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in which robs is the observed correlation between the relatives, and rexp is the expected correlation. The expected correlation is simply the proportion of the genes in common. We must point out that the observed correlation between relatives can be arti cially in ated if the environments are not random. Since we know that relatives frequently share similar (or correlated) environments, they may show a phenotypic similarity irrespective of genetic causes. It is important to keep that in mind, especially when we analyze human traits, where it may be almost impossible to rule out or quantify environmental similarity. Hence, robs may be in ated, which will in ate HN. In human beings, nger-ridge counts ( ngerprints, g. 18.15) have a very high heritability; there seems to be very little environmental interference in the embryonic development of the ridges (table 18.6). Monozygotic twins are from the same egg, which divides into two embryos at a very early stage. They have identical genotypes. Dizygotic twins result from the simultaneous fertilization of two eggs. They have the same genetic relationship as siblings. (However, environmental in uences may be different; they may be treated differently by relatives and friends.) The data therefore suggest that human nger ridges are almost completely controlled by additive genes with a negligible input from environmental and dominance variation. Few human traits are controlled this simply (table 18.7). This brief discussion should make it clear that the components of the total variance can be estimated. For a given quantitative trait, the total variance can be mea-

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R2 =

The outflow resistance is the pressure drop divided by the rate of flow of liquid As the flow through the clearance is usually laminar, the outflow resistance is only dependent on the geometrical shape of the lands, the clearance and the viscosity of the fluid 6 Ro = a b h 3 + 3096 c c Ro =

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If we assume 01 F values for both C1 and C2 in each section, we can compute the value of the resistances required to complete the design: First section: R1 = 7,723 (nearest standard 5% resistor value: 82 k ) R2 = 80,923 (nearest standard 5% resistor value: 82 k ) Second section: R1 = 6,411 (nearest standard 5% resistor value: 68 k ) R2 = 97,484 (nearest standard 5% resistor value: 100 k )

Identical (MZ) Twins (%) Hair color Eye color Blood pressure Handedness (left or right) Measles Clubfoot Tuberculosis Mammary cancer Schizophrenia Down syndrome Spina bi da Manic-depression 89 99.6 63 79 95 23 53 6 80 89 72 80 Fraternal (DZ) Twins (%) 22 28 36 77 87 2 22 3 13 7 33 20

Comments: We have chosen to x the values of the capacitors and compute the required values of the resistors because of the greater availability of resistor sizes Focus on Computer-Aided Solutions: A Matlab le that performs the computations

shown above for each quadratic section may be found in the accompanying CD-ROM Note that the m- le can be easily modi ed to design a Butterworth lter of arbitrary order

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157 Determine the order of the lter required to satisfy the requirements of Example 153 if the stop-band frequency is moved to S = 2 C 158 What is the actual attenuation of the lter of Exercise 157 at the stop-band frequency, S 159 Design a quadratic lter section with Q = 1 and a cutoff frequency of 10 rad/s Note that there can be many solutions, depending on your design

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8 10 9 6 5 10 6 775 12 2 + 2 3096

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