16 frequency and phase measurement rev 3 080524

7/27/2019 16 Frequency and Phase Measurement Rev 3 080524

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16 Measurement of Frequencyand Phase Shift

0908341 Measurements & Instrumentation

Copyright held by the author 2008: Dr. Lutfi R. Al-Sharif Page 1 of 21

Chapter 16Measurement of Frequency and Phase Shift

(Revision 3.0, 24/5/2008)

1. Introduct ionThe measurement of frequency and phase shift are closely linked. ThisChapter examines the various methods used to measure them. Many of themethods can be used to measure both variables at the same time.

2. Frequency MeasurementThere are four main methods for measuring frequency. These are discussedin the four subsections below.

2.1 Digital counter-timersThe most widely used method for measuring frequency is to use a digital

counter that is incremented by the frequency to be measured.Figure 1 shows the block diagram of digital counter used to measurefrequency. The system internally generates a pulse that is used as an enablesignal to the AND gate. The AND gates allows the signal to be measured toincrement an n bit counter for the duration of the enable pulse. The signal tobe measured is signal conditioned into a digital logic square wave that has thesame phase and frequency of the original signal.

The final count on the counter represents the frequency to bemeasured. At the start of the enable pulse the counter is reset, and at the endof the enable pulse the output of the counter is loaded onto a buffer thatdisplays the number on a digital display.

If the counter overflows, this indicates that the frequency to bemeasured is too high for this range and a higher range must be selected bychoosing a pulse with shorter time duration. In auto-ranging systems, thischange in range is done automatically.

For continuous measurement, this process is repeated so that thedisplay is updated with a new value of the frequency of the signal every cycle.

The following equation applies in this case:

( )121),(max

=n

unknownexternal

enablef

t


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Figure 1: Block diagram of the circuit ry for a digital counter (suitable for the measurement ofhigh frequencies).

The digital counter system discussed above is suitable for high frequencysignals. For low frequency signals this system fails, as it will take a long timeto measure the frequency of the signal (e.g., if the signal has a frequency of0.1 Hz, it will take it 10240 seconds (2.8 hours) to fill a 10 bit counter!).

When the frequency of the signal is low, the digital timer method ismore suitable. Figure 2 shows a block diagram of a digital timer used to

measure low frequency signals. Notice that the roles of counting and timeenable have now been reversed: whereby the signal to be measuredprovides the time enable pulse, and the counting frequency signal isgenerated internally.

Internal timereference

n bit

counter

AND

Externalsignal to bemeasured

tenable2

Digital LCDdisplay

Storagebuffer

Load (attrailing edge)tenable3

tenable1

tenable

f2 rangef1 range

f3 range

Frequency rangeselector switch

(automatic or manual)

f3>f2>f1

tenable

Reset counter

(at leadingedge)

Signalconditioning

(signalconverted to

square wave atdigital logic)

An-1 A0

Overflow(indicates the need to

change to a higher range)

Binaryto BCD

For continuous measurement,the enable signal becomes

periodic as shown

Reset Load

A zero or low countindicates the need to

change to a lower range


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Figure 2: Block diagram of the circuit ry for a digital timer (suitable for the measurement oflow frequencies).

Note on the figure above: enable signal is symmetrical, as it is based on thesignal to be measured and is not under our control.

For various ranges, different internal frequencies are used as shown on thethree ranges (f1, f2 and f3). An overflow in the counter indicates the need tomove to a lower range, while a zero or low counter indicates the need to move

to a higher range. In auto-ranging systems this is done automatically.There is one important difference in this case with the digital counter

method. The final count in this case is inversely proportional to frequencymeasured (in fact it is proportional to the period of the signal). For this reasoninversion processing is need on the final count to make it representative ofthe frequency. This is shown in the figure as a value inversion block.

For continuous measurement, the process is repeated every othercycle of the signal to be measured, whereby the counter is reset at the leadingedge of the time enable signal and the count is loaded at the trailing edge ofthe time enable signal.

