Signals and Systems - Electrical Engineering

394 C H A P T E R 6: Application to Control and Communications

we letH(S)= 1 /D(S)—that is, we assign toH(S)the poles in the left-hands-plane so that the resulting

filter is stable. The roots ofD(S)in Equation (6.38) are

S^2 kN=

ej(^2 k−^1 )π

e−jπN

=ej(^2 k−^1 +N)πfor integersk=1,..., 2N

after replacing− 1 =ej(^2 k−^1 )πand(− 1 )N=e−jπN. The 2Nroots are then

Sk=ej(^2 k−^1 +N)π/(^2 N) k=1,..., 2N (6.39)

Remarks

n Since|Sk|= 1 , the poles of the Butterworth filter are on a circle of unit radius. De Moivre’s theorem guar-

antees that the poles are also symmetrically distributed around the circle, and because of the condition that

complex poles should be complex conjugate pairs, the poles are symmetrically distributed with respect to the

σaxis. Letting S=s/hpbe the normalized Laplace variable, then s=Shp, so that the denormalized

filter H(s)has its poles in a circle of radiushp.

n No poles are on the j′axis, as can be seen by showing that the angle of the poles are not equal toπ/ 2 or

3 π/ 2. In fact, for 1 ≤k≤N, the angle of the poles are bounded below and above by letting 1 ≤k and

then k≤N to get

π

2

(

1 +

1

N

)

≤

( 2 k− 1 +N)π

2 N

≤

π

2

(

3 −

1

N

)

and for integers N≥ 1 the above indicates that the angle will not be equal to eitherπ/ 2 or 3 π/ 2 , or on

the j′axis.

n Consecutive poles are separated byπ/N radians from each other. In fact, subtracting the angles of two

consecutive poles can be shown to give±π/N.

Using the above remarks and the fact that the poles must be in conjugate pairs, since the coefficients

of the filter are real-valued, it is easy to determine the location of the poles geometrically.

nExample 6.7

A second-order low-pass Butterworth filter, normalized in magnitude and in frequency, has a

transfer function of

H(S)=

1

S^2 +

√

2 S+ 1

We would like to obtain a new filterH(s)with a dc gain of 10 and a half-power frequencyhp=

100 rad/sec.

The DC gain ofH(S)is unity—in fact, when=0,S=j0 givesH(j 0 )=1. The half-power

frequency ofH(S)is unity, indeed letting′=1, thenS=j1 and

H(j 1 )=

[

1

j^2 +j

√

2 + 1

]

=

1

j

√

2

so that|H(j 1 )|^2 =|H(j 0 )|^2 / 2 = 1 /2, or′=1 is the half-power frequency.