Correct option is (d) sgn(ω)
For explanation I would say: Let x(t) = sgn(t)
The Fourier transform of sgn(t) is X(ω) = F[sgn(t)] = \(\frac{2}{jω}\)
Replacing ω with t
–> X(t) = \(\frac{2}{jt}\)
As per duality property X(t) ↔ 2πx(-ω), we have
F\(\Big[\frac{2}{jt}\Big]\) = 2πsgn(-ω) = -2πsgn(ω)
\(\frac{2}{jt}\) ↔ -2πsgn(ω)
\(\frac{2}{πt}\) ↔ sgn(ω).