exp2 passive band pass and band-stop filter

NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGYAmafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite

EXPERIMENT # 2Passive Band-Pass and Band-Stop Filter

Cauan, Sarah Krystelle P. July 05, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score:

Eng’r. Grace RamonesInstructor

OBJECTIVES

1. Plot the gain-frequency response of an L-C series resonant and an L-C parallel

resonant band-pass filter.

2. Determine the center frequency and the bandwidth of the L-C band pass filter.

3. Determine how the circuit resistance affects the bandwidth of an L-C band-pass

filter.

4. Plot the gain-frequency response of an L-C series resonant and an L-C parallel

resonant band-stop (notch) filter.

5. Determine the center frequency and the bandwidth of the L-C band-stop filter.

6. Determine how the circuit resistance affects the bandwidth of an L-C band-stop

filter.

SAMPLE COMPUTATIONS

Solution for Step 4

AdB=20 log A

−1.637dB=20 log A

−1.637dB20

=20 log A20

10−1.637 dB

20 =10 log A

10−1.637dB

20 =A

A=0.83

Solution for Step 6:

BW=1.104 kHz−914.969Hz=189.031Hz

Solution for Step 7

f o=1

2π √LC= 1

2π √ (100×10−3H ) (0.25×10−6F )=1006.58Hz

Solution for Question in Step 7

% difference=¿

Solution for Step 8

f 0=√ f C1 f C2=√ (914.969Hz )(1104Hz )=1005.05Hz


% difference=¿

Solution for Step 9

QS=X LRT

=2 π f o L

RT=2π (1006.58Hz )(100mH )

120Ω=5.27

Solution for Step 10

BW=f OQ

=1006.58Hz5.27

=191Hz


% difference=¿


AdB=20 log A

−2.189dB=20 log A

−2.189dB20

=20 log A20

10−2.189 dB

20 =10 log A

10−2.189 dB

20 =A

A=0.78


BW=1.093kHz−929.014 Hz=163.186Hz


f o=1

2π √LC= 1

2π √ (100×10−3H )(0.25×10−6 F)=f o=1006.58Hz


% difference=¿


f 0=√ f C1 f C2=√ (929.014Hz )(1093Hz)=1007.68Hz


% difference=¿


QL=X LRW

=2π f o L

RW=2π (1006.58Hz )(100mH )

20Ω=31.62


REQ=RW (QL2+1)=20Ω (31.622+1 )=20016.488Ω


1RP

= 1R

+ 1RS

+ 1REQ

= 1200000Ω

+ 15000Ω

+ 120016.488Ω

= 13922.20Ω

RP=3922.20Ω


QP=RPX L

= 3922.20Ω2π (1006.58Hz )(100mH )

=6.2015


BW=f OQ

=1006.58Hz6.2015

=162.31Hz


% difference=¿


BW=1.103kHz−918.093Hz=184.907Hz


f o=1

2π √LC= 1

2π √ (100×10−3H )(0.25×10−6 F)=f o=1006.58Hz


% difference=¿


f 0=√ f C1 f C2=√ (918.093Hz )(1.103kHz)=1006.31Hz


% difference=¿


QS=X LRT

=2 π f o L

RT=2π (1006.58Hz )(100mH )

120Ω=5.27


BW=f OQ

=1006.58Hz5.27

=191Hz


% difference=¿


BW=1.098kHz−914.969Hz=183.031Hz


f o=1

2π √LC= 1

2π √ (100×10−3H )(0.25×10−6 F)=f o=1006.58Hz


% difference=¿


f 0=√ f C1 f C2=√ (914.969Hz )(1.098kHz)=1002.32Hz


% differenc e=¿


QL=X LRW

=2π f o L

RW=2π (1006.58Hz )(100mH )

20Ω=31.62


REQ=RW (QL2+1)=20Ω (31.622+1 )=20016.488Ω


RP=(R ) (REQ )R+REQ

=(4kΩ ) (20016.488Ω )4 kΩ+20016.488Ω

=3333.79Ω


QP=RPX L

= 3333.79Ω2π (1006.58Hz ) (100mH )

