A#15_Solutions - Assignment #15 — Solutions — p.1...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Assignment #15 — Solutions — p.1 ECSE-2410 Signals & Systems — Spring 2009 Due Tue 03/24/09 50%? +1] 1(60). H ((0) = (a) & (b) Sketch the straight—line Bode magnitude and the phase plot. . . J 60 a) +1 —— + 1 1(00 >[ 50 ) . . . . . 1 (C) Usmg the Bode phase appr0x1mat10n, determlne the approxunate phase when a) = —. J1—6 Amway. 322; +M‘(WS)~W;CJV%) * Wfi (‘93:) “ ~1_A_fl;g_3_flflfi2.oz__ ,_,° 1:. W5 W5 .... W.— H—flufi mama-{e 4% Fame apps/mi gee nah’twc) fags. Assignment #15 -— Solutions - p.2 ECSE~2410 Signals & Systems ~ Spring 2009 Due Tue 03/24/09 90 Accuracy of the Bode phase approximation {ghase iéegrees} vs. freguencv {rad/sec} 80 em E l 70 k—ww 60; i i i W 1 i 50 7: T I / tan—lwz———— 1/”, 2 ‘0 4O 30% w=logspace(~2,2, 100); x=atand(w); w 1 =10gspace(—2,0, 100); xneg=57.3*w1; w2=logspace(0,2, 100); xpos=57.3*((pi/2)-(1./w2)); p10t(10g10(w),x,10g10(w1),xneg,log10(w2),xpos); grid -.01 -.1 1 10 100 1 4 tWWW“W.NflEWWMW.MmmgmmMWWMIWWMMTWWWWM _ mommrmmwm .. W Max phase error occurs within the 12 a), two octives astride the break frequency 10 W error = tan‘1 (0—- 7: 8 r A 6 W i wwmmmmémvtWWW.»MWWWW U3 < E o 4 Wwm~,,4,,,w,w,§,.,www~nwNMNWNMEMWWMMWM 4 “in WW3 5.3 2 x 0 g _2 meéww M W WEEKme , wl:10gspace(—2,0,100); a xexactl:atand(w1); -4 WWWW§WWWWW§W WW xneg=57 .3*w1; _6 WWW 5 WWW "WW W,“ m . w2=logspace(0,2,100); 3 xexact2=atand(w2); -8 “WMWW E "'““”W“‘”‘§““”W“’WW xpos=57.3*((pi/2)-(1./w2)); _ 1 O "WWWQWMWWWWW wwwéwwmwww eneg=xexact1‘xneg; ‘ epos=xexact2-xpos; -1 2 ’ 9 ' ' plot(log10(wi),eneg,log10(w2),epos); _1 4 grid -.01 -.02 —.05 -.1 —.2 —.5 1 2 5 1O 20 50 100 frequency (rad/sec) Assignment #15 - Solutions -— p.3 ECSE-2410 Signals & Systems ~ Spring 2009 Due Tue 03/24/09 Accuracy analysis of our problem. Study the contribution of the various terms. Neither phase term is evaluated within an octive of its break frequency, so the approximation should be fairly accurate. This term contributes very little at a) = 7:, 10 because it is evaluated more than two decades 1 from its break frequency of a) = 50. Phase at a) = —: JiE Exact phase. 7r 1 0.02 AH 60 = ——- + tan’1 x/l0 -— tan"l — tan—1 : ——90° + 72.45" — 1755“ —0.36° = —35.46° Approximate phase. 7: 7r 1 1 0.02 2 2 J15 M M = —90° + 71.89° —l8.12° —-O.36° = ~36.60° % error in the approximation is = - 100 = 3.2% 35.46 Assignment #15 -— Solutions — p.4 ECSE-2410 Signals & Systems — Spring 2009 Due Tue 03/24/09 2. Continued. (d) Using the Bode phase approximation, at what frequency is the phase —- g; ? ASSN 34w 45c: Z a} “l _.1__.._‘t2,.,z..‘-l..ufi 45:19: “%"‘Ca "5)”03. w) Csvi" 72 2:; 3*; ~17 3 «ii... 32... a. ameztsrw+53=o 41‘ in 5:3 “‘7 w 3; mo‘i'S’. 46?- ???ng its WWW 7&4» W Majesty) 3w: €47.92 ragga! C(QWEGMriO “mi W (0:: Slflfl‘agg (F'er Mmké (347+) M515 5% ewwt 5-” “a 1.028 ,5 46,5”9 ’_ 4g“ (5 .‘l- Ewe; 4w; ~K+WF§ 343“ a) afar-9"“ : _%o +3333?