#Prac10_Fundamentals

HW7_Fundamentals
#prac9
HW5 principles: 10.1-4; 9.11-17
Problem 10.1
An analogue-to-digital converter has an input range of 0 to 5 V and incorporates a 12-bit encoder.
(a) Assuming a binary encoder, find:
(i) the maximum percentage quantisation error;
(ii) the digital output signals corresponding to input voltages of 0.55 V and 2.63 V.
(b) Convert the binary codes of (a) (ii) into hexadecimal form.
(c) Repeat (a) assuming a three-decade, 8:4:2:1 binary-coded decimal (b.c.d.) encoder.

Solution:
𝑉𝑖𝑛 = [0,5] 𝑉
𝑛 = 12
𝑉𝑖𝑛1 = 0.55𝑉
𝑉𝑖𝑛2 = 2.63𝑉
𝑎)
(𝑖)
100% 100%
𝑒=± =± = ± 0.0122%
2(2𝑛 −1) 2(212 −1)
(𝑖𝑖)
𝑉𝑖𝑛
𝑁= 2𝑛
𝑉𝑖𝑚
0.55 12
𝑁1 = 2 = 450
5
(450)2 = 000111000010

2.63 12
𝑁2 = 2 = 2154
5
(2154)2 = 100001101010
b)
(450)2 = (1𝐶2)16
(2154)2 = (86𝐴)16
c)
100%
𝑒= ± = ± 0.05005%
2(103 − 1)
0.55
𝑁1 = 999 = 109
5
(109)𝐵𝐶𝐷 = 000100010000

2.63
𝑁2 = 999 = 525
5
(525)𝐵𝐶𝐷 = 010100100101

2
Problem 10.2
The digital-to-analogue converter of Figure 10.6(a) is required to give an output voltage in the range 0 to 2.55
V, corresponding to an 8-bit digital input signal 00000000 to 11111111.
(a) Assuming VREF = −15 V, R = 1 kΩ, calculate the value of RF required.
(b) Find the output voltage corresponding to an input signal of 11000101.

Solution:
(a) 𝑉𝑅𝐸𝐹 = −15𝑉, 𝑅 = 1𝑘Ω
𝑅𝑟𝑒𝑓 1 𝑉𝑜𝑢𝑡 1
𝑉𝑜𝑢𝑡 = − 𝑉𝑅𝐸𝐹 ∑ 𝑖 => 𝑅𝑟𝑒𝑓 = − 𝑅 = 85.3 Ω
𝑅 2 𝑉𝑟𝑒𝑓 ∑ 1
2𝑖
(b) (𝑁)2 = 11000101
𝑅𝑟𝑒𝑓 1 85.3 1 1 1 1
𝑉𝑜𝑢𝑡 = − 𝑉𝑅𝐸𝐹 ∑ 𝑖 = 15 ( 0 + 1 + 5 + 7 ) = 1.97 𝑉
𝑅 2 1000 2 2 2 2

Problem 10.3

Solution:
1
𝑎) 𝐺(𝑠) =
1+τ𝑠
1 1
|𝐺(𝑗ω)| = | |=
1 + 𝑗τω √1 + (τω)2
1 1 1
= ⇒ 1 + (τω)2 = 2 ⇒ (τω)2 = 1 ⇒ ω =
√1 + (τω)2 √2 τ
1
𝜔 ∈ [0, 𝜏 ]

3
 2 1−𝑧 −1
b) 𝑠 = ⋅
Δ𝑇 1+𝑧 −1
1 1
𝐺(𝑠) = ⇒ 𝐺(𝑧) =
1 + τ𝑠 2 1 − 𝑧 −1
1 + τ ⋅ Δ𝑇 ⋅
1 + 𝑧 −1

1 + 𝑧 −1
𝐺(𝑧) =
2τ
(1 + 𝑧 −1 ) + (1 −1 )
Δ𝑇 − 𝑧

1 + 𝑧 −1
𝐺(𝑧) =
2τ 2τ
(1 + Δ𝑇) + (1 − Δ𝑇) 𝑧 −1

2τ
1 1−
𝑎0 = 𝑎1 = , 𝑏1 = Δ𝑇
2τ 2τ
1 + Δ𝑇 1 + Δ𝑇
𝑎0 +𝑎1 𝑧 −1
c) 𝐺(𝑧) =
1−𝑏1 𝑧 −1
𝑌(𝑧) 𝑎0 + 𝑎1 𝑧 −1
𝐺(𝑧) = =
𝑋(𝑧) 1 − 𝑏1 𝑧 −1

