Blogsheet week 9
1 Measure the resistance of
the speaker. Compare
this value with the value you would find online.
The measured value was between 8.6 and 8.7 ohms which very close to the value we found online which was 8 ohms.
The measured value was between 8.6 and 8.7 ohms which very close to the value we found online which 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)
Video1: shows what happens when we increase the frequency.
Fill the following table. Discuss your
results.
Frequency

Observation

1 Hz

no sound

10 HZ

Vibration

100 Hz

low pitch

1KHz

high pitch

2 KHz

Pitch keep increasing

The sound that is coming from the speaker will increase as the frequency increases, the sound will get higher after 100 Hz and there will be a vibration before or sound that is difficult to hear, and as we decrease the frequency the sound will decrease as well.
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.
Resistor
value

Oscilloscope
output

Observation

47 Ω
 
820 Ω

Table 3.1: shows the voltage across two 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)High pass filter works by blocking the sound before specific frequency and the sound will occur after it. To build a high pass filter we connect the speaker in series with capacitor that is connected to a resistor in serires which is connected in series with two parallel resistors with equivalent resistance of 91.6, that makes the resistance of the speaker and the resistors is 100 ohms. This make the high pass note to go through the speaker. When the frequency reaches 700 ohms, we started to hear a sound and the sound keeps increasing after it but there will be no sound before 700 ohms.
b) Fill out the following
table by adding enough (1015 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?
I think we should have taken more frequency values but it seems the range of our data wasn't enough but when we looks at the graph it seems that the cut off frequency will be after 5000 Hz because when the Vinput/ Voutput will stop increasing and it starts to decrease and as it is seen in the graph the line will start to decrease after 5000 Hz.
I think we should have taken more frequency values but it seems the range of our data wasn't enough but when we looks at the graph it seems that the cut off frequency will be after 5000 Hz because when the Vinput/ Voutput will stop increasing and it starts to decrease and as it is seen in the graph the line will start to decrease after 5000 Hz.
e) 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, ’or’)
xlabel(‘
’); %Right your xaxis label in ‘’
ylabel(‘
’); %Right your yaxis label in ‘’
Also, try loglog instead of plot.
Graph 4.2 shows the plot and code for the graph using MATLAB 
Graph 4.3 shows the loglog and code for the graph using MATLAB 
f) Calculate
the cut off frequency theoretically and compare with one that was found in c.
theoretically the cut off frequency is calculated by the equation fc= 1/(2 * pi * R * C)= 72379 Hz. We really can't compare to our graph result because what we concluded in our graphs is that the cut off frequency will be after 5000 Hz. However, from what we have seen in our graph we thought that the cut off frequency will be right after 5000 Hz, between 5000Hz and 10000Hz but it seems that our conclusion wasn't correct.
g) 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 4ag using the new designed circuit.
a)explain the operation(video)
b) Fill out the following table by adding enough (1015 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?

Graph5.1: shows what happens when we draw Vout/Vin with respect to frequency in low pass filter. 
By looking at the graph in part c I would say that the cutoff value should be around 1KHz or even a lower frequency. If we had a wider range of frequency we would be able to see the true cut off value.
e) Draw Vout/Vin with respect to frequency using MATLAB. Your code would look like this;
Graph 5.2 shows the plot and code for the graph using MATLAB 
Graph 5.3 shows the loglog and code for the graph using MATLAB 
f) Calculate the cut off frequency theoretically and compare with one that was found in c.
theoretically the cut off frequency is calculated by the equation fc= 1/(2 * pi * R * C)= 72379 Hz. We determine that the cutoff frequency is somewhere around or before 1KHz. Although we don't see this in our data in our observation we noticed that around 7 and 8 KHz the speaker made less noticeable of a sound.
g) Explain how the circuit works as a low pass filter.
Works the same way that the high pass filter but in the opposite way allowing low frequencies to pass and blocking the higher frequencies.
6Construct 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.)
Good blog this week guys. I really like the use of the tables, and how organized and easy to read it is. Our tables were not identical to what you guys have, but still pretty close. It may have been the different values we have? Overall good work this week!
ReplyDeleteThank you for the comment Angel.
ReplyDelete