# 1. What is Phase?¶

## 1.1. Objective¶

The objective of this lab activity is to understand what is meant by the phase relationship between signals and to see how well theory agrees with practice. A secondary outcome will be a preliminary understanding of the Red Pitaya STEMlab hardware and software - test & measurements applications.

## 1.2. Notes¶

In this tutorials we use the terminology taken from the user manual when referring to the connections to the Red Pitaya STEMlab board hardware.

## 1.3. Background¶

We will investigate the concept of phase by looking at sine waves and passive components that will allow us to observe phase shift with real signals. First we will look at a sin wave and the phase term in the argument. You should be familiar with the equation:

(1.1)$f(t) = \sin(\omega t + \theta)$

$$\omega$$ sets the frequency of the sinusoidal wave as $$t$$ progresses and $$\theta$$ defines an offset in time which defines a phase shift in the function.

The sine function results in values ranging from 1 to -1. First set time variable $$t$$ equal to a constant, say 1. The argument, $$\omega t$$, is now no longer a function of time. With $$\omega$$ in radians, $$\sin(\frac{\pi}{4}) \approx 0.7071$$.

$$2\pi$$ radians equals $$360^{\circ}$$, so $$\frac{\pi}{4}$$ in radians corresponds to $$45^{\circ}$$; $$\sin(45^{\circ}) = 0.7071$$, too.

Now let $$t$$ vary with time like it normally does. When the value of $$\omega t$$ changes linearly with time it yields a sine wave as shown in Fig. 1.1. As $$\omega t$$ goes from 0 to $$2 \pi$$ the sine wave goes from 0 up to 1, down to -1 and back to 0. This is one cycle or one period, $$T$$, of a sine wave. The x-axis is the time varying argument/angle, $$\omega t$$, which varies from 0 to $$2\pi$$.

The value of $$\theta$$ is 0 in the function plotted in Fig. 1.1. Since the $$sin(0) = 0$$ the plot starts at 0. This is a simple sine wave without offset in time, which means no phase offset. Note that if we are using degrees $$\omega t$$, in a range from from 0 to $$2 \pi$$ or 0 to $$360^{\circ}$$ to yield the sine wave shown in Fig. 1.1.

Fig. 1.1 2 cycles of $$\sin(t)$$

As a side note, what happens if $$\omega t > 2\pi$$?

Enter $$2.5\pi$$ in a calculator and see for yourself. As you should know, the sine function repeats every $$2\pi$$ radians or $$360^{\circ}$$. It is similar to subtracting $$N\, 2 \pi$$ rad from the argument where $$N$$ is the largest integer that yields a non-negative result.

What happens if we plot a second sine wave into Fig. 1.1 with the same $$\omega$$ value and $$\theta = 0$$?

We have another sine wave which lands on top of the first sine wave. Since $$\theta = 0$$ there is no phase difference between the sine waves and they look the same in time.

Now, alter $$\theta$$ to $$\pi/4$$ rad or $$45^{\circ}$$ for the second sine wave. We see the original sine wave and a second sine wave shifted to the left in time. Fig. 1.2 shows the original sine wave (green) and the second sine wave (orange) with an offset in time. Since the offset is constant, we see the original sine wave shifted in time by a value of $$\theta$$ which is $$1/4$$ of the period in this example.

Fig. 1.2 green - $$\sin(t)$$, orange - $$\sin(t + \pi/4)$$.

$$\theta$$ is time offset or phase portion of Eq. 1.1. The phase angle defines offset in time and vice versa. Eq. 1.2 shows the relationship. We happened to choose a particularly common offset of $$90^{\circ}$$. The phase offset between a sine and cosine wave is $$90^{\circ}$$. The offset angle is almost always not 90. As a matter of fact it is often a function of frequency ($$f$$).

## 1.4. Phase¶

When there are two sine waves, e.g. displayed on a scope, the phase angle can be calculated by measuring the time between the two waveforms (negative to positive zero crossings or “rising edges”, can be used as time measurement reference points in the waveform). One full period of the sine wave in time is the same as $$360^{\circ}$$. Taking the ratio of the time between the two waveforms as $$\Delta t$$, and the time in one period of a full sine wave as $$T$$, you can determine the angle between them. Eq. 1.2 gives the exact relationship.

(1.2)\begin{align}\begin{aligned}\theta &= \frac{\Delta t}{T} 360^{\circ}\\&= \frac{\Delta t}{T} 2\pi \, rad\\&= \Delta t f 2 \pi \, rad\end{aligned}\end{align}

where $$T$$ is the period of the sine wave.

