Example determine Linear Constant- Coefficient Difference Equations of SFG Source Sink Node b n、Mode n Solution C wn=xn +aw,n +bw,n [n]=cwna w,[n]=cx[n]+acw,n]+ bcw,[nl yln]=din]+ew , n cerl n 1-ac-bC ce y[n]=d+ 1-ac-bc 28
28 Example : determine Linear ConstantCoefficient Difference Equations of SFG x[n] w y[n] 1 [n] w2 [n] a b c d e Source Node Sink Node 1 2 2 w n x n aw n bw n [ ] [ ] [ ] [ ] = + + 2 1 w n cw n [ ] [ ] = 2 y n dx n ew n [ ] [ ] [ ] = + [ ] [ ] 1 = + − − ce y n d x n ac bc2 2 2 w n cx n acw n bcw n [ ] [ ] [ ] [ ] = + + Solution: 2 [ ] [ ] 1 cex n w n ac bc = − −
Block Diagram vS Signal Flow Graph vn yn branching point Canonic direct Form btbd H(=)= |z| 1-az Source node Win Sink node xin o W yn elay By convention, variables is branch represented as sequences rather than as z-transforms W4()=W2(z) Delay branch cannot be represented in time domain by a branch gain by Z-transform a unit delay branch has a gain of z 29
29 Block Diagram vs. Signal Flow Graph x[n] + a z −1 + b1 b0 w[n] y[n] x[n] w1 [n] w2 [n] w3 [n] a b1 b0 1 2 3 4 w4 [n] y[n] 1 0 1 1 ( ) , | | | | 1 b b z H z z a az − − − + = Delay branch cannot be represented in time domain by a branch gain z −1 Delay branch Canonic direct Form Source Node Sink Node =w2 [n-1] 2 W z( ) 1 4 2 W z z W z ( ) ( ) − = by z-transform, a unit delay branch has a gain of z -l . By convention, variables is represented as sequences rather than as z-transforms branching point
Block Diag w,[n]=x[n]+aw[nj Determine the w,In]=wIn Function n)from wiN]=bow2[n]+b, wIn w4m]=w2[n-1 Solution yIn]=wsln XIn y2h320 4w4[n] 30
Determine the difference equation (System Function) from the Flow Graph. 30 Block Diagram vs. Signal Flow Graph x[n] + a z −1 + b1 w b0 [n] y[n] x[n] w1 [n] w2 [n] w3 [n] a b1 b0 z −1 1 2 3 4 w4 [n] y[n] 1 4 w n x n aw n [ ] [ ] [ ] = + 2 1 w n w n [ ] [ ] = 3 0 2 1 4 w n b w n b w n [ ] [ ] [ ] = + 4 2 w n w n [ ] [ 1] = − 3 y n w n [ ] [ ] = Solution:
Block Di w,[n]=x[n]+aw4[n] Determine wIn=w difference wn=bw,n +b,win equation w4m]=w2n-1] yln=waln y[n]=wa[n]=bow,n]+b,w,n-1]difficult in wiN]=wIn]=x[n]+aw,n-1 time-domain (z)=(b+b2)W2(=) W,()=X(2)+az W2(a) Y(>(6+6,2) X(=) 1-az W2()=X() 1-az-y(n]=[n-1]+box[n]+b,x[n-1]
Block Diagram vs. Signal Flow Graph 31 3 y n w n [ ] [ ] = 0 2 1 2 = + − b w n b w n [ ] [ 1] 2 1 w n w n [ ] [ ] = 2 = + − x n aw n [ ] [ 1] ( ) ( ) ( ) 2 1 0 1 Y z b b z W z − = + ( ) ( ) ( ) 2 1 2 W z X z az W z − = + 2 1 1 ( ) ( ) − − = az X z W z ( ) 1 ( ) ( ) 1 1 0 1 X z az b b z Y z − − − + = 0 1 y n ay n b x n b x n [ ] [ 1] [ ] [ 1] = − + + − 1 4 w n x n aw n [ ] [ ] [ ] = + 2 1 w n w n [ ] [ ] = 3 0 2 1 4 w n b w n b w n [ ] [ ] [ ] = + 4 2 w n w n [ ] [ 1] = − 3 y n w n [ ] [ ] = Determine difference equation difficult in time-domain
Ex 6.3 Determine the System Function from Flow Graph y Solution 2=1小x: W(=)=W4(=)-X(= ]=am[ W2()=cW( [=w2[]+x[n Wy(=)=W2(=)+X(=) yn=12n+4 ] Y()=W2(=)+W4()
32 w n w n x n 1 4 = - W z W z X z 1 4 ( ) = ( )- ( ) Ex. 6.3 Determine the System Function from Flow Graph w n w n 4 3 = -1 w n w n x n 3 2 = + w n w n 2 1 = y n w n w n = + 2 4 W z W z X z 3 2 ( ) = + ( ) ( ) Y z W z W z ( ) = + 2 4 ( ) ( ) W z W z 2 1 ( ) = ( ) ( ) ( ) -1 W z W z z 4 3 = Solution: