# Lesson 35 Flashcards — Pulse-Doppler Radar

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<summary><strong>1. Define pulse width, PRF, and PRI.</strong></summary>
<div class="card-answer"><p><strong>Pulse width</strong> <span class="math notranslate nohighlight">\(\tau\)</span> is how long the transmitter is on. <strong>PRI</strong> is the time from the start of one pulse to the start of the next. <strong>PRF</strong> is pulses per second, with <span class="math notranslate nohighlight">\(PRI = 1/PRF\)</span>.</p></div>
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<summary><strong>2. How does a radar measure range?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R = c\,\Delta t/2\)</span> from the round-trip time. The factor of 2 is because the pulse travels out and back.</p></div>
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<summary><strong>3. Give the range resolution.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(\Delta R = c\tau/2\)</span> — set by <strong>pulse width</strong>. Two targets closer than <span class="math notranslate nohighlight">\(\Delta R\)</span> merge into one return.</p></div>
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<summary><strong>4. A 1 µs pulse gives what range resolution?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\((3\times10^8)(1\times10^{-6})/2 = 150\)</span> m. Two aircraft within 150 m appear as a single target.</p></div>
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<summary><strong>5. Give the maximum unambiguous range.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R_\text{unamb} = c\,PRI/2\)</span> — set by <strong>PRF</strong>. The radar must hear the echo before it sends the next pulse. It is the equation sheet's <em>Unambig Range</em> row.</p></div>
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<summary><strong>6. What happens to a target beyond the unambiguous range?</strong></summary>
<div class="card-answer"><p>Its echo arrives after the next pulse goes out, so the radar times it from the wrong pulse and reports a <strong>ghost</strong> at the true range minus <span class="math notranslate nohighlight">\(R_\text{unamb}\)</span>. With <span class="math notranslate nohighlight">\(R_\text{unamb} = 15\)</span> km, a target at 18 km is reported at 3 km.</p></div>
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<summary><strong>7. What does each waveform setting cost?</strong></summary>
<div class="card-answer"><p>A <strong>short pulse</strong> gives fine resolution but carries less energy, so detection range drops. A <strong>low PRF</strong> gives a long unambiguous range but puts fewer pulses on the target each second. They are two separate trades, and real radars change both by mode.</p></div>
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<summary><strong>8. Write the Doppler equation.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(f_\text{return} = f_\text{original}(1 \pm 2v\cos\theta/c)\)</span> — <strong>plus for a closing target, minus for an opening one</strong>. <span class="math notranslate nohighlight">\(v\)</span> is the target's speed in m/s, and <span class="math notranslate nohighlight">\(\theta\)</span> is the acute angle between its velocity and the radar's line of sight.</p></div>
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<summary><strong>9. Why does the factor of 2 appear in the Doppler equation?</strong></summary>
<div class="card-answer"><p>For the same reason as in range: the path is <strong>two-way</strong>.</p></div>
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<summary><strong>10. What Doppler shift does a target flying across the beam produce?</strong></summary>
<div class="card-answer"><p><strong>None.</strong> At <span class="math notranslate nohighlight">\(\theta = 90^\circ\)</span>, <span class="math notranslate nohighlight">\(\cos\theta = 0\)</span> and there is no shift at all.</p></div>
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<summary><strong>11. What is the Doppler shift, and how is a return frequency reported?</strong></summary>
<div class="card-answer"><p>The Doppler shift is the change in frequency, <span class="math notranslate nohighlight">\(\Delta f = 2v\cos\theta\,f_\text{original}/c\)</span>, in Hz. Add it for a closing target, subtract it for an opening one, and report <span class="math notranslate nohighlight">\(f_\text{return}\)</span> to the <strong>nearest hertz</strong> — the shift is only a few kHz on a GHz carrier.</p></div>
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<summary><strong>12. How do you find a target's speed and direction from its return frequency?</strong></summary>
<div class="card-answer"><p>A return <strong>above</strong> <span class="math notranslate nohighlight">\(f_\text{original}\)</span> means the target is closing; <strong>below</strong> means opening. The speed is <span class="math notranslate nohighlight">\(v = c\,|\Delta f|/(2f_\text{original}\cos\theta)\)</span>. The answer is a speed and a direction, such as 125 m/s, opening.</p></div>
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<summary><strong>13. A target's speed is given in km/h. What do you do first?</strong></summary>
<div class="card-answer"><p>Divide by <strong>3.6</strong> to get m/s, because 1 km/h is 1000 m in 3600 s. A target at 1440 km/h is moving at 400 m/s.</p></div>
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