Signal Processing and Linear Systems-B.P.Lathi copy

# 76a t he e rror between the actual plot and the

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Unformatted text preview: begin) as t he 20 dB line as before(see Fig. 7.9a). In a ddition, we have a real zero a t s = - 100 a nd a p air of complex conjugate poles. W hen we e xpress t he s econd-order factor in s tandard form, * ~ J :t: &quot;&quot; 10 I 0 N 0 I -10 - 20 i + 2 (w n s + w~ we have Wn = 10 and ._._--- - 30 S2 + 2s + 100 = _. . _-- I I t\ &quot;- ! ; i I I .~ I ! Ii I i I 41lO JW I ml I I 1/ I ' If ....... -&quot;'r ~ iI T LL i H:-I ~ j~- I! I i I - I .' .' ..' .. ' 'f j I I i I i ~ I~ i&quot; 10:&gt; I I ---·t~ I - ttl -t I ---t---t- -ff--i--- -t- I . i II i (a) \ I , I, . &quot; I ,,~ 90 f ----50 Ex&gt; p tp ot IJ~O 1 ,1 llillnototic pi .' .'I ..,. I ! .. -. -' I ( = 0.1 S tep 1 . D raw an a symptote o f - 40 d B/decade ( -12 d B/octave) s tarting a t W = 10 for t he c omplex c onjugate poles, a nd d raw a nother a symptote o f 20 d B/decade, s tarting a t W = 100 for t he (real) zero. S tep 2 . A dd b oth a symptotes. S tep 3 . A pply t he c orrection a t w = 100, w here t he c orrection because of t he c orner frequency w = 100 is 3 dB. T he c orrection because of t he c orner frequency w = 10 may b e ignored. Next, a pply t he c orrection a t w = 10, w here t he c orrection because o f t he c orner frequency w = 10 is 13.90 d B (see Fig. 7.7a for ( = 0.1 ). We m ay find corrections a t a few more p oints. T he r esulting p lot is i llustrated i n Fig. 7.9a. \=-= . ;' .9 iT I !\ ! 30 ~ 20 (7.26) 489 (b) I tt I T I! I I I &quot; ~ .c ~ - 50 P hase P lot T he a symptote for t he complex conjugate poles is a s tep function with a j ump of _ 900 a t w = 1 0, a nd t he a symptote for t he zero a t s = - 100 is a s traight line with a slope o f 45° / decade, leveling off a t w = 10 a nd w = 100 t o 0° a nd 90°, respectively. T he two a symptotes a dd t o give t he s awtooth s hown in Fig. 7.9b. We nOw a pply t he c orrections from Fig. 7.7b a nd Fig. 7.5b t o o btain t he e xact p lot. • o C omputer E xample C 7.2 Solve E xamples 7.3 a nd 7.4 using M -file f unctions in MATLAB. Frequency response m ay b e p lotted u sing several functions. For t he p urpose o f B ode plots, however, t he m ost suitable file is b ode.m, which c an be used as d emonstrated here. % F or E xample 7 .3 n um=[20 2 000 O J;den=[l 1 2 2 0]; b ode(nurn,den) % F or E xample 7 .4 n um=[O 1 0 1 000];den=[1 2 1 00); b ode(nurn,den) 0 Poles and Zeros in RHP In our discussion so far, we have assumed t he poles and zeros of the transfer function t o b e i n t he LHP. W hat if some of the poles a nd/or zeros of H (s) lie in the R HP? I f t here is a pole in the RHP, t he integral in Eq. (2.49) with s = jw does not - 100 F ig. 7 .9 A mplitude a nd p hase response o f a s econd-order s ystem in E xample 7.4. converge,. and H {jw) is meaningless. Anyway, such systems (unstable) are useless for any Signal processing application. For this reason, we shall consider only t he case of the R HP zero. We can readily show t hat t he amplitude function of a R HP zero a t a is identical to t hat of a L HP zero a t - a. T he reason is t hat 1+:I=11- :l= (l+::)t j j 1 Therefore, the log amplitUde plots...
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