A#13_Solutions

# A#13_Solutions - Assignment#13 «— Solutions — 13.1...

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Unformatted text preview: Assignment #13 «— Solutions — 13.1 ECSE~2410 Signals & Systems ~ Fail 2006 dyU) 1(10). Text 6.9. Given t +5y(t)=2x(t) (a) Find 1imyslep(t). Ti? £61m 9% ﬁg gawﬁgﬁm gait (b) Find I0 such that ymp (to) = ymp(oo)[1—w}3w). 39%»? W; dé. - “it A?" \$ﬂfé9m‘zm“? Fri 10/27/06 (:74? i Assignment #13 — Soiutions —— {3.2 ECSE—24I0 Signals & Systems v Fail 2006 Fri 10/27/06 2(20). Text 6.15. Determine damping for: d 2W) (Mt) (a) +4wggw+4y(t)=x0) dtz r) 2.9% 5% =4 a; «25:2. d 2W) 62%) (C) dig jomawwtw) 23w“:2w mfg} 1 = Em WW 2 W a dams) m _ gdxm (d) 5 dig +4 dz +5y(r)—7x(t)+3 d: + 5W 4%.. [E5 Assignment #13 m Solutions ~— p.3 ECSEQMO Signals & Systems — Fall 2006 Fri 10/27/06 _ w:w¢gw M @5in .. m? Mfma 3(10). Text 6.19 Assignment #13 -— Solutions m» 33.4 ECSE-2410 Signals & Systems - Fall 2006 Fri 30/27/06 4(30). Text 6.32(a). Make the compensator frequency response, H C (am), as simple as you can. This means pole-zero cancellation of terms is permissible. There are many possible solutions. Verify that your design meets the given specs in the problem. .2 \ m be: MM m} We! C(wi fire)- Hm; : (ii) Eta. a 1; osmotic/j 40\$ {Homage "Desired We) (W afoot.) Assignment #13 —— Solutions w p.5 ECSE-2410 Signals & Systems - Fall 2006 Fri 10/27/06 5(30). Text 6.56. You need only consider straight-line Bode magnitude plots. Neglect the phase plots. (3) __ __ ’_ nil} l“ I} “r A YB") l +- ‘ \w-ip so m W a 1 Has) a...” m W ._.... ir- The signal spectrum is given as: The signal is referenced to 0 dB for convenience. The important factor is the ratio of signal-to-noise. The noise spectrum is: i AB T he spectrum of the output, 15160); wouid have the following characteristics. In the frequency range, SOHZ < f < SKHZ , l . . . The signal—to~noise ratio is 0:18 to —— 40dB , i.e., in»: m m 100. Thus the Signal 18 100 times “1'66” “stronger” than the noise. But in the frequency range, lOKHz < f < ZOKHZ , The signal~to—noise ratio is 0618 to — 28dB , i.e., the signal is only 25 times “stronger” than the noise. This ratio is too sniali, because the “hiss” will be more noticeable. The solutions is to preﬁlter, H1050), the ingot signal and rnake it 4 times (12633) 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 #13 — Solutions —- p.6 ECSE~2410 Signais & Systems ~ Fail 2006 Fri 10f27/06 5(30). Text 656. Continued. We want an enhanced spectrum that looks like: The straighbline Bode magnitude pEot of H1(a)) is 12 éﬂbﬁﬁ (b) We now have 36% Where the spectrum of the output, p(t) , is: Wk Note that the high frequencies above IOKHZ are ampliﬁed Thus the output sounds as if it has lots of treble, or high frequency. Assignment #13 — Solutions — p.7 ECSE-2410 Signals & Systems - Fall 2006 Fri 10/27/06 5(30). Text 6.56. Continued (2). (c) To make the output signal be §(r) = 5(t) , we need to ﬁlter out these high frequencies introduced by . 2 i ( +—2 fail ﬁiter H1 (60). Thus H2012) 2: mmim and its Bode plot is 23-51? H1060) [1+ jco EHJwﬁ 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 piayer in our case.) ...
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A#13_Solutions - Assignment#13 «— Solutions — 13.1...

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