𝑌(𝑧)(1 − 𝑏1 𝑧 −1 ) = 𝑋(𝑧)(𝑎0 + 𝑎1 𝑧 −1 )

𝑦(𝑖) − 𝑏1 𝑦(𝑖 − 1) = 𝑎0 𝑥(𝑖) + 𝑎1 𝑥(𝑖 − 1)

𝑦(𝑖) = 𝑎0 𝑥(𝑖) + 𝑎1 𝑥(𝑖 − 1) + 𝑏1 𝑦(𝑖 − 1)

d)
2τ
= 1 ⇒ Δ𝑇 = 2
Δ𝑇
1
𝐺(𝑧) = (1 + 𝑧 −1 )
2
1 ωΔ𝑇
𝑧 = 𝑒 𝑗ωΔ𝑇 ⇒ 𝐺(𝑒 𝑗ωΔ𝑇 ) = (1 + 𝑒 −𝑗ωΔ𝑇 ) = 𝑒 −𝑗ωΔ𝑇/2 cos ( )
2 2

ωΔ𝑇 ωΔ𝑇
|𝐺(𝑗ω)| = cos ( ), arg 𝐺 (𝑗ω) = −
2 2

ωΔ𝑇 1 ωΔ𝑇 π
cos ( )= ⇒ =
2 √2 2 4
π
⇒ω=
2Δ𝑇
π
Δ𝑇 = 2τ ⇒ ω =
4τ
π
ω ∈ [0, ]
4τ
Problem 10.4
The output signal of a vortex flowmeter has a frequency range between 5 and 50 Hz. The mean period of the
signal is to be measured using a signal counter and a clock counter. The signal is input to the signal counter,
which increments until Ns cycles have been counted. The clock counter is 16-bit binary and has a clock input of
frequency 10 kHz. The clock counter increments until a count of Ns is registered on the signal counter. What is
the maximum possible value of Ns?
Solution:
𝑓𝑠 = [5, 50] 𝐻𝑧
𝑛 = 16
𝑓𝑐 = 10 𝑘𝐻𝑧

𝑁𝑠 𝑁𝑐
𝑇= =
𝑓𝑠 𝑓𝑐
𝑛
𝑁𝑐 = 2
𝑁𝑐 × 𝑓𝑠 65536 × 5
𝑁𝑠 = = = 32
𝑓𝑐 10
Problem 9.11:
Design an a.c. amplifier system which incorporates an ideal operational amplifier, to meet the following
specifications:
Given:
𝑅𝐼𝑁 = 10𝑘Ω
𝐺 = 100
𝑓1 = 100𝐻𝑧, 𝑓2 = 1000𝐻𝑧
Solution:
𝑍𝐹
𝐺= ⇒ 𝑍𝐹 = 𝐺𝑍𝐼𝑁 = 10 ∗ 103 ∗ 100 = 1𝑀Ω
𝑍𝐼𝑁
1 1 1
𝑓1 = ⇒ 𝐶𝑖𝑛 = = = 0.159µ𝐹
2π𝐶𝑖𝑛 𝑅𝑖𝑛 2π𝑓1 𝑅𝑖𝑛 2π100 ∗ 104
1 1 1
𝑓2 = ⇒ 𝐶𝐹 = = = 0.159𝑛𝐹
2π𝐶𝐹 𝑅𝐹 2π𝑓2 𝑅𝐹 2π100 ∗ 106

