# The Instrumentation And Measurement Engineering Essay

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Published: *Mon, 5 Dec 2016*

A single strain gauge having a resistance of 500â„¦, gauge factor of 2 and a temperature coefficient of 1 Ã- 10-5 per °C at room temperature is mounted on the beam and connected in the arm AB of the bridge shown in figure Q4 for measuring a strain in cantilaver beam. The other three arms BC, CD and DA of the bridge have resistance of 100â„¦, 100â„¦ and 500 â„¦ respectively. The detector connected across A and C of the bridge has a resistance (Rg) of 100 â„¦ and sensitivity 5 mm per µA. The voltage supply to the bridge is 12 V.

Determine the detector deflection for a gauge strain of 0.002.

Given:-

R1 = 500 â„¦

R2 = 100 â„¦

R3 = 500 â„¦

R4 = 100 â„¦

Resistance across A and C of the bridge, Rg = 100 â„¦

Î± = temperature coefficient of 1 x 10-5 °C

Gauge strain, = 0.002

Voltage supply = 12 V

Sensitivity 5 mm / Î¼A

Gauge factor of 2

## Solution:-

When a strain is introduced, the strain sensitivity, which is also called the gage factor (GF), and also the strain is defined as the amount of deformation per unit length of an object when a load is applied. Strain is calculated by dividing the total deformation of the original length by the original length (L).

Substitute all the value that given and find out change in resistance,

After get the change in resistance, âˆ†R1. Total resistance measured is equal to R1 and R3 in parallel and R2 and R4 in parallel. If strain gauge is changes even a little bit in value can cause the bridge unbalanced and can define that R1 = âˆ†R1 + R1.

Wheatstone resistive bridge sensors can be analyzed using Thevenin’s Theorem, where the circuit is reduced to voltage sources with series resistance

Voltage across the bridge, VAC varies change as strain gauge, R1

So we can determine the voltage across the bridge, VAC terminals by applying the Ohm’s Law.

R is the resistance that across the A and C of the bridge

Lastly, determine the detector deflection for a gauge strain and the deflection are known as below,

Deflection = Sensitivity x Current, Ig flow at VAC

Deflection = (5 mm / Î¼A) x 29.71 Î¼A

Deflection = 148.55 mm

Determine the change in strain indicated for an increase of 20 °C in room temperature.

Substitute the value that to the equivalent change in strain

## QUESTION 2

A telephone line will be used to carry measurement data as a frequency-modulated signal from 5kHz to 6kHz. The line is shared with unwanted voice data below 500Hz, and switching noise occurs above 500kHz. Design a band-pass RC filter that reduces the unwanted voice by 80% and reduces the switching noise by 90%. Assume CH is 0.05µF, and use a resistance ratio r of 0.02. What is the Vout/Vin on the passband frequency of 5.5kHz?

Given:-

Frequency – modulated signal from 5 kHz to 6 kHz

Unwanted voice data below 500Hz

Switching noise above 500 kHz

Reduced unwanted voice 80%

Reduced switching noise 90%

CH = 0.05 Î¼F

Resistance ratio, r = 0.02

First need to find out the low pass filter

Passive RC Low Pass Filter

Also know that the capacitive reactance of a capacitor in an AC circuit is given as below

The High Pass Filter is the exact opposite to the low pass filter. This filter has no output voltage from DC (0Hz), up to a specified cut-off frequency (Æ’c) point. This lower cut-off frequency point is 70.7% or -3dB (dB = -20log Vout/Vin) of the voltage gain allowed to pass.

Passive RC High Pass Filter

Also know that the capacitive reactance of a capacitor in an AC circuit is given as below

After getting value of then substituting to the

Substitute the R1 value to the formula to get the R2 value as well

Using the FCL find out the C2

## Band Pass Filter Circuit

Passive RC Band Pass Filter

Band Pass Filters passes signals within a certain “band” or “spread” of frequencies without distorting the input signal or introducing extra noise. This band of frequencies can be any width and is known as the filters Bandwidth

Band Pass Filter Bode Plot

## QUESTION 3

Describe how the sensor control works in cycle with relay in filling and draining water from the tank. Find the value of amplifier gain, K, required to open the valve when the level reached 1.5 m.

## Description:-

Input flow Q1 and Q2 fill the tank without controlled. When the level of water in tank reaches the height h = 1.5m, the level sensor sends signal’s voltage, Vh to the amplifier to amplify the voltage to relay’s voltage Vr = KVh with a gain of K, which the voltage of relay will be large enough to drive the relay closes.

As the relay’s voltage reaches Vr = 6V, the relay is closed and activates the valve to open and water in the tank is started to drain out. After some time, the water level drops to 1.1m and the level sensor will read the signal. Again, the voltage Vh is amplified to Vr = 4.8V and latch the relay to open. The open relay is then instructed the valve to close.

Even though the water stops draining out, the water tank is still filling with water. The water level will increase to h = 1.5m again. The same cycle is expected to occur continuously.

## Given:-

Level sensor’s voltage, Vh = 0.8h + 0.4V

Relay’s voltage, Vr = KVh

Relay’s voltage closes, Vr = 6V

h = 1.5m

Find the amplifier gain, K

## Solution:-

Level sensor linear static operating characteristics which is given, Vh = 0.8h + 0.4V. Apply this formula to the voltage relay to get the value of amplifier gain, K. Substitute all the info that given to the relay’s voltage, Vr = KVh

At what level does the valve close?

## Given:-

Relay’s voltage closes, Vr = 4.8V

Level sensor’s voltage, Vh = 0.8h + 0.4V

Amplifier gain, K = 3.75

## Solution:-

When the valve close, the voltage of relay, Vr = 4.8V and given that K = 3.75 and substitute to the formula relay’s voltage to get the level, h of valve close.

