week 9

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.

    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

                           Table 2.1: shows what happened when we change the frequency. 

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	wave with  448 mV pk-pk	as we increases the frequency of the speaker the frequency increases and the period decreases.
820 Ω	wave with 54mV pk-pk	it does the same thing but it does it faster because it has smaller pk-pk.

                                Table 3.1: shows the voltage across two different resistors. 

When we put the resistor in series with the speaker, the pitch sound was the the same whether with resistors or without resistor, but the change happened for the volume, when we increased the resistance the volume decreased.  

   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)

                                           Video 2: shows the operation of high pass filter. 

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 (10-15 data points) frequency measurements. Vout is measured with the DMM, thus it will be rms value.

Table 4.1: shows the measurements of Vout and Vout/Vin
for a high pass filter.

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

Graph4.1: shows what happens when we draw  Vout/Vin with respect to
frequency in high pass filter. 

    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.

   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, ’o-r’)

xlabel(‘  ’); %Right your x-axis label in ‘’

ylabel(‘  ’); %Right your y-axis 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.

High pass filter works by blocking a low signal so we can't hear any sound lower than a certain frequency and we can hear a sound above the certain frequency. in other words, it blocks low frequencies and allow high frequencies to pass. 

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)

                                                 Video3: shows the operation of 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.

Table 5.1: shows the measurements of Vout and Vout/Vin
for a low pass filter.

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

Graph5.1: shows what happens when we draw  Vout/Vin with respect to frequency 
in low pass filter. 

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

By looking at the graph in part c I would say that the cut-off 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. 

   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 3 shows the operation of the circuit.