# Lesson 36 Flashcards

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<summary><strong>1. What is SEAD, and what two kinds of effect can it have?</strong></summary>
<div class="card-answer"><p>Suppression of enemy air defenses: activity that neutralizes, destroys, or temporarily degrades surface-based air defenses. <strong>Destruction</strong> removes a site for the rest of the mission; <strong>suppression</strong> degrades it only while the suppression is applied.</p></div>
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<summary><strong>2. Lethal versus non-lethal SEAD — and can an anti-radiation missile suppress rather than destroy?</strong></summary>
<div class="card-answer"><p>Lethal: anti-radiation missiles (HARM) homing on the radar's emission, or weapons on a located site. Non-lethal: electromagnetic attack — noise jamming, including a standoff jammer. Yes: a HARM in the air gives operators reason to shut the radar down, and a silent radar is suppressed, not destroyed — it comes back when the threat passes.</p></div>
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<summary><strong>3. What is the suppression window, and why does it couple SEAD to the route?</strong></summary>
<div class="card-answer"><p>Suppression lasts only while the suppressor is on station, so it opens and closes in time. A package gains only for the part of its time inside the site's lethal zone that falls inside the window; a late package, or a long path through the zone, flies the rest exposed. Fixes: leave earlier, shorten the path through the zone, or lengthen the window.</p></div>
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<summary><strong>4. How does the beam/notch maneuver hide an aircraft?</strong></summary>
<div class="card-answer"><p>The radial velocity is <span class="math notranslate nohighlight">\(v_r = v\cos\alpha\)</span>. Fly perpendicular to the line of sight (<span class="math notranslate nohighlight">\(\alpha = 90^\circ\)</span>) and <span class="math notranslate nohighlight">\(v_r \to 0\)</span>, so <span class="math notranslate nohighlight">\(f_d = 2v_r/\lambda \to 0\)</span>: the echo lands in L5's MTI notch with the ground clutter and is canceled with it. The notch is narrow, so the heading must be held within a few degrees of the beam.</p></div>
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<summary><strong>5. What does beaming a radar cost the B-21 on its X-band pattern?</strong></summary>
<div class="card-answer"><p>The pattern's sharpest spike sits exactly at the beam: about −3 dBsm against −15 dBsm nose-on, 12 dB brighter. With <span class="math notranslate nohighlight">\(R \propto \sigma^{1/4}\)</span> that doubles the range at which a radar mode that does not rely on Doppler can see you. The notch wants exactly beam, and the signature punishes exactly beam.</p></div>
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<summary><strong>6. When does turning cold defeat a missile?</strong></summary>
<div class="card-answer"><p>Inside the MEZ but outside the LEZ (L32): with your tail on the site, the missile flies a stern chase and runs out of time of flight. Inside the LEZ, <span class="math notranslate nohighlight">\(R_{\text{LEZ}} = (v_m - v_t)t_{\max}\)</span>, it catches even a target running straight away. Dragging holds the cold heading to spend the missile's energy; the endgame belongs to chaff (L18) and flares (L24).</p></div>
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<summary><strong>7. What two things change when a jammer moves off the target to a standoff orbit?</strong></summary>
<div class="card-answer"><p><strong>Two ranges:</strong> the echo still travels the target's range <span class="math notranslate nohighlight">\(R_t\)</span> both ways, but the jamming travels the jammer's range <span class="math notranslate nohighlight">\(R_j\)</span> one way, and <span class="math notranslate nohighlight">\(R_j\)</span> stays fixed while <span class="math notranslate nohighlight">\(R_t\)</span> closes. <strong>Two gains:</strong> the radar hears the target through its mainbeam <span class="math notranslate nohighlight">\(G_r\)</span> but the jammer through the gain it points toward the jammer, <span class="math notranslate nohighlight">\(G_r'\)</span>.</p></div>
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<summary><strong>8. Write the standoff jam-to-signal ratio.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\dfrac{J}{S} = \dfrac{4\pi P_jG_jR_t^4}{P_tG_t\sigma R_j^2}\cdot\dfrac{G_r'}{G_r}\cdot\dfrac{B_r}{B_j}\)</span> — the echo's two-way loss <span class="math notranslate nohighlight">\(R_t^4\)</span> over the jammer's one-way loss <span class="math notranslate nohighlight">\(R_j^2\)</span>, times the gain ratio and L17's bandwidth factor.</p></div>
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<summary><strong>9. How does the standoff equation reduce to L17's self-protect equation?</strong></summary>
<div class="card-answer"><p>Put the jammer on the target: <span class="math notranslate nohighlight">\(R_j = R_t = R\)</span> and <span class="math notranslate nohighlight">\(G_r' = G_r\)</span>. Then <span class="math notranslate nohighlight">\(R_t^4/R_j^2 = R^2\)</span> and the gain ratio is 1, leaving <span class="math notranslate nohighlight">\(J/S = 4\pi P_jG_jR^2 B_r/(P_tG_t\sigma B_j)\)</span>. With L17's Quick Problem numbers both forms give +14.5 dB at 30 km and a 17.8 km burn-through.</p></div>
