butterworth lowpass: this filter is characterized by the property that its magnitude response is...

Butterworth Lowpass:This filter is characterized by the property that its magnitude response is flat in both passband and stopband. The magnitude-squared response of an N-th order lowpass filter is given by:

|𝐻 𝑎 ( 𝑗Ω )|2= 1

1+( ΩΩ𝑐 )2𝑁

Here, N is the order of the filter and c is the cutoff frequency in rad/sec

Plot of the magnitude-squared response is as depicted below:

|𝐻 𝑎 ( 𝑗Ω )|

( ΩΩ𝑐 )

At = c , for all N.

This implies a 3dB attenuation at c.

To determine the system function s= j =

= =

= = =

The roots of the denominator polynomial are given as:

= where k = 0,1,……,(2N-1)

Interpretation of

For there are 2N poles which are equally distributed on a circle of radius with angular spacing of radians.

For odd values of N, poles are given by where k = 0,1,……,(2N-1)

For even values of N, poles are given by where k = 0,1,……,(2N-1)

Poles are symmetric with respect to axis

If N is odd a pole never falls on the imaginary axis. Instead it is on the real axis.

Example:Given that determine the analog filter’s system function Ha

that =

N= 3=0.5

Hence =

=

=

N=3;gc=0.5;k=0:5;

pk = gc*exp(i*(pi/(2*N)*(2*k+N+1));

pk =

Columns 1 through 5

-0.2500 + 0.4330i -0.5000 + 0.0000i -0.2500 - 0.4330i 0.2500 - 0.4330i 0.5000 - 0.0000i

Column 6

0.2500 + 0.4330i

Design Procedure:

Given a digital lowpass filter with specifications wp, ws, Rp, As we want toDetermine H(z) by first designing an equivalent analog filter and then mapping it into the desired digital filter.

Steps to follow: Choose T and determine the analog frequencies

and Design an analog filter Ha using specifications p, s, Rp, As

Using PFE expand Ha into the form Ha

Now transform the analog poles {pk} into digital poles {} to obtain the digital filter:

Ha =

After taking inverse laplace the corresponding impulse response is

Impulse response of the discrete time filter is obtained by sampling T and is

h[n] =T =u[n]

= =u[n]

=

=

=

The system function of the discrete-time filter is therefore

H(z) =

Note: a pole at s = pk transform into a pole z= in the z-plane.

Suppose that our Ha is as follows:

Ha =

Transform this analog filter into a digital filter H(z) using the impulse invariance method.Assume T = 0.1.

Ha =

The poles are at p1 = -3 and p2 = -2

H(z) =

Ex2:

Find the digital IIR filter for the second-order lowpass analog filter with frequencyresponse Ha as depicted below. Assuming a sampling period of T = 0.05 secs.

Ha(s) = First we must apply PFE to Ha

Ha(s) = =

= 15

+ = 015 = 15 =1 and =-1

Ha(s) = =

H(z) = *T=