05/27 - Lab 26: Signals with Multiple Frequency Components

Lab 26: Signals with Multiple Frequency Components

In this lab, we used Waveforms to create complex voltage signals which were applied to a simple RC circuit. The first signal composed of multiple sinusoidal waves of different frequencies, and the second signal was a sinusoidal wave with a time-varying frequency. The response of the circuit was measured in order to see how the "shape" of the input is affected. The following circuit was used in this lab:

The values used in the circuit were R = 1 kOhm, and C = 100 nF, where V_in(t) is the complex sinusoidal signal, and V_out is the voltage measured across the "load" resistor. The actual circuit is pictured below:

For the pre-lab, we calculated the magnitude response of the circuit, which is simply the ratio of the amplitude of the input sinusoid to the amplitude of the output sinusoid, in order to determine the zeros and poles of the circuit.

The function used for our first input sinusoid was a function containing multiple sine functions at various frequencies:

V_in(t) = 20[sin(1000πt) + sin(2000πt) + sin(20,000πt)]

The frequency of the overall sinusoid was adjusted throughout the experiment. We used frequencies of 500 Hz, 1000 Hz, and 10000 Hz. Pictured below is the oscilloscope depicting the input and output voltages for each case:

500 Hz

Yellow: Input Voltage  |  Blue: Output Voltage

1000 Hz

Yellow: Input Voltage  |  Blue: Output Voltage

10000 Hz

Yellow: Input Voltage  |  Blue: Output Voltage

The second input function used in this lab was a sinusoidal sweep function that contained a time-varying frequency. We did not need a mathematical equation for this, as Waveforms has a built-in sweep function. The input wave using this built-in feature looks like this:

Pictured below is the Waveforms oscilloscope containing the input voltage, as well as the voltage across the "load" resistor.

In Class Examples

1. For the circuit shown at right, calculate the gain I o (ω)/I i (ω) and its poles and zeros.

2. Given the circuit shown below and i(t)= I cos ωt amps, find the transfer function H(ω) = Vo / I and sketch the frequency response.

3. Given the circuit shown at right and vin = Vin cos ωt volts, find the transfer function H(ω ) = Vout / Vin where Vout is the voltage across the inductor and sketch the frequency response.

05/24 - Lab 25: Apparent Power and Power Factor

Lab 25: Apparent Power and Power Factor

In this lab, we used a basic LR circuit to gain understanding of apparent power and power factor in order to determine the power delivered to a load. Apparent power, average power, and the power factor associated with the circuit were calculated on our white boards, and then measured during the experiment to compare our expectations with the actual values. The circuit used in this experiment is pictured below:

For this experiment, R_T = 10 Ohms, L = 1 mH, and the load resistance R_L will vary throughout the experiment from 10 Ohms to 47 Ohms, and finally 100 Ohms. The input signal V(t) had an amplitude of 1V @ 5 kHz. The circuit we built in the breadboard is pictured below (with R_L = 10 Ohms).

Our first set of calculations (with R_L = 10 Ohms) is pictured on our white board below:

We predicted average power P = 7.2 mW, apparent power S = 13.4 mVA, V_rms = 0.707, 0° V, and I_rms = 19, -57.61° mA. The actual results from measuring values on our circuit are pictured below:

The actual V_rms turned out to be 591 mV, which is slightly off but still fairly close to our predicted result (only off by ~0.1V), and our measured I_rms current turned out to be 18.3 mA which was only ~0.7 mA from our predicted result. The phase shift of the current from the input voltage can be seen as being ~-60°, which correlates with our predictions.

Next, we changed out R_L from the 10 Ohm resistor and replaced it with a 47 Ohm resistor. Our white board calculations are pictured below:

Our calculations yielded an average power P = 6.8 mW, apparent power S = 7.8 mVA, V_rms = .707, 0° V, and I_rms = 11, -28.8° mA. The actual results of our circuit are pictured below:

The actual voltage obtained from measuring the circuit was V_rms = 608 mV (again, ~0.1V off from our prediction), and our I_rms from measurements turned out to be 10.9 mA, which was extremely close to our prediction of 11 mA. The phase shift of the current can be seen as being slightly negative, which makes sense with our prediction of -28.8°.

Finally, we replaced R_L with a 100 Ohm resistor. Our white board calculations are shown below:

Our calculations yielded an average power P = 4.0 mW, apparent power S = 4.2 mVA, V_rms = .707, 0° V, and I_rms = 5.9, -15.9° mA. The actual measurements from our circuit are pictured below:

The actual voltave V_rms turned out to be 645 mV (~ .05V from our prediction), and the actual current I_rms was slightly off from our prediction at 7.1 mA (predicted 5.9 mA). The phase shift of the current from the input voltage can be seen as very small, and negative, which corresponds nicely to the value we predicted of -15.9°.

In Class Examples

1. Calculate the RMS value of the signal shown in below.

2. A relay coil is connected to a 210-V, 50-Hz supply. If it has a resistance of 30 Ω and an inductance of 0.5 H, calculate the apparent power and the power factor.

3. The voltage across a load is v(t) = 60 cos(ωt − 10◦) V and the current through the element in the direction of the voltage drop is i(t) =1.5 cos(ωt + 50◦). Find complex and apparent powers.

4. When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.

05/19 - Lab 24: Inverting Voltage Amplifier

Lab 24: Inverting Voltage Amplifier

In this lab, we explored the use of op amps with capacitors in order to gain an understanding of the steady-state circuit response to sinusoidal inputs. By comparing input and output signals, the circuit can be seen to provide some amplitude gain between the input and output signals, as well as a phase shift between the phasors of input and output. We measured the gain and phase change response of an inverting op amp and compared them to our theoretical expectations.

