Blog sheet Week 11: Strain Gauges
Part A: Strain Gauges:
Strain gauges are used to measure the strain or stress levels on the materials. Alternatively, pressure on the strain gauge causes a generated voltage and it can be used as an energy harvester. You will be given either the flapping or tapping type gauge. When you test the circle buzzer type gauge, you will lay it flat on the table and tap on it. If it is the long rectangle one, you will flap the piece to generate voltage.
1. Connect the oscilloscope probes to the strain gauge. Record the peak voltage values (positive and negative) by flipping/tapping the gauge with low and high pressure. Make sure to set the oscilloscope horizontal and vertical scales appropriately so you can read the values. DO NOT USE the measure tool of the oscilloscope. Adjust your oscilloscope so you can read the values from the screen. Fill out Table 1 and provide photos of the oscilloscope.
Table 1: Strain gauge characteristics
2. Press the “Single” button below the Autoscale button on the oscilloscope. This mode will allow you to capture a single change at the output. Adjust your time and amplitude scales so you have the best resolution for your signal when you flip/tap your strain gauge. Provide a photo of the oscilloscope graph.
Part B: HalfWave Rectifiers
1. Construct the following halfwave rectifier. Measure the input and the output using the oscilloscope and provide a snapshot of the outputs.
2. Calculate the effective voltage of the input and output and compare the values with the measured ones by completing the following table.
3. Explain how you calculated the rms values. Do calculated and measured values match?
4. Construct the following circuit and record the output voltage using both DMM and the oscilloscope.
5. Replace the 1 µF capacitor with 100 µF and repeat the previous step. What has changed?
Part C: Energy Harvesters
1. Construct the halfwave rectifier circuit without the resistor but with the 1 µF capacitor. Instead of the function generator, use the strain gauge. Discharge the capacitor every time you start a new measurement. Flip/tap your strain gauge and observe the output voltage. Fill out the table below:
2. Briefly explain your results.
3. If we do not use the diode in the circuit (i.e. using only strain gauge to charge the capacitor), what would you observe at the output? Why?
4. Write a MATLAB code to plot the date in table of Part C1.
Flick plot:
x=[10 20 30];
y1=[26 102 61];
y2=[103 346 603];
y3=[375 430 350];
y4=[2800 2300 3000];
plot(x,y1,'o',x,y2,'s')
xlabel('Duration (s)')
ylabel('Output Voltage (mV)')
legend('1 Flick per second','4 Flicks per second')
Tap plot:
plot(x,y3,'o',x,y4,'s')
xlabel('Duration (s)')
ylabel('Output Voltage (mV)')
legend('1 Tap per second','4 Taps per second')
Wednesday, March 29, 2017
Monday, March 20, 2017
Lab 10
In this week’s lab, you will collect more data on low pass and high pass filters and “process” them using MATLAB.
PART A: MATLAB practice.
1. Open MATLAB. Open the editor and copy paste the following code. Name your code as FirstCode.m
Save the resulting plot as a JPEG image and put it here.
clear all;
close all;
x = [1 2 3 4 5];
y = 2.^x;
plot(x, y, 'LineWidth', 6)
xlabel('Numbers', 'FontSize', 12)
ylabel('Results', 'FontSize', 12)
2. What does clear all do?
Clear all simply clears the command window.
3. What does close all do?
Close all clears the command window and closes all MATLAB figures and graphs etc.
4. In the command line, type x and press enter. This is a matrix. How many rows and columns are there in the matrix?
There is one row and 5 columns in the matrix x.
x =
1 2 3 4 5
5. Why is there a semicolon at the end of the line of x and y?
