�1

Biosensors and Instrumentation: Tutorial 1 Solutions

1. Two of the most important characteristics of a sensor system are accuracy and precision.1.1. Explain these terms with the help of a diagram.

The accuracy of a measurement defines how close the measured value is to the true value, while the precision describes how the measured value changes with repeated measurements. You could just use a graph like the one in the lecture to show this but a nice analogy involves a comparison with hits on a target in archery:

�In the figure on the left the hits are accurate but imprecise, i.e. they are centred on the bullseye but are well spread out. In the figure on the right the hits are more precise but less accurate, i.e. they are grouped closely together but are not well aimed at the centre of the target.

1.2. Think about how you might attempt to determine them for a particular sensor. What methods would you use?

Determining the accuracy and precision of a sensor will both require multiple measurements, typically made on a known “standard” which will obviously change depending on what the sensor is measuring. For example with a pH sensor you would need one or more solutions of known pH. Measurements need to be made with the same conditions and you need to be reasonably sure that the parameter being measured is not changing with time. Ideally the measurements would be made quickly to avoid any problems with drift in the sensor. The actual value of the standard isn’t really important for measurement of precision, you just need to make enough measurements of an otherwise unchanging variable to determine the variance or standard deviation of the results. Determination of the accuracy will also require multiple measurements, unless of course you know for sure that the measurement is extremely precise, but the important figure here is the mean value of the parameter. The accuracy will be defined by the offset between the known value of the standard and the mean of the set of measurements. Of course these measurements will often only give you an idea of the accuracy and precision at one measurement point and it would be a good idea to have two or more standards to measure which are are spread out along the range of values you expect to be using the sensor to measure.

1.3. How could you deal with the effects of poor accuracy and precision in a real sensor?

Basically if the precision is poor then you need to make multiple measurements and take the average of these as your answer. The way to deal with an offset or inaccuracy is to measure this offset against a standard (or multiple standards) and calibrate the sensor

X

XX

X

X

X

X XX XX

Biosensors and Instrumentation

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instrumentation to remove this. If the offset is not a constant this might need to be done at several points. Systematic offsets may be caused by other factors, related to the selectivity of the sensor as in the next question.

2. Selectivity is another important characteristic of a sensor system2.1. Explain the meaning and importance of sensor selectivity.

Ideally a sensor will only detect the parameter it is designed to measure and will be insensitive to any interfering factors. However in the real world there are any number of environmental parameters (temperature, pH, other chemicals) that can affect a biosensor. The selectivity can be defined as the sensitivity of a sensor to the targeted parameter divided by the sensitivity of the confounding factor.

2.2. Why might a biosensor have a poor selectivity and what methods might there be to improve this?

Biosensors typically detect molecular binding events. These can be quite specific by there is always the possibility of non-specific binding of other molecules. There can also be effects due to the local pH and temperature of the environment which is harder to control. The sensors could be covered with a membrane which tends to only let through the chemical we’re trying to measure or we could have a “control” sensor which will measure the level of non-specific binding which can then be subtracted from the result from the real sensor.

2.3. Based on the graph, figure 1, would you expect temperature to have an adverse effect on an implanted glucose sensor with this characteristic? Give reasons for your answer.

Temperature is unlikely to have a huge effect on the sensor if it’s being implanted as the operating temperature is likely to be fairly constant at around human body temperature of 37℃. The graph suggests that the temperature sensitivity is quite low at this point which makes sense as the GOx enzyme presumably evolved to operate around this temperature. Small variations in temperature will therefore not affect the sensor.

�Figure 1. Glucose sensor response vs. temperature.

Sensors 2002, 2 132

Selection of pH value

Actually, the effect of pH on the sensor involves two factors: (1) the effect of pH on the electrocatalystic property of the PB layer (Fig.4 (a)); (2) the effect of pH on the activity of GOD (Fig. 4(b)). Fig. 4(c) shows the combined relation between the response current of the sensor and the pH values of the glucose solution. The sensor exhibits that larger response current can be obtained within the range of pH 6.0 to 7.0. However, the higher value of pH in the glucose solution, the lower stability of the PB. Thus, pH6.8 phosphate buffer solution was chosen as bulk solution.