=5.27


BW=f OQP

=1006.58Hz5.27

=191Hz

Solution for Step Question in 50

% difference=¿

DATA SHEET

MATERIALS

One function generator

One dual-trace oscilloscope

Capacitors: one 0.1µF, one 0.25 µF

Inductors: one 50 mH, one 100 mH

Resistors: 100 Ω, 200 Ω, 2 kΩ, 4 kΩ, 5kΩ, 200kΩ

THEORY

In electronic communications systems, it is often necessary to separate a specific range of frequency from the total frequency spectrum. This is normally accomplished with filters. A filter is circuit that passes a specific range of frequencies while rejecting other frequencies. A passive filter consists of passive circuit elements, such as capacitors, inductors, and resistors. There are four basic types of filters, low-pass, high-pass, band-pass, and band-stop. A low-pass filter is designed to pass all frequencies below the cutoff frequency and reject all frequencies above the cutoff frequency. A high-pass filter is designed to pass all frequencies above the cutoff frequency. A high-pass filter is designed to pass all frequencies above the cutoff frequency and reject all frequencies below the cutoff frequency. A band-pass filter passes all frequencies within a band of frequencies and rejects all other frequencies outside the band. A band-stop filter rejects all frequencies within a band of frequencies outside the band. A band-stop filter is often referred to as a notch filter. In this experiment, you will study band-pass and band-stop (notch) filters.

The most common way to describe the frequency response characteristics of a filter is to plot the filter voltage gain (Vo/Vi) in db as function of frequency (f). The frequency at which the output power gain drops is 50% of the maximum value is called the cutoff frequency (fC). When the output power gain drops to 50%, the voltage gain drops 3 db (0.707 of the maximum value). When the filter dB voltage gain is plotted as a function of frequency on semi log graph using straight lines to approximately the actual frequency response, it is called a Bode plot. A bode plot is an ideal plot of filter frequency response because it assumes that the voltage gain remains constant in the passband until the cutoff frequency is reached, and then drops in straight line. The filter network voltage gain in dB is calculated from the actual voltage gain (A) using the equation

AdB = 20 log A

Where A = Vo/Vi

An L-C series resonant band-pass filter is shown in Figure 2-1. The impedance of the series L-C circuit is lowest at the resonant frequency and increases on both sides of the resonant frequency. This will cause the output voltage to be highest at the resonant frequency and decrease on both sides of the resonant frequency. At L-C parallel resonant band-pas filter is shown in Figure 2-2. The impedance of the parallel L-C circuit is highest at the resonant frequency and decreases on both sides of the resonant frequency. This will

also cause the output voltage to be highest at the resonant frequency and decrease on both sides of the resonant.

Figure 2-1 L-C Series Resonant Band-Pass Filter

`

Figure 2-2 L-C Parallel Resonant Band-Pass Filter

An L-C series resonant band-stop (notch) filter is shown in Figure 2-3. The impedance at the series L-C circuit is lowest at the resonant frequency and increase on both sides of the resonant frequency. This will cause the output voltage to be lowest at the resonant frequency and increase on both side of the resonant frequency. An L-C parallel resonant band-stop (notch) filter is shown in Figure 2-4. The impedance of the parallel L-C

circuit is highest at the resonant frequency and decreases on both sides of the resonant frequency. This will also cause the output voltage to the lowest at the resonant frequency and increase on both sides of the resonant frequency.

Figure 2-3 L-C Series Resonant Band-Stop (Notch) Filter

Figure 2-4 L-C Parallel Resonant Band-Stop (Notch) Filter

The center frequency (fO) for the L-C series resonant and the L-C parallel resonant band-pass and band-stop (notch) filter is equal to the resonant frequency of the L-C circuit, which can be calculated from

f o=1

2π √LCFor an L-C parallel resonant filter, the equation is accurate only for a high Q inductor

coil (Qf ≥ 10) where QL is calculated from

QL=X LRW

and XL is the inductive reactance at the resonant frequency (center frequency, fO) and Rw is the inductor coil resistance.

