Sunday, March 19, 2017

Week 9: High and Low Pass Filters

Problem 1: Measure the resistance of the speaker. Compare this value with the value you would find online.

Between 8.6 and 8.7 Ohms, the listed resistance online is 8 Ohms.

Problem 2: Build the following circuit using a function generator setting the amplitude to 5V (0V offset). What happens when you change the frequency?

The pitch of the speaker changes.

Fig. 1: Speaker directly connected to function generator.


Problem 3: Add one resistor to the circuit in series with the speaker (first 47 Ω, then 820 Ω). Measure the voltage across the speaker. Briefly explain your observations. 

Resistor value Oscilloscope output  Observation
47 Ω 448 mV peak to peak Period of the wave decreases as the freq. increases
820 Ω 54 mV peak to peak Same as the other resistor


Problem 4: Build the following circuit. Add a resistor in series to the speaker to have an equivalent resistance of 100 Ω. Note that this circuit is a high pass filter. Set the amplitude of the input signal to 8 V. Change the frequency from low to high to observe the speaker sound. You should not hear anything at the beginning and start hearing the sound after a certain frequency. Use 22 nF for the capacitor. 

a.) Explain the operation. 


Fig. 2: High pass filter explanation.

b.) Fill out the following table by adding enough (10-15 data points) frequency measurements. Vout is measured with the DMM, thus it will be rms value. 

Frequency Vout (mV)(rms) Vout (rms) / Vin (rms)
10 Hz 15.4 0.002570952
50 Hz 15.6 0.002604341
100 Hz 15.4 0.002570952
200 Hz 15.3 0.002554257
300 Hz 15.6 0.002604341
500 Hz 15.6 0.002604341
600 Hz 15.8 0.00263773
700 Hz 15.9 0.002654424
800 Hz 16.6 0.002771285
900 Hz 17.5 0.002921536
1 KHz 16.8 0.002804674
1.2 kHz 17.2 0.002871452
1.4 kHz 17.7 0.002954925
1.6 kHz 18.4 0.003071786
1.8 kHz 18.9 0.003155259
2 kHz 20.8 0.003472454
2.5 kHz 22.1 0.003689482
3 kHz 24.3 0.004056761
3.5 kHz 25.6 0.00427379
4 kHz 28.3 0.004724541
5 kHz 34.2 0.005709516
6 kHz 41.2 0.00687813
8 kHz 56.1 0.009365609
10 kHz 71.5 0.011936561
12 kHz 87.9 0.014674457
14 kHz 102 0.017028381
16 kHz 116 0.019365609
18 kHz 130 0.021702838
20 kHz 145 0.024207012
40 kHz 294 0.049081803
80 kHz 551 0.091986644
100 kHz 630 0.105175292
200 kHz 1120 0.186978297
400 kHz 1900 0.317195326
600 kHz 1920 0.320534224
800 kHz 2320 0.387312187
1000 kHz 2660 0.444073456

c.) Draw Vout/Vin with respect to frequency using Excel. 

d.) What is the cut off frequency by looking at the plot in b?

Between 80 and 40 kHz.

e.) Draw Vout/Vin with respect to frequency using MATLAB.


f.) Calculate the cut off frequency theoretically and compare with one that was found in c. 

V(Max)*1/sqrt(2)=.44*.707=0.311mV which is about 400 Hz, which does not coincide with what was found in part c at all.

g.) Explain how the circuit works as a high pass filter.

The circuit works as a high pass filter because low frequencies are blocked by the capacitor in series due to its impedance. As the frequency goes lower, the harder it is for the voltage to pass through the capacitor.

Problem 5: Design the circuit in 4 to act as a low pass filter and show its operation. Where would you put the speaker? Repeat 4a-g using the new designed circuit.

a.) Explain the operation.



b.) Fill out the following table by adding enough (10-15 data points) frequency measurements. Vout is measured with the DMM, thus it will be rms value. 

Frequency Vout (mV)(rms) Vout (rms) / Vin (rms)
5 Hz 330 0.05509182
10 Hz 330 0.05509182
50 Hz 331 0.055258765
100 Hz 332 0.05542571
200 Hz 334 0.055759599
300 Hz 337 0.056260434
500 Hz 348 0.058096828
1 KHz 373 0.062270451
5 kHz 338 0.056427379
10 kHz 343 0.057262104
50 kHz 582 0.097161937
100 kHz 835 0.139398998
200 kHz 1200 0.20033389
300 kHz 1810 0.302170284
400 kHz 1530 0.25542571
500 kHZ 1100 0.183639399
1000 kHz 429 0.071619366
2000 kHz 182 0.030383973
3000 kHz 105 0.017529215
4000 kHz 62 0.010350584

c.) Draw Vout/Vin with respect to frequency using Excel.

d.) What is the cut off frequency by looking at the plot in b? 

300 kHz.

e.) Draw Vout/Vin with respect to frequency using MATLAB.


f.) Calculate the cut off frequency theoretically and compare with one that was found in c. 

V(Max)*1/sqrt(2)=0.3*.707=0.21mV = 300 Hz which is was similar to what was found earlier.

g.) Explain how the circuit works as a low pass filter.

The circuit works as a low pass filter by having the capacitor in parallel with the output, and having said capacitor then grounded. This means that when given a low frequency, the capacitor will offer a higher impedance to the flow of electricity, allowing the voltage to flow unhindered. However, as the frequency gets higher, the capacitor allows for the voltage to flow across it easier, and since electricity follows the path of least resistance, it will flow through the capacitor, and not to the output, thus "filtering" out the high frequencies.

Problem 6: Construct the following circuit and test the speaker with headsets. Connect the amplifier output directly to the headphone jack (without the potentiometer). Load is the headphone jack in the schematic. “Speculate” the operation of the circuit with a video.

6 comments:

  1. Your cut off frequencies were much higher than our high and low pass filters, but the graphs do look good. Also your #6 video appears to have different connection on your headphone jack than other groups. We tried your set up and it seemed to work perfectly.

    ReplyDelete
  2. Very clear explanation in your videos. Comparing your cut off frequency to ours, yours is higher than ours which is make more sense. Thanks for you explanation about the headphone type in question 6.

    Good Job!

    ReplyDelete
  3. I liked how you guys took a lot of measurements for the high pass frequency and low pass frequency, that will help to find the cut off frequency. For your graphs, the data in your plot looks correct, it isn't completely as our graphs but if we had taken more measurements, i think it will be close to your graphs. However, you guys are missing captions for videos, graphs and pictures. Make sure to add more explanation for Q4 and Q5. Good Job!

    ReplyDelete
  4. It seemed that you guys used a much broader range for your measurements then us for question 4 and 5. I wonder if that will give you better results than us? Other than adding some more explaining here there and adding captions for the videos and picture good blog.

    ReplyDelete
  5. You guys did excellent in taking more than the required 15 measurements in questions 4 and 5. I feel that by doing so you would get a more accurate representation on the graph. Your graphs do look somewhat different than ours but at the end of the day its still the same. Good job!

    ReplyDelete