5.0 5.5 6.0 6.5 7.0 7.5 8.0

0.7

0.8

0.9

1.0

Rel

ativ

e re

spon

se

pH

5.0 5.5 6.0 6.5 7.0 7.5 8.0

0.6

0.7

0.8

0.9

1.0

Rel

ativ

e re

spon

se

pH

5.0 5.5 6.0 6.5 7.0 7.5 8.00.4

0.5

0.6

0.7

0.8

0.9

1.0

Rel

ativ

e re

spon

se

pH

(a) (b) (c)

Figure 4. Effect of pH on the responses for (a) PB/Pt electrode, (b) GOD/Pt electrode and (c) GOD/PB/Pt electrode.

Effect of temperature on response

Fig. 5 shows the effect of temperature on the response current of the sensor (concentration of glucose, 3.0mM). The current at certain concentration increased with temperature until 40oC and then decreased. The lower temperature, the lower activity of GOD is. When the temperature is higher than 40oC, the GOD on the WE will partially lost its activity. Hence, it is necessary to control seriously environmental temperature during measurement or to make compensation for temperature.

10 20 30 40 50

160

200

240

280

320

Res

pons

e/nA

Temperature/ oC

Figure 5. Effect of temperature on response current of the sensor. Biosensors and Instrumentation

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3. An amperometric sensor for glucose has a characteristic response defined by the calibration curve shown in figure 2.

�Figure 2. Amperometric glucose sensor output characteristic.

3.1. Estimate the linear range of this sensor and the sensitivity in this region.The linear region appears to run from about 0mM of glucose up to about 7mM. In the same range the output current goes from 0 to about 700nA. Therefore we can estimate the sensitivity of this glucose sensor to be about 100 nA mM-1 or 0.1 μA M-1.

3.2. The sensor will be connected to a digital readout which is basically a voltmeter with a resolution of 0.1V. Assuming that a glucose measurement resolution of 0.1mM is required design a circuit using an operational amplifier to send a suitable signal to the digital readout system.

As stated in the answer to the previous question the sensor sensitivity is 100 nA mM -1, so in the linear range a change in glucose concentration of 0.1 mM means a change in output current of 10 nA. The amplifier used for this sensor should be a current follower, with the circuit shown below in figure T1:

�Figure T1. Current Follower

The characteristic equation of a current amplifier is: Vout = -iR and we require ∂Vout for a change in current (∂i) of 10 nA to be equal to 0.1 V. Therefore R = 10 MΩ! This may be too high in some situations. There is also the problem of the input bias current of the operational amplifier. This will need to be very low, requiring an expensive, high quality component to avoid losing some of the current signal. How could we improve on this circuit to allow resistors which have a more sensible range to be used?

�

+outV

R

I

Biosensors and Instrumentation

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4. The relationship between the concentration (activity) of the oxidised and reduced components of a redox couple and the potential E at the working (sensing) electrode is given by the Nernst Equation:

� (at 25 ℃)

Identify which of the following equations correctly interprets the Nernst Equation for the reaction:

!

(a) ! (b) !

(c) ! (d) !

The Nernst Equation applies to the reaction: Ox + ne- ↔ R

in which Ox is the oxidised species and R the reduced species of the redox couple. Thus, for the reaction:

!

equations (c) and (d) are correct.

5. A glass electrode based potentiometric pH sensor operating in its linear range has a response to pH defined by the following equation:

�5.1. Explain the terms of this equation.

E0 - Characteristic reduction potential for the particular combination of glass electrode and reference electrode usedR - Universal gas constant = 8.314472(15) J K−1 mol−1

T - Absolute temperature (K)F - Faraday constant = 9.64853399(24)×104 C mol−1

Supplementary questions: Where does the figure “2.303” come from and can you simplify the equation for T=298K (25℃)?

