Digital Signal Processing Reference
In-Depth Information
Putting the output y[n] all in terms of the input, x[n]:
y[n]
= afx[n] + bfx[n1] + cfx[n2] + dfx[n3]
+ agx[n1] + bgx[n2] + cgx[n3] + dgx[n4]
+ akx[n2] + bkx[n3] + ckx[n4] + dkx[n5]
+ amx[n3] + bmx[n4] + cmx[n5] + dmx[n6]:
Simplifying:
y[n]
= afx[n] + (ag + bf)x[n1] + (ak + bg + cf)x[n2]
+
(am + bk + cg + df)x[n3] + (bm + ck + dg)x[n4]
+
(cm + dk)x[n5] + dmx[n6]:
We see that our expression for y[n] now contains only a linear combination of the
input (and delayed versions of it), multiplied by constants. The resulting coecients
are, therefore,faf, (ag + bf), (ak + bg + cf), (am + bk + cg + df), (bm + ck + dg),
(cm + dk), dmg. This pattern is the convolution offa;b;c;dgwithff;g;k;mg.
We have found a resulting lter using convolution. In the next section, we will
see how this analysis could be done with the z-transform.
8.3
Revisiting Sequential Filter Combination with z
Let's return to the example in Figure 8.1. We want to replace the two lters h
1
[n] =
fa;b;c;dgand h
2
[n] =ff;g;k;mgwith a single equivalent lter. First, we can write
the z-transform for these two lters directly, just by examining their coecients.
H
1
(z) = az
0
+ bz
1
+ cz
2
+ dz
3
H
2
(z) = fz
0
+ gz
1
+ kz
2
+ mz
3
To nd the equivalent lter, we multiply the z-transforms H
1
(z) and H
2
(z) together.
We do this because multiplication in the frequency-domain is the same as convolu-
tion in the time-domain, as demonstrated in section 8.8. We will call the z-transform
for our third lter H
3
(z),
H
3
(z)
= H
1
(z)H
2
(z)
