# Lesson 21 Flashcards — Signal Conditioning

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<summary><strong>1. Name the three pitfalls signal conditioning addresses.</strong></summary>
<div class="card-answer"><p><strong>Aliasing</strong> (a frequency problem), <strong>clipping</strong> (signal too big), and <strong>wasted dynamic range</strong> (signal too small).</p></div>
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<summary><strong>2. Where must conditioning happen relative to the ADC?</strong></summary>
<div class="card-answer"><p><strong>Before</strong> it, always. Both aliasing and clipping destroy information at the moment of conversion; afterward is too late.</p></div>
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<summary><strong>3. How do you specify an anti-aliasing filter?</strong></summary>
<div class="card-answer"><p>A low-pass filter with <span class="math notranslate nohighlight">\(f_\text{cutoff}\)</span> just above the highest frequency you care about, then choose <span class="math notranslate nohighlight">\(f_s \geq 2f_\text{cutoff}\)</span> — with margin. A real first-order R-C still passes <span class="math notranslate nohighlight">\(0.707\)</span> at the cutoff, so <strong>oversample</strong> rather than sitting exactly at <span class="math notranslate nohighlight">\(2f_\text{cutoff}\)</span>.</p></div>
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<summary><strong>4. What does clipping do beyond losing the peaks?</strong></summary>
<div class="card-answer"><p>It <strong>creates new frequency content</strong> that was never in the original signal — so it corrupts the spectrum as well as the waveform.</p></div>
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<summary><strong>5. What is dynamic range?</strong></summary>
<div class="card-answer"><p>The span between a device's smallest and largest usable signal — for an ADC, <span class="math notranslate nohighlight">\(V_\text{max} - V_\text{min}\)</span>. Conditioning exists to make the transducer's dynamic range and the converter's line up.</p></div>
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<summary><strong>6. Why is wasted dynamic range a problem if nothing looks broken?</strong></summary>
<div class="card-answer"><p>Because every unused level is resolution you paid for. A ±2 V signal into a ±5 V, 8-bit ADC uses only 40% of the range — 102 of 256 levels, about <strong>6.7 effective bits</strong> out of the 8 you paid for.</p></div>
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<summary><strong>7. Write the signal conditioning equation.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(v_\text{out}(t) = K\,v_\text{in}(t) + B\)</span> — gain <span class="math notranslate nohighlight">\(K\)</span> scales the span, bias <span class="math notranslate nohighlight">\(B\)</span> shifts the center.</p></div>
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<summary><strong>8. How do you solve for K and B?</strong></summary>
<div class="card-answer"><p>Map <strong>min to min and max to max</strong>. That gives two equations in two unknowns; subtract them to eliminate <span class="math notranslate nohighlight">\(B\)</span> and solve for <span class="math notranslate nohighlight">\(K\)</span>, then back-substitute.</p></div>
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<summary><strong>9. A transducer outputs −2 mV to +8 mV into an ADC spanning −5 V to +10 V. Find K.</strong></summary>
<div class="card-answer"><p>Subtracting the endpoint equations: <span class="math notranslate nohighlight">\(-15\ \text{V} = K(-10\ \text{mV})\)</span>, so <span class="math notranslate nohighlight">\(K = 1500\)</span>. Then <span class="math notranslate nohighlight">\(B = -2\ \text{V}\)</span>.</p></div>
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<summary><strong>10. Must the gain K always be greater than 1?</strong></summary>
<div class="card-answer"><p>No. If the transducer's span is <em>larger</em> than the ADC's, <span class="math notranslate nohighlight">\(K < 1\)</span> and the conditioning attenuates. Assuming conditioning always amplifies is a common error.</p></div>
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<summary><strong>11. What is the last step of any conditioning calculation?</strong></summary>
<div class="card-answer"><p><strong>Check both endpoints</strong> in <span class="math notranslate nohighlight">\(v_\text{out} = Kv_\text{in}+B\)</span>. It takes seconds and catches sign errors.</p></div>
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