Newsgroups: sci.electronics.design
Subject: Re: Anyone know of a Squarewave to Sinewave converter IC?
Date: Mon, 06 May 1996 17:35:32 -0400
Organization: Lockheed Sanders
>
> That's what I'm looking for--a chip that accepts a squarewave input of,
> say, .1 to 1.0 Mhz and outputs a sine wave of the same frequency...
>
> Dave
Generating a true sinewave from an input square wave over an octave in
frequency is not simple. A sinewave oscillator would be easy. Filtering
to the fundimental would be easy for a frequency range less than an
octive.
The best I was able to come up with is the attached circuit which
approximately generates the sine of the input voltage. Integrating the
squarewave to a triangle wave and inputting it to the circuit below
should work.
There is another way to approximate a sine wave using zener diodes
in an op amp circuit to form a straight line approximation, but I
couldn't find that circuit.
Anyway, here is one method:
The following is taken from "Functional Circuits" , Yu Jen Wong and
William Ott, McGraw-Hill (c) 1976 pp207-209
+----------------------------------------------+
| |
R1 |
| +-R6-+-----R5------+
+-----R3--+ v | |
| | \ | +--------+ | | \ |
e-in -+-R2+--|- \ | A | mult | +--|- \ |
| |op >--+-------| x xy| C |op >------+-e-out
| +-|+ / | --|---R4-+-|+ /
| | | / +-| y 10| | | /
| v | +--------+ |
| +--------+ | |
| | mult | | |
+-----| x xy| B | |
| | --|----+ +-R7---------+
+-----| y 10| | |
| +--------+ | R8
| | |
+------------------------+ v
e-in = triangle wave
A = -5.212*ein - 4.520*eout
B = ein^2/10
C = (-ein^2/100)* (5.212*ein+4.520*eout)
e-out = (1.0287*ein - 0.10423*ein^3) / (1 + 0.0904*ein^2)
= 1.0287ein - ein^2(0.10423*ein+0.0904*eout)
R1 = 11.06 K
R2 = 9.594 K
R3 = 50 K
R4 = 4.857 K
R5 = 10 K
R6 = 3.333 K
R7 = 9.442 K
R8 = 10 K
op - operational amplifier
mult - four quadrant multiplier chip
v - ground
IMPLICIT FEEDBACK AND FOUR QUADRANT SINE GENERATOR
The Tchebysheff polynomials method .. can be applied to approximate a
sine function in all four quadrents. Two terms are generally not enough
to achieve reasonable accuracies. If three terms are retained, the
approximation, as given in [the equation below], will yield less than
0.75 percent error from -180 degrees to +180 degrees.
eo=sin(ei) ~= 0.98402ei - 0.15328ei^3 + 0.0054523ei^5^Ô
However, if implicit feedback is utilized, the four quadrent sine
function can be approximated by:
eo= sin(ei)
1.0287ei - 0.10423ei^3
~= -------------------------
1 + 0.0904ei^2
The approxmation has less than 1.25 percent error in four quadrents. In
order to model [the above equation], rewrite it as
eo = 1.0287ei - ei^2(0.10423ei + 0.0904eo)
The circuit...requires two multipliers and two operational amplifiers.