The following equation applies to the digital timer mode:

Internalfrequencyreference

n bit

counter

AND

Externalsignal to bemeasured

fx

Digital LCDdisplay

Storagebuffer

Load (attrailing edge)

tenable=1/fx

f2 range

f1 range

f3 range

Frequency rangeselector switch

(automatic or manual)

f3>f2>f1

tenable

Reset counter

(at leadingedge)

Signal conditioning(signal converted to

square wave at digitallogic)

Width of pulse equalsperiod of signal

An-1 A0

Overflow(indicates the need to

change to a lower range)

For continuous measurement, the enablesignal becomes periodic as shown

Measurement takes place every other period

Reset Load

A zero or low countindicates the need to

change to a higher range

Valueinversion*

Binary toBCD

*Digital processing toconvert from periodtofrequency


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n

internalunknownexternal

enable

fft 2

11

),min(

==

Figure 3 below shows a multi-function 3.5 GHz Counter that can measurefrequency, period and total number of events.

Figure 3: 3.5 GHz Mutli-funct ion Counter (Frequency, Period, Totalise) Model # 1856D(courtesy of: B+K Precision Instrument, CA, USA).

2.2 OscilloscopeAnother method of measuring frequency is to use an oscilloscope in the so-called X-Y mode. If a signal with an unknown frequency is entered on the Xchannel and another signal from a function generator is entered on the Y

channel, and the frequency of the signal from the function generator iscontinuously changed until we get a stable pattern, it is possible to deduce thefrequency of the unknown signal from the resultant pattern.

The resultant pattern is called a Lissajous Pattern, examples of whichare shown in Figure 4. Note that the shape of a Lissajous Pattern depends onthe ratio of the two frequencies as well as the phase shift between them.

Some basic examples of Lissajous patterns are:

1. A straight line indicates that the frequencies of the two signals areequal and that the phase shift between them is zero degrees.


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2. A circle indicates that the frequencies of the two signals are equaland that the phase shift between them is 90 degrees.

3. An ellipse the major axis of which is pointing towards the top righthand corner, means that the frequencies of the two signals areequal and that the phase shift is somewhere between zero and 90degrees

1.

Figure 4 shows six graphs for the cases where the x and y signals have equalfrequency but are out of phase by a certain angle (0, 30, 45, 90, 135 and180). Note that the x signal is applied to the left and right hand side plates(i.e., horizontal deflection plates) and that the y signal is applied to the top andbottom plates (i.e., vertical deflection plates) of the oscilloscope.

1Circular and elliptical wave polarisation has applications in finding the resultant electrical

field of two or more electromagnetic waves. For waves that are travelling in the z direction,the resultant the electrical fileds in the x and y directions depends on the magnitude and

phase shift of the components of the electrical components in the x and y directions. Thisresults in a rotating electrical field component. The magnitude of this rotating electrical filedfollows a circular or ellipticall path.


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16 Measurement of Frequency and Phase Shift 0908341 Measurem

Copyright held by the author 2007: Dr. Lutfi R. Al-Sharif

Figure 4: Lissajous patterns for two sinusoidal signals with equal frequencies and various phase shift angles.


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16 Measurement of Frequency andPhase Shift

0908341 Measurements &Instrumentation


In case where the frequencies of the two signals are not equal, the figures getmore complicated than just a line, ellipse or circle.

Figure 5 shows the Lissajous pattern that would result from twosinusoidal signals applied to the x and y axes, with the frequency of the y-axissignal twice that of the x-axis and the phase shift between them of 90degrees.

Figure 5: Lissajous pattern caused by two s inusoidal signals with a frequency ratio of1:2 (x:y) and 90 degrees phase shi ft.

In general it is meaningless to talk about a phase shift between two signals ofdifferent frequencies. However, it does become possible to do this forfrequencies that are whole multiples of each other. In this example the 90degrees phase shift represents 90 degrees phase shift for the lower frequencysignal and 180 degrees for the higher frequency signal that has twice thefrequency. This is shown in Figure 6.


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Figure 6: Phase shi ft between two signals that do not have the same frequency.

