Lissajous figures - Ted Pavlic

1 Lissajous figures Lab 1: Introduction to InstrumentationECE 209:Circuits and Electronics LaboratoryALissajous ( LEE-suh-zhoo ) figureis aparametric plotof theharmonic system(x(t) =Axsin( xt+ ),y(t) =Aysin( yt+ + )( ,y(x) =Aysin y x arcsin(xAx) + where|x| Ax). ytIn our case, we plot an inputx(t) and outputy(t) of alinear time-invariant (LTI) system. Becausecomplexexponentialsareeigenfuncti onsof LTI systems and sinusoids aresums of complex exponentials, the outputfrequency will match the input frequency ( , x= y= = 2 f). Our LTI system ( , the phase-shiftercircuit) is anall-pass filter, and so it ensures thatAy=Ax=A. So we are consider the simpler system(x(t) =Asin(2 f t+ ),y(t) =Asin(2 f t+ + )( ,y(x) =Asin arcsin(xA) |{z}x ygraph has no dependence on .where|x| A),(1)and we use a Lissajous figure to find thephase shift . We obtain the Lissajous figure with theoscilloscopein itsX Ymodewith the input of our system tied to theXchannel and the output tied to each instant, the scope plots a dot with the inputXsample as the horizontal coordinate and the outputYsample as the vertical coordinate.))

2 Because the dotspersiston the screen for a short time, their ghosts form a Lissajous figure on the screen. To see the rotation direction, we can slow down the input (t)y(t)(x(t) = sin(2 t)y(t) = sin(2 t 4)|{z} = 45 = 45 x(t)y(t)(x(t) = sin(2 t)y(t) = sin(2 t)|{z} =0 (in phase) = 0 x(t)y(t)(x(t) = sin(2 t)y(t) = sin(2 t 2)|{z} = 90 (in quadrature ) = 90 x(t)y(t)(x(t) = sin(2 t)y(t) = sin(2 t 3 4)|{z} = 135 = 135 x(t)y(t)(x(t) = sin(2 t)y(t) = sin(2 t )|{z} = 180 (inverted) = 180 x(t)y(t)(x(t) = sin(2 t)y(t) = sin(2 t 5 4)|{z} = 225 = 225 x(t)y(t)(x(t) = sin(2 t)y(t) = sin(2 t 3 2)|{z} = 270 (in quadrature ) = 270 x(t)y(t)(x(t) = sin(2 t)y(t) = sin(2 t 7 4)|{z} = 315 = 315 IncreasingphaseshiftNegative slope (II & III)Positive slope (I & IV)Clockwise (I & II)Counter clockwise (III & IV)LTI Lissajous figures areovalswitheccentricityanddirection of rotationdetermined by phase shift . Document Source code 2007 2009 by Theodore P.))))))))

3 PavlicCreative Commons Attribution-Noncommercial LicensePage 1 of 2 ECE 209 [Lab 1: Introduction to Instrumentation] Lissajous figuresSo if we knowboththe angle of themajor axisof the Lissajous curveandthe direction of the curve srotation, then we can determine thequadrantof the phase shift . That is, = 0 iflinewithpositive slope0 > > 90 ifcounter clockwiseandpositive slope = 90 ifcounter clockwisecircle 90 > > 180 ifcounter clockwiseandnegative slope = 180 iflinewithnegative slope 180 > > 270 ifclockwiseandnegative slope = 270 ifclockwisecircle 270 > > 360 ifclockwiseandpositive slope(2)where we consider onlynegative becausephysical systemsarecasualand will only = 0 , the input and output are said to be in phase. Alternatively, when = 180 , the input andoutput are inverted copies of each other and are said to be out of phase or simply inverted. In the othertwo cases, when = 90 or = 270 , the input and output are said to be inquadrature ( , theyare a quarter wavelength away from being in phase).

4 Quadrature motion is perfectly circular and has a widerange of applications throughout phase shift from measurements:To determine the precise phase shift from measurements,we must useEquation (1). If we know the sinusoidal amplitudeAand a measurement (x(t0), y(t0)) fromtimet0, then we can usex(t0) to solve for 2 f t0+ , and then we can usey(t0) to solve for . That is, = arcsin x(t0)A arcsin y(t0)A .(3)Because each arcsin can match as many astwoangles in any 360 range, there are four possible onefor each of the four quadrants. So we useEquation (2)to pick the correct out of the method for the laboratory:The following procedure helps prevent measurement errors fromnonzeroDC offset. If a measurement at timet0hasx(t0) = 0, thenEquation (3)becomes = arcsin(0) arcsin(y(t0)/A). This case corresponds to finding the point where the Lissajous figure intersects with thevertical (0, y(t0))(0, y(t0))2y(t0)Yoscilloscope channelXoscilloscope channel2 AAAYour calculator givesyou ,Arcsin 2y(t0)2A.

5 Then 0 90 , and = 360 + (Q-I), or 180 (Q-II), or 180 + (Q-III), or (Q-IV).(note:inthis example,it mustbe that is in quadrantII or III because the majoraxis has a negative slope)1. UseXcursorsandXposition knob tohorizontallycenter the Lissajous figure on the on-screen UseYcursorsto measure thedistance( , Y) between two intersection points ( , find 2y(t0)).3. UseYcursorsto measure the maximum vertical span ( , 2A).4. Let ,Arcsin( 2y(t0)/(2A) ) and choose { , 180 + , 180 , 360 + }usingEquation (2). Our phase-shifting circuit delays by no more than 180 , and so is in quadrant III or IV. Further,the major axis in this example has a negative slope, and so is quadrant III ( , = 180 + ).Copyrightc 2007 2009 by Theodore P. PavlicCreative Commons Attribution-Noncommercial LicensePage 2 of 2