Answer: 𝑅𝐼𝑁 = 10𝑘Ω, 𝑅𝐹 = 1𝑀Ω, 𝐶𝐼𝑁 = 0.159µ𝐹, 𝐶𝐹 = 0.159𝑛𝐹

Problem 9.12:
The electronic torque balance D/P transmitter of Figure 9.20(a) is to have an output range of 4 to 20 mA for
input ranges between 0 to 0.5m of water and 0 to 5.0 m of water. The diaphragm has a diameter of 10 cm, and
the maximum value of a/b is 2.0. Take g = 9.81 m s−2, and the density of water = 103 kg m−3.
(a) Complete the design by calculating the electromagnetic force constant and the zero spring force.
(b) Find the values of a/b corresponding to input ranges of 0 to 1 m and 0 to 5m.
Given:
𝐼 = [4,20]𝑚𝐴
𝑥1 = [0,0.5]𝑚
𝑥2 = [0,5]𝑚
 𝜌 = 103 𝑘𝑔 𝑚−3
𝑎/𝑏 = 2
d = 10cm
𝑔 = 9.81𝑚 𝑠 −2
Solution:
𝑎 𝐴𝐷 𝐹0 𝑎 𝐴𝐷 𝐹0
𝑖= (𝑃1 − 𝑃2 ) + = 𝜌𝑔(𝑥𝑖1 − 𝑥𝑖2 ) +
𝑏 𝐾𝑀 𝐾𝑀 𝑏 𝐾𝑀 𝐾𝑀
(a)
𝐾𝑀 𝑖𝑚𝑖𝑛 = 𝐹0 , (1)
𝑎
𝐾𝑀 𝑖𝑚𝑎𝑥 = 𝐴𝐷 𝜌𝑔(𝑥11 − 𝑥12 ) + 𝐹0 , (2)
𝑏
(2) − (1):
𝑎 π𝑑 2
𝐾𝑀 (𝑖𝑚𝑎𝑥 − 𝑖𝑚𝑖𝑛 ) = (𝑥11 − 𝑥12 )𝜌𝑔
𝑏 4
𝑎 π𝑑2 (𝑥11 − 𝑥12 ) π10−2
𝐾𝑀 = 𝜌𝑔 = 2.0 ∗ ∗ 0.5 ∗ 103 ∗ 9.81 = 4.8 ∗ 103 𝑁/𝐴
𝑏 4 (𝑖𝑚𝑎𝑥 − 𝑖𝑚𝑖𝑛 ) 4
𝐹0 = 𝐾𝑀 𝑖𝑚𝑖𝑛 = 4.8 ∗ 4 = 19.3 𝑁
(b) 𝑥1 = [0,1]𝑚, 𝑥2 = [0,5]𝑚
𝑎 𝐾𝑀 𝑖𝑚𝑎𝑥 − 𝐹0 4.8 ∗ 103 ∗ 20 ∗ 10−3 − 19.3
= = =1
𝑏 𝐴𝐷 𝜌𝑔(𝑥11 − 𝑥12 ) π10−2
4 1000 ∗ 9.81 ∗ 1
𝑎 𝐾𝑀 𝑖𝑚𝑎𝑥 − 𝐹0 4.8 ∗ 103 ∗ 20 ∗ 10−3 − 19.3
= = = 0.2
𝑏 𝐴𝐷 𝜌𝑔(𝑥21 − 𝑥22 ) π10−2
4 1000 ∗ 9.81 ∗ 5
Answer: (𝑎)4.8 × 103 𝑁/𝐴, 19.3𝑁(𝑏) 1.0, 0.2

Problem 9.13:
The variable reluctance displacement transducer of Problem 8.4 is incorporated into an electrical oscillator
circuit with a fixed capacitance of 500 pF. Find the variation in frequency of the oscillator output signal
corresponding to a variation in air gap of between 1 and 3 mm.
𝐶 = 500𝑝𝐹
𝐿1 = 7.6𝑚𝐻
𝐿2 = 3.4𝑚𝐻
Solution:
1 1
𝑓1 = = = 81.6𝑘𝐻𝑧
2𝜋√𝐿1 𝐶 2π√7.6 ∗ 10−3 500 ∗ 10−12
1 1
𝑓2 = = = 122.4𝑘𝐻𝑧
2𝜋√𝐿2 𝐶 2π√3.4 ∗ 10−3 500 ∗ 10−12
Problem 9.14:
The natural frequency fn Hz of a thin-walled tube executing circumferential vibrations in a fluid of density ρ kg
m−3 is given by:
2
1.25 ∗ 1010
𝑓𝑛 =
𝜌 + 350
The stiffness of the tube is 109 N m−1 and the damping ratio ξ = 0.1. The tube is incorporated in a closed-loop
system which also includes a drive coil of sensitivity 104 N V−1 and a displacement sensing coil of sensitivity
103 V m−1. The system is required to give a sinusoidal voltage output signal whose frequency changes with
fluid density in the range ρ = 250 to 1500 kg m−3. Find the gain and phase characteristics of the maintaining
amplifier. What is the effect of changes in fluid viscosity on the system?
Given:
𝑘 = 109 𝑁𝑚−1
𝜉 = 0.1
𝐾𝑆 = 104 𝑁 𝑉 −1
𝐾𝐷 = 103 𝑉𝑚−1
𝜌 = [250,1500] 𝑘𝑔 𝑚−3
Solution:
1 2𝜉𝑘 2 ∗ 0.1 ∗ 109
|𝐺(𝑗𝜔𝑛 )| = = = = 20
|𝐻(𝑗𝜔𝑛 )| 𝐾𝑆 𝐾𝐷 107
𝑤ℎ𝑒𝑛 𝑎𝑟𝑔(𝐺(𝑗𝜔𝑛 )) = −90°