5.2. Draw the equivalent circuit of a glass electrode with resistance RSee figure T2.

E = Eo +RT

nFln

✓[Ox]

[R]

◆= Eo +

0.059

nlog10

✓[Ox]

[R]

◆

Fe3+ + e� ! Fe2+

E = Eo � 0.059

nlog10

✓[Fe

3+]

[Fe

2+]

◆E = Eo

+

0.059

nlog10

✓[Fe

2+]

[Fe

3+]

◆

E = Eo

+

0.059

nlog10

✓[Fe

3+]

[Fe

2+]

◆E = Eo � 0.059

nlog10

✓[Fe

2+]

[Fe

3+]

◆

Fe3+ + e� ! Fe2+

E = E0 � 2.303RT

FpH (V)

Biosensors and Instrumentation

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�T2. Equivalent circuit of a potentiometric pH sensing electrode

5.3. With a typical reference electrode the characteristic response of a glass electrode pH meter looks something the graph in figure 3. The linear region helpfully includes the most useful range of pH values. Assuming that we will connect the pH sensor to a voltmeter and would like the readout to be equal to the actual pH, design an amplifier for this electrode using an operational amplifier. The output resistance of a typical pH sensor is 500 MΩ. What is the output voltage of your amplifier for pH=7?

�Figure 3. Potentiometric pH sensor output characteristic.

In the linear region of this graph the output voltage varies by about 0.75 V over a pH range of 13, with a negative slope. Therefore the sensitivity is around -0.06 V pH-1

(Nernst type response). In order to get an output resolution of 1 V pH-1 we need an amplification factor G of -16.67. Because the output resistance of the pH sensor is very high there could be a problem with using an inverting amplifier which has a finite input impedance. One solution is to use a voltage follower or buffer between the sensor and the inverting amplifier as shown in figure T3. The gain of this circuit is given by the following equation:

�

R

E

G =V

out

Vin

=�R

f

Rin

Biosensors and Instrumentation

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�Figure T3. pH amplifier circuit.

For a gain of -16.67 using standard resistor values you could use:Rf = 20 kΩ and Rin = 1.2 kΩ

but there are obviously many other possible solutions. For a pH of 7 we can estimate the output voltage of the sensor to be around -0.42V. The output of the amplifier will therefore be ~7V which is a good confirmation that it’s operating correctly.As a supplementary problem, imagine that the op-amp in the buffer has an input bias current of 1pA. What is the voltage error introduced by this at the amplifier input and what error in the measured pH does this lead to? What would it be if the input current was 1nA?

6. A fluorine ion selective probe gives an output reading of 0.112 V when calibrated against a solution containing 0.01 M fluorine ions at room temperature (25 ℃). The probe is then inserted into a test solution (at 25 ℃) and the voltage reading changes to 0.203 V. It is assumed that there is a linear relationship between the probe voltage reading and the fluorine ion concentration [F–] of the form:

Eprobe = K + 0.059 log[F–]6.1. Based on the voltage readings obtained, calculate the fluorine ion concentration of the

test solution.Based on the linear relationship:

Eprobe = K + 0.059 log[F–]

If Eprobe = 0.112 V at [F–] = 0.01 M, then K = 0.112 – 0.059 log(0.01) = 0.230 VThen for Eprobe = 0.203 V:

log[F–] = (0.203 - 0.230)/0.059 = –0.458, to give [F–] = 0.349 M

6.2. Would it be advisable to calibrate the probe using more than one calibration solution? Justify your answer.

It would be advisable to calibrate the probe using two more fluorine ion solutions (e.g., 250 mM and 500 mM) to check the validity of the assumed linear relationship in the [F–] region that includes that (~350 mM) of the test sample.

7. An amperometric sensor for the measurement of partial pressure of dissolved oxygen in a solution has an output response as shown in figure 4.

R

V

Sensor

−

+

−

+outV

R

Rin

f

Biosensors and Instrumentation

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�Figure 4. Output characteristic of an amperometric oxygen sensor.

7.1. Estimate the sensitivity of this sensorThe linear fit to the sensor output shows a change of 10 nA for a change in oxygen partial pressure of 50 mmHg. The sensitivity is therefore 0.2 nA/mmHg.

7.2.Assuming that the sensor will be connected to an analogue to digital convertor with an input range of 0 to 5 V, design a circuit to convert and amplify the sensor current output to get the best voltage sensitivity.

