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Musical-Applications-of-Microprocessors-2ed-Chamberlin-H-1987

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596 MUSICAL ApPLICATIONS OF MICROPROCESSORS<br />

N<br />

-c_L<br />

74161<br />

00 ~OIVIOEO<br />

UTPUT<br />

ClK<br />

OC<br />

10<br />

[7<br />

CET OB<br />

CEP OA<br />

-----3.<br />

0<br />

~B3 CO QQ ~<br />

~ B2<br />

L3 15 13 0 00<br />

~ 81 OC ~<br />

~ BO ZZ2 12 C oc<br />

MS8 7483A 74/75 OB ~<br />

,,~ ~B OS<br />

I<br />

A3<br />

3 A2 OA ~<br />

-- ~ AI L0 9 4 A QA<br />

10 AD<br />

CI<br />

~ ClK<br />

~ B3<br />

B2<br />

~<br />

13<br />

1<br />

14<br />

CO<br />

00 ~<br />

15 13 0 Q5<br />

L3<br />

v-!---2- BI ~ I~<br />

2<br />

~ BO L2<br />

12 C OC<br />

7483A 74175 OB ~<br />

1 6 5<br />

A3 LI<br />

B OS<br />

3 A2 OA ~<br />

--~ AI L0 9 4 A OA<br />

LSB<br />

10 A0 9 >--- ClK<br />

Cl<br />

J3<br />

MASTER<br />

CLOCK<br />

Fig. 17-6. Experimental accumulator divider<br />

As an example, let's assume that an oscillator with I-cent (0.06%)<br />

resolution over a 20-Hz to 20-kHz frequency range and inaudible jitter is<br />

desired. Under normal conditions, 0.5% frequency shift is about the<br />

minimum audible for single tones <strong>of</strong> moderate frequency. It is reasonable to<br />

assume that this much jitter at 20 kHz would be completely inaudible, while<br />

at midfrequencies the jitter is far less anyway. At 1 kHz, for example, it<br />

would be a mere 0.025%, which is much lower than the wow and flutter<br />

figures for most audio equipment.<br />

The first step is to determine the clock frequency from the jitter requirement.<br />

The period <strong>of</strong> the highest frequency is 50 J.Lsec and 0.5% <strong>of</strong> this<br />

is 250 nsec. Thus, the clock frequency must be at least 4 MHz to meet the<br />

jitter specification. To meet the resolution requirement at 20 Hz, N must be<br />

1/0.0006 or about 1,600. If the output frequency equation is rewritten as<br />

F=6.4 X 10 9 /2 M , M can be found by applying M=10g2(6.4 X 10 9 /20),<br />

which yields a figure <strong>of</strong> 28.2. Thus, the adder and register must be 28 bits<br />

long. Note, however, that N will never be greater than 1,342,178, which<br />

means that only about 20 bits are needed to represent N. The remaining

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