## 1.5. Naturally occurring time offsets in sine waves¶

Some passive components yield a time offset between the voltage across them and the current through them. In class we showed that the voltage across and the current through a resistor was a simple time independent relationship. $$V / I = R$$, where $$R$$ is real and in ohms. So the voltage across and current through a resistor are always in phase.

For capacitors and inductors the equation relating voltage $$V$$ to current $$I$$ is similar. $$V / I = Z$$, where $$Z$$ is an complex impedance with real and imaginary parts. We are only looking at a capacitors in this lab.

Generally, capacitors are made of two conductive plates separated by a dielectric material. When a potential difference is applied across the plates, hence an electric field is created between the plates. Capacitor dielectrics can be made of many materials, including thin insulating films and ceramic. A capacitor’s distinguishing characteristic is its capacitance (C), measured in Farads (F), which measures the ratio between voltage and charge buildup.

The basic rule for capacitors is that the voltage across the capacitor will not change unless there is current flowing into the capacitor. The rate of change of the voltage ($$dv_C/dt$$) depends on the magnitude of the current. For an ideal capacitor the current $$i_C(t)$$ is related to the voltage by the following formula:

(1.3)$i_C(t) = C \frac{dv_C(t)}{dt}$

Right now, the full implications of this is beyond the scope of this lab. You will observe this behavior in later labs. The impedance of a capacitor is a function of frequency. The impedance goes down with frequency conversely the lower the frequency the higher the impedance.

(1.4)$Z_C = \frac{1}{j \omega C},$

where $$\omega = 2 \pi f$$ is defined as the angular velocity.

One subtle thing about Eq. 1.4 is the imaginary operator $$j$$. When we looked at a resistor, i.e., there is no imaginary operator in the equation for the impedance. The sinusoidal current through a resistor and the voltage across a resistor have no time offset between them, as the relationship is completely real. The only difference is in amplitude. The voltage is sinusoidal and is in phase with the current sinusoid. This is not the case with a capacitor. When we look at the waveform of a sinusoidal voltage across a capacitor it will be time shifted compared to the current through the capacitor. The imaginary operator $$j$$ is responsible for this. Looking at Fig. 1.3, we can observe that the current waveform has a peak (maximum) if the slope of the voltage waveform ($$dv/dt$$) is maximal.

The time difference can be expressed as a phase angle between the two waveforms as defined in Eq. 1.2.

Fig. 1.3 Phase angle determination between voltage (V) and current (I).

You probably have seen circuits made entirely from resistors. These circuits have only real impedance, which means that voltages throughout the circuit will all be in phase (i.e. $$\theta = 0$$ deg.) as it is the complex impedance that shifts the current in time with respect to the voltage. Note that the impedance of a capacitor is pure imaginary. Resistors have real impedances, so circuits that contain both, resistors and capacitors, will have complex impedances.

In order to calculate the theoretical phase angle between voltage (V) and current (I) in an RC circuit:

$i(t) = \frac{v(t)}{Z_{tot}},$

where $$Z_{tot}$$ is the total circuit impedance.

Rearrange the equation until it looks like $$Z_{tot} = a + jb$$, where $$a$$ and $$b$$ are real numbers. The phase relationship of the current relative to the voltage is then:

$\theta = \arctan\left(\frac{b}{a}\right).$

## 1.6. Materials¶

• Red Pitaya STEMlab 125-14 or STEMlab 125-10
• $$2 \times 470\Omega$$ resistors
• $$1 \times 1 \mu F$$ capacitor

You are going to use Red Pitaya’s STEMlab board and the Oscilloscope & Signal generator application. User guide for starting the Red Pitaya STEMlab board can be found at quickstart, while Oscilloscope & Signal generator application is explained here.

## 1.7. Instructional Objectives¶

1. Explore the phase relationship of voltages in a resistive circuit.
2. Explore the phase relationship of voltages in an RC circuit.