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<summary><strong>10. Write the standoff burn-through range. Why a fourth root?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R_{bt} = \left[\dfrac{10\,P_tG_t\sigma R_j^2 B_j}{4\pi P_jG_j(G_r'/G_r)B_r}\right]^{1/4}\)</span>, the target range where <span class="math notranslate nohighlight">\(J/S\)</span> falls to 10 (+10 dB). A fourth root because only the echo is two-way; the jammer's <span class="math notranslate nohighlight">\(R_j^2\)</span> is a constant, fixed by the orbit.</p></div>
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<summary><strong>11. When is a standoff jammer in the radar's mainlobe?</strong></summary>
<div class="card-answer"><p>When it sits on the target's azimuth, within half a beamwidth — on the jammer's side of the radar. Then <span class="math notranslate nohighlight">\(G_r'/G_r = 1\)</span> (0 dB). Anywhere else the jamming enters through a sidelobe and <span class="math notranslate nohighlight">\(G_r'/G_r\)</span> is the sidelobe level, typically −25 to −35 dB. A target on the far side of the radar from the SOJ is on the same line but not in the mainlobe.</p></div>
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<summary><strong>12. What is the jammed detection contour, and what is the strobe?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R(\phi) = \min(R_{det}, R_{bt}(\phi))\)</span>: the sidelobe burn-through range around most of the circle, and a deep, narrow notch one beamwidth wide on the jammer's azimuth, where it falls to the mainlobe burn-through range. That notch is the strobe — on the radar's display, a bright radial line of jamming pointing at the jammer.</p></div>
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<summary><strong>13. What is the protected corridor, and how wide is it?</strong></summary>
<div class="card-answer"><p>The part of the strobe between the mainlobe and sidelobe burn-through ranges: a striker there is hidden, while a few degrees to either side the same radar sees it. Its width at range <span class="math notranslate nohighlight">\(R\)</span> is about <span class="math notranslate nohighlight">\(R\,\theta_B\)</span> — for a 5° beam, about 2.6 km at 30 km. A corridor, not a sky.</p></div>
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<summary><strong>14. How does burn-through scale with the jammer's standoff range?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R_{bt} \propto R_j^{1/2}\)</span>: the one-way <span class="math notranslate nohighlight">\(R_j^2\)</span> inside a fourth root. Four times the standoff only doubles burn-through, so backing a jammer off to clear the missile zones is cheap — and pulling it in buys little.</p></div>
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<summary><strong>15. How does burn-through scale with the radar's sidelobe level?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R_{bt} \propto (G_r'/G_r)^{-1/4}\)</span>: every 10 dB of sidelobe is a factor of <span class="math notranslate nohighlight">\(10^{10/40} = 1.78\)</span> in burn-through range, and 30 dB is 5.6. Lower sidelobes push the radar's burn-through out; flying the jammer's azimuth wins that whole factor back for free.</p></div>
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<summary><strong>16. What EP defeats sidelobe jamming, and why is the mainlobe position prized?</strong></summary>
<div class="card-answer"><p>L15's sidelobe cancelers (an auxiliary antenna nulls the jammer), sidelobe blankers, and ultra-low sidelobes all attack the sidelobe path. None of them can remove a jammer sitting in the beam the radar is looking through — so the mainlobe is the one door EP cannot close, and the route rides the strobe.</p></div>
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<summary><strong>17. What are the four deconfliction rules for one standoff jammer?</strong></summary>
<div class="card-answer"><p>(1) Orbit outside every MEZ plus a margin — the SOJ radiates, so it will be found. (2) One jammer beam, one victim: a directional SOJ holds its transmit mainlobe on one radar, and the others are (to first order) unjammed. (3) The route rides the victim's strobe. (4) The other sites' zones are crossed inside the SEAD window, or avoided.</p></div>
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<summary><strong>18. Standoff jamming has two different "mainlobes." Which is which?</strong></summary>
<div class="card-answer"><p>The <strong>radar's receive</strong> lobes, <span class="math notranslate nohighlight">\(G_r'/G_r\)</span>, decide <em>where</em> around a jammed radar the jamming bites hardest: the mainlobe on the jammer's azimuth (the strobe), a sidelobe elsewhere. The <strong>jammer's transmit</strong> beam, <span class="math notranslate nohighlight">\(G_j\)</span>, decides <em>which</em> radar is jammed at all: a directional SOJ aims its ERP at one victim at a time. A radar outside that beam still hears the jammer faintly through its sidelobes; the Final Project treats it as unjammed, a stated simplification.</p></div>
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