For the circuit above, at zero input condition( (v1 = v2 = 0), VD1_0 = VD2_0 = 6 V, the amplification matrix is defined as[vod voc] = [Add Acd Adc Acc][vid vic]where Vid = v1-v2, Vic = (v1+v2)/2, Vod = vD1-vD2, Voc = (vD1+vD2)/2. Given Add = -89, Acd = 0.2, Adc = 0.1, Acc = −0.20, and v1 = −1.01, v2 = −0.98 determine the potential difference between VD1 and ground. (VD1 = VD1_0+vD1 and VD2 = VD2_0+vD2)

For the circuit above, at zero input condition( (v1 = v2 = 0), VD1_0 = VD2_0 = 6 V, the amplification matrix is defined as[vod voc] = [Add Acd Adc Acc][vid vic]where Vid = v1-v2, Vic = (v1+v2)/2, Vod = vD1-vD2, Voc = (vD1+vD2)/2. Given Add = -89, Acd = 0.2, Adc = 0.1, Acc = −0.20, and v1 = −1.01, v2 = −0.98 determine the potential difference between VD1 and ground. (VD1 = VD1_0+vD1 and VD2 = VD2_0+vD2)

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For the circuit above, at zero input condition( ( v 1 = v 2 = 0 ) , V D 1 _ 0 = V D 2 _ 0 = 6 V , the amplification matrix is defined as
[ v o d v o c ] = [ A d d A c d A d c A c c ] [ v i d v i c ]
where Vid=v1-v2, Vic=(v1+v2)/2, Vod=vD1-vD2, Voc=(vD1+vD2)/2. Given Add=-89, A c d = 0.2 , A d c = 0.1 , A c c = 0.20 , and v 1 = 1.01 , v 2 = 0.98 determine the potential difference between VD1 and ground. (VD1=VD1_0+vD1 and VD2=VD2_0+vD2)

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