Question

Increasing the resolution of an ideal N-bit ADC at constant reference voltage will:

• A. Increase quantization noise power linearly
• B. Decrease quantization noise power linearly
• C. Not affect quantization noise power
• D. Decrease quantization noise power exponentially

Hint:

Each additional bit added to an ADC cuts its quantization step size in half. Since noise power is proportional to the square of the step size (\(\Delta^2/12\)), every extra bit reduces the quantization noise power by a factor of 4 (\(6\text{ dB}\) SNR improvement).

Answer 1

The Correct Option is D

Concept: Quantization error is an inherent limitation when mapping a continuous analog signal into discrete digital levels. For an ideal \(n\)-bit Analog-to-Digital Converter (ADC) with a reference voltage range \(V_{\text{ref}}\), the step size (also known as the voltage resolution width or Least Significant Bit value, \(\Delta\)) is defined as: \[ \Delta = \frac{V_{\text{ref}}}{2^n} \] Assuming the input signal varies actively across multiple quantization steps, the resulting quantization error can be modeled as a uniform random variable distributed evenly between \(\pm \frac{\Delta}{2}\). The mean-squared value of this error represents the average theoretical quantization noise power (\(P_q\)), which is calculated as: \[ P_q = \frac{\Delta^2}{12} \]

Step 1: Finding the mathematical relationship between noise power and bit count.
Let us substitute our expression for the step size \(\Delta\) directly into the quantization noise power equation: \[ P_q = \frac{1}{12} \left( \frac{V_{\text{ref}}}{2^n} \right)^2 \] Expanding the squared terms inside the brackets: \[ P_q = \frac{1}{12} \cdot \frac{V_{\text{ref}}^2}{(2^n)^2} \] Using index laws, we can rewrite the denominator term \((2^n)^2\) as \((2^2)^n = 4^n\): \[ P_q = \frac{V_{\text{ref}}^2}{12 \cdot 4^n} \] To see the mathematical relationship clearly, we can pull the variable \(n\) out as a negative exponent base: \[ P_q = \left( \frac{V_{\text{ref}}^2}{12} \right) \cdot 4^{-n} = \left( \frac{V_{\text{ref}}^2}{12} \right) \cdot 2^{-2n} \]

Step 2: Analyzing the impact of changing the resolution.
In this expression, the reference voltage \(V_{\text{ref}}\) remains constant. Therefore, the term \(\frac{V_{\text{ref}}^2}{12}\) behaves as a fixed constant multiplier. The quantization noise power \(P_q\) depends entirely on the exponential term: \[ P_q \propto 4^{-n} \quad \text{or} \quad P_q \propto \left(\frac{1}{4}\right)^n \] Because the resolution variable \(n\) is located within a negative exponent, any linear increase in the number of bits \(n\) causes the overall quantization noise power to drop exponentially. Specifically, adding a single bit halves the step size \(\Delta\), which reduces the quantization noise power to one-quarter (\(\frac{1}{4}\)) of its previous value. This corresponds perfectly to Option (D).