# Lesson 5 Flashcards

Click a question to reveal the answer.

<div data-flashcards data-deck="b1l5"></div>

<details>
<summary><strong>1. Write the Doppler shift formula.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(f_d = 2 v_r / \lambda\)</span>. The factor of 2 is the two-way path; <span class="math notranslate nohighlight">\(v_r\)</span> is the radial velocity.</p></div>
</details>

<details>
<summary><strong>2. At X-band, what Doppler shift does 300 m/s produce?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(f_d = 2(300)/0.03 = 20\)</span> kHz. At 30 m/s it is 2 kHz.</p></div>
</details>

<details>
<summary><strong>3. Why can't a radar reject clutter by amplitude?</strong></summary>
<div class="card-answer"><p>Ground clutter has an enormous RCS — far stronger than the target — so the aircraft return is buried on amplitude alone. Clutter must be rejected by Doppler instead.</p></div>
</details>

<details>
<summary><strong>4. Where does clutter sit in Doppler, and why?</strong></summary>
<div class="card-answer"><p>At <span class="math notranslate nohighlight">\(f_d \approx 0\)</span>, because stationary terrain has no radial velocity. Movers sit away from zero Doppler, which is what lets them be separated.</p></div>
</details>

<details>
<summary><strong>5. Write the single delay-line MTI canceller.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(y[n] = x[n] - x[n-1]\)</span>. Stationary returns cancel; movers, whose phase advanced between pulses, survive.</p></div>
</details>

<details>
<summary><strong>6. What is the MTI frequency response?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(|H(f)| = 2|\sin(\pi f T)|\)</span> with <span class="math notranslate nohighlight">\(T = \text{PRI}\)</span>. It has notches at <span class="math notranslate nohighlight">\(f = 0, \text{PRF}, 2\,\text{PRF}, \dots\)</span></p></div>
</details>

<details>
<summary><strong>7. What are blind speeds?</strong></summary>
<div class="card-answer"><p>Target velocities whose Doppler lands on an MTI notch, so the target is cancelled with the clutter: <span class="math notranslate nohighlight">\(v_{\text{blind},n} = n\lambda\,\text{PRF}/2\)</span>.</p></div>
</details>

<details>
<summary><strong>8. How does PRF move the blind speeds?</strong></summary>
<div class="card-answer"><p>Raising PRF raises the blind speeds. At X-band, 1 kHz gives a first blind speed of 15 m/s; 10 kHz pushes it to 150 m/s, usually out of the clutter band — at the cost of range ambiguity.</p></div>
</details>

<details>
<summary><strong>9. What is pulse-Doppler processing?</strong></summary>
<div class="card-answer"><p>Collect <span class="math notranslate nohighlight">\(N\)</span> pulses in a coherent dwell and take an <span class="math notranslate nohighlight">\(N\)</span>-point FFT across pulses at each range bin, giving a Doppler spectrum per range bin with bin width <span class="math notranslate nohighlight">\(\text{PRF}/N\)</span>.</p></div>
</details>

<details>
<summary><strong>10. What does a range-Doppler map look like?</strong></summary>
<div class="card-answer"><p>A 2D image of range vs. Doppler. Clutter forms a vertical ridge at zero Doppler across all ranges; movers land off the ridge in their own cells, easy to threshold.</p></div>
</details>

<details>
<summary><strong>11. How could a jammer make a target hide on the R-D map?</strong></summary>
<div class="card-answer"><p>Match the platform's own ground-clutter Doppler line so the false return sits on the zero-Doppler ridge and is cancelled with the clutter. EP counters include geometry checks and inverse-cancellation logic.</p></div>
</details>

<details>
<summary><strong>12. What SNR benefit does coherent integration give?</strong></summary>
<div class="card-answer"><p>Integrating <span class="math notranslate nohighlight">\(N\)</span> pulses coherently gives an SNR gain proportional to <span class="math notranslate nohighlight">\(N\)</span>, while finer Doppler resolution comes from the larger dwell.</p></div>
</details>
