# Lesson 17 Flashcards

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<summary><strong>1. Define electromagnetic attack (EA).</strong></summary>
<div class="card-answer"><p>The use of EM energy to deny, degrade, or disrupt the enemy's use of the spectrum. It is offensive EW — where ES listens and EP shields, EA reaches out and breaks the threat's picture. Its three reaches are jamming, deception (L18), and directed energy.</p></div>
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<summary><strong>2. Cover jamming vs deception — what is the difference?</strong></summary>
<div class="card-answer"><p>Cover (noise) <em>hides</em> the real return by raising the noise floor, so the radar sees nothing clearly — it <strong>denies</strong> information. Deception <em>creates</em> false returns by mimicking or replaying the signal, so the radar sees the wrong thing — it <strong>corrupts</strong> information. This lesson is the noise half.</p></div>
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<summary><strong>3. What single trade-off governs all noise-jamming types?</strong></summary>
<div class="card-answer"><p>A jammer's power is fixed; only its distribution in frequency is free. Spread it wide for coverage and you lose density per hertz; concentrate it for density and you lose coverage. Spot, barrage, and swept-spot are three answers to that trade.</p></div>
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<summary><strong>4. Describe spot jamming and its weakness.</strong></summary>
<div class="card-answer"><p>All of the jammer's power on one narrow band matched to the threat frequency — very high power density. Its weakness: a frequency-agile radar simply hops out from under the spot, leaving the jammer shouting into an empty channel.</p></div>
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<summary><strong>5. Describe barrage jamming and the dilution penalty.</strong></summary>
<div class="card-answer"><p>Power spread across the whole band at once — immune to hopping, since the radar cannot escape a band jammed everywhere. The penalty: finite power divided over a wide bandwidth drops the power <em>per hertz</em> in the threat's passband. Coverage costs density.</p></div>
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<summary><strong>6. Describe swept-spot jamming.</strong></summary>
<div class="card-answer"><p>A high-density spot that sweeps across the band. At any instant it is a full-density spot; over time it covers the whole band. It trades dwell time for coverage — any given channel is jammed only a fraction of the time.</p></div>
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<summary><strong>7. Self-protect (SPJ) vs stand-off (SOJ) jamming geometry?</strong></summary>
<div class="card-answer"><p>SPJ rides on the target it defends, so jammer and skin echo share one range to the radar. SOJ is a dedicated platform jamming from a safe range outside the threat ring while strikers ingress, so its range to the radar stays roughly fixed. Geometry sets <span class="math notranslate nohighlight">\(R\)</span> in the J/S equation.</p></div>
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<summary><strong>8. Define J/S and give the self-protect formula.</strong></summary>
<div class="card-answer"><p>The jam-to-signal ratio: effective jamming power in the receiver passband over the signal (skin echo) power. For a self-protect jammer, <span class="math notranslate nohighlight">\(\frac{J}{S} = \frac{4\pi\,P_j G_j\,R^2}{P_t G_t\,\sigma}\)</span>. Only power <em>in the passband</em> counts.</p></div>
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<summary><strong>9. Why does self-protect J/S scale as <span class="math notranslate nohighlight">\(R^2\)</span>?</strong></summary>
<div class="card-answer"><p>The skin echo is two-way (round trip), so <span class="math notranslate nohighlight">\(S \sim 1/R^4\)</span>; the jammer is one-way, so <span class="math notranslate nohighlight">\(J \sim 1/R^2\)</span>. The ratio is <span class="math notranslate nohighlight">\((1/R^2)/(1/R^4) = R^2\)</span>. The jammer wins far out and loses as the target closes.</p></div>
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<summary><strong>10. The target halves its range. What happens to J/S?</strong></summary>
<div class="card-answer"><p>Since <span class="math notranslate nohighlight">\(J/S \sim R^2\)</span>, halving <span class="math notranslate nohighlight">\(R\)</span> scales J/S by <span class="math notranslate nohighlight">\((1/2)^2 = 1/4\)</span> — a factor-of-four drop, or <span class="math notranslate nohighlight">\(-6\)</span> dB. Closing the range helps the radar.</p></div>
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<summary><strong>11. What is the burn-through range <span class="math notranslate nohighlight">\(R_{bt}\)</span>?</strong></summary>
<div class="card-answer"><p>The range at which the closing skin echo overtakes the jam and J/S drops back through the radar's tolerable threshold — inside <span class="math notranslate nohighlight">\(R_{bt}\)</span> the radar "burns through" and re-detects. A modern pulse-Doppler radar tolerates J/S up to about <span class="math notranslate nohighlight">\(+10\)</span> dB, so burn-through occurs while the jammer still nominally "wins."</p></div>
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<summary><strong>12. How does lowering RCS affect burn-through?</strong></summary>
<div class="card-answer"><p>RCS <span class="math notranslate nohighlight">\(\sigma\)</span> is in the denominator of J/S, so a smaller RCS means a weaker echo and a higher J/S at every range. The radar must get closer to recover the track, so burn-through moves <strong>inward</strong>. LO and jamming reinforce each other.</p></div>
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<summary><strong>13. Why doesn't a stand-off jammer get the same range relief as it closes?</strong></summary>
<div class="card-answer"><p>An SOJ sits at a roughly fixed standoff range; its <span class="math notranslate nohighlight">\(R\)</span> to the radar barely changes as the strike package presses in, so its J/S does not fall the way an SPJ's does. It also typically jams through the radar's side lobes rather than the main beam.</p></div>
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