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.)))

2 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. 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))))))))

3 LTI Lissajous figures areovalswitheccentricityanddirection of rotationdetermined by phase shift . Document Source code 2007 2009 by Theodore P. 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.

4 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). 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.

5 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 .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)).

6 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