The semicolon keeps the x and y from being displayed in the command window when the code is ran.
6. Remove the dot on the y = 2.^x; line and execute the code again. What does the error message mean?
It means that with out the period multiple values of x and y cannot be calculated.
7. How does the LineWidth affect the plot? Explain.
This is how thick the line on the plot is going to be.
8. Type help plot on the command line and study the options for plot command. Provide how you would change the line for plot command to obtain the following figure (Hint: Like ‘LineWidth’, there is another property called ‘MarkerSize’)
x = [1 2 3 4 5];
y = 2.^x;
plot(x,y,'ro','Linewidth',5,'markersize',15)
xlabel('Numbers', 'FontSize', 12)
ylabel('Results', 'FontSize', 12)
9. What happens if you change the line for x to x = [1; 2; 3; 4; 5]; ? Explain.
For this code, the array for "x" becomes a vertical one instead of a horizontal one; ie., there is one column with 5 rows instead of 5 columns with one row. The corresponding output is shown below:
x =
1
2
3
4
5
However, this change in the x array does not change the values for "y" and thus the plot remains the same.
10. Provide the code for the following figure. You need to figure out the function for y. Notice there are grids on the plot.
plot(x,y,'k:s','Linewidth',5,'markersize',15)
xlabel('Numbers', 'FontSize', 12)
ylabel('Results', 'FontSize', 12)
x
GridLineStyle = ''
grid on
11. Degree vs. radian in MATLAB:
a. Calculate sinus of 30 degrees using a calculator or internet.
The calculated value of sin 30 is .5.
b. Type sin(30) in the command line of the MATLAB. Why is this number different? (Hint: MATLAB treats angles as radians).
MATLAB uses radians as a default measurement and if the user wants degrees they just have to change the code a bit. Below, the code and the output are displayed for "sin(30)."
>> sin(30)
ans =
0.9880
c. How can you modify sin(30) so we get the correct number?
Using just "sin" will be accurate for values in radians, but to get values for degrees, "sind" can be used instead. The MATLAB input and output are displayed here:
>> sind(30)
ans =
0.5000
12. Plot y = 10 sin (100 t) using Matlab with two different resolutions on the same plot: 10 points per period and 1000 points per period. The plot needs to show only two periods. Commands you might need to use are linspace, plot, hold on, legend, xlabel, and ylabel. Provide your code and resulting figure. The output figure should look like the following:
t=linspace(0,4*pi/100,10);
y=10.*sin(100.*t);
plot(t,y,'ro')
hold on
t2=linspace(0,4*pi/100,1000);
y2=10.*sin(100.*t2);
plot(t2,y2,'k')
xlabel('Time (s)')
ylabel('y function')
legend('Coarse','Fine')
Note that the upper value of the linspace command for the two different t values is 4*pi/100. this is because the graph is of the first 2 periods and the period of a sinusoidal function is 2pi/b where "b" is the coeffecient of "t" in the sin function (y=Asin(bt)). So, to show two periods, 2pi/ b was multiplied by 2 to get 4pi/100.
13. Explain what is changed in the following plot comparing to the previous one.
So for the updated graph, the fine value has a set maximum of 5 now, as opposed to the original function that had a maximum value of 10. Because there are values that would normally be about the 5, the new graph just flattens out at the maximum.
14. The command find was used to create this code. Study the use of find (help find) and try to replicate the plot above. Provide your code.
t=linspace(0,4*pi/100,10);
y=10.*sin(100.*t);
plot(t,y,'ro')
hold on
t2=linspace(0,4*pi/100,1000);
y2=10.*sin(100.*t2);
%plot(t2,y2,'k')
xlabel('Time (s)')
ylabel('y function')
legend('Coarse','Fine')
I=find(y2<5);
