Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Comp-B5, Computer Communications — National Exams, May 2015. Closed-book, 3 hours; seven questions of equal value (20% each); ANY FIVE constitute a complete exam (all seven answered below as a complete study resource).
Reference texts: Stallings, Data and Computer Communications, 10th ed. — sampling and aliasing (Ch.5, Q1), channel capacity/Shannon-Hartley (Ch.3, Q2), PCM and uniform quantization (Ch.5, Q3), Manchester/Differential Manchester line coding (Ch.5, Q4), LAN/network topologies (Ch.16, Q5), spread spectrum (Ch.9, Q6), and physical/link/network-layer terminology (Ch.3, 6, 9, 16, 17, Q7); Kurose & Ross, Computer Networking: A Top-Down Approach, 8th ed. — application-layer protocols and transport-layer terminology (Ch.1–3, Q7).
Check: assumes the "10 V full-scale" is the full input range (FSR) covered by the quantizer's $2^{12}$ codes, i.e. the quantizer maps $[0,\text{FSR}]$ (or an equivalent $10\text{ V}$ span) onto codes $0$ to $2^{12}-1$ — the paper does not state whether the range is unipolar ($0$–$10\text{ V}$) or bipolar ($-5$ to $+5\text{ V}$); the step-size arithmetic is identical either way.
Given.
Quantity
Value
Full-scale range, FSR
10 V
Code length
12 bits
Quantization
Uniform
Find. (a) normalized step size, (b) actual step size (V), (c) maximum quantized level (V), (d) actual voltage resolution.
Approach. With $n=12$ bits there are $L=2^{12}=4096$ quantization levels spanning the full-scale range; the normalized step is $1/L$ of full scale, and every other quantity follows by scaling that fraction by the $10\text{ V}$ range.
Number of quantization levels.
$$L = 2^{n} = 2^{12} = 4096$$
(a) Normalized step size (fraction of full scale per code):
$$\Delta_{\text{norm}} = \frac{1}{L} = \frac{1}{4096} = \boxed{2.4414\times10^{-4}}$$
(b) Actual step size in volts:
$$\Delta = \frac{\text{FSR}}{L} = \frac{10\text{ V}}{4096} = \boxed{2.4414\text{ mV}}$$
(c) Maximum quantized level in volts (the top code, $L-1$, one step below full scale):
$$V_{\max} = (L-1)\,\Delta = 4095 \times 2.4414\text{ mV} = \boxed{9.9976\text{ V}}$$
(d) Actual voltage resolution: the resolution — the smallest voltage change the 12-bit code can distinguish — is by definition the same quantity as the actual step size found in (b):
$$\text{resolution} = \Delta = \boxed{2.4414\text{ mV}}$$