Signals and Systems - Electrical Engineering

6.5 Analog Filtering 395

Thus, the desired filter with a dc gain of 10 is obtained by multiplyingH(S)by 10. Furthermore,

if we letS=s/100 be the normalized Laplace variable whenS=j′hp=j1, we get thats=jhp=

j100, orhp=100, the desired half-power frequency. Thus, the denormalized filter in frequency

H(s)is obtained by replacingS=s/100. The denormalized filter in magnitude and frequency

is then

H(s)=

10

(s/ 100 )^2 +

√

2 (s/ 100 )+ 1

=

105

s^2 + 100

√

2 s+ (^104) n
Design
For the Butterworth low-pass filter, the design consists in finding the parametersN, the minimum
order, andhp, the half-power frequency, of the filter from the constrains in the passband and in the
stopband.
The loss function for the low-pass Butterworth is
α()=−10 log 10 |HN(/hp)|^2 =10 log 10 ( 1 +

(

/hp

) 2 N

)

The loss specifications are

0 ≤α()≤αmax 0 ≤≤p

αmin≤α() <∞ ≥s

At=p, we have that

10 log 10 ( 1 +

(

p/hp

) 2 N

)≤αmax

so that
(
p
hp

) 2 N

≤ 10 0.1αmax− 1 (6.40)

and similarly for=s, we have that

10 0.1αmin− 1 ≤

(

s

hp

) 2 N

(6.41)

We then have that from Equation (6.40) and (6.41), the half-power frequency is in the range

p

( 10 0.1αmax− 1 )^1 /^2 N

≤hp≤

s

( 10 0.1αmin− 1 )^1 /^2 N

(6.42)

and from the log of the two extremes of Equation (6.42), we have that

N≥

log 10 [( 10 0.1αmin− 1 )/( 10 0.1αmax− 1 )]

2 log 10 (s/p)

(6.43)