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)

Figure 1: Test setup for the speaker.

Fill the following table. Discuss your results.

Table 1: Write your caption here…

Frequency	Observation
660 Hz	Low pitch sound at E note when using square wave
800 Hz	Low pitch sound
1000 Hz	Higher than before but still a lower pitched sound
1300 Hz	Pitch continues to increase with the 300 Hz added
1600 Hz	With another 300 Hz added the pitch continued to get higher and louder

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.

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 1.58kHz	We 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 added	With 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

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.

Explain the operation.  (video)

High Pass Filter

High Pass Filter

3. 
   

   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

3. 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. 

3. 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 ‘’

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

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

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 (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.