Blogsheet week 9

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

 The resistance we measured on the speaker using an ohmeter was about 8.1 ohms, it did fluctuate. The value we found online was 8 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)

Video 2.1 Shows frequent changes that cause the speaker to emit higher and lower pitches

Fill the following table. Discuss your results.

Table 2.1 Frequencies and corresponding pitch of sounds from speaker

Frequency (KHz)	Observation
.5	Very Low pitch
1.5	Low pitch
3.5	Medium pitch
8	High pitch
16	Very high pitch

When the frequency of the function generator was low the speaker would emit a low pitch sound. Around 500 Hz we were able to barely hear and audible sound. As the frequency was increased the speaker's pitch would become higher all the way up to about 16 KHz we were able to hear. Anything higher than that was hard to hear anything.

   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.

Fill the following table. Discuss your results.

Table 3.1 Show resistor values and corresponding noise output of speaker

Resistor value	Oscilloscope output (mV rms)	Observation
47 Ω	46.1	Medium Low pitch Loud
820 Ω	5.6	Medium Low pitch Quiet

When we put the 47 ohm resistor in series with the speaker, the speaker was slightly quieter. After putting the 820 ohm resistor in series the speaker was much quieter than the 47 ohm resistor setup.

   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.

Figure 2: Test setup for the high pass filter.

    a.     Explain the operation.  (video)

Video 4.1 High pass filter (passive) 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.

Table 4.1 High pass filter data table

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

Graph 4.1 High pass filter output line graph

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

By looking at our graph we see that our cut off frequency was about 2KHz.

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

Frequency = [0.01 0.03 0.06 0.1 0.4 0.6 1 2 3 4 5 6];

Output = [0.000125 0.00025 0.004 0.004625 0.005625 0.007125 0.00775 0.00825 0.00825 0.007375 0.00625 0.005375];

plot(Frequency, Output, 'o-r')

xlabel('Frequency');

ylabel('Vout/Vin');

Graph 4.2 High pass filter output (MATLAB 'plot' graph)

Also, try loglog instead of plot.

Frequency = [0.01 0.03 0.06 0.1 0.4 0.6 1 2 3 4 5 6];

Output = [0.000125 0.00025 0.004 0.004625 0.005625 0.007125 0.00775 0.00825 0.00825 0.007375 0.00625 0.005375];

loglog(Frequency, Output, 'o-r')

xlabel('Frequency');

ylabel('Vout/Vin');

Graph 4.3 High pass filter output (MATLAB 'loglog' graph)

    f.      Calculate the cut off frequency theoretically and compare with one that was found in c.
   fc=Zc=(1/(2pi*F*C))= 7KHz                                                                                                                                                            Theoretically our cut off frequency was much higher, I am not sure as to why we saw only 2KHz when testing our filter's cut off frequency. We may have used a resistor with a higher value than was called for perhaps.

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

                  The operation of a high-pass filter is that the capacitor controls the frequency by allowing the high frequency to get through and blocking the low frequency.

   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.  (video)

Video 5.1 Low pass filter (Passive) 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. 

Table 5.1 Low pass filter data table

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

Graph 5.1 Low Pass filter output line graph

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

  As the graph shows the cut off frequency is about 1.6KHz. Also our theoretical cut off frequency was much different than what we tested it for. But the frequency value was very close to the value we achieved with the high pass filter.

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

Frequency = [0.01 0.04 0.08 0.1 0.4 0.8 1 1.5 2 3 5 6];

Output = [0.047125 0.05425 0.054125 0.054 0.05525 0.06 0.05875 0.06125 0.060125 0.05625 0.05 0.0475];

plot(Frequency, Output, 'o-r')

xlabel('Frequency');

ylabel('Vout/Vin');

Graph 5.2 Low Pass filter output (MATLAB 'plot' graph)

Also, try loglog instead of plot.

Frequency = [0.01 0.04 0.08 0.1 0.4 0.8 1 1.5 2 3 5 6];

Output = [0.047125 0.05425 0.054125 0.054 0.05525 0.06 0.05875 0.06125 0.060125 0.05625 0.05 0.0475];

loglog(Frequency, Output, 'o-r')

xlabel('Frequency');

ylabel('Vout/Vin');

Graph 5.3 Low Pass filter output (MATLAB 'plot' graph)

    f.      Calculate the cut off frequency theoretically and compare with one that was found in c.
    fc=Zc=(1/(2pi*F*C))= 7KHz             The value calculated was much higher than found in c here as well in the low pass filter. But it was close to the value we had in the high pass filter.                                                                                                                                               

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

The circuit for the low pass filter will work just the opposite which the capacitor will be allowing the low frequency to get through and blocking the high frequency. 

   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.

Video 2.1 shows the setup for the headphone jack circuit. 

We "speculated" that the microphone's output would be amplified and be directed to the audio jack so we could hear the noises emitted by the microphone.