## 1.8. Procedure¶

• Be sure the STEMlab is plugged into a local network and start up the web interface using web browser.
• Start the Oscilloscope & Signal generator application. The main screen should look like a scope display with adjustable range, position and measurement parameters.
• On the left bottom of the screen be sure that OUT1 V/div and OUT2 V/div are both set to 200 mV/div. You can set V/div by selecting the desired channel and using vertical +/- controls.
• In the OUT1 controls menu, set the frequency of OUT1 to 1000 Hz with $$0^{\circ}$$ phase and 0.9 V amplitude. Select SINE waveform shape and enable output.
• In the OUT2 controls menu, set the frequency of OUT2 to 1000 Hz and 0.9 V amplitude. Select SINE waveform shape and enable output.
• Set t/div to 200 us/div (using horizontal +/- controls).

### 1.8.1. Measure the phase angle between two generated waveforms¶

From the previous settings you should see what looks like 1 sine wave. There are two just one is on top of the other - zero phase angle!

• In the OUT1 control menu, change the phase to $$90^{\circ}$$.
• In the OUT2 control menu, change the phase to $$135^{\circ}$$.
• Which channel looks like the sine is occurring before the other?

The OUT2 signal should look like it is leading (happening before) the OUT1 signal. The OUT2 signal crosses the 0 V axis from below to above before the OUT1 signal. It turns out a positive $$\theta$$ is called a phase lead. The low to high crossing time reference point is arbitrary. The high to low crossing could also be used.

Fig. 1.5 Oscilloscope application showing two sine signal with phase difference.

• Change the phase of OUT2 to $$45^{\circ}$$. Now it looks like the CHB signal lags the CHA signal.
• Press the red STOP button to pause the Oscilloscope acquisition.
• Select “CURSOR” menu and enable X1 and X2 cursors
• Using horizontal +/- controls set Time to 100 us/div.
• Using mouse and left press+hold on the cursor marker(white arrow on the end of the cursor line) set one cursor position so that cursor line going through point where OUT1 is crossing 0V line.

Repeat the step for the second cursor and OUT2 signal.

• Readout the time difference between cursors.
• What is $$\Delta t$$?
• Use the measured $$\Delta t$$ and Eq. 1.2 to calculate the phase offset $$\theta$$ in degrees.

Note you cannot measure the frequency of a signal that does not have at least one full period displayed on the screen. Usually you need more than two cycles to get consistent results. You are generating the frequency so you already know what it is. You don’t need to measure it in this part of the lab.

### 1.8.2. Measuring magnitude using a real circuit¶

Fig. 1.6 R-R circuit.

Build the circuit shown in Fig. 1.5 on your solderless breadboard using two $$470 \Omega$$ resistors, oscilloscope probes and Red Pitaya STEMlab board.

NOTICE: For ground pin use probes ground leads (crocodile connectors).

Fig. 1.7 R-R circuit on the breadboard.

We have connected OUT1 directly to IN1 so we can observe a real voltage signal across resistors R1and R2.

• In the OUT1 controls menu, set the Frequency to 200 Hz with 0° Phase and 0.9 V amplitude. Deselect “Show” button, select SINE waveform shape and select “ON” button.
• Set the horizontal time scale to 1.0 mS/Div to display two cycles of the waveform.
• Click on the scope Start button if it is not already running.
• Using vertical +/- controls set 200 mV/div for IN1 and IN2

The voltage waveform displayed in IN1(yellow) is the voltage across both resistors (VR1+VR2). The voltage waveform displayed in IN2 is the voltage across just R2(VR2). To display the voltage across R1we use the Math waveform display options. Under the math menu for Signal1 select IN1, select operator “-“, for Signal2 select IN2 then select enable. You should now see a third waveform for the voltage across R1(VR1).

• Using vertical +/- controls set 200 mV/div (0.2 V/div) for MATH trace.

With this settings you are observing:

• IN1- Input excitation signal

• IN2- Voltage drop signal across R2

• MATH - Voltage drop signal across R1

• Record VR1and VR2.

• VR1_______Vpp.
• VR2_______Vpp.
• VR1+VR2_______Vpp.
• Can you see any difference between the zero crossings of VR1and VR2?

• Can you even see two distinct sine waves?

Probably not. There should be no observable time offset and thus no phase shift.

You can see that MATH (purple) and IN2 (green) trace are overlapping. To see both traces you can adjust the vertical position of a channel to separate them.

This can be done by selecting trace marker (on the left side of the grid) using mouse left button and moving trace up-down. Make sure to set the vertical position back to 0 to realign the signals.

Here we don’t have phase shift and value of R1= R2so the signal amplitudes for VR1and VR2will be the same. The result is that we have two identical signals (IN2=VR2, MATH=VR1) on the Oscilloscope.

