Transfer function h in factored form h s k s z1

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: · (s − zm) (s − p1) · · · (s − pn) Bode plots: 20 log10 |H (j ω )| = 20 log10 |k | + − ￿ H (j ω ) = ￿ n ￿ i=1 k+ m ￿ i=1 20 log10 |j ω − zi| 20 log10 |j ω − pi| m ￿ i=1 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly ￿ (j ω − zi) − n ￿ i=1 ￿ (j ω − pi) 10 Graphical interpretation: |H (j ω )| ! × • ￿m dist(j ω , zi) i |H (j ω )| = |k | ￿n=1 i=1 dist(j ω , pi ) s = j! ◦ × ! where dist(u, v ) = |u − v | × This means that • |H (j ω )| gets big when j ω is near a pole • |H (j ω )| gets small when j ω is near a zero EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 11 Graphical interpretation: ￿ H (j ω ) ! × • s = j! ◦ × ! × ￿ H (j ω ) = k + ￿ m ￿ i=1 ￿ (j ω − zi) − n ￿ i=1 ￿ (j ω − pi) This means that ￿ H (j ω ) changes rapidly when a pole or zero is near j ω EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 12 Example: Single Pole Lowpass Filter H (s) = 1 ; s+a a>0 From distance interpretation, corner frequency (3dB point) is ωc = ±a Phase shift goes from 0 to -90◦ Bode Plot 0 Magnitude (dB) ! ω=a −20 dB/decade !20 !30 !40 !50 !2 10 !1 0 10 10 ω/a 1 2 10 10 0 × −a !20 ! Phase (deg) √ a2 −3 dB !10 ω = −a !40 !60 !80 !100 !2 10 !1 0 10 10 Frequency (rad/s) 1 2 10 10 ω/a EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 13 Gain Enhancement by Poles • A pole at s = σ + j ω increases the frequency response at j ω • As the pole gets closer to the j ω axis: – The increase in the frequency response gets larger – The range of frequencies effected gets narrower Bode Plot Bode Plot 20 20 10 ! × Magnitude (dB) Magnitude (dB) 10 0 !10 !20 0 10 !40 !1 10 1 10 Phase (deg) !100 !100 !150 !150 !200 !1 10 1 10 !50 !50 Phase (deg) × 0 10 0 0 ! !20 !30 !30 !40 !1 10 0 !10 0 10 Frequency (rad/s) s = −0.1 ± j EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 1 10 !200 !1 10 0 10 Frequency (rad/s) 1 10 s = −0.05 ± j 14 Gain Suppression by Zeros • A zero at s = σ + j ω decreases the frequency response at j ω • As the zero gets closer to the j ω axis: – The decrease in the frequency response gets larger – The range of frequencies effected gets narrower Bode Plot Bode Plot 40 40 30 ! ◦ Magnitude (dB) Magnitude (dB) 30 20 10 0 10 0 !10 !10 !20 !1 10 20 0 10 !20 !1 10 1 10 Phase (deg) ◦ 1 10 200 150 150 Phase (deg) 200 ! 0 10 100 0 !1 10 100 50 50 0 10 Frequency (rad/s) 1 10 s = −0.1 ± j EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 0 !1 10 0 10 Frequency (rad/s) 1 10 s = −0.05 ± j 15 Filter Design Idea • Basic components are real zeros and poles and complex pairs of zeros and poles. • Approximate the desired response by combinations of these components • Many simple filters can be designed directly • Properties of more complex filters can be analyzed from this perspective EE102A:Signal Processing and Linear Sy...
View Full Document

Ask a homework question - tutors are online