0 is represented by p(t) And 1 is represented by q(t) And ψ1(t) and ψ2(t) are orthogonal signal set (i) p(t)=ψ1(t) and q(t)=−ψ1(t) So signal space diagram will be,
∴dmin1=2 (ii) p(t)=ψ1(t) and q(t)=√Eψ2(t) So signal space diagram will be
∴dmin2=√E+1 Now bit error probability is same in both cases ∴dmin1=dmin2 √1+E=2 ∴E=3