What happens if you use $$220 \Omega$$ value for R2?

### 1.8.3. Measuring RC circuit¶

• Replace R2with a 1 uF capacitor C1.

NOTICE: For $$1\, \mu F$$ capacitor you will be probably using an electrolytic capacitor.

This capacitors are polarity sensitive i.e on the positive capacitor pin the voltage should never go negative and on negative pin (GND) voltage should never go positive.

From previous example (RR circuit) and Oscilloscope & Signal generator settings we are generating sine wave which is going from -0.9 V to 0.9 V, causing a wrong polarization of capacitor (it can damage a capacitor) we need to adjust our output signal so we generate a sine signal which is always positive (sine signal with an offset).

• In the OUT1 settings menu set Amplitude and Offset values to 0.45 V (Now we are generating sine signal which is oscillating around 0.45 V of DC offset value i.e sinusoidal signal is going from 0 V to 0.9 V)

Because there is no DC current through the capacitor, we are not interested in this DC value. In order to re-center our signals on the grid, we need to shift signals in vertical direction using negative offset values.

• In the IN1 and IN2 settings menu set the value of Vertical Offset to -450 mV
• For the stable acquisition set the trigger level in TRIGGER menu to 0.45 V
• Measure IN1, IN2 and Math P2P (peak to peak) value. What signal is the Math waveform?
• Record VR1, VC1and VR1+VC1.
• VR1____________VPP.
• VC1_______________VPP.
• VR1+VC1____________VPP.

Now something to do with phase. Hopefully you see a few sine waves with time offsets or phase differences displayed on the grid. Let’s measure the time offsets and calculate the phase differences.

• Measure the time difference between VR1and VC1. and calculate the phase offsets.

Use Eq. 1.2 and the measured $$\Delta t$$ to calculate the phase angle $$\theta$$.

The CURSORS are useful for determining $$\Delta t$$; here’s how:

• Display at least 2 cycles of the sine waves.
• Set the horizontal time/div to 500 us/div. Note the Delta cursor display keeps track of the sign of the difference.

You can use the measurement display to get frequency. Since you set the frequency of the source you don’t really need to depend on the measurement window for this value.

Assume $$\Delta t$$ is 0 if you really can’t see any difference with 1 or 2 cycles of the sine wave on the screen.

• Put a first cursor at the neg. to pos. zero crossing location for the IN1 ( VR1+ VC1) signal. Put a second cursor at the nearest neg. to pos. zero crossing location for the math ( VR1) signal. Record the time difference and calculate the phase angle. Note $$\Delta t$$ maybe a negative number. Does this mean the phase angle leads or lags?

$$\Delta t$$ _________, $$\theta$$ _________

• Put a first cursor at the neg. to pos. zero crossing location for the IN1 ( VR1+ VC1) signal. Put a second cursor at the nearest neg. to pos. zero crossing location for the IN2 ( VC1) signal. Record the time difference and calculate the phase angle.

$$\Delta t$$ _________, $$\theta$$ _________

• Put a first cursor at the neg. to pos. zero crossing location for the Math ( VR1) signal. Put a second cursor at the nearest neg. to pos. zero crossing location for the IN2 (VC1) signal. Record the time difference and calculate the phase angle.

$$\Delta t$$ _________, $$\theta$$ _________

• Measure the time difference and calculate the phase $$\theta$$ offset at a different frequency.

• Set OUT1 frequency to 1000 Hz and the time / div to 200 us/div.

• Put a first cursor at the neg. to pos. zero crossing location for the IN1 ( VR1+ VC1) signal. Put a second cursor at the nearest neg. to pos. zero crossing location for the math (VR1) signal. Record the time difference and calculate the phase angle. Note $$\Delta t$$ maybe a negative number. Does this mean the phase angle leads or lags?

$$\Delta t$$ _________, $$\theta$$ _________

• Put a first cursor at the neg. to pos. zero crossing location for the IN1 ( VR1+ VC1) signal. Put a second cursor at the nearest neg. to pos. zero crossing location for the IN2 ( VC1) signal. Record the time difference and calculate the phase angle.

$$\Delta t$$ _________, $$\theta$$ _________

• Put a first cursor at the neg. to pos. zero crossing location for the math ( VR1) signal. Put a second cursor at the nearest neg. to pos. zero crossing location for the IN2 (VC1) signal. Record the time difference and calculate the phase angle.

$$\Delta t$$ _________, $$\theta$$ _________