Lab Report Day 24 - Signals with Multiple Frequency Components

Transfer Function：

We first talk about transfer function. 

Transfer function is ratio of numerator polynomial N(omega) to denominator polynomial D(omega) that has frequency as its variable. The previously learned voltage gain, current gain, transfer impedance, and transfer admittance are all transfer functions. 

We learn how to use FreeMat to graph a transfer function, and later in semi-log domain.

Those are the codes. 

This is what the graph looks like.

Signals with Multiple Frequency Components

In this lab, we will measure the response of a circuit that is applied with multiple input signals, in other words, multiple sinusoidal waves of different frequency. Another response we look at is a sinusoidal signal with a time-varying frequency, known as a sinusoidal sweep. We are going to find the magnitude response of an electrical circuit and use this information to infer the effect of the circuit on some relatively complex input signals. 

In low frequency, our circuit acts like DC, which the capacitor is open, the voltage across R2 is half of Vin. In high frequency, the circuit acts like AC, which the capacitor acts like short, and voltage across R2 is 0. This means that as omega goes to 0, Vout also goes to 0. It is a low-pass filter, meaning low frequencies flows through while high frequencies are blocked.

We first calculate the theoretical phase shift at different frequency.


This is our input graph. We create a custom waveform, 20(sin(1000πt)+sin(2000πt)+sin(20,000πt)) and sinusoidal sweep function. We are going to using Analog Discovery  to measure Vin and Vout from the circuit.

This is the Sweep wave.

The wave at 500 Hz.

The wave at 1000 Hz.

The wave at 10K Hz.

This shows that the output voltage and the input voltage has some phase shift.


For second part of lab, the input voltage is a sinusoidal sweep from 100Hz to 10 Khz in 20 msec:

The graph of a sweep wave. Vin and Vout rise and fall in the same rate and has no phase shift. Both output graphs are what we expected.

Summary:

Today, we learn about frequency response. The frequency response of a circuit is determined by the variation in its behavior with change in signal frequency. We can use the transfer function H(omega) to find the frequency response of a circuit. The transfer function of a circuit is the frequency dependent ratio of a phasor output N(omega) (voltage or current) to a phasor input D(omega) . We can use these equations to analyze the transfer function.

Lab Report Day 23 - Apparent Power and Power Factor

This lab assignment emphasizes the use of apparent power and power factor to quantity the AC power delivered to a load and power dissipated by the process of transmitting this power. Our goal in is to provide some necessary average power to load with minimum power lost during the power delivery process. 

To do that, we will used a 10 ohm resistor, and a series combination of 1mH inductor and resistor (10 ohm, 47 ohm, or 100 ohm) resistor to be the load. We use a sinusoidal wave with (1cos(2pi*5000t) in the circuit and calculate the power dissipated by the resistor. 

This is the set up of the lab. 

We measure the resistors to be 11.9 ohm (10), 48.6 ohm (47) and 99.8 ohm (100). Number in "()" are theoretical values. 

Those are results and calculations. The resistance is series with a load that changes each time. We measure the voltage of the load and the voltage across the resistor in series with the load.

The oscilloscope screen with 11.9 ohm resistor

The oscilloscope screen with 48.6 ohm resistor

The oscilloscope screen with 99.8 ohm resistor

This is our measured value. The phase difference are very big. 

Then we include a capacitor in parallel with the load. This can make the voltages measured more in phase.

The oscilloscope screen with 11.9 ohm resistor with a capacitor

The oscilloscope screen with 48.6 ohm resistor with a capacitor

The oscilloscope screen with 99.8 ohm resistor with a capacitor

We can see that the phase angles are very small, less than 3 degrees for each.


Summary:

Today we are told that the electric company charge us with peak voltage instead of average power, and later we are told how to avoid it. All we need to make it more efficiency and draw less current from the power source is to add a capacitor parallel to the load. Therefore, that will. And we do a lab to prove this.

Lab Report Day Twenty Two - AC Average and Maximum Power

Today, we do not actually do any labs. But professor Mason uses a demo to show us how AC power works visually. 

This is a picture that when professor Mason uses AC power and DC power to light the bulb at the same voltage. We can clearly see that the one with AC is dimmer. It shows that the bulb with AC has less power. 

When professor Mason turns the voltage for AC to about 1.4 times of the DC power, they seems the same bright. This proves that Vrms = 1/sqrt(2) Vmax. The voltage we read is Vmax instead of Vrms. 

It proves that power of AC is P=Vrms^2/R. 

Summary:
Today we review inductor and average and maximum power calculation in a  AC circuits, and see a demonstration to see how rms values dictate maximum power of AC circuits.

Lab Report Day Twenty-one - Inverting Voltage Amplifier, Op-Amp Relaxation Oscillator

Inverting Voltage Amplifier

In the first lab of the lab, we build an inverting voltage amplifier using an op-amp.  We first calculate the theoretical voltage gain and the phase shift.

We first derive the formula we are going to use. 