t3= t2(I);
y3= y2(I);
plot(t3,y3,'k')
To obtain the graph from 13, the above code was entered. the "find" code finds the indices of the values of a vector that are above or below a certain value. For this function, we used "find" to find the values for y2 that were less than 5, where y2 was the smooth function for 10sin(100t). With these indices found, we called these values to the new matrices of t3 for time and y3 for the y values of the function. Then we plotted y3 with respect to t3 to get the graph displayed above.
PART B: Filters and MATLAB
1. Build a low pass filter using a resistor and capacitor in which the cut off frequency is 1 kHz.
Observe the output signal using the oscilloscope. Collect several data points particularly around the cut off frequency. Provide your data in a table.
MATLAB Code:
F=[10 50 100 200 300 400 500 600 700 800 850 900 950 1000 1050 1100 1150 ...
1200 1300 1400 1500]
Vout=[.838 .847 .849 .838 .820 .797 0.768 .737 .709 .680 .668 .648 .637 ...
.625 .607 .599 .583 .573 .545 .516 .493]
plot(F,Vout,'LineWidth', 3)
xlabel('Frequency (Hz)')
ylabel('rms Output (V)')
title ('Output Voltage Vs Frequency: Low Pass Filter')
Graph: Low pass filter and resulting output voltage
3. Calculate the cut off frequency using MATLAB. find command will be used. Provide your code.
MATLAB Code:
F=[10 50 100 200 300 400 500 600 700 800 850 900 950 1000 1050 1100 1150 ...
1200 1300 1400 1500];
Vout=[.838 .847 .849 .838 .820 .797 0.768 .737 .709 .680 .668 .648 .637 ...
.625 .607 .599 .583 .573 .545 .516 .493];
plot(F,Vout,'LineWidth', 3)
xlabel('Frequency (Hz)')
ylabel('rms Output (V)')
title ('Output Voltage Vs Frequency: Low Pass Filter')
hold on
c=find(Vout<.707*.838);
d=c(1);
CutoffFrequency=F(d)
MATLAB Output:
CutoffFrequency =
1150
4. Put a horizontal dashed line on the previous plot that passes through the cutoff frequency.
5.1.
Table: High pass filter frequency and the resulting output voltage
5.2.
PART A: MATLAB practice.
1. Open MATLAB. Open the editor and copy paste the following code. Name your code as FirstCode.m
Save the resulting plot as a JPEG image and put it here.
clear all;
close all;
x = [1 2 3 4 5];
y = 2.^x;
plot(x, y, 'LineWidth', 6)
xlabel('Numbers', 'FontSize', 12)
ylabel('Results', 'FontSize', 12)
2. What does clear all do?
Clear all simply clears the command window.
3. What does close all do?
Close all clears the command window and closes all MATLAB figures and graphs etc.
4. In the command line, type x and press enter. This is a matrix. How many rows and columns are there in the matrix?
There is one row and 5 columns in the matrix x.
x =
1 2 3 4 5
5. Why is there a semicolon at the end of the line of x and y?
The semicolon keeps the x and y from being displayed in the command window when the code is ran.
6. Remove the dot on the y = 2.^x; line and execute the code again. What does the error message mean?
It means that with out the period multiple values of x and y cannot be calculated.
7. How does the LineWidth affect the plot? Explain.
This is how thick the line on the plot is going to be.
8. Type help plot on the command line and study the options for plot command. Provide how you would change the line for plot command to obtain the following figure (Hint: Like ‘LineWidth’, there is another property called ‘MarkerSize’)
x = [1 2 3 4 5];
y = 2.^x;
plot(x,y,'ro','Linewidth',5,'markersize',15)
xlabel('Numbers', 'FontSize', 12)
ylabel('Results', 'FontSize', 12)
9. What happens if you change the line for x to x = [1; 2; 3; 4; 5]; ? Explain.
For this code, the array for "x" becomes a vertical one instead of a horizontal one; ie., there is one column with 5 rows instead of 5 columns with one row. The corresponding output is shown below:
