Day 24 05/28/2015 Signals with Multiple Frequency Components

Class Note

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Circuit build

Pre-Lab

 This lab today is to aply different signals with multiple frequency components. The circuit consisted of 1kohm and series with and capacitor and also parallel to another 1kohm resistors

In the prelab we calculate the magnitude that response to the circuit. we know the s = jw and we simply it to get 1 / (RCS +2)

We first created a custom waveform 20(sin(1000πt)+sin(2000πt)+sin(20,000πt)) and sinusoidal sweep function.

20(sin(1000πt)+sin(2000πt)+sin(20,000πt)) function

At this graph we can see the high frequency part is more suppressed than the low frequency part.

500 Hz

1000 Hz

10,000 Hz

sinusoidal sweep

This one we see that frequency is being suppressed more .

Also being suppressed a lot here

As what we know from handout when the lower frequencies, the capacitor has infinite impedance, the circuit becomes a simple voltage divider which will simply divide income voltage half.

w = 0 => Vout/ vin = 1/2

However right now we were playing a high frequencies, but we didn't get our Independence to zero . our capacitor didn't act like short circuit. 

w = infinite => Vout/Vin =0

Summary: what I learned today is how frequency response. The frequency response the circuit is determined by the depend on the transfer function H(w) . The frequency dependent ratio of a phasor output to phasor input . 

Important function:

Day 23 05/26/2015 (Missed day the only day miss the class...) ( Data from classmate)

Today they supplied the AC supply to a resistor in series to inductor and resistor.

They do this lab to show the use of apparent power and power factor to quantify the AC power delivers to the load.

They first detail list all the components and the result of their pre-lab

The result of their resistance in series is changed each time.

10 ohm

47 ohm.

1000 ohm

Circuit.

This lab new required to include a capacitor in parallel with load. This can get the voltage being measured to be more in phase.

10Ω

47Ω

100Ω

This lab I learned the apparent power and power factor. we can see that the power is product the rms values of voltage and current and is in unit of voltage amperes. I see the power are cosine phase different between voltage and current. No cosine seems to be found. This lab kind help me able to calculate th AC power delivers to the load. 

Day 22 05/19/2015

Class Note

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Mason's tiny little power lab

The result is above which means DC is not exactly twice of AC

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Note

Day 21 05/14/2015 Inverting Voltage Amplifier & Op Relaxation Oscillator

Class Note

Nodal analysis and we all know at inverting voltage amplifier the node at (-) & (+) is 0

Key point at star ~!!

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Inverting Voltage Amplifier

In this lab we were required to build an inverting voltage amplifier using an op-amp

We then asked to calculate the theoretical voltage gain and the phase shift.

In this lab we were using R=10k C=10nF sinusoidal signal amplitude 2V and offset 0V with 

frequency: 100 Hz, 1kHz, and 5kHz

100Hz Vin 2.01 Vout 1.704

1kHz Vin 2.01 Vout 0.312

5kHz Vin 2.01 Vout 0.312

We calculate the Amp gain By Vout/ Vin

Phase Shift by Delta T / T(period) * 360

The result of this lab is frequency increase will decrease the Amp gain because Vout decrease as frequency increase.

Unlike the Amp gain, the phase shift is increase when frequency increase but quickly reach a shift of maximum.

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Setup of Circuit

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Our next lab required us to build an op-amp relaxation oscillator and manipulate the resistance to obtain a frequency of 178

we use digit number 155 and observe 178 

We then calculate period then plug into the formula theat period = 2RCln[(1+B)/(1-B)]

then we assume R1=R2=1k ohm

Then we solved R = 29428 ohm

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Second Lab set up

Day 20 05/12/2015 Phasors:Passive RL Circuit Response

Class Note

This one is the nodal analysis

this one is mesh analysis

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Phasor:Passive RL Circuir Response

In Pre-Lab we calculate the phasors shift and frequency

The Result end up not doable because the frequency is way too high

Next we keep our circuit setup

Then change to W= 10kHz and 20kHz

then we have the result above

At this result we solved the phase shift at

10 kHz = 0.77

20 kHz = 0.73

Another way:

For 10 kHz we observe amplitude Vin = 1.002 V at -442 us ; amplitude In = 1.008 A at -416 us

delta T b/w peak = 26us

Period V = 100 us

Period I = 99.996 us

We then solved by delta T/ Period * 360

Period and Amplitude is observed here

The lab today we were able to find the amplitude gain by comparing the phasors representing by Vin and Vout.

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Class Note