We use 2 10K ohm resistors, 1 1 uF Capacitor, and 1 Op Amp to build our circuit. 

Input and output voltage at 100 Hz. 

Input and output voltage at 1k Hz. 

Input and output voltage at 5k Hz. 

This is our result table including the theoretical value and experimental value of gain and phase different at f=100 Hz, 1K Hz, and 5k Hz. 

This is a brief summary table.

 Op-Amp Relaxation Oscillator:

In this lab, we first construct a relaxation oscillator on EveryCircuit.

We are going to generate a frequency of 99 Hz. We calculate the theoretical value for the pot to make this work to be 8371 ohm.

 The circuit consists of 1 10K resistor, 1 100 ohm resistor, 1 10k POT, 1 OP 27, and 1 0.1 uF capacitor.

This is the out put of the circuit. The frequency is 99.8 Hz. It has a percent difference less than 1%. We do this lab successfully.

Summary:

Today we go over how OP Amps work in AC circuit, and do labs to see how to see how they can apply in real life. We learn how op amp relaxation oscillator works and how we can make our desired frequency. 

Lab Report Day Twenty - Phasors: Passive RL Circuit Response

Phasors: Passive RL Circuit Response

 

In this lab assignment, we measure the gain and phase responses of a passive RL circuit and compare these measurements with expectations based on analysis.

 

In the pre-lab, we predict the amplitude gain and phase difference with 3 different frequency: Wc, Wc/10, and 10Wc, where Wc is 2.2*10^5 Hz. We also find the cutoff frequency for the circuit. When W= Wc, we expect amplitude gain = 0.0321 and expect phase difference = -45degree. When frequency = Wc/10 expected amplitude gain = 0.0452 and expected phase difference = -5.711 degree. When frequency = 10Wc expected amplitude gain =4.52*10^-3 and expected phase difference = -84.3 degree. 

 

We use a 22 ohm resistor and a 100 uH inductor to build the circuit.

At f=35k Hz, Vin=989.4 mV, and I=31.25 mA. phase difference is -37.8 degree. Gain is 0.0316. 

At f=3.5k Hz, Vin=973.4 mV, and I=38.55 mA. phase difference is -6.3 degree. Gain is 0.0452. 

 

At f=350k Hz, Vin=994.8 mV, and I= 5.665 mA. phase difference is -78.12 degree. Gain is 5.695*10^-3. 

 

This is a summary table of our experiment including our percent difference, experimental value, and theoretical value for phase difference and gain.

 

Summary:

Today, we analyze the steady-state response of electrical circuits from sinusoidal inputs. We can find the amplitude gain by comparing the Vin and Vout. We also learn how to analyze AC circuits using different methods of analysis including nodal analysis, mesh analysis, superposition, Thevenin and norton analysis. 

Lab Report Day Nineteen - Impedence

Impedance

In this lab assignment, we measure impedes of resistors, capacitors, and inductors. The measured values will be compared with our expectations based on analyses. 

In the pre-lab, we calculate the impedance for a RR circuit, RL circuit, and RC circuit.

 

This is the set up of our RR circuit. 

 

This is the RR circuit at 1k frequency. Vt= 1.354 V, I=13.2 mA. 

 

This is the RR circuit at 5k frequency. Vt= 1.354 V, I=13.2 mA. 

 

This is the RR circuit at 10k frequency. Vt= 1.354 V, I=13.2 mA. 

 

We determined the resistor impedance = 47 + R 

the real resistor values came out 46.6 ohm and 99.9 ohm, and therefore the experimental impedance for resistors = 146.5 ohm. 

We get our percent difference to be 0.34%. 

 

 

This is the set up for the RL circuit. 

 

This is the RL circuit at 1k frequency. Vt= 0.2652 V, I=40.05 mA. 

 

This is the RL circuit at 5k frequency. Vt= 1.0584 V, I=34.2 mA. 

 

This is the RL circuit at 10k frequency. Vt= 1.546 V, I =25.4 mA. 

 

The circuit is supplied with a 2V sine wave with different frequencies of 1kHz, 5kHz, and 10 kHz frequencies. In the inductor circuit the voltage leads the current by 90°. 

 

This is a RC circuit we build.

This is the RC circuit at 1k frequency. Vt= 1.999 V, I =1.05 mA. 

 

This is the RC circuit at 5k frequency. Vt= 1.9774 V, I =4.955 mA. 

 

 

This is the RC circuit at 10k frequency. Vt= 1.924 V, I =9.475 mA. 

 

The circuit is supplied with a 2V sine wave with different frequencies of 1kHz, 5kHz, and 10 kHz frequencies. In the capacitor circuit the current leads voltage by 90°

 

Summary:

Today, we analyze AC circuits. We find that a resistor AC circuit has no phase change between the voltage and current. The voltage leads the current by 90° in a circuit with an inductor and the current leads the voltage by 90° in a circuit with a capacitor. We learn that circuit elements can be represented as impedance.