0195136047.pdf

176 TIME-DEPENDENT CIRCUIT ANALYSIS

In this so-called Butterworth bandpass filter, the center frequency and the bandwidth of the

bandpass filter are given by

ω 0 =

√

ωC 1 ωC 2 and B=ωC 2 −ωC 1

and

ω 0

B

=Q

√

ωC 1

ωC 2

<< 1

whereωCHP=ωC 1 is the cutoff frequency of the high-pass filter andωCLP=ωC 2 is the cutoff

frequency of the low-pass filter.

In order to design a bandpass filter with a passband gain of 0 dB and cutoff frequencies of

ωC 1 =10 rad/s andωC 2 =50 rad/s, with the stopband gains of less than−20 dB at 2 rad/s and

250 rad/s, one has chosen the following low-pass and high-pass transfer functions:

TLP(s)=

1

(s/ 50 )^2 +

√

2 (s/ 50 )+ 1

THP(s)=

(s/ 10 )^2

(s/ 10 )^2 +

√

2 (s/ 10 )+ 1

When circuits realizing these two functions are connected in cascade, the overall transfer function

is given by

TBP(s)=THP(s)×TLP(s)

Use MATLAB to illustrate the specified bandpass response with the center frequency√ ω 0 =

10 × 50 = 22 .4 rad/s, the bandwidthB= 50 − 10 =40 rad/s, and the filter quality factor

Q=ω 0 /B= 0. 56 .Note thatQbeing less than 1 indicates a broad-band response.

Solution

The M-file and answers are as follows:

function example363

clc

% 2nd order Butterworth low pass wc = 50

TLPs = tf(1,[1/50∧2 sqrt (2)/50 1])

% 2nd order Butterworth high pass wc = 10

THPs = tf([1/10∧2 0 0], [1/10∧2 sqrt(2)/10 1])

% 4th order Butterworth bandpass wc1 = 10 wc2 = 50

TBPS = TLPs * THPs

bode (TBPs,{1,1e3});

Transfer function:

1

--------------------------

0.0004 s∧2 + 0.02828s+1

Transfer function:

0.01 s∧ 2

-----------------------

0.01 s∧2 + 0.1414s+1