The circuit used in this lab is shown in the diagram below.

Our pre-lab assignment was to determine the value of R such that the angular frequency obtained was 864 rad/s.

Next, we built the circuit from the diagram above. Because there were no resistors near 526 Ohms, we decided to place 2 1k resistors in parallel to achieve a resistance of 500 Ohms. The actual circuit we built is pictured below.

We were able to obtain a square wave graph from the output voltage with an amplitude of 4V. This means we were able to take a sinusoidal voltage input and get a voltage output that looks like a square wave, where voltage is essentially switching between 4V and -4V.

In Class Examples:

1. Determine vo (t) for the op amp circuit below if vs = 3 cos 1000t V.

2. Refer to the circuit depicted below. Find the average power absorbed by each element.

3. Find the value of ZL in the circuit of below for maximum power transfer.

05/17 - Lab 23: Passive RL Circuit Response

Lab 23: Passive RL Circuit Response

In this lab, we calculated the theoretical phase and gain response of an RL circuit and compared the results to actual measurements taken from a passive RL circuit. The circuit used was very simple and only contains a 47 Ohm resistor in series with a 1 mH inductor.

The cutoff frequency was given as R/L, and we were instructed to perform our calculations and measurements of the circuit for the cutoff frequency, 10 * cutoff frequency, and 1/10 * cutoff frequency. The actual resistance of the 47 Ohm resistor was 46.9 Ohms, and the true inductance of the inductor was 0.963 mH. Using these values, we calculated the cutoff frequency to be w = 48182.8 rad/s, and thus a frequency of f = 7669 Hz. The actual circuit used for this experiment is pictured below.

First, we used the cutoff frequency as our voltage input, with an amplitude of 2V. White board calculations and measurements from the actual circuit are pictured below.

The impedance of the inductor was calculated as jwL = 46.4j Ohms. Since the resistor is 46.4 Ohms, the total impedance was calculated to be sqrt(46.4^2 + 46.4^2) = 65.6 Ohms, with a phase shift of atan(1) = 45°, thus Z_tot = 65.6 Ohms, 45°. The current in the circuit can be calculated as V/Z, which is 2, 0°/65.6, 45° = 30.5, -45° mA. The voltage across the inductor was calculated by multiplying the current by the impedance of the inductor, which yielded V_L = 1.42, 45° V. The results obtained from measuring the actual circuit were nearly spot on to our calculations.

Oscilloscope at f = 7669 Hz (cutoff frequency)

Blue = Input Voltage

Red = Current

Oscilloscope at f = 7669 Hz (cutoff frequency)

Blue = Input Voltage

Yellow = Inductor Voltage

Red = Current

The wrong scale was used in the second set of graphs (it should have shown the yellow scale to show voltage across the inductor); however, looking closely it can be seen that the scale used for both voltages are 0.5V / division. The input voltage was 2 V, and the voltage across the inductor is slightly less than 1 division below the input voltage, thus the inductor voltage we calculated was correct at ~1.4 V. It can also be seen that the inductor voltage is out of phase with the input voltage by ~45°.

Next, we changed the input voltage frequency to be 1/10 the cutoff frequency; thus, w = 4818.3 rad/s, and f = 766.9 Hz. White board calculations as well as measurements taken from the circuit are pictured below.

Changing the input frequency affects the impedance of the inductor (by a factor of 1/10), and thus the total impedance was much smaller than the previous setup at Z_tot = 46.6, 5.71° Ohms. The current and voltage were calculated using the same method as before, yielding values of 43.0, -5.71° mA, and 0.2, 34.4°

Oscilloscope at f = 766.9 Hz (1/10 cutoff frequency) 

Blue = Input Voltage

Yellow = Inductor Voltage

Red = Current

Oscilloscope at f = 766.9 Hz (1/10 cutoff frequency)

Blue = Input Voltage

Yellow = Inductor Voltage

Red = Current

The first set of graphs shows the vertical scale in terms of mA, and it can be seen that the peak current measured across the circuit was just over ~42 mA with a very slight phase shift from the input voltage, which was nearly identical to our white board calculation. The second graph shows the same image but with a scale indicating the voltage. The voltage across the inductor was measured to be ~0.25 V with a slightly lower phase shift than the first setup.

Finally, we tested the same RL circuit setup but with the input frequency at 10 * the cutoff frequency, or w = 481,828 rad/s, and f = 76,695 Hz. The white board calculations as well as measurements taken from the actual circuit are pictured below.

Since w was increased by a factor of 10, the impedance of the inductor was also increased by the same factor, making Z_L = 464j Ohms, and thus Z_tot = 466.3, 84.3° Ohms. Using the same calculations as previous setups, the current and voltage were calculated to be 4.29, -84.3° mA

Oscilloscope at f = 76695 Hz (10 * cutoff frequency)

Blue = Input Voltage

Yellow = Inductor Voltage

Red = Current

Oscilloscope at f = 76695 Hz (10 * cutoff frequency)

Blue = Input Voltage

Yellow = Inductor Voltage

Red = Current

In the first series of graphs, the vertical scale indicates the voltage across the inductor. Just as we predicted, the voltage across the inductor is nearly the same as the input voltage, with a very slight phase shift of ~5°. The second set of graphs shows the same thing but with the scale depicting the current through the circuit in mA. We predicted a current with an amplitude of 4.3 mA, and the measurements taken show this result exactly. The phase shift of the current was calculated to be -84.3° from the input voltage, which is precisely what our measurements indicate.

In Class Examples

1. Find i x in the circuit shown using nodal analysis.

2. Determine current Io in the circuit below using mesh analysis.

3. We were asked to use mesh currents again for the previous example, but with only a current source that delivers I = -2.353 + j2.353 A