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42 Some Tools and Examples <strong>of</strong> Filter Synthesis<br />

Repeated roots are usually determined to within two or three significant<br />

figures; this may be adequate for most but certainly not all engineering work.<br />

3.1.5. Polynomial Scaling. By the initial guess z=O.1 +jl for the root location,<br />

there is an assumption that the roots are not too far from the origin.<br />

Some polynomials may require scaling <strong>of</strong> coefficients to obtain the assumed<br />

condition, and the roots will require subsequent rescaling to correspond to the<br />

original problem. Two methods will be described, as given by Turnbull (1952):<br />

(1) decreasing all roots by the factor 10; (2) decreasing all roots by subtracting<br />

some fixed amount. The choice <strong>of</strong> method and amount depends on the<br />

problem being solved; there is usually adequate information to make those<br />

choices.<br />

To reduce all root real and imaginary components by a factor <strong>of</strong> 10, reduce<br />

all polynomial coefficients <strong>of</strong> the kth-power terms by lOexp(n - k), where the<br />

polynomial degree is n. The following example clarifies the procedure.<br />

Example 3.5.<br />

Consider the polynomial<br />

f(z) = 19404-394z+2z', (3.33)<br />

which has roots 98 + jO and 99 + jO (available from the root-finder program).<br />

Rewrite the polynomial with revised coefficients using the rule given above:<br />

f,(z)= 194.04-39.4z+2z'. (3.34)<br />

The root-finder program will show that the roots <strong>of</strong> (3.34) are 9.8 + jO and<br />

9.9 + jO. Similarly, the roots <strong>of</strong><br />

are 0.98 + jO and 0.99 + jO.<br />

f,(z)=1.9404-3.94z+2z' (3.35)<br />

The method for shifting the roots by a given amount is somewhat more<br />

involved but uses synthetic division in an interesting way. Again, consider the<br />

degree-3 polynomial in (3.12) without loss <strong>of</strong> generality. Suppose that variable<br />

z is decreased by amount h:<br />

z=s+h or s=z- h. (3.36)<br />

Making that substitution in (3.12), there must be an equivalent polynomial,<br />

F(s), in the new variable s:<br />

This is rewritten two more ways:<br />

f(z) = F(s)=bo+ b,s+b,s' + bJs'.<br />

(3.37)<br />

F(s) = bo+ s(co+ c,z + c,z'), (3.38)<br />

F(s)=bo+s[b,+s(co+zc;>J. (3.39)<br />

Note that (3.39) is nested in the same fashion as (3.25). Now (3.13) is written<br />

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