Monday, March 13, 2017

Blog week 9

Blog Week 9

1. Measure the resistance of the speaker. Compare this value with the value you would find online.
~~Our speaker: 9 ohms
~~Online speaker: 6 to 7 ohms (for an 8 ohm speaker)

2. Build the following circuit using a function generator setting the amplitude to 5V (0V offset). What happens when you change the frequency? (video)
Frequency
Observation
1 Khz
Moderate tone (pitch)
3 Khz
Moderate-high tone (pitch)
5 Khz
High tone (pitch)
7 Khz
Higher tone (pitch)
9 Khz
Loud squeal



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
.49 V (Peak to Peak)
.3 V
820 Ω
.992 V (Peak to Peak)
.002 V
~~ The observations that we saw during this segment was that the frequency was constant throughout the segment. The voltage and Peak to Peak values were the only varying measurements. The output of the 47 ohm was barely audible however the output of the 820 ohm speaker was not audible to us, as group members.

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.

Picture obtained from Dr. Kaya's blogsheet
a. Explain the operation. (video)
~~
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.

c. Draw Vout/Vin with respect to frequency using Excel.
d. What is the cut off frequency by looking at the plot in b?
    The filter starts attenuating around 9 kHz, which seems to be our cutoff frequency.


Table and a graph of the measurements obtained during the High Pass filter testing. It shows the Vout/Vin (RMS) on the y-axis and the frequency in kHz on the x-axis.



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


Graph using MatLab where Frequency is in Kilo Hertz and Vout/Vin is in Volts
f. Calculate the cut off frequency theoretically and compare with one that was found in c.
R=100; C=22*10^-9;
LH=1/(2*pi*R*C)~~~~~~~~~~~~~~~~~ 7.2343*10^4 Hz
This differs from the what the cutoff frequency appears to be on our graph (9 kHz), we believe this is because the value calculated is just theoretical, and doesn't have very much merit on our real life circuit. It is odd that it is such a big discrepancy though.

g. Explain how the circuit works as a high pass filter.
~~A high pass filter decreases the ability for low frequency tones to pass through easily and high frequency tones go through much easier.

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.
~~The speaker would go in parallel with the resistor and the capacitor.

~a. Explain the operation. (video)

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

~d. What is the cut off frequency by looking at the plot in b?
~~ Approximately 2KHz
~e. Draw Vout/Vin with respect to frequency using MATLAB. Your code would look like this;

~f. Calculate the cut off frequency theoretically and compare with one that was found in c.
R=100; C=22*10^-9;
LH=1/(2*pi*R*C)~~~~~~~~~~~~~~~~~ 7.2343*10^4 Hz

~g. Explain how the circuit works as a Low pass filter.
~~A low pass filter decreases the ability for high frequency tones to pass through easily and low frequency tones go through much easier.

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.















5 comments:

  1. For numbers 4 and 5 part g you guys didn't really explain why the filters work and just said what they do. You guys had you cut off frequency be lower the your theoretical value when which is funny because ours was much higher for our low pass filter at least it seems like.

    ReplyDelete
  2. There seems to be some key sections you are missing in your blog. Our group also had issues with our graphs not really matching exactly with the theoretical ones as well.

    ReplyDelete
  3. Number 4 could use some more explanation. Good job overall though

    ReplyDelete
  4. Number 4 could use some more explanation. Good job overall though

    ReplyDelete