ISRO EC 2025 Q.41: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. A synchroniser fails when a flip-flop stays metastable for longer than the resolution time $t_r$ allowed before the next stage samples it. The standard result is $$\text{MTBF}=\frac{e^{t_r/\tau }}{T_0{f}_{clk}{f}_{data}}$$ where $\tau$ is the flip-flop's regeneration time constant, $T_0$ is proportional to the width of the metastability window (setup plus hold aperture), $f_{clk}$ is the sampling clock frequency and $f_{data}$ is the rate at which the data input changes.

2. (a) Decreasing the setup time narrows the window, so $T_0$ falls and MTBF rises. This does not reduce MTBF.

3. (b) Increasing the resolution time $t_r$ raises the exponent $t_r/\tau$, so MTBF rises exponentially. This does not reduce MTBF.

4. (c) MTBF is inversely proportional to the product $f_{clk}{f}_{data}$. A faster clock and a higher data arrival rate give more chances per second for a data edge to fall inside the window, so MTBF falls. This option reduces MTBF.

5. (d) 'All of the above' would need (a) and (b) to reduce MTBF as well, and they do not. So only (c) is correct.