86.1.3 Analysis of Block Diagrams Combining the last two equations we get 3÷8+-1 W3 1+δz Substituting the above equation in Wi=X-az-W3, Y=BW1+r2-W3 we finally arrive at Yβ+(β6+y8)z-1+yz 2 H(2)X1+(6+8)2+§6.1.3 Analysis of Block Diagrams • Combining the last two equations we get 1 1 3 1 1 W W z z − − + + = , 3 1 W1 = X −z − W 3 1 Y = W1 +  z − W 1 2 1 2 1 ( ) ( ) ( ) − − − − +  +  +   +   +  +  = = z z z z X Y H z we finally arrive at • Substituting the above equation in