In the band-pass and band-stop (notch) filters, the low-cutoff frequency (fC1) and the high-cutoff frequency (fC2) on the gain-frequency plot are the frequencies where the whole voltage gain has dropped 3dB (0.707) from the highest dB gain. The filter bandwidth (BW) is the difference between the cutoff frequency (fC2) and the low-cutoff frequency (fC1) Therefore,

BW = fC2 – fC1

The center frequency (f0) is the geometric mean of the low-cutoff frequency and the high-cutoff frequency. Therefore,

f 0=√ f C1 f C2The quality factor (Q) of the band-pass and the band-stop (notch) filters is the ratio

of the center frequency (fO) and the bandwidth (BW), and it is an indication of the activity of the filter. Therefore,

Q=f OBW

A higher value of Q means a narrower bandwidth and a more selective filter.

The quality factor (QS) of a series resonant filter is determined by first calculating the inductive reactance (XL) of the inductor at the resonant frequency (center frequency, fO), and then dividing the inductive reactance by total series resistance (RT). Therefore,

QS=X LRT

Where X L=2π f oL

The quality factor (QP) of a parallel resonant filter is determined by first calculating the inductive reactance (XL) of the inductor at the resonant frequency (center frequency, fO), and then dividing the total parallel resistance (RP) by the inductive reactance (XL). Therefore,

QP=RPX L

Because the inductor wire resistance (RW) is in series with inductor L, the circuits in Figures 2-2 and 2-4 are not exactly parallel resonant circuits; the series combination of inductance (L) and resistance (RW) must first be converted into an equivalent parallel network with resistance REQ in parallel with inductance L. In Figure 2-2, the parallel equivalent resistance (REQ) will also be in parallel with resistor r and resistor RS making the total resistance of the parallel resonant circuit (RP), equal to the parallel equivalent of resistors r, RS, and REQ. Therefore, RP can be solved from

1RP

= 1R

+ 1RS

+ 1REQ

In Figure 2-4, the parallel equivalent resistance (REQ) will be in parallel with resistor R, making the total resistance of the parallel resonant circuit (RP) equal to the parallel equivalent of resistor R and REQ. Therefore, RP can be solved from

RP=(R )(REQ)R+REQ

The equation for converting resistance RW to the equivalent parallel resistance (REQ) is

REQ=RW (QL2+1)

The parallel equivalent inductance LEQ is calculated from

REQ=L(QL

2+1)QL2

This equivalent inductance can be considered equal to the original inductance (L) for a high Q coil (QL≥10)

PROCEDURE

Band-Pass Filters

Step 1 Open circuit file Fig 2-1. Make sure that the following Bode plotter settings are selected. Magnitude, Vertical (Log, F=0 dB, I=–20dB), Horizontal (Log, F=2 kHz, I=500Hz)

Step 2 Run the simulation. Notice that the voltage gain in db has been plotted between the frequencies of 50Hz and 2Kz by the Bode plotter. Sketch the curve plot in the space provided.

Question: Is the frequency response curve that of a band-pass filters? Explain why.

Yes. A band-pass is a filtering device that permits only the frequencies within a certain band and rejects all other band. The plot curve shown above only allows the frequencies from 501.697 Hz to 1.993 kHz.

Step 3 Move the cursor to the center of the curve at its peak point. Record the center frequency (f0) and the voltage gain in dB on the curve plot.

fO= 996.84 Hz; AdB= – 1.637 dB

Step 4 Based on the dB voltage gain, calculate the actual voltage gain (A) of the series resonant band-pass filter at the center frequency.

A= 0.83

Step 5 Move the cursor as close as possible to a point on the left side of the curve that is 3 dB down from dB gain measured at frequency fO. Record the approximate frequency (low-cutoff frequency, fC1) on the curve plot. Next, move the cursor as close as possible to a point on the right side of the curve that is 3 dB down from the dB gain measured at frequency fO. Record the approximate frequency (high-cutoff frequency, fC2) on the curve plot.

fC1= 914.969 Hz fC2= 1.104 kHz

Step 6 Based on the values of fC1 and fC2 measured on the curve plot, determine the bandwidth (BW) of the series resonant band-pass filter.

