I/Q mismatch

hi
I do not know how to compute the I/Q mismatch?
for example
IP IN QP QN
amplitude(mv) 193 193 190 190

phase(°) 0 92 182 270

the amplitude mismathc is 3mv, is 3db?
the phase error is 2°?

Usually (if your device is differential, as I suppose) you have to consider the two I&Q differential signals instead of 4, obtained as I=IP-IN and Q=QP-QN.
No mismatch means that amp(I)=amp(Q) and ph(I)=ph(Q)-90deg.
The amplitude mismatch is simply amp(I)-amp(Q) that can be expressed in mV or in dB (20log).
The phase mismatch is the deviation respect the expected 90deg of phase between I&Q.
I hope it can help.
Mazz

if QP QN IP IN
amplitude(mv) 192.3 193.1 192.7 192.6

phase(°) 0 182 92 270

the amp(I) =amp(IP)-amp(IN) = 0.1mv
amp(Q) =amp(QP)-amp(QN) = 0.8mv
then the amp mismatch is 0.7mv or 20log(0.7)=-3db?
the phase is
phase(I)=178
phase(Q)=182
phase(Q)-90=92
phase(I)=phase(Q)-90?