Let's solve the problem step by step. --- ### **Given Data** - **Thermocouple voltage range (input):** From 2.585 mV (at 50°C) to 3.649 mV (at 70°C) - **A/D Converter:** - Resolution: 12 bits - Input range: ±5 V (i.e., total range = 10 V) - **Tasks:** 1. Estimate percent quantization error. 2. Compute amplifier gain needed to reduce quantization error to 5% or less. 3. Given SNR = 40 dB, compute noise after amplification and discuss implications. --- ## **a. Percent Quantization Error in the Digital Value** ### **Step 1: Find the input voltage range** \[ \text{Signal range} = 3.649~\text{mV} - 2.585~\text{mV} = 1.064~\text{mV} \] ### **Step 2: Find A/D converter's voltage step size (quantization level)** - **Number of steps:** \(2^{12} = 4096\) - **Total A/D input range:** 10 V (\(\pm 5\) V) - **Step size (LSB):** \[ \text{LSB} = \frac{10~\text{V}}{4096} = 0.002441~\text{V} = 2.441~\text{mV} \] ### **Step 3: Quantization error as a percentage of signal range** - **Maximum quantization error:** \(\pm \dfrac{\text{LSB}}{2} = \pm 1.2205~\text{mV}\) - **Signal range:** \(1.064~\text{mV}\) \[ \text{Percent quantization error} = \frac{\text{Max quantization error}}{\text{Signal range}} \times 100\% = \frac{1.2205~\text{mV}}{1.064~\text{mV}} \times 100\% = 114.7\% \] > **Interpretation:** The quantization error is **greater than the entire signal range**—the A/D converter is not able to resolve the temperature changes at all with this direct input. --- ## **b. Amplifier Gain to Reduce Quantization Error to 5% or Less** We want the **quantization error** to be **≤ 5% of the signal range** after amplification. ### **Step 1: Define variables** Let \( G \) be the amplifier gain. - **Amplified signal range:** \( G \times 1.064~\text{mV} \) - **A/D LSB:** \(2.441~\text{mV}\) - **Quantization error after amplification:** \( \pm 1.2205~\text{mV} \) - **Requirement:** \[ \frac{1.2205}{G \times 1.064} \leq 0.05 \] ### **Step 2: Solve for G** \[ \frac{1.2205}{G \times 1.064} \leq 0.05 \implies G \geq \frac{1.2205}{0.05 \times 1.064} \implies G \geq \frac{1.2205}{0.0532} \implies G \geq 22.95 \] ### **Step 3: Check if A/D will saturate** - **Amplified high voltage:** \(3.649~\text{mV} \times 22.95 = 83.8~\text{mV} \) - **Amplified low voltage:** \(2.585~\text{mV} \times 22.95 = 59.4~\text{mV} \) - **Amplified signal range:** \(83.8 - 59.4 = 24.4~\text{mV}\), which is much less than 10 V, so it's safe. > **Final Answer for (b):** \[ \boxed{G \geq 22.95} \] So, an amplifier gain of at least **22.95** is required to bring the quantization error to 5% or less of the signal range. --- ## **c. SNR, Amplification, and Noise** ### **Step 1: SNR and Noise Definition** Given: - **SNR:** 40 dB - **SNR (dB):** \( 20 \log_{10} \left( \frac{V_\text{signal}}{V_\text{noise}} \right) \) ### **Step 2: Find RMS Noise Voltage (before amplification)** - **Peak-to-peak signal:** \(1.064~\text{mV}\) (assuming this is peak-to-peak, for RMS divide by \(2\sqrt{2}\) or use directly for SNR calculation) - **Assume SNR refers to RMS values.** \[ 40 = 20 \log_{10} \left( \frac{V_\text{signal}}{V_\text{noise}} \right) \implies \frac{V_\text{signal}}{V_\text{noise}} = 10^{2} = 100 \implies V_\text{noise} = \frac{V_\text{signal}}{100} \] Assume \(V_\text{signal} = 1.064~\text{mV (RMS)}\): \[ V_\text{noise} = \frac{1.064~\text{mV}}{100} = 0.01064~\text{mV} = 10.64~\mu\text{V} \] ### **Step 3: Noise after Amplification** Amplifier gain: \(G = 22.95\) - **Signal after amp:** \(G \times V_\text{signal}\) - **Noise after amp:** \(G \times V_\text{noise}\) \[ V_\text{noise,amp} = 22.95 \times 0.01064~\text{mV} = 0.244~\text{mV} = 244~\mu\text{V} \] ### **Step 4: Discuss implications** - **A/D LSB:** \(2.441~\text{mV}\) - **Amplified noise:** \(0.244~\text{mV}\) - **Ratio:** \(0.244 / 2.441 \approx 0.1\) (i.e., noise is about 10% of one LSB). > **Implication:** After amplification, the quantization error is reduced to 5% of the signal range, but the **amplified noise is about 10% of one LSB**. This means the noise could cause the A/D output to fluctuate by 1 count, so the effective resolution is limited by the noise, not the quantization. In practice, **further increasing gain won't help unless the noise is reduced**. --- ## **Final Answers (Summary)** **a.** The percent quantization error without amplification is \[ \boxed{114.7\%} \] (The error is greater than the signal range.) **b.** The minimum amplifier gain required to reduce quantization error to 5% or less is \[ \boxed{22.95} \] **c.** After amplification, the noise magnitude is \[ \boxed{0.244~\text{mV}~(244~\mu\text{V})} \] This is about 10% of one A/D LSB, so the effective accuracy is limited by the noise, not quantization. **Further amplification won't improve digital resolution unless the noise is reduced.**