BW= 189.031 Hz

Step 7 Based on the circuit component values in Figure 2-1, calculated the expected center frequency (fO) of the series resonant band-pass filter.

fO(COMPONENT VALUE)= 1006.58 Hz

Question: How did the calculated value of the center frequency (fO) based on the circuit component values compare with the measured value on the curve plot?

The calculated fO based on the circuit component values and the measured value is almost equal. Their percentage difference is 0.97%.

Step 8 Based on the values of fC1 and fC2, calculate the center frequency (fO).

fO(fC1/fC2)= 1005.05 Hz

Question: How did the calculated value of the center frequency (fO) based on fC1 and fC2

compare with the measured value on the curve plot?

The percentage difference between the two is 0.82%.

Step 9 Based on the circuit component values, calculate the quality factor (QS) of the series resonant band-pass filter.

QS= 5.27

Step 10 Based on the circuit quality factor (QS) and the center frequency (fO), calculate the expected bandwidth (BW) of the series resonant band-pass filter.

BW(EXPECTED)= 191.00 Hz

Question: How did the expected bandwidth calculated from the value of QS and the center frequency compare with the bandwidth measured on the curve plot?


Step 11 Change the resistance of R to 200 Ω. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) from the curve plot and record the values.

fO=996.84 Hz BW= 350.136 Hz

Questions: What effect did changing the resistance R have on the center frequency of the series resonant band-pass filter? What effect did changing the resistance of R have on the bandwidth of the series resonant band-pass filter? Explain.

The center frequency remain constant or did not change its value while the bandwidth changed, it increased as the resistor increase. Therefore, the center frequency fO of a series resonant band-pass filter is constant with change in the resistance R while the bandwidth of the series resonant band-pass filter is directly proportional to the resistance R.

Step 12 Change the capacitance of C to 0.1 µF. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) on the curve plot and record the values Change the Bode plotter settings as needed.

fO=1.583 kHz BW= 350 Hz

Questions: What effect did changing the capacitance of C have on the center frequency of the series resonant band-pass filter? What effect did changing the capacitance of C have on the bandwidth of the series resonant band-pass filter? Explain.

The bandwidth remains constant or did not change its value. The center frequency however changed, it increased as the resistor increase. Therefore, the bandwidth of a series resonant band-pass filter is affected by the change of capacitance C while the center frequency of the series resonant band-pass filter is inversely proportional to the capacitance C.

Step 13 Change the inductance of L to 50 mH. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) on the curve plot and record the values. Change the Bode plotter settings as needed.

fO=1.423 kHz BW= 386 Hz

Questions: What effect did the changing the inductance of L has on the center frequency of the series resonant band-pass filter? What effect did changing the inductance of L have on the bandwidth of the series resonant band-pass filter? Explain.

The center frequency and bandwidth was compared to the original circuit. By changing the inductance, bandwidth BW and center frequency fO increased. Therefore, bandwidth and the center frequency of the series resonant band-pass filter are inversely proportional to the inductance of L

Step 14 Open circuit file Fig 2-2. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–20dB), Horizontal (Log, F=2 kHz, I=500Hz)

Step 15 Run the simulation. Notice that the voltage gain in dB has been plotted between the frequencies of 500 Hz and 2 kHz by the Bode plotter. Sketch the curve plot in the space provided.

Question: Is the frequency response curve that of a band-pass filters? Explain why.

Yes. The curve plot above is a band-pass because it permits only the frequencies within bandpass from 501.697 Hz to 1.993 kHz and rejects all other band.

Step 16 Move the cursor to the center of the curve at its peak. Record the center frequency (fO) and the voltage gain in dB on the curve plot.

fO= 1.003 kHz AdB= – 2.189 dB

Step 17 Based on the dB voltage gain, calculate the actual voltage gain (A) of the parallel resonant band-pass filter at the center frequency.

A= 0.78

Step 18 Move the cursor at its close as possible to a point on the left side of the curve that is 3dB down from the dB gain measured at frequency fO. Record the approximate frequency (low-cutoff frequency, fC1) on the curve plot. Next, move the cursor as close as possible to a point on the right side of the curve that is 3 dB down from the dB gain measured at frequency fO. Record the approximate frequency (high-cutoff frequency fC2) on the curve plot.

fC1= 929.014 Hz fC2= 1.093 kHz

Step 19 Based on the values of fC1 and fC2, determine the bandwidth (BW) of the parallel resonant band-pass filter.