Making an assumption that the useful pO2 range for the sensor is 0-200 mmHg and that the sensor output is linear then the current output range will be approximately 35-85nA. We want a voltage range of 5V from this current variation of 50nA. A current follower (figure T1) can convert the current into voltage with Vout=-IR. If we want 5V from 50nA then R will be 100MΩ. If this is an issue we could instead use a 100kΩ resistor and the output voltage range will be -5mV. Then use an inverting amplifier with a gain of -1000, that would be Rin =100Ω and Rf=100kΩ. This will have a low input impedance so use a voltage follower to buffer the current follower output. The output of this amplifier would vary from 3.5V to 8.5V so we also need to shift the input level. This can be done with a summing inverting amplifier with one input, which has a gain of -1, set to a DC voltage of 3.5V. This will lower the output to the range 0-5V as shown in the circuit diagram in figure T4

!3

5. An amperometric sensor for the measurement of partial pressure of dissolved oxygen in a solution has an output response as shown in figure 4.

5.1. Estimate the sensitivity of this sensor"5.2.Assuming that the sensor will be connected to an analogue to digital convertor with an

input range of 0 to 5 V, design a circuit to convert and amplify the sensor current output to get the best voltage sensitivity."

!

! "Figure 4. Output characteristic of an amperometric oxygen sensor."!!

6. The potentiostat is a basic instrument for electrochemistry experiments or the measurement of electrochemical sensors. Explain the purpose of a potentiostat with reference to the experimental setup of a three-terminal electrochemical cell."!7. Figure 5 shows a potentiostat circuit; refer to this when answering the following questions:"

7.1. What is Vout in terms of Vin, and why?"7.2. What is the potential of the working electrode (WE) vs. the reference electrode (RE)?"7.3. How is the working electrode current measured?"7.4. Assuming the current through the cell is 100 pA, design a circuit to measure this and

produce a voltage of 1 V for this current level. Assume that resistors in the range of 100 Ω to 220 kΩ are acceptable."

Biosensors and Instrumentation

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" Figure T4: buffered amplifier with level shifting.

8. The potentiostat is a basic instrument for electrochemistry experiments or the measurement of electrochemical sensors. Explain the purpose of a potentiostat with reference to the experimental setup of a three-terminal electrochemical cell.

Refer to the notes for a full explanation but the basic idea is to control the potential difference between the reference and working electrodes of the electrochemical cell, while measuring the resulting current flow. Electronic feedback is used to supply current through a counter/auxilliary electrode. The working electrode is typically connected to an external point at ground (either virtual or direct) and the reference electrode potential vs. ground is controlled. Depending on the potential on the working electrode (vs. RE) an electrochemical reaction (reduction/oxidation) may occur leading to a current flow into or out of the working electrode.

9. Figure 5 shows a potentiostat circuit; refer to this when answering the following questions:

9.1. What is Vout in terms of Vin, and why?Vout = –Vin. Assuming there is no current flow into the op-amp inputs and the inverting input of OA-1 is at a virtual earth then a current I=Vin/R will flow through both the resistors R, and the Vout terminal will be at a potential Vin lower than ground.

9.2. What is the potential of the working electrode (WE) vs. the reference electrode (RE)?VWE (vs. RE) = Vin

9.3. How is the working electrode current measured?The WE current is supplied by the counter electrode from the control amplifier OA-1, this flows through Rf and can be measured as VRf = IWE Rf.

9.4. Assuming the current through the cell is 100 pA, design a circuit to measure this and produce a voltage of 1 V for this current level. Assume that resistors in the range of 100 Ω to 220 kΩ are acceptable.

Obviously there are many answers to this. If Rf is 100kΩ then the voltage VRf = 10µV. Then we need further amplification of 105. The first step needs to be a buffered

Biosensors and Instrumentation

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differential amplifier or possibly an instrumentation amp. The output voltage of this will be:

If R2 = 1kΩ, R3 = 100kΩ, R1=100kΩ, then setting Rgain=200Ω will give roughly the correct gain.

�Figure 5. Potentiostat Circuit

10. Amino acids in aqueous solution are oxidised by the enzyme amino acid oxidase (AAO) according to the reaction:

�

Identify three different designs (in terms of their mode of detection) of an amino acid sensor.