The Lissajous pattern can also be generated for signal that have morecomplicated ratios of frequencies (e.g., 2:3, 3:4). An example of this for twosignals that have a ratio of 3 to 2 and phase shift of 90 is shown in Figure 7below.

Note that a simple way of finding out the ratio of the frequencies of thetwo signals is to count the number of edge-crossings that the pattern makeswith the x boundary and the y boundary as shown in Figure 7.

Figure 7: Lissajous pattern caused by two s inusoidal signals having a 3:2 (x:y)frequency ratio and a 90 degrees phase shift.


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Problem 2 at the end of this Chapter contains more details on the informationcontained in a Lissajous pattern and how to interpret the data, especially tofind out which of the signals leads the other in case of a phase shift betweenthem.

All the discussions so far have assumed the following:

1. The two signals have equal amplitude.2. The oscilloscope sensitivities for both channels are equal.

This has resulted in a square type of display where the x deviation and the ydeviations are equal.

If the two signals do not have equal amplitudes, or the oscilloscopesensitivities are not set to equal settings, the display will be rectangular ratherthan square. When this is the case, two signal of equal frequency and 0

phase shift would show a line that is NOT at 45 to the horizontal (e.g., asshown in Figure 8 where the amplitude of one signal is 1.5 times the other).This causes the line to be inclined at 33.7 rather than 45.

It is very important to note that this angle of incline has nothing to dowith the phase shift between the two signals and is solely caused by thedifference in amplitude.

Figure 8: The inclination of the Lissajous figure when the amplitudes of the twosignals are unequal.

Another method of measuring the frequency with the oscilloscope (but that iscrude and inaccurate) is to display the signal on the oscilloscope in the Y-tmode against the oscilloscopes time-base, and to measure the duration ofone period.

2.3 Phase Locked Loop (PLL)This is a method of measuring the frequency of signal that is submerged in

noise to an extent that it is difficult to distinguish the signal from the noise.


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Using a PLL allows regeneration of a new signal that is the same phase andfrequency as the original signal. PLL was discussed in the Noise Chapter.

The 4046 is a phase-locked loop circuit (of the 4000 CMOS logic family) thatconsists of a linear voltage controlled oscillator (VCO) and two different phasecomparators with a common signal input amplifier and a common comparatorinput [4].

2.4 Wien Br idgeThe fourth method is to use an AC bridge, called the Wien Bridge. This wasdiscussed in detail in the Chapter on AC Bridges.

3. Phase Shift MeasurementThere are four methods to measure the phase shift between two signals.These are discussed below. It is important to remember that in order to

compare the phase shift of any two signals, their frequencies must first beequal; it is meaningless to compare the phase of two signals if therefrequencies are different.

3.1 Digital TimerThe same device discussed earlier (digital counter-timer) can be used tomeasure the time elapsed between the zero crossing of one signal and thezero crossing of another. An example of a digital timer based system thatmeasures phase angle is shown in Figure 9.

Figure 9: A digital timer based, phase angle metre that can be used to synchronise

generators (Courtesy of Laureate Electronics, Inc., USA).

3.2 OscilloscopeThis has been discussed in detail earlier in this Chapter.

3.3 XY PlotterLissajous Patterns discussed earlier can be used in the same way on an XYplotter. The main limitation of an XY plotter is the limited frequency response(around 10 Hz maximum).

3.4 Phase sensitive DetectorAnother method of finding the phase shift between two signals is to multiplythem by each other and then retrieve the DC content of the resultant signal.


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It can be shown that the DC content is proportional to the cosine of thephase shift angle between the two signals, and hence achieves its peak whenthe phase shift is zero.

References & Bibliography[1] Measurement & Instrumentation Principles, Alan S. Morris, Elsevier,

2001.[2] Modern Electronic Instrumentation and Measurement Techniques,

Albert D. Helfrick and William D. Cooper, Prentice Hall InternationalEditions, 1990.

[3] Elements of Electronic Instrumentation and Measurement, Joseph J.Carr, Third Edition, Prentice Hall, 1996.