BW= 163.186 Hz

Step 20 Based on the circuit component values in Fig 2-2, calculate the expected center frequency (fO) of the parallel resonant band-pass filter.


Question: How did the calculated values of the center frequency (fO) base on the circuit component values compare with measured values recorded on the curve plot?

The percentage difference between the calculated fO based on the circuit component values and the measured value is only 0.37%.


fO(fC1/fC2)= 1007.68 Hz

Question: How did the calculated value of the center frequency (fO) base on the fC1 and fC2



Step 22 Based on the value of L and RW, calculate the quality factor (QL) of the inductor.

QL=31.62

Step 23 Based on the quality factor (QL) of the inductor, calculate the equivalent parallel inductor resistance (REQ) across the tank circuit.

REQ= 20016.488Ω

Step 24 Based on the value of REQ, RS, and R, calculate the total parallel resistance (RP) across the tank circuit.

RP= 3922.20Ω

Step 25 Based on the value of RP, calculate the quality factor (QP) of the parallel resonant band-pass filter.

QP= 6.2015

Step 26 Based on the filter quality (QP) and the center frequency (fO), calculate the expected bandwidth (BW) of the parallel resonant band-pass filter.

BW= 162.31 Hz

Question: How did the expected bandwidth calculated from the value of QP and the center frequency compare with the bandwidth measured on the curve plot?


Step 27 Change the resistance of R to 5kΩ. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) from the curve plot and record the values.

fO=1.003 kHz BW= 289.456 Hz

Questions: What effect did changing the resistance of R have on the center frequency of the parallel resonant band-pass filter? What effect did changing the resistance R have on the bandwidth of the parallel resonant band-pass filter? Explain why.

The center frequency remain constant or did not change its value while the bandwidth changed, it increased as the resistor decrease. Therefore, the center frequency fO of a series resonant band-pass filter is constant with change in the resistance R while the bandwidth of the series resonant band-pass filter is inversely proportional to the resistance R.

Band-Stop (Notch) Filters

Step 28 Open the circuit file Fig 2-3. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–20dB), Horizontal (Log, F=2 kHz, I=500Hz)


Question: Is the frequency response curve that of a band-stop (notch) filters? Explain why.

Yes. What was shown above is a band-stop filter because it rejects all frequencies that are within the bandpass, and giving easy passage only to frequencies outside the bandpass.

Step 30 Move the cursor to the center of the curve at its lowest point. Record the center frequency (fO) on the curve plot.

fO= 1.003 kHz;

Step 31 Move the cursor to the highest point on the flat part of the curve and record the db gain on the curve plot.

AdB= – 0.066 dB

Step 32 Move the cursor as close as possible to a point on the left side of the curve that is 3 dB down from the highest dB gain on the flat part of the curve. Record the approximate frequency (low-cutoff frequency fC1) on the curve plot. Next, move the cursor as close as possible to a point on the right side of the curve that is 3 dB down from the highest dB gain on the flat part of the curve. Record the approximate frequency (high-cutoff frequency fC1) on the curve plot.

fC1= 918.093 Hz fC2= 1.103 kHz

Step 33 Based on the values of fC1 and fC2, determine the bandwidth (BW) of the series resonant band-stop (notch) filter.

BW= 184.907 Hz

Step 34 Based on the circuit component values in Fig 2-3, calculate the expected center frequency (fO) of the series resonant band-stop (notch) filter.


Question: How did the calculated value of the center frequency (fO) base on circuit component values compare with the measured value recorded on the curve plot?


Step 35 Based on the values of fC1 and fC2, calculate the center frequency (fO)

fO(fC1/fC2)= 1006.31 Hz




Step 36 Based on the circuit component values calculate the quality factor (Qs) of the series resonant band-stop (notch) filter.

QS= 5.27

Step 37 Based on the circuit quality factor (QS) and the center frequency (fO), calculate the expected bandwidth (BW) of the series resonant band-stop (notch) filter.

BW(EXPECTED)= 191.00 Hz

Question: How did the expected bandwidth calculated from the value of QS and the center frequency compare with the bandwidth measured on the curve plot?


