Poles and Zeros in z-plane

Transcription

1 M58 Mixed Signal Processors page of 6 Poles and Zeros in z-plane z-plane Response of discrete-time system (i.e. digital filter at a particular frequency ω is determined by the distance between its poles and zeros to a point on the unit circle corresponding to that frequency ω. The distance from a pole to a point on the unit circle is inversely proportional to the gain of the magnitude response at that frequency. Therefore a pole placed close to the unit circle has a small distance to point corresponding to frequency ω and therefore the magnitude response of the filter at ω will be high (peak. The distance from a zero to appoint on the unit circle is directly proportional to the gain of the magnitude response at that frequency. Therefore a zero placed close to the unit circle has a small distance to point corresponding to frequency ω and therefore the magnitude response of the filter at ω will be will be low (null or notch. For a stable operation of the system poles should not be placed on the unit circle since the length of the corresponding vector will be zero, which corresponds to a gain of infinity. This causes the filter to be unstable. Notice that we have done exactly that in the previous lecture in order to produce filter resonator for an impulse type input filter oscillates/resonates at one particular frequency ω. Zeros however can be placed on the unit circle and cause the gain to be zero at the frequency where the zero is placed. Poles and zeros placed on the origin have no effect on the magnitude response of the filter but contribute to the phase response of the system.

2 M58 Mixed Signal Processors page of 6 Poles and zeros that are not real (i.e. they are not on the real axis must appear as complex conjugate pair. This is important in order to ensure that the system is real and therefore realizable. Two examples will demonstrate this theory. EXAMPLE: Design a two-pole band-pass filter with a peak at f Hz. Place poles at R.99 distance from the centre of the unit circle in the z-plane. Find the transfer function, frequency response and difference equation for this filter. What is the bandwidth of this filter? Two pole band-pass filter will have a transfer function of the form: G * ( pz ( p z jω jω ( R e z ( Re z G + az + az G where p and p* represent the complex conjugate pair of filter poles. To design this filter we need to determine filter parameters G (filter gain, a and a (filter coefficients. Equations for filter coefficients can be derived from obtained filter transfer function as: a Rcosω a R Filter gain is derived from filter normalisation requirement, i.e. from H ( ω it can be derived that: ( cos( ω G R R + R Using filter specification given in the question and normalising filter peak frequency, i.e. π f.63 ω filter parameters are obtained to be: G.58 a.956 a.96 Expression for filter transfer function is therefore: H( z Expression for filter frequency response: H ( ω z +.96z.58 j ω.956e +.96e jω

3 M58 Mixed Signal Processors page 3 of 6 Filter frequency response will have a single peak at specified pass frequency of the filter ω. Filter Magnitude Rersponse H( ω.. Filter Frequency Response 3.. ω ω Angular Frequency Starting from transfer function for this filter we have: Y z G X z + a z + a z ( + + Y z a z a z GX z Y z GX z a z Y z a z Y z Applying the inverse z-transform to the equation above we obtain a difference equation which needs to be implemented in the real-time, signal processing program. ( (.58 x( n +.956y( n.96y( n y n Gx n a y n a y n Bandwidth of the filter can be obtained from the relation: Δ ω ( R, i.e. ω Δ.99.6 Δω f which corresponds to Δ f s 5.5 Hz π

4 M58 Mixed Signal Processors page 4 of 6 EXAMPLE: Design a notch filter with the following specification: Notch frequency: 5 Hz 3 db notch width: ±5 Hz Sampling frequency: 5 Hz To reject a 5 Hz component we place a pair of complex zeros at points on the unit circle corresponding to 5 Hz, i.e. at angles: π fn π 5 ω ± ±.π 36 5 To achieve a sharp notch filter and improve amplitude response on either side of the notch frequency, a pair of complex conjugate poles can be placed at a distance/radius R from the centre of the unit circle. The width of the notch, i.e. filter bandwidth is determined by the location of the poles and given by relation: Δ ω R Δf π Δω Therefore, the pole magnitude can be obtained as: R Diagram below shows the filter pole-zero plot in the z-plane. Im z R p ω Re p z Transfer function of the filter will be of the form: ( z z( z z ( z p ( z p jω jω ( z e ( z e jω jω ( z R e ( z R e

5 M58 Mixed Signal Processors page 5 of 6 Replacing with obtained poles and zeros and doing some calculations we get: j36 j36 ( z e ( z e j36 j36 ( z.937 e ( z.937 e z.68z+ z.564z z + z.564z z This leads to a difference equation for this filter: Y z.68z z X z +.564z z ( (.68 + Y z z z X z z z.68 ( + ( (.8783 ( y n x n x n x n y n y n

6 M58 Mixed Signal Processors page 6 of 6 Slightly more complex example. EXAMPLE: A DSP system operating at a sampling rate of 5 Hz is plagued by 5 Hz frequency interference noise from power lines and its harmonics. Design a notch filter that removes all of these harmonics but remains flat at other frequencies. The fundamental harmonic is: ω π f π 5 5.π The other harmonics at fiif correspond to ωiiω. There are harmonics that lie within the Nyquist interval [,]; namely, fi for i,,,9. Because f all the harmonics that lie outside the Nyquist interval (if they have not been filtered out by the aliasing prefilter will be aliased onto harmonics inside the interval. For example the harmonic ff gets aliased with f- f-ff and so on. Our digital notch filter therefore must be designed to have notches at the frequencies within the Nyquist interval: π i ω iω i,,...,9 These frequencies correspond to roots fo the polynomial: The notch filter is then obtained by: where R ρ. 9 i ( j z N z z e ω ( ρ i N z z z N z ρ z R z Choosing the radius ρ of the poles of the above transfer function close to the unit circle, e.g. ρ.998 will result in very sharp notches at the desired frequencies. At other frequencies the magnitude response will essentially remain flat.