1.25 ∗ 1010
𝑓1 = √ = 4.56𝑘𝐻𝑧
250 + 350

1.25 ∗ 1010
𝑓2 = √ = 2.60𝑘𝐻𝑧
1500 + 350

Answer: |𝐺(𝑗𝜔𝑛 )| = 20, 𝑎𝑟𝑔𝐺(𝑗𝜔𝑛 ) = −90°, 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 [2.60, 4.56] 𝑘𝐻𝑧

Problem 9.16:
Figure Prob. 16 shows a four-lead bridge circuit; Rc is the resistance of the leads connecting the sensor to the
bridge circuit. Show that ETh ≈ VS (R0/R3) αT, i.e. the bridge output voltage is unaffected by changes in Rc.
Solution:
2𝑅𝑐 + 𝑅0 (1 + α𝑇) 2𝑅𝑐 + 𝑅0
𝐸𝑇ℎ = 𝑉𝑠 ( − )
2𝑅𝑐 + 𝑅0 (1 + α𝑇) + 𝑅4 𝑅3 + 2𝑅𝑐 + 𝑅0
Assume 𝑅3 , 𝑅4 ≫ 𝑅0 + 2𝑅𝑐 , 𝑅0 α𝑇 𝑎𝑛𝑑 𝑅3 = 𝑅4
0 + 𝑅0 α𝑇 𝑅0 α𝑇
𝐸𝑇ℎ ≈ 𝑉𝑆 ( − 0) ≈ 𝑉𝑆
𝑅4 + 𝑅0 α𝑇 𝑅4 + 𝑅0 α𝑇
As 𝑅3 , 𝑅4 ≫ 𝑅0 α𝑇 and 𝑅4 = 𝑅3 :
𝑅0
𝐸𝑇ℎ ≈ 𝑉𝑆 α𝑇
𝑅3
Problem 9.17:
A resonant pressure sensor consists of a vibrating circular silicon plate. The natural frequency fn Hz of the plate
is given by: 𝑃 = 104 − 30𝑓𝑛 + 0.12𝑓𝑛2
where P Pa is the input pressure. The stiffness of the plate is 107 N/m, and the damping ratio is 0.01. The
sensitivity of the capacitance displacement sensor is 2 × 102 V/m, and the sensitivity of the drive element is
5N/V.
(a) If the sensor is to have an output frequency range of 1.0 to 3.0 kHz, what is the input pressure range?
(b) What are the gain and phase characteristics of maintaining amplifier?
Given:
𝑘 = 107 𝑁/𝑚
𝜉 = 0.01
𝐾𝑆 = 2 ∗ 102 𝑉/𝑚, 𝐾𝐷 = 5𝑁/𝑉
Solution:
(a) 𝑓𝑛 = [1, 3]𝑘𝐻𝑧
𝑃1 = 104 − 30 ∗ 103 + 0.12 ∗ 106 = 105 𝑃𝑎
𝑃2 = 104 − 30 ∗ 3 ∗ 103 + 0.12 ∗ 9 ∗ 106 = 106 𝑃𝑎
(b)
2𝜉𝑘 2 ∗ 0.01 ∗ 107
|𝐺(𝑗𝜔𝑛 )| = = = 200
𝐾𝑆 𝐾𝐷 103
𝑎𝑟𝑔(𝐺(𝑗𝜔𝑛 )) = −90°

Answer: (𝑎) 105 𝑡𝑜 106 𝑃𝑎 (𝑏) |𝐺(𝑗𝜔𝑛 )| = 200, 𝑎𝑟𝑔𝐺(𝑗𝜔𝑛 ) = −90° 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 1.0 𝑎𝑛𝑑 3.0𝑘𝐻𝑧
TEST
1. Temperature problems can be limited by
(a) selection of proper materials
(b) providing temperature compensation
(c) proper control of temperature
(d) all of these

2. Total absolute pressure is

3. What output voltage is developed in the circuit below?

A. - 0.625 V B. 0.625 V C. - 1.25 V D. 1.25 V E. 2.5 V

4. When pressure is applied onto the diaphragm, the distance between the two metal plates changes, which
in turn changes the
(a) capacitance
(b) inductance
(c) resistance
(d) reluctance

5. The measurable property that varies with temperature in a thermocouple is

6. Zero shift is the result of

(a) temperature variations

7. If a temperature sensor gives readings of 24.0°C, 24.5°C, and 24.2°C, find the average temperature.
A) 24.1°C
B) 24.3°C
C) 24.2°C
D) 24.0°C
8. The scale of a pressure gauge reads 0.5 bar per division. What is the accuracy of the scale?
A) ±0.5 bar
B) ±0.25 bar
C) ±1.0 bar
D) ±0.1 bar

the accuracy is half the smallest division