Step 38 Open circuit file Fig 2-4. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–20dB), Horizontal (Log, F=2 kHz, I=500Hz)


Question: Is the frequency response curve that of a band-stop (notch) filters? Explain why.

Yes. The above curve plot is a band-stop filter because it rejects all frequencies that are within the bandpass, and giving easy passage only to frequencies outside the bandpass.

Step 42 Move the cursor to the center of the curve at its lowest point. Record the center frequency (fO) on the curve plot.

fO= 1.003 kHz;

Step 41 Move the cursor to the highest point on the flat part of the curve and record the dB gain on the curve plot.

AdB= – 0.055 dB

Step 42 Move the cursor as close as possible to a point on the left side of the curve that is 3 dB down from the highest dB gain on the flat part of the curve. Record the approximate frequency (low-cutoff frequency fC1) on the curve plot. Next, move the cursor as close as possible to a point on the right side of the curve that is 3 dB down from the highest dB gain on the flat part of the curve. Record the approximate frequency (high-cutoff frequency fC1) on the curve plot.

fC1= 914.969 Hz; fC2= 1.098 kHz

Step 43 Based on the values of fC1 and fC2, determine the bandwidth (BW) of the parallel resonant band-stop (notch) filter.

BW= 183.031 Hz

Step 44 Based on the circuit component values in Fig 2-4, calculate the expected center frequency (fO) of the parallel resonant band-stop (notch) filter.


Question: How did the calculated values of the center frequency (fO) base on the circuit component values compare with measured values recorded on the curve plot?



fO(fC1/fC2)= 1002.32 Hz




Step 46 Based on the value of L and RW, calculate the quality factor (QL) of the inductor.

QL=31.62

Step 47 Based on the quality factor (QL) of the inductor, calculate the equivalent parallel inductor resistance (REQ) across the tank circuit.

REQ= 20016.48Ω

Step 48 Based on the value of REQ and R, calculate the total parallel resistance (RP) across the tank circuit.

RP= 3333.79Ω

Step 49 Based on the value of RP, calculate the quality factor (QP) of the parallel resonant band-stop (notch) filter.

QP= 5.27

Step 50 Based on the filter quality (QP) and the center frequency (fO), calculate the expected bandwidth (BW) of the parallel resonant band-stop (notch) filter.

BW= 191 Hz

Question: How did the expected bandwidth calculated from the value of QP and the center frequency compare with the bandwidth measured on the curve plot?


Step 51 Change the resistance of R to 2kΩ. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) from the curve plot and record the values.

fO=1.003 kHz BW= 337.136 Hz

Questions: What effect did changing the resistance of R have on the center frequency of the parallel resonant band-stop (notch) filter? What effect did changing the resistance R have on the bandwidth of the parallel resonant band-stop (notch) filter? Explain.

The center frequency remain constant or did not change its value while the bandwidth changed, it increases as the resistor decreases. Therefore, the center frequency fO of a series resonant band-pass filter is constant with the change of the value of the resistance of resistor R while the bandwidth of the series resonant band-pass filter is inversely proportional to the value of the resistance of resistor R.

CONCLUSIONS

Based on the output signal in the bode plotter the band-pass filter only allows frequencies within the band and blocks all frequencies outside the band. Its counterpart is the band-stop (notch) filter which blocks all frequencies within the band and allows all other frequencies outside the band.

In addition to that, the voltage gain and the center frequency of the band-pass filter are at the peak point of the curve plot. On the other hand, the voltage gain and the center frequency of the band-stop filter are at the lowest point of the curve plot. Bandwidth can be determine from subtracting the frequencies 3 dB down and up from dB gain measured at center frequency fO.

Lastly, I concluded that the center frequency of the curve plot was not affected by the resistance and is inversely proportional to the capacitance and inductance. Bandwidth is not affected by capacitance, but inversely proportional to the inductance. The bandwidth in L-C series resonant band-pass/stop filter is directly proportional to the resistance, while bandwidth in L-C series resonant band-pass/stop filter is inversely proportional to the resistance.

exp2 passive band pass and band-stop filter

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