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 Posted: Tue 0:44, 25 Jan 2011 Post subject: air jordans With a digital FM signal EPROM design |
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Digital FM with EPROM design source
Ⅲ was the first pulses, the frequency of F: l / T: / (NL-m +1) (m = l, 2, ...,[link widoczny dla zalogowanych], M) a 35 - under the theory of linear frequency modulation formula. When you hear too often should be f (t) fI + kffNL-m / 2 +1 / 2) m / fs]. Where 1 is the first pulse frequency, k is a constant. Thus, the digital FM frequency with the theoretical value can be defined as poor dishes: IF (m) a Rt /) X100% = f (m A 1) flfs-km (Nl-m / 2 +1 / 2)】 / {( N1 a l'll +1) X 【f. f + km (N 【a m / 2 +1 / 2)]} X100% (normal N,>> m) ≈ 【(m ~ 1) fIfs-kmN ~. 】 / FlfsNLX100% _ factory where l =, = fo a /. (2) EPROM Digital FM nonlinearity Figure 2 shows the linear frequency modulation and digital FM characteristic curve, where A is the linear chirp of the theoretical value, curve 2 the amount of linear frequency modulation and digital modulation curve q: △ F / Af × 100% B is composed of digital FM EPRoM the actual value of the frequency change between two consecutive pulses was the first step c m a drunk red frequency shift is △ F1 / T a 1 / T a I = / f 【_ Ⅳ l-m) (_ Ⅳ 【a m +1)] FM signal bandwidth △ f1/TH a 1/TL (, a l / INL (_ Ⅳ L-M +1) 1 this definition digital frequency modulation circuit signal frequency nonlinear error is = Ⅳ L (_ Ⅳ L-M +1) / f (_ Ⅳ L-m) (_ Ⅳ L-m +1) (M-1)】 × 100% ≈ 1 / M × 100% (general NL> > M ≥ m) 4. For example assumes that the design of a frequency modulation signal source,[link widoczny dla zalogowanych], the indicators required for the signal width △ t = 10ms, the center frequency fo = 19kHz, the bandwidth △,[link widoczny dla zalogowanych], = 2kHz. by (1), (2),[link widoczny dla zalogowanych], ( 3) presents the equation:. f △ tINL-M / 2 + I / 2) M / fs a one I 』10 × 10 T. One I / (to-Af / 2) = 1 / [(i9 a J) lo] I 【T: 1 / (fo + Af / 2) l / [(19 + D10] solving equations have NL: 1890M = 190 = 34MHz calculation results show the signal from the l90 cycles followed by a difference of 29ns (1 / = 1 / (34 × 106) s) of the pulse signal composed of write EPROM of the frequency factor of 1 cycle N1: _ Ⅳ L = 1890 Ⅳ name = 945N-I =. 945 - 2 cycles N2 = 189t BU 1 = 1889N ~ 2 = 945N I, = 944 s cycle NI901890-190 = 1701I90851Ll9o = 850-36 ~ In order to use less memory and the series divider , we set the number of single-shot set of output pulse width TR: N / js set the example for N = 800, then = 800 × 29ns. So the set by the number of one-shot timer and counter preset time, the number of points to achieve comprehensive frequency divider circuit can achieve the purpose of simplifying the example of the principle of the circuit shown in Figure 3 ∞ 3 deep principle circuit set the frequency source to be written into EPROM's frequency coefficient N = N ~ N = N-N. then write AEPROM the procedures are as follows: number Address (H) NL? =: N Ⅳ _ + one -880,000 script (H) 379l7B380l7C145145145l445l509lH919l9033335. discussion (1) the capacity of the memory by the J lot, a signal which occupies only one point, so an amount of memory can store several http source frequency coefficients. This design acoustic warfare in the ever-changing} divided Li FM signal system is kept, Ill img 0 (2): * peak occasions. the r conditions restrictions within the traitor afternoon bite with poor Fana stepped off the bad linear frequency modulation circuit. The first m undecided on a ladder calculated as .= (a, bamboo +1) / (Xi = 1,2. A. acridine J set each. a step has a pulse period C, the signal width, = f (7 ++。++ ten 7) 'CM (N-M / 2 +1 / 2) / If,,. (4 unknown parameters. which has a parameter under section f1: set,[link widoczny dla zalogowanych], can be calculated} r the other three parameters. Bin Shi functional block circuit shown in Figure 4. on the sub-leading coefficient of programming methods and programming method of the previous example {the same consumption. May J kind of method, but also can analyze the amount of the other functions of the wail source. find the parameters of order 4 H {lemon; Run by the amount of circuit sites help swine lack} ∞ l234 37 iv
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