[4] HEF4046B, MSI, Phase-locked loop, Philips Semiconductors, January1995.

Problems1. We want to measure the frequency of a signal that varies between 20 kHzand 200 kHz. We shall use a 16 bit binary counter. The signal to bemeasured is converted to a square wave and directed to the 16 bit binarycounter for a period of 100 ms (i.e., 0.1 second), as shown in the figure below.

Figure 10: Block diagram of the circuitry for frequency meter.

Answer the following questions:

(a) What is the more suitable method for measuring the frequency ofthe signal in this case: Frequency measurement or periodmeasurement?

(b) What are the maximum and minimum possible counts based on theexpected range of the input frequency?

(c) What is the maximum possible frequency that this device canmeasure?

(d) What is the minimum possible frequency that this device canmeasure?

Solution

(a) As the frequency of the input signal is relatively high, the moresuitable method is the frequency method.

Internal timereference

16 bitcounter

AND Gate

Output to digitaldisplay

External signalto be measured

100 ms


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(b) The maximum possible count occurs when the input signal has afrequency of 200 kHz. As the gate opens for 100 ms, the number ofpulses will be:

0.1x200kHz= 20,000 pulses.

This is the maximum possible count.The minimum possible counter occurs when the input signal the frequency

of 20kHz.Minimum count= 0.1x 20 kHz= 2000 pulses.

(c) The maximum possible frequency is the frequency that causes thecounter to overflow. The maximum count on the counter is 2^16-1=65535.

As the gate is only open for 0.1 s, the maximum signal

frequency will be 10 times this number:Maximum frequency = 65535/0.1= 655350 Hz.

(d) The minimum frequency is the frequency that causes only onepulse to be counted. So the minimum frequency is 10 Hz. To beprecise, it has to be slightly more than 10 Hz to cause a count.

2. The diagram below shows an elliptic figure, as would be seen on thescreen of an oscilloscope with an X-Y mode setting. Assume thatChannel 1 is the X signal and Channel 2 is the Y signal. The sensitivity(Volts/Division) settings of both channels on the oscilloscope are equal.

Based on the diagram find the following (in terms of A, B, C and D ifneeded):

a) The phase shift between the two signals.b) The ratio of the frequencies of the two signals.c) The ratio of the amplitude of the two signals.d) The phase lead/lag between the two signals (i.e., which is

leading and which is lagging).


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SolutionBased on the figure above:

a) The phase shift between the two signals is equal to

arcsin(A/B)=arcsin(C/D). Note that we cannot tell solely from thevalues of A and B (or C and D) which signal is leading which.

b) As the shape is an ellipse then the frequencies of the two signalsare equal.

c) B represents the amplitude of Ch1 (x); D represents the amplitudeof Ch2 (y). Provided that the sensitivities of both channels (i.e.,volts/division) are equal, then the ratio of the amplitude of x to thatof y is B:D.

d) The relative phase of the two signals can only be decided fromLissajous Figure if we know the sense of rotation (i.e., clockwise oranticlockwise). If the rotation is clockwise, this means that Y ispeaking before X and hence that Y is leading X. If the sense ofrotation is anti-clockwise, this means that X is peaking before Y andhence that X is leading Y. [This assumes that positive X valueslead to a deflection towards the right and the positive Y values leadto a deflection towards the top].

Sense of Rotation: In practice, when using the oscilloscope with highfrequencies we cannot detect the sense of rotation and hence cannot decidefrom the figure which signals is leading. However, when using theoscilloscope with low frequency signals or when using an x-y plotter it is

possible to detect the sense of rotation.

X axis

Y axis

A

B

CD


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3. The output signal of a vortex flow-meter has frequency range between5 and 50 Hz. The mean period of the signal is to be measured usingan internal frequency of 10 kHz and a 16-bit binary.

a) What are the maximum and minimum possible counts?b) What is the minimum possible frequency that this device can

measure?

