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Circuit Response to Time Varying Signals Lab


Note: Do to the length of this laboratory, the lab Write‑Up section has been moved to its own page located at, Lab Writeup, Circuit Response to Time Varying Signals ‑ Next Page


Objectives:

Part A:
1. Measure, in terms of average value and RMS value, meter response to various combinations of DC and time varying waveforms.

Part B:
1. Become familiar with the response of capacitors and inductors to time rates of change of forcing functions.

2. Experimentally verify the analogous behavior of capacitors and inductors in terms of ic = C dvc/dt and VL = L diL/dt


Equipment and Materials:
1. Oscilloscope
2. Function Generator
3. Digital Multimeter
4. Proto Board
5. 0.01uF Capacitor
6. 100 Ohm Resistor
7. Test Leads
8. 100mH Inductor



Procedures:

Part A ‑ Meters

The concepts of average or DC value and effective or RMS (Root Mean Square) value of a time varying waveform have been discussed in class. However, in using typical laboratory meters, such as the digital Multimeter (DMM), some additional points must be made.


With the DMM set on DC volts, the meter should accurately record the average or DC value of the waveform. There is no problem here. So, what exactly does an AC meter measure? By definition, an AC waveform has a DC or average value of zero. So, the first thing the DMM does on its AC setting is to remove (i.e. ignore) any DC or average value of the waveform. Thus, a 0 to 10V sine wave that has been DC offset by +5V, looks exactly the same to the meter (set on AC volts) as a 10Vpp sine wave swinging ±5V. Therefore, the meter responds only to the AC component of the waveform, completely disregarding any DC offset the waveform may have. This voltmeter reading is often referred to as AC RMS, to differentiate it from True RMS. (True RMS does take into account the DC offset of the waveform.)

It would be normal to expect that the DMM set on AC volts would accurately record the RMS value of the AC component of a waveform. Indeed, it will indicate the AC RMS value if (and only if) the waveform is a sine wave. Unfortunately, it won't do this for any arbitrary periodic waveform.


In order to determine what the DMM will indicate on AC volts for any arbitrary waveform, it is necessary to better understand how the meter actually works. When used on AC volts, most DMMs will perform what is called a full‑wave rectification on the input waveform, which essentially results in taking the absolute value of the waveform by inverting the negative portions.

The meter then computes the average value of this rectified waveform, and from the average value it then computes the RMS value. For a sine wave input, the average value is 2Vp/π for the rectified waveform, where Vp is the peak value. Of course, the effective or RMS value is Vp/√2 = 0.707Vp (for a sine wave only), therefore,

RMS value / average value of rectified waveform = (Vp / √2) / (2Vp / π) = 1.11 (sine wave only)

The DMM on AC volts will compute the average value of the rectified input waveform and then multiply by 1.11 to indicate the effective value. Since the 1.11 factor is unique to sine waves, the indicated value is accurate only for sine waves. However, by knowing the formula that the meter uses, we should be able to predict what the meter would read for any arbitrary waveform, such as a square wave.

In the experiment that follows, you are asked to predict the DC value, the True RMS value, the AC RMS value, and the expected meter AC value for a square wave, triangle wave, and sine wave. Using the technique learned in lecture, the sine wave should be easy.

The square wave will be a little more difficult, and will require using the information discussed above about AC meters. The triangle wave, however, presents a major challenge. The technique presented in terms of calculating the RMS value are valid for a triangle wave, but in order to calculate the area under the squared function, integral calculus is required since the squared waveform takes on a parabolic shape. Therefore, the result is given to you here. The RMS value (i.e. the True RMS value) of a triangle wave (with zero average value) is Vp / √3

In the experiment that follows, you will also be asked to calculate values for DC offset waveforms. A useful simplification can be derived for calculating the RMS value. Consider a DC offset periodic waveform. This waveform can be expressed as the sum of the DC component and an AC component as follows: v(t) = VDC + vAC

To determine the RMS value, we use the fundamental definition:

First, square the function.
Second, calculate the mean of the squared function (i.e. the average value)
Third, take the square root of the average value

The square of the function is:

v2 (t) = (VDC + vAC)2

v2 (t) = V2DC + 2VDC ∙ vAC + v2AC

The mean or average value is:

area under v2 (t) / T = [ area under V2DC + area under (2VDC ∙ vAC) + area under v2AC ] / T

but the area under 2VDC ∙ vAC is zero since the area under vAC is zero, by definition.