x =
1
2
3
4
5
However, this change in the x array does not change the values for "y" and thus the plot remains the same.
10. Provide the code for the following figure. You need to figure out the function for y. Notice there are grids on the plot.
plot(x,y,'k:s','Linewidth',5,'markersize',15)
xlabel('Numbers', 'FontSize', 12)
ylabel('Results', 'FontSize', 12)
x
GridLineStyle = ''
grid on
11. Degree vs. radian in MATLAB:
a. Calculate sinus of 30 degrees using a calculator or internet.
The calculated value of sin 30 is .5.
b. Type sin(30) in the command line of the MATLAB. Why is this number different? (Hint: MATLAB treats angles as radians).
MATLAB uses radians as a default measurement and if the user wants degrees they just have to change the code a bit. Below, the code and the output are displayed for "sin(30)."
>> sin(30)
ans =
0.9880
c. How can you modify sin(30) so we get the correct number?
Using just "sin" will be accurate for values in radians, but to get values for degrees, "sind" can be used instead. The MATLAB input and output are displayed here:
>> sind(30)
ans =
0.5000
12. Plot y = 10 sin (100 t) using Matlab with two different resolutions on the same plot: 10 points per period and 1000 points per period. The plot needs to show only two periods. Commands you might need to use are linspace, plot, hold on, legend, xlabel, and ylabel. Provide your code and resulting figure. The output figure should look like the following:
Graph: Sinusoidal curve of how the coarse would normally appear
To obtain this graph, the following code was entered into MATLAB:t=linspace(0,4*pi/100,10);
y=10.*sin(100.*t);
plot(t,y,'ro')
hold on
t2=linspace(0,4*pi/100,1000);
y2=10.*sin(100.*t2);
plot(t2,y2,'k')
xlabel('Time (s)')
ylabel('y function')
legend('Coarse','Fine')
Note that the upper value of the linspace command for the two different t values is 4*pi/100. this is because the graph is of the first 2 periods and the period of a sinusoidal function is 2pi/b where "b" is the coeffecient of "t" in the sin function (y=Asin(bt)). So, to show two periods, 2pi/ b was multiplied by 2 to get 4pi/100.
13. Explain what is changed in the following plot comparing to the previous one.
Graph: Function with a maximum value of 5, and what it would normally look like
So for the updated graph, the fine value has a set maximum of 5 now, as opposed to the original function that had a maximum value of 10. Because there are values that would normally be about the 5, the new graph just flattens out at the maximum.
14. The command find was used to create this code. Study the use of find (help find) and try to replicate the plot above. Provide your code.
t=linspace(0,4*pi/100,10);
y=10.*sin(100.*t);
plot(t,y,'ro')
hold on
t2=linspace(0,4*pi/100,1000);
y2=10.*sin(100.*t2);
%plot(t2,y2,'k')
xlabel('Time (s)')
ylabel('y function')
legend('Coarse','Fine')
I=find(y2<5);
t3= t2(I);
y3= y2(I);
plot(t3,y3,'k')
To obtain the graph from 13, the above code was entered. the "find" code finds the indices of the values of a vector that are above or below a certain value. For this function, we used "find" to find the values for y2 that were less than 5, where y2 was the smooth function for 10sin(100t). With these indices found, we called these values to the new matrices of t3 for time and y3 for the y values of the function. Then we plotted y3 with respect to t3 to get the graph displayed above.
PART B: Filters and MATLAB
1. Build a low pass filter using a resistor and capacitor in which the cut off frequency is 1 kHz.
Observe the output signal using the oscilloscope. Collect several data points particularly around the cut off frequency. Provide your data in a table.
Frequency (Hz)