Based on the enzyme reaction described, the following amino acid sensor designs could be applied:(i) Amperometric: Use a Clark oxygen electrode to monitor depletion of oxygen in

the reaction - or to detect generation of hydrogen peroxide by electrochemical oxidation of the peroxide.

(ii) Potentiometric: Use an ammonium ion selective electrode to determine generation of ammonium ions (NH4+).

(iii) Conductometric: Detect the increase of conductivity of an immobilised enzyme arising from generation of the ammonium ions (NH4+). The enzyme can be immobilised in a paste or gel formed onto an interdigitated electrode array.

A full answer would also summarise the relative advantages and disadvantages of these three sensor types.

11.You have been given a conductometric biosensor consisting of interdigitated microelectrodes coated with a enzyme that reduces in resistivity when exposed to a specific analyte.

11.1. Suggest a suitable circuit to convert the resistance to a voltage. Explain your choice with reference to the characteristics of the sensor and the resistance to voltage converter.

You could use an inverting op-amp circuit with the sensor replacing the input resistor with it driven by a voltage source but there are advantages to using a Wheatstone

Vout

= VRf

✓1 +

2R1

Rgain

◆R

3

R2

WE

RE

CE

−

+

−

+OA-1

OA-2

R

R

Vin

Rf

Vout

VRf

Biosensors and Instrumentation

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bridge with the sensor as one of the resistors, particularly if the resistance change is small, there is also the possibility of using the bridge to compensate for cross correlation effects like temperature. A typical setup is shown in figure T5.

� Figure T5. Wheatstone bridge circuit for resistance measurement

11.2. Assuming the sensor has an initial resistance R0 of 1 kΩ that varies with ∂R of up to 10 Ω in the presence of the analyte, design a circuit to produce a voltage signal with a maximum output that does not go higher than 5V when the sensor is detecting the maximum level. If you also assume the resistance can increase with the same maximum change make sure that the output voltage does not go negative.

Using a Wheatstone bridge with the sensor taking the place of R1, balance the bridge by making all the other resistors 1 kΩ. If we make Vs = 10V and ∂R1 is ±10 Ω then we can calculate the range of values for the output Vo:

�

�We can measure Vout with a buffered differential amplifier with some little gain, lets say G = 10. The circuit is shown connected to our Wheatstone bridge in figure T6, remember that the equation for the output of the differential amplifier is:

�If Rx = 1 kΩ and Ry = 10 kΩ we will get the output voltage Vamp ranging from ~0.251 V when the analyte is at the highest level (R1 = 990Ω) to –0.249 V when at opposite end of the scale, (R1 = 1010Ω).

A

B

C

DVV

RR

R R

1

os

24

3

I1I2

Vout

= Vin

✓R2

R1 + �R1 +R1� R4

R3 +R4

◆= 10

✓1000

2010� 1

2

◆= �24.9mV

Vout

= Vin

✓R2

R1 + �R1 +R1� R4

R3 +R4

◆= 10

✓1000

1990� 1

2

◆= 25.1mV

Vamp

=R

y

Rx

(V2 � V1) =R

y

Rx

Vout

Biosensors and Instrumentation

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� Figure T6. Wheatstone bridge with differential amplifier output.

The final stage would be to amplify by ~10 and level shift the signal, the circuit I’ve shown for doing that before is an inverting amplifier so you would need two stages, either with gains of –10 and –1 or maybe –2 and –5? One amplifier also adds in a voltage which could be tuned to give the correct scale of 0-5V. However, there is a better way to do this in one stage with the operational amplifier circuit as shown in figure T4. If we assume that R1=R3 and R2=R4, then the output of this non-inverting level shifter circuit is:

�Therefore, setting R1=R3=10 kΩ, R2=R4=100 kΩ, and Vref =2.5 V will give us roughly the output that we’re looking for. If you want to assure yourself of this, you can perform the nodal analysis for the op-amp input terminals.

�

Figure T6. Non-inverting level shifter circuit.

12. A system consisting of resistance R and capacitance C will exhibit an impedance of the form:

!

