# Lesson 37 Flashcards — Electronic Warfare and Jamming

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<summary><strong>1. Define ES, EP, and EA.</strong></summary>
<div class="card-answer"><p><strong>ES</strong> (Electromagnetic Support) senses and characterizes emissions. <strong>EP</strong> (Electromagnetic Protection) makes our own systems resistant. <strong>EA</strong> (Electromagnetic Attack) jams, deceives, or destroys threat sensors.</p></div>
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<summary><strong>2. Why must ES come before EA?</strong></summary>
<div class="card-answer"><p>You cannot jam intelligently a signal you have not first intercepted and characterized.</p></div>
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<summary><strong>3. What does a jammer actually do to a receiver?</strong></summary>
<div class="card-answer"><p>It does not break it. It <strong>adds its own received power to the noise</strong>, so the receiver works only when <span class="math notranslate nohighlight">\(S/(N + J) \ge SNR_{\min}\)</span> — the same <span class="math notranslate nohighlight">\(SNR_{\min}\)</span> as with the jammer off. The jamming dwarfs the thermal noise <span class="math notranslate nohighlight">\(N\)</span>, so in practice the test is <span class="math notranslate nohighlight">\(S/J \ge SNR_{\min}\)</span>.</p></div>
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<summary><strong>4. Write the jamming SNR of a communications link in full.</strong></summary>
<div class="card-answer"><p>The ratio of two Friis calculations into the same receiver: <span class="math notranslate nohighlight">\(SNR_\text{comm} = \dfrac{P_{T,\text{sig}}G_{T,\text{sig}}G_R\lambda^2/(4\pi R_\text{sig})^2}{P_{T,\text{jam}}G_{T,\text{jam}}G_R\lambda^2/(4\pi R_\text{jam})^2}\)</span>.</p></div>
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<summary><strong>5. What cancels when you simplify the jamming SNR, and what is left?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(G_R\)</span>, <span class="math notranslate nohighlight">\(\lambda^2\)</span>, and <span class="math notranslate nohighlight">\((4\pi)^2\)</span> all cancel, leaving the equation sheet's row, <span class="math notranslate nohighlight">\(SNR_\text{comm} = P_{T,\text{sig}}G_{T,\text{sig}}R_\text{jam}^2/(P_{T,\text{jam}}G_{T,\text{jam}}R_\text{sig}^2)\)</span>. The reading writes <span class="math notranslate nohighlight">\(R_\text{txrx}\)</span> for <span class="math notranslate nohighlight">\(R_\text{sig}\)</span>.</p></div>
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<summary><strong>6. What does that cancellation assume?</strong></summary>
<div class="card-answer"><p>That the receive antenna has the <strong>same gain toward both</strong> the transmitter and the jammer, and that the jammer is <strong>on the victim's frequency</strong>, inside its bandwidth. A receive antenna that nulls the jammer's bearing breaks it. What is left to fight over is power, gain, and above all the <strong>ratio of ranges, squared</strong>.</p></div>
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<summary><strong>7. In a jamming problem, where is every range measured to?</strong></summary>
<div class="card-answer"><p>The <strong>victim receiver</strong>. <span class="math notranslate nohighlight">\(R_\text{sig}\)</span> runs from the transmitter to it and <span class="math notranslate nohighlight">\(R_\text{jam}\)</span> from the jammer to it — never from the jammer to the transmitter.</p></div>
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<summary><strong>8. Why does a stand-in jammer beat a stand-off jammer of the same power?</strong></summary>
<div class="card-answer"><p>Because <span class="math notranslate nohighlight">\(SNR_\text{comm} \propto R_\text{jam}^2\)</span>. A jammer twice as far away is <strong>four times</strong> less effective. Geometry beats power.</p></div>
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<summary><strong>9. Write the SNR of a radar whose target carries the jammer.</strong></summary>
<div class="card-answer"><p>The echo (radar range equation) over the jamming (Friis into the radar's own antenna): <span class="math notranslate nohighlight">\(SNR_\text{rad} = \dfrac{P_T\,G_T\,RCS}{P_{T,\text{jam}}\,G_{T,\text{jam}}\,4\pi R^2}\)</span>. It assumes the jammer is <strong>at the target's range</strong>: <strong>self-screening</strong> (the target carries it) or an <strong>escort</strong> flying with the target, so one <span class="math notranslate nohighlight">\(R\)</span> serves both paths. True stand-off radar jamming needs two ranges, which is outside this course.</p></div>
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<summary><strong>10. Why does jamming favor the jammer against radar specifically?</strong></summary>
<div class="card-answer"><p>The jammer's signal travels <strong>one way</strong> (<span class="math notranslate nohighlight">\(1/R^2\)</span>) while the radar's echo travels <strong>two</strong> (<span class="math notranslate nohighlight">\(1/R^4\)</span>). So <span class="math notranslate nohighlight">\(SNR_\text{rad} \propto 1/R^2\)</span>: far out the jamming wins, and the echo gains on it only as the target closes.</p></div>
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<summary><strong>11. What is the burn-through range?</strong></summary>
<div class="card-answer"><p>The range at which <span class="math notranslate nohighlight">\(SNR_\text{rad}\)</span> reaches the radar's <span class="math notranslate nohighlight">\(SNR_{\min}\)</span>: <span class="math notranslate nohighlight">\(R_{BT} = \sqrt{\dfrac{P_T\,G_T\,RCS}{P_{T,\text{jam}}\,G_{T,\text{jam}}\,4\pi\,SNR_{\min}}}\)</span>, in meters because <span class="math notranslate nohighlight">\(RCS\)</span> is in m². Inside it the radar sees through the jamming.</p></div>
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<summary><strong>12. How does jamming change a radar's detection range?</strong></summary>
<div class="card-answer"><p>It does not move the radar equation's detection range. It adds a second, tighter limit, the <strong>burn-through range</strong>: a jammed radar detects only inside it.</p></div>
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<summary><strong>13. A burn-through range comes out smaller than the aircraft's altitude. What does that mean?</strong></summary>
<div class="card-answer"><p>The aircraft is never closer to the radar than its own altitude (30,000 ft is <span class="math notranslate nohighlight">\(30{,}000 \times 0.305 = 9{,}150\)</span> m), so that range is never reached: the radar <strong>never burns through</strong>. A small burn-through range is a jammer that works, not a mistake.</p></div>
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<summary><strong>14. Name four low-probability-of-intercept techniques.</strong></summary>
<div class="card-answer"><p><strong>Spread spectrum</strong> (power density below the noise floor), <strong>frequency hopping</strong>, <strong>power management</strong> (transmit the least that closes the link), and <strong>narrow beams</strong>.</p></div>
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<summary><strong>15. Why is brute transmit power usually the worst answer to jamming?</strong></summary>
<div class="card-answer"><p>Because <span class="math notranslate nohighlight">\(SNR_\text{comm}\)</span> is only <em>linear</em> in <span class="math notranslate nohighlight">\(P_{T,\text{sig}}\)</span> but goes as the <em>square</em> of range. Shortening the link, or using EP so the jammer cannot concentrate power in your band, buys far more.</p></div>
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<summary><strong>16. Name four electronic-attack countermeasures beyond noise jamming.</strong></summary>
<div class="card-answer"><p><strong>Chaff</strong> (false returns), <strong>flares</strong> (IR decoys), <strong>towed and active decoys</strong>, and <strong>anti-radiation missiles</strong> that home on the radar's own emissions.</p></div>
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