EEL 6502 - DIGITAL SIGNAL PROCESSING I COMPUTER ASSIGNMENT #4 (See Course Homepage) I. FREQUENCY RESPONSE FOR RATIONAL SYSTEM FUNCTIONS Background Reading: Oppenheim & Schafer: Section 5.3. Overview: A simple geometric construction can be used to approximately sketch the frequency response function directly from the pole-zero plot. The procedure is based on the following facts: a) The frequency response corresponds to the system function evaluated on the unit circle. b) The complex value of each pole and zero can be drawn as a vector in the z-plane from the pole or zero to a point on the unit circle. The frequency response for a system with a single zero is shown in Figure 5.8 in the text and the corresponding geometric construction is shown in Figure 5.9. PROBLEM I Consider the second order system, H ( z) 1 1 j 1 (1 re z )(1 re z ) (1 2r cos z 1 r 2 z 2 ) j 1 This system function has a pole at z re j and another at the conjugate location, z re j . a) Remember that the log magnitude of the frequency response in dB is given by, -10 log10[1+ r2 - 2r cos( - )] - 10 log10[1+ r2 - 2r cos(+)] and the phase angle is given by, -arctan(r sin( - )/(1-r cos( - ))) - arctan(r sin( + )/(1-r cos( + ))). Plot these functions for r = 0.95 and = /3. b) If v1, v2 and v3 represent vectors from the first pole, from the second pole and from the origin respectively to a point on the unit circle, respectively, show that, |( )| = |v3|2 / (|v1|.|v2|) = 1 / (|v1|.|v2|) ( ) = 2 - v1 - v2 where, v1 e j re j and v2 e j re j Plot the magnitude response using these results for r = 0.95 and = /3 and compare with that obtained earlier. 1 Sankar/DSP-I c) Add a zero at z = - 0.5 for the system, H(z) given above and plot the new frequency response. Observe the effect of zero on the frequency response to that obtained in part (a). Identify how the filter changed by adding this new zero to the pole-zero plot. PROBLEM II Consider the following system function (See Example 5.10 on page 268 in 2nd Edition or Example 5.8 on page 299 in 3rd Edition), 0.05634(1 z 1 )(1 1.0166z 1 z 2 ) H(z ) (1 0.683z 1 )(1 1.4461z 1 0.7957z 2 ) The zeros of this system function are at, Radius Angle rad 1 1 1.0376 rad (59.45 deg) and the poles are at, Radius Angle 0.683 0 rad 0.892 0.6257 rad (35.85 deg) a) Draw the pole-zero plot for this system, and plot the log magnitude and phase of its response. Note the effect of the zeros that are placed on the unit circle at 1.0376 and radians. Observe that the poles are placed such that rather than peaking for frequencies close to their angles, the total log magnitude remains close to 0 dB over a band from 0 to 0.2 (and by symmetry, from 1.8 to 2 ), and then it drops abruptly and remains below -25 dB from 0.3 to 1.7 This system is a low pass filter designed by an approximation method. The example suggests that useful approximations to frequency selective filter responses can be achieved using poles to build up the magnitude response and zeros to suppress it. b) See how you can move the poles and zeros such that the cutoff frequency of the filter can be increased or decreased? (Hint: Try moving the complex conjugate poles and zeroes by = /6 radians.) 2 Sankar/DSP-I