Vin

R1

−

+

−

+

−

+Vamp

+

−

R1

R2

R3

R4

Rx

Rx

Ry

Ry

Vout

Vout

= Vin

R2

R1+ V

ref

−

+

Vout

R3

Vin

R4

R1

R2

Vref

Biosensors and Instrumentation

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where Ro and R∞ are the limiting low- and high-frequency values of the equivalent series resistance, and τ is a characteristic time constant (τ = RC). Representing the complex impedance in the form Z = R – jX, the equivalent series components are:

! , !

By eliminating (ωτ)2 from both of these equations, derive the following relationship:

!

to indicate that the locus of all values of the effective resistance R and reactance X, as we proceed from low to high frequencies, will take the form of a semicircle of radius (Ro – R∞)/2 with its centre at (Ro + R∞)/2.

Full Solution will be added at the end of this document. There’s too much mathematics so it’s easier to do it in LaTeX!

Biosensors and Instrumentation

Bisoensors and Intrumentation

Tutorial 1

1 Tutorial 1 - Question 12 Solution

Z(!) = R1 +

R0 �R1

1 + j!⌧

(1)

Can be written in the form: u+ v = R0 �R1

Where u = Z(!)�R1 and v = j!⌧(Z(!)�R1).

In the complex plane, the vectors u and v are oriented at 90

�to each other, with a vector

sum equal to the constant and real quantity (R0 � R1). The right angle included by u

and v is thus inscribed in a semicircle of radius (R0 � R1)/2, which represents the locus

of all values of the e↵ective resistance R and reactance X as ! varies from a very low to a

high frequency. The locus is a semicircle, and not a circle, because the capacitive reactance

is only a negative quantity. A semicircle of this form, centred at x1, is described by the

equation:

(x� x1)2+ y

2=

✓R0 �R1

2

◆2

(2)

The first step is to eliminate (!⌧)

2from the following two equations:

R = R1 +

R0 �R1

1 + !

2⌧

2(3)

X =

(R0 �R1)!⌧

1 + !

2⌧

2(4)

Which happens as follows, from equation (3):

1 + !

2⌧

2=

R0 �R1

R�R1(5)

) !

2⌧

2=

R0 �R1

R�R1� 1 (6)

1

Tutorial 1 S. Smith

) !

2⌧

2=

R0 �R1 � (R�R1)

R�R1(7)

) !

2⌧

2=

R0 �R

R�R1(8)

Substitute equation (5) into the bottom line of equation (4):

X =

(R0 �R1)!⌧

R0�R1R�R1

(9)

and cancel to leave:

X = (R�R1)!⌧ (10)

Then square both sides and substitute with equation (8):

X

2=

(R�R1)

2(R0 �R)

R�R1(11)

) X

2= (R�R1)(R0 �R) (12)

Which gives us the following relationship:

X

2+ (R�R0)(R�R1) = 0 (13)

The next stage is to add

�R0�R1

2

�2to both sides of (13) and aim to achieve an equation

with the same form for a semicircle as equation (2). Firstly:

(R�R0)(R�R1) +

✓R0 �R1

2

◆2

+X

2=

✓R0 �R1

2

◆2

(14)

then in the first bracket, (R�R0), we’re going to add (R1 �R1) and then rearrange:

[R�R0 + (R1 �R1)](R�R1) +

✓R0 �R1

2

◆2

+X

2=

✓R0 �R1

2

◆2

(15)

) [(R�R1)� (R0 �R1)](R�R1) +

✓R0 �R1

2

◆2

+X

2=

✓R0 �R1

2

◆2

(16)

)"(R�R1)

2 � (R0 �R1)(R�R1) +

✓R0 �R1

2

◆2#+X

2=

✓R0 �R1

2

◆2

(17)

Finally, it should be reasonably obvious that the section enclosed in the square brackets in

equation (17) can be factored or rewritten to give this final equation:

(R�R1)� R0 �R1

2

�2+X

2=

✓R0 �R1

2

◆2

(18)

This describes a semicircle with radius r =

(R0�R1)2 centred at x =

(R0+R1)2

2