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Unformatted text preview: Assignment #15 — Solutions — p.1 ECSE2410 Signals & Systems — Spring 2009 Due Tue 03/24/09
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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=xexact2xpos;
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
ECSE2410 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 ECSE2410 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}
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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)
afar9"“ : _%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 straightline 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
ECSE24lO Signals & Systems  Spring 2009 Due Tue 03/24/09 3(20). Text 6.56. You need only consider straightline 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 signalto~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 signalto—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 signaltonoise 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 signalto—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
ECSE2410 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.) ...
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