Monday, March 13, 2017

Week Nine

Not Shocking People After Spring Break

Week Nine

1. Measure the resistance of the speaker.

We measured 8.6 Ohms, while online is says this model of speaker offers 32 Ohms.

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


Figure 1: Frequency of AC Voltage applied to a speaker.

Figure 2: Circuit Layout

Figure 3: Frequencies compared speaker output.

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.



Figure 4: Observations while placing resistors in parallel with speaker. The higher the resistor, the lower the amplitude. Therefore, the lower the volume.

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 shouldnot hear anything at the beginning and start hearing the sound after a certain frequency. Use 22 nF for the capacitor.

a.
Figure 5: Video explaining the operation of the high pass  filter connected to the speaker.

b. & c.
Figure 6: V_out measured and V_out/V_in Calculated with respect to frequency increasing.

d. It looks like the cutoff frequency is 5000 with the voltage V_out capping off at 0.014V.
e. 
Figure 7: Plot of Frequency v. Output. The climax of the curve shows the critical frequency of the circuit.

f. The Cutoff Frequency according to our plot and theoretical calculations is 5000 Hz.
g. As the frequency increases, to a high 2000 Hz, the voltage amplitude takes on a significant enough value to be audible to human ears. However, when the frequency reaches 5000 Hz, the voltage reaches its cutoff point and begins to decrease. 

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.

We would place the speaker in parallel with a capacitor, the pair of which would be in series with a resistor. 

a.
Figure 8: Video explaining the operation of the low pass filter.

b. & c.
Figure 9:V_out measured and V_out/V_in Calculated with respect to frequency increasing in a low pass circuit.

d. The cutoff frequency is at 1000, capping off at .596 V.
e.
Figure 10: Plot of Frequency v. Output of a low pass circuit. The climax of the curve shows the critical frequency of the circuit.

f. According to the plot, our cutoff frequency is at 1000.
g. The low pass filter works in a way where the capacitor works as an extremely high resistance to low frequency signals, making them pass to the speaker, whereas high frequency will pass through the capacitor, and not to the speaker, as can be seen above. 

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.


Figure 11: Video with speculation of the operation of the circuit.


11 comments:

  1. Comparing your data with ours I think yours make more sense. This lab was really confusing for me. by looking to your blogs it gives me a better understanding. For the cut off frequency, don't you guys think it should be the same value for the high pass and the low pass? because that is what we got by using the same equation twice (fc=(1/(2pi*F*C)).
    Well done!

    ReplyDelete
    Replies
    1. Thank you very much. That equation is one for capacitive reactance of a low pass filter, not the cutoff.

      Delete
  2. I feel like your highness and lowpass filter data is better then my groups. We had a lot of trouble and couldn't really find the cutoff frequency very easy and it seemed to be much higher then what it should of been. Yours seems to be much more of what ours should of looked like data wise.

    ReplyDelete
    Replies
    1. Thank you, you may want to check your circuits resistors, which may stifle your frequency amplitude.

      Delete
  3. This comment has been removed by the author.

    ReplyDelete
  4. Good Job with your blog, I actually used it to help confirm that the data from both our tables were similar. Don't forget the Excel Plots of your tables, also the videos are only 240p on youtube and are difficult to observe (Check to see if there is a setting on your camera, youtube app, or try using a different phone). Nice post!

    ReplyDelete
    Replies
    1. Thank you very much, we'll take a look at our video resolutions and see if we can do anything about them.

      Delete
  5. Your plot of both Questions 4 and 5 look different from ours, our graph never peaked. I wonder what was different for our groups? Other than that its a well detailed and well done blog.

    ReplyDelete
    Replies
    1. If your graph never peaked, that might mean you need a larger range of input values, or you have broken components. The peak is what represents the cutoff frequencies for the filter.

      Delete
  6. Very well done data! Your graphs correlate a lot more with the theoretical ones than most groups had, including ours. Good job as always you two.

    ReplyDelete