SolutionAs the frequency of the input signal is relatively low, we will measure theperiod of the signal. The period of the 5 Hz signal is 0.2 seconds (or 200 ms)and the period of the 50 Hz signal is 0.02 seconds (or 20 ms). As we areusing a 10 kHz signal, the total count during one period of the input signal willbe:

(a) For the 5 Hz signal: 10k x 0.2 = 2000 (maximum count)For the 50 Hz signal: 10k x 0.02= 200 (minimum count)

(b) The minimum possible frequency corresponds to the maximum possibleperiod length. The maxim possible count of the counter is 2

16-1= 65535.

At a pulse rate of 10 kHz, it will take a signal with the period length below to fillit before overflow:

65535/10000= 6.5535 seconds

A period of 6.5535 s corresponds to a frequency of 0.1526 Hz. This is lowest

frequency that this counter can measure.


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4. The diagram below shows an elliptic figure, as would be seen on thescreen of an oscilloscope. Answer the following questions:

a. What mode do you need to set the oscilloscope to in order tosee this screen?

b. What is the relationship between the frequency of the twosignals X and Y?

c. What is the purpose of this test?d. If the value of A is 5 divisions and B 6 divisions, what can you

say about X and Y signals.

Solved Advanced Problem5. Figure 11 below shows an RC circuit fed from a function generator.

The outputs V1 and V2are connected to an oscilloscope. The resultantscreen of the oscilloscope can be seen in Figure 12. Answer thefollowing questions a) to e).

X axis

Y axis

A

B


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Figure 11: RC circui t fed from a function generator.

Figure 12: Figure seen on the screen of the oscilloscope.

a) What mode do you need to set the oscilloscope to in order to see thisscreen?

AnswerXY mode.

b) If the value of A is 3.5 divisions and the value of B is 4.95 divisions, what is

the value of the capacitor C.

C

V1 x-axis

COM

1 kHzSinusoidal

FG

R = 100

V2 y-axis

x axis

y axis

A

B


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AnswerThe ratio of A to B is the arcsin of the phase shaft angle between the twosignals. So the angle is equal to arcsin 0.707=45.

The voltage V2 is proportional to the current throught he RC network.The fact that the angle is 45 means that the phase angle of the combinationof R and C has a phase angle of 45. This means that at this frequency, theimpedance of the capacitor it equal to 100 .Thus: 1/(C)=100, or C=1.59 F

c) If the value of the frequency from the function generator is changed to 500Hz, draw what you would see on the screen? (Show your answer in Error!Reference source not found.).

-What are the relative values of A and B in this case?-Explain your answer.

AnswerThe impedance of the capacitor at this frequency will be(1/C)=1/(25001.59 F)=200

So the new phase angle between V1 and V2 will now become equal toarctan (200/100)= arctan (2)= 63.5

So the relative values of A and B in the Lissajous pattern will become =sin(63.5)=0.894 (A:B = 0.894:1).

Figure 13: Answer to part c).


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d) If the frequency of the function generator is changed to 1 MHz, draw whatyou would see on the oscilloscope screen? (Show your answer in Error!Reference source not found.).-Explain your answer.

AnswerIn this case the impedance of the capacitor becomes very small (0.1 ) whichis negligible compared to the resistance. So in effect the two signals V1 andV2become identical and the Lissajous pattern becomes a straight line inclinedat 45 degrees.

Figure 14: Answer to part d).

e) Based on the circuit in Error! Reference source not found. below, drawwhat you would see now on the screen of the oscilloscope (show your answerin figure 6 below).

-Explain your answer.


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Figure 15: Setup for question e).

AnswerThe two signals now represent the voltage across the capacitor and thecurrent flowing through it, which are always at 90 by definition. Theirmagnitude is also equal at 1 kHz, so they will show as a circle on theLissajous pattern.

Figure 16: Answer to part e).

C

V1-V2 x-axis

COM

1 kHzSinusoidal

FG

R = 100

V2 y-axis

Subtractor+

-

V1

V2


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Example Datasheets on Counter/Timers (Courtesy of BK Precision)


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16 frequency and phase measurement rev 3 080524

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