Circuit containing two-port The transfer function from
Vin to
V2 is desired. There is only one forward path: :*
Vin to
V1 to
I2 to
V2 with gain G_1 = -y_{21} R_L \, There are three loops: :*
V1 to
I1 to
V1 with gain L_1 = -R_\text{in} y_{11} \, :*
V2 to
I2 to
V2 with gain L_2 = -R_L y_{22} \, :*
V1 to
I2 to
V2 to
I1 to
V1 with gain L_3 = y_{21} R_L y_{12} R_\text{in} \, : \Delta = 1 - ( L_1 + L_2 + L_3 ) + ( L_1 L_2 ) \, note:
L1 and
L2 do not touch each other whereas
L3 touches both of the other loops. : \Delta_1 = 1 \, note: the forward path touches all the loops so all that is left is
1. : G = \frac { G_1 \Delta_1 } { \Delta } = \frac { -y_{21} R_L } {1 + R_\text{in}y_{11} + R_L y_{22} - y_{21} R_L y_{12} R_\text{in} + R_\text{in}y_{11} R_L y_{22} } \,
Digital IIR biquad filter Digital filters are often diagramed as signal flow graphs. : There are two loops :* L_1 = -a_1 Z^{-1} \, :* L_2 = -a_2 Z^{-2} \, : \Delta = 1 - ( L_1 + L_2 ) \, Note, the two loops touch so there is no term for their product. :There are three forward paths :* G_0 = b_0 \, :* G_1 = b_1 Z^{-1} \, :* G_2 = b_2 Z^{-2} \, : All the forward paths touch all the loops so \Delta_0 = \Delta_1 = \Delta_2 = 1 \, : G = \frac { G_0 \Delta_0 +G_1 \Delta_1 + G_2 \Delta_2 } {\Delta} \, : G = \frac { b_0 + b_1 Z^{-1} + b_2 Z^{-2} } {1 +a_1 Z^{-1} + a_2 Z^{-2} } \,
Servo The signal flow graph has six loops. They are: :* L_0 = - \frac {\beta} {sM} \, :* L_1 = \frac{-( R_M + R_S)} {sL_M} \, :* L_2 = \, \frac {-G_M K_M} {s^2 L_M M} :* L_3 = \frac {-K_C R_S} {sL_M} \, :* L_4 = \frac {-K_V K_C K_M G_T} {s^2 L_M M} \, :* L_5 = \frac {-K_P K_V K_C K_M } {s^3 L_M M} \, : \Delta = 1 - (L_0+L_1+L_2+L_3+L_4+L_5) + (L_0 L_1 + L_0 L_3 )\, There is one forward path: :* g_0 = \frac {K_P K_V K_C K_M } {s^3 L_M M} \, The forward path touches all the loops therefore the co-factor \Delta_0 = 1 And the gain from input to output is \frac {\theta_L} {\theta_C} = \frac {g_0 \Delta_0} {\Delta} \, == Equivalent matrix form ==