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548 Chapter 10 ROUND-OFF EFFECTS IN DIGITAL FILTERS<br />

We can also show that the variance of q(n), for both the truncation or<br />

rounding, is given by<br />

∞∑<br />

∫ π<br />

σq 2 = σe<br />

2 |h(n)| 2 = σ2 e<br />

|H(e jω )| 2 dω (10.15)<br />

2π −π<br />

−∞<br />

The variance gain from the input to the output (also known as the normalized<br />

output variance) isthe ratio<br />

σq<br />

2 ∞∑<br />

= |h(n)| 2 = 1 ∫ π<br />

|H(e jω )| 2 dω (10.16)<br />

2π<br />

σ 2 e<br />

−∞<br />

Forareal and stable filter, using the substitution z =e jω , the integral in<br />

(10.16) can be further expressed as a complex contour integral<br />

∫ π<br />

|H(e jω )| 2 dω = 1 ∮<br />

H(z)H(z −1 )z −1 dz (10.17)<br />

2πj<br />

−π<br />

UC<br />

where UC is the unit circle and can be computed using residues (or the<br />

inverse Z-transform) as<br />

∫ π<br />

−π<br />

|H(e jω )| 2 dω = ∑ [Residues of H(z)H(z −1 )z −1 inside UC](10.18a)<br />

−π<br />

= Z −1 [ H(z)H(z −1 ) ]∣ ∣<br />

n=0<br />

(10.18b)<br />

10.1.7 MATLAB IMPLEMENTATION<br />

Computation of the variance-gain for the A/D quantization noise can be<br />

carried out in MATLAB using (10.16) and (10.18). For FIR filters, we can<br />

perform exact calculations using the time-domain expression in (10.16).<br />

In the case of IIR filters, exact calculations can only be done using (10.18)<br />

in special cases, as we shall see (fortunately, this works for most practical<br />

filters). The approximate computations can always be done using the<br />

time-domain expression.<br />

Let the FIR filter be given by the coefficients {b k } M−1<br />

0 . Then using<br />

the time-domain expression in (10.16), the variance-gain is given by<br />

σq<br />

2<br />

σe<br />

2<br />

=<br />

M−1<br />

∑<br />

k=0<br />

|b k | 2 (10.19)<br />

Let an IIR filter be given by the system function<br />

∑ N−1<br />

l=0<br />

H(z) =<br />

b lz −l<br />

1+ ∑ N−1<br />

k=1 a (10.20)<br />

kz −k<br />

with impulse response h(n). If we assume that the filter is real, causal, and<br />

stable and has only simple poles, then using the partial fraction expansion,<br />

we can write<br />

N−1<br />

∑ R k<br />

H(z) =R 0 +<br />

(10.21)<br />

z − p k<br />

k=1<br />

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