Output rms Voltage (V)

10

0.838

50

0.847

100

0.849

200

0.838

300

0.820

400

0.797

500

0.768

600

0.737

700

0.709

800

0.680

850

0.668

900

0.648

950

0.637

1000

0.625

1050

0.607

1100

0.599

1150

0.583

1200

0.573

1300

0.545

1400

0.516

1500

0.493

Table: Low pass filter frequency and resulting output voltage
*Input is 1 V peak to peak, or 0.707 rms V.
2. Plot your data using MATLAB. Make sure to use proper labels for the plot and make your plot line and fonts readable. Provide your code and the plot.MATLAB Code:
F=[10 50 100 200 300 400 500 600 700 800 850 900 950 1000 1050 1100 1150 ...
1200 1300 1400 1500]
Vout=[.838 .847 .849 .838 .820 .797 0.768 .737 .709 .680 .668 .648 .637 ...
.625 .607 .599 .583 .573 .545 .516 .493]
plot(F,Vout,'LineWidth', 3)
xlabel('Frequency (Hz)')
ylabel('rms Output (V)')
title ('Output Voltage Vs Frequency: Low Pass Filter')
Graph: Low pass filter and resulting output voltage
3. Calculate the cut off frequency using MATLAB. find command will be used. Provide your code.
MATLAB Code:
F=[10 50 100 200 300 400 500 600 700 800 850 900 950 1000 1050 1100 1150 ...
1200 1300 1400 1500];
Vout=[.838 .847 .849 .838 .820 .797 0.768 .737 .709 .680 .668 .648 .637 ...
.625 .607 .599 .583 .573 .545 .516 .493];
plot(F,Vout,'LineWidth', 3)
xlabel('Frequency (Hz)')
ylabel('rms Output (V)')
title ('Output Voltage Vs Frequency: Low Pass Filter')
hold on
c=find(Vout<.707*.838);
d=c(1);
CutoffFrequency=F(d)
MATLAB Output:
CutoffFrequency =
1150
4. Put a horizontal dashed line on the previous plot that passes through the cutoff frequency.
Graph: Low pass filter frequency with respect to voltage
5. Repeat 13 by modifying the circuit to a high pass filter.
Frequency (Hz)

Output rms Voltage (V)

10

0.030

50

0.048

100

0.072

200

0.163

300

0.235

400

0.301

500

0.369

600

0.423

700

0.467

800

0.516

850

0.532

900

0.549

950

0.568

1000

0.582

1050

0.596

1100

0.610

1150

0.622

1200

0.634

1300

0.655

1400

0.678

1500

0.689

5.2.
MATLAB Code:
F=[10 50 100 200 300 400 500 600 700 800 850 900 950 1000 1050 1100 1150 ...
1200 1300 1400 1500];
Vout=[0.030 .048 .072 .162 .235 .301 .369 .423 .467 .516 .532 .544 .568 ...
.582 .596 .610 .622 .634 .655 .678 .689];
plot(F,Vout,'LineWidth', 3)
xlabel('Frequency (Hz)')
ylabel('rms Output (V)')
title ('Output Voltage Vs Frequency: High Pass Filter')
Graph: High pass filter frequency with respect to voltage
5.3 Calculate the cut off frequency using MATLAB. find command will be used.
F=[10 50 100 200 300 400 500 600 700 800 850 900 950 1000 1050 1100 1150 ...
1200 1300 1400 1500];
Vout=[.030 .048 .072 .162 .235 .301 .369 .423 .467 .516 .532 .544 .568 ...
.582 .596 .610 .622 .634 .655 .678 .689];
plot(F,Vout,'LineWidth', 3);
xlabel('Frequency (Hz)');
ylabel('rms Output (V)');
title ('Output Voltage Vs Frequency: Low Pass Filter');
x=find(Vout>.549);
c=F(s);
cutoff= min(c)
cutoff =
950
Thursday, March 16, 2017
Lab 9
1. Measure the resistance of the speaker. Compare this value with the value you would find online.
The resistance of our speaker was about 18.8 Ohms, how ever we noticed it was moved even a little it would change the resistance, so our DMM ended up varying from about 1320 ohms. Looking online, most speakers have a resistance of either 4, 8 or 16 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)
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.
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)
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.
d. What is the cut off frequency by looking at the plot in b?
The cutoff frequency from this set of data appears to be about 8000 Hz. This can be seen in the graph above and the two graphs below because the cutoff frequency is where the Vin/Vout stops increasing with respect to frequency.
e. Draw Vout/Vin with respect to frequency using MATLAB.
f. Calculate the cut off frequency theoretically and compare with one that was found in c.
Our calculated value for the cutoff frequency was 72kHz, while our observed cutoff frequency was about 8000 Hz. This large discrepancy could have been avoided if we had tested larger values for Frequency instead of assuming that this apparent downward trend after 8000 Hz would continue.
g. Explain how the circuit works as a high pass filter.
A high pass filter blocks waves of lower frequencies from passing while allowing higher frequency waves, depending on the setting of the filter, to pass. This means that the the input and output voltages for the higher frequency waves should be about the same. the result of this is that the higher frequencies are audible, and the lower ones are silenced.
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.
5a.
5b. Low pass filter
5c.
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 resistance of our speaker was about 18.8 Ohms, how ever we noticed it was moved even a little it would change the resistance, so our DMM ended up varying from about 1320 ohms. Looking online, most speakers have a resistance of either 4, 8 or 16 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)
Image: Simple Speaker Circuit Diagram
Frequency (Hz)

