.</p></div>
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<summary><strong>4. What is effective aperture?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(A_e = \lambda^2 G_R/4\pi\)</span> — the effective area over which a receiving antenna collects power. Multiplying power density by <span class="math notranslate nohighlight">\(A_e\)</span> gives received power.</p></div>
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<summary><strong>5. What happens to P_R if you double the transmit power? The gain?</strong></summary>
<div class="card-answer"><p>Both <strong>double</strong> <span class="math notranslate nohighlight">\(P_R\)</span> — these terms enter linearly.</p></div>
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<summary><strong>6. What happens to P_R if you double the wavelength?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(P_R\)</span> goes up by <strong>four times</strong>, since <span class="math notranslate nohighlight">\(\lambda\)</span> is squared. Lower frequencies travel farther for the same power.</p></div>
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<summary><strong>7. What happens to P_R if you double the range?</strong></summary>
<div class="card-answer"><p>It drops to <strong>one quarter</strong> — range enters squared.</p></div>
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<summary><strong>8. Higher frequency hurts Friis but helps antennas. Explain the tension.</strong></summary>
<div class="card-answer"><p>Free-space loss goes as <span class="math notranslate nohighlight">\(1/\lambda^2\)</span>, but dish gain goes as <span class="math notranslate nohighlight">\(1/\lambda^2\)</span> too. Raising frequency costs you in propagation and pays you back in antenna gain — a real design tradeoff.</p></div>
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<summary><strong>9. Give the decibel conversions worth memorizing.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\times 2 = +3\)</span> dB, <span class="math notranslate nohighlight">\(\times 4 = +6\)</span> dB, <span class="math notranslate nohighlight">\(\times 10 = +10\)</span> dB, <span class="math notranslate nohighlight">\(\times 100 = +20\)</span> dB, <span class="math notranslate nohighlight">\(\div 2 = -3\)</span> dB. These are <strong>power</strong> ratios, <span class="math notranslate nohighlight">\(10\log_{10}\)</span>; the equation sheet's <span class="math notranslate nohighlight">\(20\log_{10}\)</span> row is for voltage gain (Lesson 17).</p></div>
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<summary><strong>10. Why do link budgets use decibels at all?</strong></summary>
<div class="card-answer"><p>Because <strong>multiplying becomes adding</strong>, and the dynamic range involved is enormous. In this course, convert a dB value to a linear ratio before it goes into an equation.</p></div>
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<summary><strong>11. How do you find the power-limited range of a link?</strong></summary>
<div class="card-answer"><p>Solve Friis for <span class="math notranslate nohighlight">\(R\)</span> with <span class="math notranslate nohighlight">\(P_R = P_{R,\min}\)</span>: <span class="math notranslate nohighlight">\(R = \frac{\lambda}{4\pi}\sqrt{P_T G_T G_R/P_{R,\min}}\)</span>, in meters. It is the second Friis row on the equation sheet.</p></div>
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<summary><strong>12. The LOS range comes out in miles and the Friis range in meters. How do you compare them?</strong></summary>
<div class="card-answer"><p>Convert first. <span class="math notranslate nohighlight">\(R_{LOS,mi}\)</span> takes heights in feet and gives miles; multiply by <strong>1.61</strong> for kilometers. Compare the two limits only in the same unit.</p></div>
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<summary><strong>13. A satellite is directly overhead. What range goes into Friis?</strong></summary>
<div class="card-answer"><p>The <strong>difference of the two altitudes</strong>, <span class="math notranslate nohighlight">\(R = h_\text{sat} - h_\text{aircraft}\)</span>, in meters (feet × 0.305). Straight up there is no horizon in the way, so there is no LOS limit to compute.</p></div>
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<summary><strong>14. What two tests must a communication link pass?</strong></summary>
<div class="card-answer"><p>It needs <strong>line of sight</strong> and <strong>adequate power</strong>. The maximum range is the <em>smaller</em> of the two limits — and <strong>either one can bind</strong>, so compute both.</p></div>
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