Next,

area under V2DC / T = V2DC

and from the definition,

√(area under V2AC / T) = VAC RMS

V2AC RMS = area under v2AC / T

We get,

VRMS = √( V2DC + V2AC RMS)

We can compute the RMS value of a DC offset waveform by squaring the DC or average value and adding to that the square of the RMS value of the AC component, and taking the square root of the sum.

1. We first consider periodic waveforms with no DC offset. Using the information discussed above, fill in the predicted values asked for in Table 1 assuming a 10V peak‑to‑peak waveform (i.e. ±5V). Then, set your function generator to produce the desired waveforms at a frequency of 100Hz. Use your oscilloscope to verify the magnitude and frequency as well as the 0V average value of the waveform. Then, measure the waveform using your DMM set on either DC volts or AC (whichever is appropriate) volts and record the meter readings in Table 1. Compare your predictions with the experimental meter readings.

Table 1A: DMM Readings for 10Vpp AC Waveforms


2. Now consider a DC offset waveform that varies between 0V and 10V. Predict the values called for in Table 2, and set your function generator to produce the desired waveforms at 100Hz. Again, use your oscilloscope to verify the correctness of the waveform, and use your DMM to measure the DC and AC values. Enter data in Table 2 and compare with your predictions.

Table 2A: DMM Readings of 0 - 10V DC Offset Waveforms



Part B: C an L Response to Time Varying Waveforms

From the fundamental relationships between the terminal current through and voltage across resistors, capacitors, and inductors we have:



Resistor: iR = vR / R

Capacitor: iC = C dvC/dt

Inductor: vL = L diL/dt

In calculus you will learn that dv/dt is the derivative of the voltage across the capacitor with respect to time. However, in the absence of calculus, dv/dt is simply the rate‑of‑change of the capacitor voltage with time, or, the slope of the voltage versus time waveform. We can determine the slope of a waveform experimentally by,

slope = rise / run = ∆VC/∆t ≈ dvC/dt


The units for dvC/dt are (volts/second), and for diL/dt they are (amp/second). If the voltage is increasing with time, the slope is positive. If the voltage is decreasing with time, the slope is negative. Thus, the above fundamental equation dictates that if there is to be any current flowing through a capacitor, the voltage across the capacitor must be changing with time.

If the voltage is increasing, the current will be positive; but if the voltage is decreasing, the current will be negative. If the voltage is constant, the current will be zero. Likewise, the inductor has an analogous behavior, except the voltage and current exchange roles. In order for there to be a voltage across the inductor, the current through it must be changing with time. If the current is increasing, the voltage is positive, if the current is decreasing, the voltage is negative, and if the current is constant, the voltage is zero.

1. One of the simplest waveforms to examine experimentally is the triangle wave in which the slope (i.e. dv/dt) is constant for each half cycle. Consider the circuit shown in Figure 1. Since the oscilloscope cannot measure current directly, a 100 ohm resistor has been included in the circuit as a current viewing device. The current waveform can be viewed indirectly by viewing the voltage across the resistor, since Ohm's Law dictates that the current and voltage will always have the same time‑dependent shape for a resistor. The 100 ohm value has been chosen so that the resistor drops a very small fraction of the input voltage, so in your calculations of the current through the capacitor you should assume that essentially all of the input voltage is dropped across the capacitor.

Begin by calculating the values called for in Table 1 for a frequency of 100Hz. Then draw the input voltage waveform and the expected current through the capacitor waveform on graph paper for a frequency of 100Hz. Assemble the circuit and measure the voltage across the resistor with your oscilloscope. Determine the measured current by dividing the measured voltage across the resistor by the 100 ohm resistance value. Superimpose this measured current waveform on your expected current waveform as a dotted line on graph paper.