1.28 = 0.8h + 0.4

0.8h = 0.88

h = 1.1 m

Suppose Q1 = 5 m3/min, Q2 = 2 m3/min, and Qout = 9 m3/min (when open). Determine the time for water level to rise from 1.1 to 1.5 meters and the time to drain out. Find the total time of cycle.

## Given:-

Input flow rate (velocity), Q1 = 5m3/min and Q2 = 2m3/min

Output flow rate (velocity), Qout = 9m3/min (when valve open)

## Solution:-

Time for water level to rise from 1.1m to 1.5m

Velocity shows how fast an object is moving to which direction. Average velocity can be calculated by dividing displacement over time.

Where the t1 is the time when water start rise at height 1.1m. Assume the t1 = 0 (Initial time)

Time for water level to drain out from 1.5m to 1.1m

Output flow rate (velocity), Qout = 9m3/min (when valve open)

t2 is the time when water reaches 1.1m, water drains out is stopped. Assume t2 = 0

## QUESTION 4

A measurement of temperature using a sensor that outputs 6.5 mV/ËšC must measure to 100ËšC. A 6-bit ADC with a 10V reference is used. Develop a circuit to interface the sensor and the ADC. Find the temperature resolution.

## Given:-

Output of the sensor = 6.5 mV/ ËšC measure to 100ËšC

6-bit ADC = 10Vref

## Solution:-

Find the output sensor during 100ËšC where the output sensor 6.5 mV/ ËšC measure 1ËšC is given.

Resolution can define electrically, and expressed in volts. The minimum change in voltage required to guarantee a change in the output code level is called the LSB (least significant bit, since this is the voltage represented by a change in the LSB).

The resolution Q of the ADC is equal to the LSB voltage. The voltage resolution of an ADC is equal to its overall voltage measurement range divided by the number of discrete voltage intervals:

N is the number of voltage intervals,

EFSR is the full scale voltage range, 10 V

Normally, the number of voltage intervals is given by,

Where the M is the ADC’s resolution in bits.

## Solution:-

Develop a circuit to interface the sensor

Block Diagram

Figure 4.1 Interfacing an Analog Output Temperature Sensor to an ADC

At first sensor consists of a band gap reference circuit that produces a voltage.

A switched capacitor op amp amplifier is used to amplify the temperature coefficient to a voltage mV/°C because of the ease of building capacitors that are a ratio of each other.

Lowpass filter is used to remove the switching noise of the amplified signal. The output signal is then driven by a buffer amplifier.

The temperature sensor’s output pin is driven by an op amp that has output impedance (ROUT). The input of the ADC consists of a simple sample and hold circuit.

A switch is used to connect the signal source with a sampling capacitor, while the ADC measures the CSAMPLE capacitor’s voltage in order to determine the temperature. The ROUT and RSWITCH resistances and the CSAMPLE capacitor form a time constant that must be less than the sampling rate (TSAMPLE) of the ADC as shown.

An external capacitor can be added to the output pin to provide additional filtering and to form an anti-aliasing filter for the ADC. This capacitor may impact the time response of the sensor and the designer must allow time for the capacitor to charge sufficiently between ADC conversions.

Also, the sensor amplifier may oscillate if the filter capacitor is too large. A small resistor of approximately 10 to 100Î© can be added between the output pin of the sensor and CFILTER to isolate the sensor’s amplifier from the capacitive load.

The output impedance of the sensor (ROUT) varies as a function of frequency. Thus, a series resistor should be added to the effective ROUT resistance if CFILTER is intended to serve as the ADC’s anti-aliasing filter. The output impedance of the TC1047A is less than 1Î© because operational amplifier A2 functions as a voltage buffer.

The output impedance of the sensor is low due to the negative feedback of the buffer circuit topology. The negative feedback results in an output impedance that is equal to the impedance of the amplifier divided by the open-loop gain of the amplifier. The open-loop gain of the op amp is relatively large which, in turn, forces the output impedance to be small.

## QUESTION 5

A pressure sensor has a resistance that changes with pressure according to R = (0.15 kâ„¦/psi)p + 2.5 kâ„¦. This resistance is then converted to a voltage with the transfer function,

The sensor time constant is 350 ms. At t = 0, the pressure changes suddenly from 40 psi to 150 psi.

What is the voltage output at 0.5 s? What is the indicated pressure at this time?

## Given:-

Pressure sensor has resistance changes with pressure, R = (0.15 kâ„¦/psi)P + 2.5 kâ„¦

Transfer function of voltage,

Sensor time constant, Ï„ = 350 ms

At t = 0, Pressure, P change suddenly from 40 psi â†’ 150 psi

## Solution:-

Pressure changes suddenly from 40 psi to 150 psi and we can assume that the initial pressure, Po = 150 psi

Voltage output after t = 0.5 sec, find pressure after 0.5 sec first;

Basic formula a quantity pressure depends exponentially on time t if.

During the resistance 0.5 sec, P = 35.95 psi. Substitute the indicated pressure at 0.5s, P in resistance changes with pressure according to to find out the resistance during 0.5 sec

Voltage output at 0.5 s,

2. At what time does the output reach 5.0 V?

Find out the resistance by substitute the output voltage at the formula that giving

Indicated pressure at output 5.0 V

50 psi

A quantity pressure depends exponentially on time t if, substitute the value of initial pressure, indicated pressure output at 5.0 V and time constant is giving 350 ms. Finally we can get the time does the output reach 5.0 V.

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