Let's solve the problem step by step. --- ### **Given Data** - **Thermocouple voltage range:** 2.585 mV to 3.649 mV (for 50°C to 70°C) - **Change in voltage (ΔV):** \(3.649 - 2.585 = 1.064 \text{ mV}\) - **A/D converter:** 12-bit, ±5 V (i.e., 10 V total range) - **SNR (signal-to-noise ratio):** 40 dB --- ## **a. Percent Quantization Error (without amplification)** ### **Step 1: Calculate A/D converter resolution** Total number of possible counts for a 12-bit converter: \[ 2^{12} = 4096 \text{ levels} \] Voltage range of A/D converter: \[ V_{\text{range}} = 10~\text{V} \text{ (from -5V to +5V)} \] Smallest voltage change (1 LSB): \[ \text{Resolution (LSB)} = \frac{V_{\text{range}}}{2^{12}} = \frac{10~\text{V}}{4096} \approx 0.00244~\text{V} = 2.44~\text{mV} \] ### **Step 2: Compare signal variation to 1 LSB** The thermocouple signal varies by: \[ \Delta V = 1.064~\text{mV} \] But the A/D converter's 1 LSB is: \[ 2.44~\text{mV} \] So, the entire signal range is **less than 1 LSB**. This means the A/D converter cannot resolve any change in the thermocouple voltage over this range—**all values will map to the same digital code**! ### **Step 3: Calculate percent quantization error** Quantization error is usually defined as the maximum possible error relative to the signal span. For an n-bit converter, the maximum quantization error is ±0.5 LSB. - Maximum quantization error: \(0.5 \times 2.44~\text{mV} = 1.22~\text{mV}\) - Signal range: \(1.064~\text{mV}\) \[ \text{Percent quantization error} = \frac{1.22~\text{mV}}{1.064~\text{mV}} \times 100\% \approx 115\% \] **This means the error is greater than the signal itself!** --- ### **Final Answer (a):** > **Percent quantization error is approximately 115% (i.e., the quantization step is larger than the entire signal range).** --- ## **b. Required Amplifier Gain for Quantization Error ≤ 5%** ### **Step 1: Find the maximum allowable quantization error** Allowable error = 5% of signal range: \[ \text{Allowed error} = 0.05 \times 1.064~\text{mV} = 0.0532~\text{mV} \] Maximum quantization error after amplification should be ≤ 0.0532 mV. ### **Step 2: LSB after amplification** After amplification by gain \(A\): - Signal range: \(1.064 \times A\) mV - Should fit within 10 V (\(A \leq \frac{10}{1.064 \times 10^{-3}}\)) 1 LSB of A/D converter: 2.44 mV (remains the same) But after amplification, 1 LSB (in terms of input mV) corresponds to \(2.44/A\) mV \[ 0.5~\text{LSB} = 1.22~\text{mV (after amp)} = \frac{1.22}{A}~\text{mV (in input range)} \] Set this less than or equal to 0.0532 mV: \[ \frac{1.22}{A} \leq 0.0532 \] \[ A \geq \frac{1.22}{0.0532} \approx 22.94 \] So, **minimum gain required is approximately 23**. ### **Step 3: Check that amplified signal doesn't exceed A/D input range** Amplified signal range: \[ 1.064~\text{mV} \times 23 = 24.47~\text{mV} \] This is much less than 10 V, so no risk of overdriving the A/D. --- ### **Final Answer (b):** > **Amplifier gain required: at least 23 (rounded up to next integer, so Gain = 23).** --- ## **c. Noise after Amplification (SNR = 40 dB)** ### **Step 1: Find noise level before amplification** SNR (in dB): \[ \text{SNR} = 20 \log_{10}\left(\frac{V_{signal}}{V_{noise}}\right) \] \[ 40 = 20 \log_{10}\left(\frac{1.064~\text{mV}}{V_{noise}}\right) \] \[ 2 = \log_{10}\left(\frac{1.064}{V_{noise}}\right) \] \[ 10^2 = \frac{1.064}{V_{noise}} \] \[ V_{noise} = \frac{1.064}{100} = 0.01064~\text{mV} = 10.64~\mu\text{V} \] ### **Step 2: Noise after amplification** \[ V_{noise,amp} = V_{noise} \times \text{Gain} = 10.64~\mu\text{V} \times 23 = 244.7~\mu\text{V} = 0.245~\text{mV} \] ### **Step 3: Discussion** - After amplification, noise is 0.245 mV. - The **quantization step** (after amplification) is still 2.44 mV. - The **noise is about 10% of 1 LSB**, and about 23% of the maximum quantization error (0.5 LSB = 1.22 mV). - In this setup, the noise level is still **less than the quantization step**, but it is not negligible. Further increasing the gain to use more of the A/D range would amplify the noise proportionally. #### **Conclusion** - Increasing gain to reduce quantization error also increases the noise presented to the A/D, potentially making noise a significant part of the signal. There's a tradeoff between quantization error and noise: improving one can worsen the other. --- ## **Summary Table** | Part | Calculation | Final Answer | |------|-------------|-------------| | a | Quantization error without amplification | **115%** (quantization step is larger than signal range) | | b | Required gain for ≤5% quant error | **Gain = 23** | | c | Noise after amplification | **0.245 mV** (noise becomes a significant fraction of 1 LSB; further increasing gain will further amplify noise) | --- **If you need the equations or reasoning for each step, let me know!**