“ @”‘ “13%.? x W’”‘ I;le 4% 52+ CZ» 5‘ )“ L%"§%‘+)“L%" 753:5) (e) Use the bode command in MATLAB to obtain the magnitude and phase plots. Submit your MATLAB code with the two graphs. Your name should appear in the title of each graph. (f) By hand, draw the Bode straight~line approximations on top of the MATLAB graphs. Submit these two graphs. 25 , ,, , See next page. Assignment #15 — Solutions - p.6 ECSE—2410 Signals & Systems « Spring 2009 Due Tue 03/24/09 2(20) (a)(lO). Use the bode command in MATLAB to obtain the magnitude and phase plots for . j a) H (0)) = . Submit your MATLAB code with the two graphs. Your name (ja) + 10(fl + 1] should appear in the title of each graph. (b)(10). By hand, draw the Bode straight-line approximation (magnitude) and approximation of phase on top of the MATLAB graphs. Submit graphs. See. aw~i~ pang/Q - Smog; 324 25000 3 +0 ~H(w) A $g+ no 32%— 351203 + 2.5000 ESmmE mvom (69p) eseqd (ap) epnuufiew Assignment #15 — Solutions — p.8 ECSE-24lO Signals & Systems - Spring 2009 Due Tue 03/24/09 3(20). Text 6.56. You need only consider straight-line Bode magnitude plots. Neglect the phase plots. (a) far) The signal is referenced to 0 dB for convenience. The important factor is the ratio of signal-to~noise. The noise spectrum is: SKHZ lOKHz w The spectrum of the output, ((0), would have the following characteristics. In the frequency range, 50Hz < f < SKHz , 1 (£35) The signal-to—noise ratio is OdB to — 40dB, i.e., = 100. Thus the signal is 100 times S 7x? “stronger” than the noise. But in the frequency range, lOKHz < f < ZOKHZ , The signal-tonoise ratio is 0618 to - 28dB , i.e., the signal is only 25 times “stronger” than the noise. This ratio is too small, because the “hiss” will be more noticeable. The solution is to prefilter, H 1 (0)) , the input signal and make it 4 times (12dB) larger in the frequency range lOKHz < f < ZOKHZ where the noise is strongest. Thus we keep the signal-to—noise ratio at 40dB over the entire frequency range of the signal (boost the signal in the frequency range most affected by noise.) Assignment #15 - Solutions - [2.9 ECSE-2410 Signals & Systems - Spring 2009 Due Tue 03/24/09 3(20). Text 6.56. Continued. We want an enhanced spectrum that looks like: SKHZ lOKHz ZOKHZ SKHZ IOKHz ZOKHZ Note that the high frequencies above lOKHz are amplified. Thus the output sounds as if it has lots of treble, or high frequency. Assignment #15 - Solutions —- p.10 ECSE—2410 Signals & Systems - Spring 2009 Due Tue 03/24/09 3(20). Text 6.56. Continued (2). (c) To make the output signal be §(t) 2 s(t) , we need to filter out these high frequencies introduced by a) 2 1 [HQ—J ] filter H,(a)). Thus [12(60): = ' 2 ,and its Bode plot is H100) [1+ ja) ] 1H2 (a2)[dB SKHZ lOKHz a) We have now investigated a practical method for reducing noise. The strategy is to boost the input (desired) signal at the frequencies where the noise is strongest. Thus the noise becomes a smaller “percentage or ratio” of the contaminated signal being processed (amplified) in the follow—on stages of the system (tape player in our case.) ...
View Full Document

Page1 / 10

A#15_Solutions - Assignment #15 — Solutions — p.1...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online