# Lesson 2 Flashcards

Click a question to reveal the answer.

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<summary><strong>1. Relate wavelength and frequency.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\lambda f = c\)</span>, with <span class="math notranslate nohighlight">\(c \approx 3\times10^{8}\)</span> m/s. Radar shortcut: <span class="math notranslate nohighlight">\(\lambda \approx 30/f\)</span> cm for <span class="math notranslate nohighlight">\(f\)</span> in GHz.</p></div>
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<summary><strong>2. What is the wavelength at X-band (10 GHz)?</strong></summary>
<div class="card-answer"><p>About 3 cm (<span class="math notranslate nohighlight">\(30/10\)</span>). X-band is the dominant fire-control band.</p></div>
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<summary><strong>3. List the IEEE radar bands from L to Ka with rough frequencies.</strong></summary>
<div class="card-answer"><p>L (1–2 GHz), S (2–4), C (4–8), X (8–12), Ku (12–18), K (18–27), Ka (27–40 GHz).</p></div>
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<summary><strong>4. What is the tradeoff between low and high radar frequency?</strong></summary>
<div class="card-answer"><p>Lower frequency: longer range, less weather loss, but large antennas and coarse resolution. Higher frequency: fine resolution and small apertures, but more atmospheric attenuation.</p></div>
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<summary><strong>5. Why do fire-control radars cluster around X-band?</strong></summary>
<div class="card-answer"><p>X-band balances good resolution against manageable antenna size and acceptable atmospheric loss — the sweet spot for tracking and engaging aircraft.</p></div>
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<summary><strong>6. Write the free-space path loss formula (km, GHz).</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(L_\text{fs,dB} = 20\log_{10}(R_\text{km}) + 20\log_{10}(f_\text{GHz}) + 92.45\)</span>.</p></div>
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<summary><strong>7. State the two 6 dB rules.</strong></summary>
<div class="card-answer"><p>Doubling the range adds 6 dB of loss; doubling the frequency adds 6 dB of loss. Both come from <span class="math notranslate nohighlight">\(20\log_{10}(2)\approx 6\)</span> dB.</p></div>
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<summary><strong>8. How much loss does a target add by flying from 50 km to 200 km?</strong></summary>
<div class="card-answer"><p>50→200 km is two doublings (×4), so +12 dB of one-way path loss.</p></div>
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<summary><strong>9. What does polarization mismatch cost?</strong></summary>
<div class="card-answer"><p>A full cross-polarization mismatch (e.g., horizontal into a vertical antenna) can cost 20–30 dB. Polarization is both a target-RCS factor and an EP/EA lever.</p></div>
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<summary><strong>10. Relate antenna gain to aperture and wavelength.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(G \approx 4\pi A_e / \lambda^2\)</span>. For a fixed aperture, shorter wavelength gives higher gain and a narrower beam.</p></div>
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<summary><strong>11. Estimate the half-power beamwidth of an aperture.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\theta_\text{BW} \approx \lambda / D\)</span>, where <span class="math notranslate nohighlight">\(D\)</span> is the aperture dimension. Higher frequency → tighter beam from the same dish.</p></div>
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<summary><strong>12. Why do missile seekers push into Ku/Ka band?</strong></summary>
<div class="card-answer"><p>High frequency buys high gain and fine angular resolution from a <em>small</em> aperture — exactly what a size-limited seeker needs, accepting the higher atmospheric loss over short ranges.</p></div>
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