GATE 2026 EC Q.22: solution and answer

Official answer: CSource: Official answer key, GATE 2026 EC (IIT Guwahati)

1. The source has four symbols, mapped one-to-one to the four amplitude levels $-1,0,+1,+2$, so $s(t)$ is a 4-level baseband PAM signal.

2. The symbol rate is $R_s=10^4\text{symbols/s}$, so the symbol period is $T_s=10^{-4}\text{s}$.

3. The number of levels only sets the bits per symbol, ${\log }_{2}4=2$, giving a bit rate of $2\times 10^4\text{bit/s}$. It does not enter the Nyquist bandwidth, which depends only on the symbol rate.

4. For ISI-free baseband signalling at $R_s$ symbols per second, the minimum (equivalent Nyquist) bandwidth is $B_N=\frac{R_s}{2}$ (one-sided, in Hz).

5. $$B_N=\frac{10^4}{2}\text{Hz}=5\text{kHz}$$

6. This is option (C).