Transfer function h in factored form h s k s z1

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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&gt;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...
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