Question 3 of 5: Short-Answer Justify Questions

Question 3: Short-Answer Justify Questions (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Five independent short-answer items on number representation, Boolean evaluation, flip-flop characterization and memory addressing.

Find. The correct option (with justification) for each of (a), (b), (c), (d), (e).

Approach. Compute each item directly from its definition — two's-complement conversion, binary-fraction-to-decimal conversion, direct Boolean substitution, characteristic-table pattern matching against the standard flip-flop tables, and address-line counting from memory capacity — verifying every arithmetic step in Python.

1. Part (a) — 2's complement of $-49$, 8-bit. $49 = (00110001)_2$; inverting every bit gives $(11001110)_2$, and adding 1 gives $(11001111)_2$. Checking each option directly: i) $(10110001)_2=-121+48=-79$ (using $-2^7$ for the sign bit); ii) $(11001111)_2 = -128+64+8+4+2+1=-49$; iii) $(00110001)_2=+49$ (positive, since the sign bit is 0). $\boxed{\text{Option ii) } (11001111)_2 = -49}$.
2. Part (b) — closest binary value to $1.6_{10}$. Converting every option to decimal: i) $1.1_2=1.5$ (diff $0.100$); ii) $1.011_2=1.375$ (diff $0.225$); iii) $1.110001_2=1.765625$ (diff $0.166$); iv) $1.101_2=1.625$ (diff $\mathbf{0.025}$). $\boxed{\text{Option iv) } 1.101_2=1.625}$ has the smallest absolute difference from 1.6 of the four options.
3. Part (c) — evaluate $X=\overline{(A\oplus B)}+C$. With $A{=}1,B{=}0$: $A\oplus B = 1\oplus 0 = 1$, so $\overline{A\oplus B}=0$ (this is the XNOR of $A,B$). Then $X = 0 + C = 0+0=\boxed{0}$.
4. Part (d) — identify the flip-flop from its characteristic table. Writing the table as a Boolean function of $(A,B,Q_n)$ and minimizing gives $Q_{n+1}=B\cdot\overline{Q_n}+\overline{A}\cdot Q_n$ — which is exactly the standard JK flip-flop characteristic equation $Q_{n+1}=J\overline{Q_n}+\overline{K}Q_n$ with $\boxed{J=B,\ K=A}$: $(A,B){=}(0,0)\Rightarrow(J,K){=}(0,0)\to$ hold $Q_n$; $(1,0)\Rightarrow(J,K){=}(0,1)\to$ reset to 0; $(0,1)\Rightarrow(J,K){=}(1,0)\to$ set to 1; $(1,1)\Rightarrow(J,K){=}(1,1)\to$ toggle to $\overline{Q_n}$ — matching every row of the given table exactly.
5. Part (e) — address/data lines for a 2 KB byte-wide memory. $2\text{ KB} = 2\times1024 = 2048 = 2^{11}$ bytes, so exactly 11 address lines uniquely select every byte, and "byte-wide read/write" means 8 data lines (one per bit of a byte). $\boxed{\text{Option iv) 11 address lines, 8 data lines}}$; options ii)/iii) would address 4 KB/8 KB respectively, and option i) is not a power of two.