week 9

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

• The speaker has a resistance of 9ohm

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: Video showing speaker function.

Figure 2:Table displaying speaker functions at different frequencies.

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.

Figure 3: Table showing differences with different resistors.

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. (video)

Figure 4: Video showing high pass filter.

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.

Figure 5: Table displaying values for high pass filter.

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

Figure 6: Graph of high pass filter Vout/Vin vs. Frequency

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

• The cutoff frequency occurs when the Vout/Vin ratio is 1/2 of the max Vout. In our case this occurred between 8 and 10 KHz. 

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

Figure 7: Vout/Vin vs Frequency graph using MATLAB

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

• The max theoretical cutoff frequency gives us a value of 9KHz. We got this value by taking half the max of Vout and finding when Vout/Vin is equal to it. The one we found in part c was between 8 and 10KHz, which is close to the theoretical value. 

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

• Its a high pass filter because it allows high frequency signals and cuts off low frequency signals. This happens as a result of the capacitor's inductance being dependent on frequencies and its relationship when in series. The capacitor has high resistance to low frequencies and low resistance to high frequencies.

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 is in parallel to the capacitor rather than in series.

a. Explain the operation. (video)

Figure 8: Video explaining low pass filter.

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.

Figure 9: Table displaying low pass filter values.

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

Figure 10: Graph of low pass filter Vout/Vin vs Frequency.

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

• The cutoff frequency is 8KHz.

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.

The video above explains the operation.