EE 205 Dr. A. Zidouri. Electric Circuits II. Frequency Selective Circuits (Filters) Bode Plots: Complex Poles and Zeros.

Transcription

1 EE 5 Dr. A. Zidouri Electric Circuits II Frequecy Selective Circuits (Filters) Bode Plots: Complex Poles ad Zeros Lecture #4 - -

2 EE 5 Dr. A. Zidouri The material to be covered i this lecture is as follows: o Bode Diagrams for Complex Poles ad Zeros o Straight-Lie Amplitude Plots o Correctig Straight-Lie Amplitude Plots o Phase Agle Plots - -

3 EE 5 Dr. A. Zidouri After fiishig this lecture you should be able to: Use Bode Diagrams for Complex Poles ad Zeros Draw Straight-Lie Approximatio of the Amplitude Plot for a Pair of Complex Poles (Zeros) Draw Straight-Lie Approximatio of the Phase Agle Plot Correct the plots for a Pair of Complex Poles (Zeros) - 3 -

4 EE 5 Dr. A. Zidouri Bode Diagrams for Complex Poles ad Zeros The complex poles ad zeros of H(s) always appear i cojugate pairs Always combie the cojugate pair ito a sigle quadratic term Oce the rules for hadlig poles is uderstood, their applicatio to zeros becomes apparet. Let s cosider Hs K ( ) = s + α jβ s + α + jβ (4-) ( )( ) The product ( s α jβ )( s α jβ ) ca be writte as: ( s α ) β s αs α β + + = (4-) s + α s+ α + β = s + ζ s+ (4-3) = α + β (4-4) ad ζ α or where = (4-5) The term is the corer frequecy of the quadratic factor The term ζ is the dampig coefficiet of the quadratic factor, if ζ < the roots are complex if ζ > the roots are real, if ζ = this is the critical value. For real roots we treat them as we have see i previous lecture (Lec#4). Assume K (4-5) Hs ζ < the ( ) = s+ ζ s

5 EE 5 Dr. A. Zidouri Hs H j Let ( ) = K (4-6) i stadard form will be Hs ( ) = K ζ s s + + s s + + ζ ( ) = K (4-8) where K = K j ζ + u = K the H ( j ) = (4-9) i polar form u + H jζ u ( j K ) = u + jζ u β (4-) = log = log K log u ( ζ u + ) (4-) From which: AdB H ( jw) u 4 ( ζu ) u u ζ log + = log + + (4-) θ= β= ta ζ u u ad ( ) (4-3) (4-7) - 5 -

6 EE 5 Dr. A. Zidouri Approximate Amplitude Plots db log log 4 A = K u + u ζ + Because u = therefore u as ad u as log u4 u ζ log 4 ζ 4log + + as u u + u + u as u We coclude that the approximate amplitude plot cosists of two straight lies. o For frequecies < the straight lie has a slope of db o For frequecies > the straight lie has a slope of -4dB/decade. These two straight lies joi o the db axis at u= or =. Fig. 4- shows the straight lie approximatio for a quadratic factor with ζ <. thus: - 6 -

7 EE 5 Dr. A. Zidouri Correctig Straight-Lie Amplitude Plots: Correctig the straight-lie amplitude plot for a pair of complex poles is ot as easy as correctig a first-order real pole, because the correctios deped o the dampig coefficiet ζ The straight-lie amplitude plot ca be corrected by locatig four poits o the actual curve: o Poit at ½ the corer frequecy = = u = the u4 u ζ ζ ζ = C A db ( ) = log ζ o Poit at peak amplitude frequecy The amplitude peaks at p ζ u AdB p = o Poit 3 at the corer frequecy u = = db ( ) = log 4ζ or A db ( ) = A o Poit 4 at zero amplitude frequecy = ζ = p = + 8 = =, thus ( ) log 4ζ ( ζ ) log ζ - 7 -

8 EE 5 Dr. A. Zidouri 3 A db A db / p Fig. 4- Four poits o the corrected amplitude plot for a pair of complex poles / (rad/s) p Fig. 4-4 the amplitude plot for Example

9 EE 5 Dr. A. Zidouri Example 4- Compute the trasfer fuctio for the circuit show i Fig. 4-3 a) Fid the corer frequecy, b) The value of K, c) The dampig coefficiet ζ d) Make a straight lie amplitude plot ragig from to 5 rad/s. e) Sketch the actual corrected amplitude i db at, p,,ad. f) Describe the type of filter represeted by the circuit i Fig. 4-3 Ω V i V o Fig. 4-3 The circuit for Example 4- Solutio: Hs Trasformig to s domai ( ) = a) from the expressio of H(s), 5 b) By defiitio K 5 = or = LC 5 R s + s s s L LC =, therefore = 5rad s - 9 -

10 EE 5 Dr. A. Zidouri c) The coefficiet of s is ζ therefore ζ = =. d) See Fig. 4-4 e) The actual amplitudes are AdB ( ) = log (.65) =.db = 5.9 = 47.96rad s AdB p ( p ) ( )( ) AdB ( ) = log (.4) = 7.96dB = 67.8 p = db AdB ( ) = db = log.6.96 = 8.4dB Fig. 4-4 shows the corrected plot. f) From the plot we see that this filter is a LPF, its cutoff frequecy appears to be 55rad/s almost the same as that predicted by the straight-lie approximatio. - -

11 EE 5 Dr. A. Zidouri Phase Agle Plots The phase agle is zero at zero frequecy The phase agle is -9 o at corer frequecy The phase agle is -8 o at large frequecy as For small values of ζ, the phase agle chages rapidly i the viciity of the corer frequecy. The lie taget to the phase agle curve at -9 o has a slope of -.3/ζ rad/decade or -3/ζ degree/decade ad it itersects the o ad -8 o ζ ζ lies at u = 4.8 ad u = 4.8 respectively. Fig. 4-5 depicts the straight-lie approximatio for.3.6 = 4.8 ζ ζ = ad shows the actual phase agle plot..6 = 4.8 ζ - -

12 EE 5 Dr. A. Zidouri Self Test 4: The umerical expressio for a trasfer fuctio is Hs ( ) Compute the corer frequecy, the dampig coefficiet, frequecies whe H ( j ) is uity, peak amplitude of H ( j ) i db, frequecy at which the peak occurs, ad amplitude of H ( j ) at half the corer frequecy. Aswer: = 5krad s, C ζ =.,, 67.8krad s 8.4dB, 47.96krad s.db 5 s 5 8 = s - -