GATE 2024 IN Q.58: solution and answer

Official answer: 3Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Number of quantisation steps. The full-scale range divided by the step size is $\frac{V_{FS}}{\Delta }=\frac{1.4\text{V}}{0.2\text{V}}=7$ steps. The output levels are $0,0.2,0.4,\dots ,1.4\text{V}$, which is $7+1=8$ distinct levels.

2. Number of bits. With the convention $\Delta =\frac{V_{FS}}{2^n-1}$ we get $2^n-1=7$, so $n=3$ exactly. With the other convention, $\Delta =\frac{V_{FS}}{2^n}$, we need $\frac{1.4}{2^n}\le 0.2$, so $2^n\ge 7$. $n=2$ gives $1.4/4=0.35\text{V}$, which is too coarse, and $n=3$ gives $1.4/8=0.175\text{V}$, which meets the requirement. So the smallest whole number of bits is $n=3$. Both conventions give $n=3$ bits per sample.

3. Nyquist sampling rate. $f_s=2f_{max}=2\times 500\text{MHz}=1\text{GHz}$, which is $10^9$ samples per second.

4. Serial bit rate. Each sample produces an $n$-bit parallel word. The parallel-to-serial converter has to send all $n$ bits one after another within one sample period $T_s=1/f_s$, so $R_b=n{f}_s$.

5. $$R_b=3\times 10^9\text{bits/s}=3\text{Gbps}$$