Observation

1

No sound

15

Quiet chirping sound

222.22

Quiet low pitched hum

1000

High pitched hum

23000

No sound

Table: Sound observations from the varying frequency
Video: Explanation of the Effect of Varying frequency.
As stated in the table above, very low frequency and very
high frequencies were inaudible from the speaker. In the range of audible
frequencies, the lower ones had a lower pitch, and as the frequency was
gradually increased, the pitch of the sound from the speaker increased. This increase
was not gradual, but occurred in steplike increments.
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.
Resistance (Ω)

Oscilloscope output over speaker (V)

Observation

47

1.2

Same pitch as without resistance, but it is quieter

820

0.16

The pitch still remains the same, and it is even quieter.

Table: Voltage across certain resistances and what was observed
This data shows that the resistance values of a circuit
involving a speaker do not have an apparent impact on the pitch of the
resulting sound. This means that the resistance does not affect the frequency of
the voltage. Resistance, does, however, affect volume i.e.; the higher the resistance
the lower the volume.
not hear anything at the beginning and start hearing the sound after a certain frequency. Use 22 nF for the capacitor.
Image: Speaker Circuit Diagram with a Capacitor
a. Explain the operation. (video)
Video: Operation of the High pass filter
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.
Frequency (Hz)

Vout (rms) (mV)

Vout (rms)/ Vin (rms) (mV)

0

1

.18

20

1

.18

100

2

.3536

500

2

.35

1000

5

.884

1500

6

1.061

2000

7

1.238

2500

8

1.45

3000

9

1.59

3500

10

1.77

4000

11

1.95

5000

12

2.12

10000

12

2.12

15000

8

1.45

20000

6

1.061

Table: Varying frequency and the measured output
c. Draw Vout/Vin with respect to frequency using Excel.
Graph: Vout/Vin and frequency
d. What is the cut off frequency by looking at the plot in b?
The cutoff frequency from this set of data appears to be about 8000 Hz. This can be seen in the graph above and the two graphs below because the cutoff frequency is where the Vin/Vout stops increasing with respect to frequency.
e. Draw Vout/Vin with respect to frequency using MATLAB.
Vout/Vin Vs Frequency
Graphs: Top one is Vout/Vin with respect to frequency, Bottom is loglog graph with same values
f. Calculate the cut off frequency theoretically and compare with one that was found in c.
Our calculated value for the cutoff frequency was 72kHz, while our observed cutoff frequency was about 8000 Hz. This large discrepancy could have been avoided if we had tested larger values for Frequency instead of assuming that this apparent downward trend after 8000 Hz would continue.
g. Explain how the circuit works as a high pass filter.
A high pass filter blocks waves of lower frequencies from passing while allowing higher frequency waves, depending on the setting of the filter, to pass. This means that the the input and output voltages for the higher frequency waves should be about the same. the result of this is that the higher frequencies are audible, and the lower ones are silenced.
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.
5a.
5b. Low pass filter
Frequency (Hz)

Vout (rms) (V)

Vout (rms)/ Vin (rms) (V)

0

.001

.001707

20

.286

.0505

100

.287

.0507

220

.29

.0512

500

.297

.05249

650

.32

.0565

700

.326

.0576

750

.331

.05848

830

.329

.058

1000

.315

.0556

1500

.293

.0518

3000

.239

.0472

5000

.197

.0348

10000

.116

.0205

15000

.069

.012

20000

.042

.0074

Graph: Low pass filter with varying frequency
Graph: The x axis is the frequency in Hz, and the yaxis is the Vout/Vin voltage
e.
Graphs: Top one is Vout/Vin with respect to frequency, Bottom is loglog graph with same values
Image: Circuit Diagram with a Microphone and Headphone Jack
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