Monday, March 14, 2016

Week 9

IMPORTANT NOTE:
Starting this week, the course format will slightly change:
  • Blogsheet information entry to the blog as group (blogG) will be reduced. You will only require to put short sentences as comments. The blogsheet will still be due Monday. The points for this part will be reduced to 10 points from 20 points.
  • Post-quiz will only be 10 points and include experiment related questions. Blog related questions will be removed from Friday’s quiz. However;
  • Blog related post-quiz questions (10 points) will be individually submitted (blogI) along with detailed blogsheet information submission (10 points) that are typed in red text in the blogsheet. This assignment will be due Wednesday the following week.
  • Commenting and responding to comments will be due the following Wednesday.


Blogsheet week 9
  1. Measure the resistance of the speaker. Compare this value with the value you would find online.
resistance- 8.1 Ohms Online- Around 8 Ohms



  1. Build the following circuit using a function generator setting the amplitude to 5V (0V offset). What happens when you change the frequency? (video)
C:\Users\kolar1km.CENTRAL\Desktop\AC1.PNG
Figure 1: Test setup for the speaker.




Fill the following table. Discuss your results.
Table 1: Write your caption here…
Frequency
Observation
660 HzLow pitch sound at E note when using square wave
800 HzLow pitch sound 
1000 HzHigher than before but still a lower pitched sound
1300 HzPitch continues to increase with the 300 Hz added
1600 HzWith another 300 Hz added the pitch continued to get higher and louder


  1. 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.
With the resistor added to the circuit in series with the speaker we were able to observe the circuit. With the frequency low the sound was more of a low pitch and as we increased our frequency the pitch began to rise and get louder.



Fill the following table. Discuss your results.
Table 1: Write your caption here…
Resistor value
Oscilloscope output
Observation
47 Ω
The output shows a sine wave with a frequency of 1.58kHzWe hooked both the input and the output up on the oscilloscope and the output has a smaller amplitude  
820 Ω
The output of the sine wave got much smaller with the higher resistor addedWith both of the inputs and outputs hooked up on the scope it is shown that with the higher resistor added it lowers the amplitude greatly 


  1. 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.
Figure 2: Test setup for the high pass filter.

Explain the operation.  (video)
High Pass Filter

High Pass Filter


  1. Frequency(Hz)      
    Vout(V)
              Vout/Vin
    500
    0.003
              0.00053
    1000
    0.005
              0.000884
    2000
    0.009
              0.001591
    3000
    0.012
              0.002122
    4000
    0.015
              0.002652
    5000
    0.016
              0.002829
    6000
    0.017
              0.003006
    7000
    0.019
              0.003359
    8000
    0.02
              0.003536
    9000
    0.017
              0.003006
    10000
    0.017
              0.003006
    11000
    0.017
              0.003006
    12000
    0.015
              0.002652
    13000
    0.015
              0.002652
    14000
    0.014
              0.002475
    16000
    0.011
              0.001945
    20000
    0.007
              0.001238


Graph 1




  1. What is the cut off frequency by looking at the plot in b

Looking at our graph from excel it is shown that the cutoff frequency of our filter is around 8000 Hz. 






  1. Draw Vout/Vin with respect to frequency using MATLAB. Your code would look like this;


Frequency = [ ]; % data points will be in the brackets
Output = [ ]; % Vout/Vin data points would be in the brackets.
plot(Frequency, Output, ’o-r’)
xlabel(‘  ’); %Right your x-axis label in ‘’
ylabel(‘  ’); %Right your y-axis label in ‘’


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

  1. Explain how the circuit works as a high pass filter.


  1. 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 (e, f, and g are for blogI).
Low Pass Filter
Low Pass Filter

For creating the low pass filter the speaker will end up in parallel with the capacitor in order to have the low pass filter properties.

     Low Pass


Frequency
                   Vout                     
Vout/Vin
        1000
                   0.336
0.059406
        3000
                   0.259
0.045792
        5000
                   0.214
0.037836
        7000
                   0.172
0.03041
        10000
                   0.125
0.0221
        13000 
                   0.093
0.016443
        15000
                   0.077
0.013614
        17000
                   0.064
0.011315
        20000
                   0.048
0.008487
        23000
                   0.035
0.006188
        25000
                   0.028
0.00495
        27000
                   0.021
0.003713
        30000
                   0.013
0.002298
        33000
                   0.007
0.001238
        35000
                   0.005
0.000884




6.
1.     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.

    With the circuit set up with an amplifier it was possible to put audio into the microphone and with our headphones hooked up we could somewhat hear the music. The music did not come through very clear but it was possible to hear the song. 






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




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

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

6 comments:

  1. This comment has been removed by the author.

    ReplyDelete
  2. I noticed that your low pass graph didn't show any dramatic changes in shape. What was different about your data points from high pass to low pass? Good work otherwise.

    ReplyDelete
    Replies
    1. honestly I am not sure why the low pass graph looks so much better than the high pass graph wise. Only thing that I can think of is that we took better data points vs the high pass

      Delete
  3. Why do you think Vout began to drop after about 8 kHz for the high pass filter, rather than staying consistent?

    ReplyDelete
    Replies
    1. The high pass filter didnt seem to work well for us. I agree that the Vout should remain the same as we see in every other high pass graph

      Delete
  4. I am puzzled with this data as well.

    ReplyDelete