1.

1 The following table shows % reading and % full scale for each data
point. There is no need to do a least squares fit.
Inputs 0.50 1.50 2.00 5.00 10.00
Outputs 0.90 3.05 4.00 9.90 20.50
Ideal Output 1.00 3.00 4.00 10.00 20.00
Difference – 0.10 +0.05 0.0 – 0.10 +0.50
% Reading 11.1 1.6 % 0% – 1.0% +2.4 % = Difference/Output × 100
Full Scale – 0.5% 0.25% 0% – 0.50% + 2.5% = Difference/20 × 100
Inspection of these data reveals that all data points are within the
"funnel" (Fig. 1.4b) given by the following independent nonlinearity = ±2.4% reading or
±0.5% of full scale, whichever is greater. Signs are not important because a symmetrical
result is required. Note that simple % reading = ±11.1% and simple % full scale = 2.5%.

1.2 The following table shows calculations using equation (1.8).

Inputs Xi 0.50 1.50 2.00 5.00 10.00 ! = 3.8
Outputs Yi 0.90 3.05 4.00 9.90 20.50 !  = 7.6
Xi –! – 3.3 – 2.3 – 1.8 1.2 6.2
Yi – ! – 6.7 – 4.55 – 3.6 2.3 12.9
(Xi – !)(Yi – !) 22.11 10.465 6.48 2.76 79.98 Σ = 121.795
(Xi – !)2 10.89 5.29 3.24 1.44 38.44 Σ = 59.3 (59.3)1/2 = 7.701
(Yi – !)2 44.89 20.7 12.96 5.29 166.41 Σ = 250.25 (250.25)1/2 = 15.8
2
121.795
r = (7.701)(15.82) = 0.9997

1.3 The simple RC high-pass filter:

I C

+ +
Vin R Vout
- -
The first order differential equation is:
d x(t) œy(t) y(t)
C =
dt R
1
CD + y(t) = (CD) x(t)
R
y(D) D
= Operational transfer function
x(D) 1
D+
RC
Y(jω) jωRC ωRC 1
= = ∠ φ = Arctan
X(jω) jωRC + 1 (ωRC)2 + 1 ωRC
;where 1/RC is the corner frequency in rad/s.

1.4 For sinusoidal wing motion the low-pass sinusoidal transfer function is
Y(jω) K
=
X(jω) (jωτ + 1)
For 5% error the magnitude must not drop below 0.95 K or
K K
= = 0.95 K
jωτ + 1 2 2
ω τ +1
Solve for τ with ω = 2πf = 2π(100)
2 2
(ω τ + 1) (.95)2 = 1
1/2
1 œ(0.95)2
τ= = 0.52 ms
(0.95)2(2π100)2
Phase angle φ = tan–1 (–ωτ) at 50 Ηz
φ50 = tanœ1 (œ2π × 50 × 0.0005) = œ9.3 °
at 100 Hz
φ100 = tanœ1 (œ2π × 100 × 0.0005) = œ18.2 °

1.6 Find the spring scale (Fig. 1.11a) transfer function when the mass is
negligible. Equation 1.24 becomes
dy(t)
B + Ks y(t) = x(t)
dt
When M = 0. This is a first order system with
1
K = static sensitivity = K
s
B
τ = time constant = K
s
 Thus the operational transfer function is
y(D) 1/Ks 1
= =
x(D) B Ks + BD
1+ D
Ks
and the sinusoidal transfer function becomes
y(jω) 1/Ks 1/Ks B
= = ∠ φ = tanœ1 œω
x(jω) B 2 Ks
1 +jω D ω B2
Ks 1+
2
Ks

1.8 For a first order instrument

Y(jω ) K
=
X (jω ) (jωτ + 1)
K K
= = 0.93 K
(jωτ + 1) 2 2
ω τ +1
2 2
ω τ + 1 (0.93)2 = 1
1 1 1 œ(0.93)2
f = ω = = 3.15 Hz
2π 2π (0.93)2 (.02)2
φ = tanœ1 (œωτ) = œ21.6°

1.9
Keœζωnt 2 2
y(t) = K sin 1 œζ ωnt + φ where φ = sinœ1 1 œζ
2
1 œζ
ζ = 0.4; fn = 85 Hz
3π 7π
œφ œφ
2 2
tn = = 7.26 ms tn+1 = = 20.1 ms
2 2
ωn 1 œζ ωn 1 œζ

10 10
y(tn) = 10 + eœζω ntn y(tn+1) = 10 + eœζω ntn+1
2 2
1 œζ 1 œζ
= 12.31 = 10.15
2.4 From (2.21)
E = 38.7T + (0.082/2)T2 = 38.7T + 0.041T2
T 38.7T 0.41T2 E
°C µV µV µV

0 0 0 0
10 387 4 391
20 774 16 790
30 1161 37 1196
40 1548 66 1614
50 1935 102 2037
The second term is small. The curve is almost linear but slightly concave
upward.

2.5 From (2.22)

α = dE/dT = a + bt = 38.7 + 0.082T µV/˚C
= 38.7 + 0.082(37) = 41.7 µV/˚C

2.6 From (2.24)

–4000
α = – β/T2 = = –4.4%/K
(300)2
2.8 In Example 2.3 C = 500 pF for the piezoelectric transducer. The amplifier
input impedance = 5 MΩ.
1
F = 0.05 Hz =
2πRCequivalent
Thus Cequivalent = 0.637 × 10–6 = Cpiezoelectric + Cshunt
Cshunt = 0.636 µF = 636 nF
Q = CV, where charge Q is fixed, capacitance C increases by 636 nF/0.5
nF = 1272 times. Voltage V (sensitivity) decreases by 1/1272.
The sensitivity will be decreased by a factor of 1272 due to increase in the
equivalent capacitance.

2.9 Select a feedback Cf = 100 nF (much larger than 500 pF). To achieve low
corner frequency, add Rf = 1/(2πfcCf) = 1/(2π·0.05·100 nF) = 32 MΩ. To achieve high
corner frequency add separate passive filter or active filter with Ro = 10 kΩ and Co =
1/(2πfcRo) = 1/(2π·100·10 kΩ) = 160 nF.