86.1.3 Analysis of Block Diagrams Eliminating e(z from the previous two equations we arrive at l1-G(2G2(ZY(2-G(zx(z which leads to Y(z) G1(z) H(z)=X(z)1-G(z)G2(2z)
§6.1.3 Analysis of Block Diagrams • Eliminating E(z) from the previous two equations we arrive at [1-G1 (z)G2 (z)]Y(z)=G1 (z)X(z) which leads to 1 ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 G z G z G z X z Y z H z − = =
86.1.3 Analysis of Block Diagrams Example- Analyze the cascaded lattice structure shown below where the dependence of signal varia bles are not shown for brevity
§6.1.3 Analysis of Block Diagrams • Example - Analyze the cascaded lattice structure shown below where the zdependence of signal variables are not shown for brevity
86.1.3 Analysis of Block Diagrams The output signals of the four adders are given by WI=X-as 2=W1-8S w3=S1-EW2Y=βW1-yS2 From the figure we observe
§6.1.3 Analysis of Block Diagrams • The output signals of the four adders are given by W1= X - S2 W2= W1 - S1 W3= S1 - W2 Y= W1 - S2 • From the figure we observe S2= z-1 W3 S1= z-1 W2
86.1.3 Analysis of Block Diagrams Substituting the last two relations in the first four equations we get WiX-azW3 W2=W1-8TW2 W3=zW2+EW2 Y=BWI+yTW3 From the second equation we get W2=W1/(1+8z) and from the third equation we get W3=(E+z)W2
§6.1.3 Analysis of Block Diagrams • Substituting the last two relations in the first four equations we get W1=X- z -1 W3 W2= W1 - z -1 W2 W3= z-1 W2 + W2 Y= W1 + z -1 W3 • From the second equation we get W2= W1 /(1+ z -1 ) and from the third equation we get W3=( + z -1 )W2
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