2. Repeat the process in step 1 above for a 10Vpp triangle wave at a frequency of 1kHz.


Figure 1: Current Through a Capacitor Driven by a Triangle Wave


Table 1B: Current Through a Capacitor Driven by a Triangle Wave


3. An inductor exhibits an analogous behavior to a capacitor, but with the roles of current and voltage reversed. From vL = LdiL/dt we can see that a constant voltage across an inductor requires a constant slope or time rate of change of the current through the inductor. Consider an inductor driven by a square wave of voltage. Replace the capacitor in Figure 1 with a 100mH inductor, and set the function generator for a 10Vpp square wave. Again, the 100 ohm resistor is used as a current viewing device and should be ignored in calculating the voltage across the inductor. First, calculate the values called for in Table 2 for a frequency of 5 kHz. Sketch the input waveform and the expected waveform for the current through the inductor on graph paper. Next, assemble the circuit and measure the voltage across the current sensing resistor. Determine the measured current by dividing the resistor voltage by the 100 ohm value of the resistor. Superimpose the measured current onto the same graph as used to sketch the expected value, again using a dotted line.



4. Repeat step 3 above using a frequency of 50kHz.

Table 2B: Inductor Response to a Square Wave of Voltage


5. Unlike a triangle wave, a sine wave does not have a constant slope. The slope changes with time, sometimes being positive, sometimes negative, and sometimes zero. In fact, the derivative or slope of a sine wave is a cosine wave, which in turn is just another sine wave phase shifted by 90 degrees.

Assuming a sine wave of current through the inductor iL = Ip sin ωt,using calculus, it can be shown that,

diL/dt = slope = ω IP cos ωt = ω IP sin (ωt + 90)

Thus,

vL = L diL/dt = ω L IP sin (ωt + 90)

The ratio of the peak voltage across the inductor to the peak current through it is,

VP / IP = ω L IP / IP = ω L = XL Ω

and the voltage will be a sine wave that leads the current sine wave by 90 degrees.

Using the above information, consider the circuit in Figure 1, but with the capacitor replaced with a 100mH inductor and the function generator input voltage a 10Vpp sine wave. Enter the calculated values called for in Table 3 for a frequency of 2kHz. Sketch the input waveform and the expected current through the inductor on graph paper. Then, construct the circuit and measure the voltage across the current sensing resistor on the oscilloscope, keeping track of the phase angle relative to the input voltage. Determine the current waveform from vR/R and sketch the measured current on the same graph paper as you drew the expected waveform, but show it as a dotted line.

6. Repeat step 5 for a sine wave frequency of 20kHz and 10Vpp.

Table 3B: Sine Wave Response of Inductor


7. Similarly, we can determine the sine wave response of a capacitor. Assuming a sinusoidal input voltage


vC = VP sin ωt

dvC/dt = slope = ωVP cos ωt = ωt VP sin(ωt + 90)

iC = CdvC/dt = ωCVP sin (ωt + 90)

VP / IP = VP / ω C VP = 1/ωC = XC Ω



Consider the capacitor in Figure 1 being driven by a 10Vpp sine wave instead of a triangle wave. Enter the calculated values called for in Table 4 for a frequency of 500Hz. As before, calculate and sketch your predictions and verify your predictions experimentally.

8. Repeat step 7 using a 5kHz sine wave.

Table 4B: Sine Wave Response of Capacitor


Lab Writeup, Circuit Response to Time Varying Signals ‑ Next Page


Electrical Engineering lab key words: RMS, root mean square, Thevenin, Norton, superposition, time varying signals, DC average, capacitor, inductor, specifications, timing, angular velocity, function generator, peak to peak, derivative, DMM, meter, waveform, sin, freq, slope, offset voltage, phase, RC, RL, series, delta, parallel, AC circuit, rectifier, triangle wave, theorem, transient analysis, average, cycle, frequency, sine wave, phase angle, measured, simulation, pspice.

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