ADS功放管直流分析

ADS进行功放管直流分析时，出现WARNING告警，Mutual inductor `FET2FSL1.T4.Mutual1' has an unphysical mutual inductance M=-1.244e-11. L1=0.000e+00 L2=0.000e+00; limit for M is +/-0.000e+00.，该如何解决？