Signals and Systems - Electrical Engineering

718 CHAPTER 12: Applications of Discrete-Time Signals and Systems

FIGURE 12.2

Lattice forn 0 =0, 1andn 1 =0,..., 3(the values in

parentheses are the indices of the samples). Notice the

ordering in the two columns.

n 1

n 0

(^01)
2
3
1
(0)
(2)
(4)
(6)
(1)
(3)
(5)
(7)
Figure 12.2 displays a lattice forn 0 andn 1 and the indices of the samples are in parentheses. By
replacingk= 4 k 1 +k 0 , we get the first step of the decimation-in-time.
If we lety[n]=x[2n] andz[n]=x[2n+1],n=0,..., 3, we can then repeat the above procedure
by factoring 4=pq= 2 ×2 and expressingY[k] andZ[k] as we did forX[k]. Thus, we have
Y[k 0 ,k 1 ]=

∑^1

n 0 = 0

W

n 0 (k 1 p+k 0 )

4

∑^1

n 1 = 0

y[n 0 ,n 1 ]W

n 1 (k 1 p+k 0 )

2

=

∑^1

n 1 = 0

y[0,n 1 ]W

n 1 (k 1 p+k 0 )

2 +W

(k 1 p+k 0 )

4

∑^1

n 1 = 0

y[1,n 1 ]W

n 1 (k 1 p+k 0 )

2

and

Z[k 0 ,k 1 ]=

∑^1

n 1 = 0

z[0,n 1 ]W

n 1 (k 1 p+k 0 )

2 +W

(k 1 p+k 0 )

4

∑^1

n 1 = 0

z[1,n 1 ]W

n 1 (k 1 p+k 0 )

2

where now

k= 2 k 1 +k 0 fork 0 =0, 1,k 1 =0, 1

n= 2 n 1 +n 0 forn 0 =0, 1,n 1 =0, 1

If we replacek= 2 k 1 +k 0 , we obtain

Y[k]=I[k]+W 4 kG[k]

Z[k]=H[k]+W 4 kF[k]

where

I[k]=

∑^1

n 1 = 0

y[0,n 1 ]Wn 21 k=

∑^1

n 1 = 0

y[2n 1 ]W 2 n^1 k=

∑^1

n 1 = 0

x[4n 1 ]W 2 n^1 k