Homework #6 Key Assignment solution
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School
University of Texas, Arlington *
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Course
3360
Subject
Aerospace Engineering
Date
Dec 6, 2023
Type
Pages
8
Uploaded by matthew02crash
1
Homework #6 Key Assignment Solution MAE3360
Due 11:59 Nov. 21, 2023
The schematic for a suspension system is shown below.
The input is the road irregularities r(t).
w
z
r
The stroke of a suspension is the maximum possible decrease in the gap between the two masses
without the suspension bottoming out. Thus, for this problem, the decrease in the gap size will be
the output-of-interest, i.e. y(t) = z(t) – w(t).
It can be shown that the transfer function relating the
input road irregularities r(t) to y(t) is
𝑌(?) = [
1200?
2
?
4
+ 48?
3
+ 1776?
2
+ 14400? + 172800
] 𝑅(?)
1. (5%) What are the eigenvalues of this suspension?
−3.35 ± 𝑗10.7
and
−20.6 ± 𝑗30.8
>> G=tf([1200
0
0],[1
48
1776
14400
172800]);
>> damp(G)
Pole
Damping
Frequency
Time Constant
(rad/seconds)
(seconds)
-3.35e+00 + 1.07e+01i
2.99e-01
1.12e+01
2.98e-01
-3.35e+00 - 1.07e+01i
2.99e-01
1.12e+01
2.98e-01
-2.06e+01 + 3.08e+01i
5.57e-01
3.71e+01
4.84e-02
-2.06e+01 - 3.08e+01i
5.57e-01
3.71e+01
4.84e-02
2.
(7%) What are the damping ratios and time constants of these eigenvalues?
Damping ratios: 0.299 & 0.557
time constants: 0.298 and 0.0484
3.
(6%) What is the DC gain of this suspension? Explain in words what the significance of this
DC gain is regarding the performance of the suspension and the output-of-interest.
Setting s = 0 in the transfer function gives: DC gain = 0
The DC gain is used for a step input since the final value of a step input is a constant which
means that its frequency is zero. When a vehicle goes over a bump in the road, the space
between the two masses shown in the schematic changes. These changes are the output of
2
interest y(t).
If the bump in the road is a step which has a constant final value, then the
final value of y(t) will be the DC gain times the amplitude of the step.
So, for a DC gain of
zero, the size of the space between the masses will eventually return to its value before the
bump.
Thus, the final value of the change will be zero.
This is shown in the figure below for the step response of the change in gap size y(t); that is,
the change is zero to start with and returns to zero eventually.
>> G=tf([1200
0
0],[1
48
1776
14400
172800]);
>> step(G)
4. Generate the frequency response of the transfer function for Y(s) using ‘bode’ in MATLAB.
3
(4.a) (6%) Does the DC gain obtained from your frequency response agree with your answer to
part 3? Explain:
Yes, because the magnitude plot is headed to
??
−∞
= 0 as the frequency gets
very small.
Note, zero frequency does not exist on a log scale for frequency; the trend of
the magnitude is definitely downward as the frequency decreases.
(4.b) (10%) Notice that the frequency response shows that the transfer function represents a
notch filter; that is, low input frequencies (smooth road) and high input frequencies (tiny pot
holes) are filtered out meaning that the gap experiences minimal change for input low and high
input frequencies. Using data tips on the frequency response, determine the frequency range
where the dynamic gain is greater than 0.1 (-20 dB); assume that the magnitude of the gap
changes can be ignored for input frequencies above and below this range.
Low: 3.62 rad/s (0.58 hz)
High: 112 rad/s (17.8 hz).
(4.c) (5%) Use a data tip on the frequency response to determine the dynamic gain at the first
resonant frequency.
The resonant frequencies are the imaginary parts of the eigenvalues. As
observed in the eigenvalues, the resonant frequencies are 10.7 rad/s and 30.8 rad/s.
At the
first resonant frequency, 10.7, the dynamic gain = 10
(3/20)
= 1.41.
(4.d) (8%) We know that the road profile has spatial
frequencies that only are converted to rad/s
or hz once the vehicle speed is specified. For a speed of 10 m/s, what is the road irregularity
wavelength associated with the first mode resonant frequency?
The oscillation period for 10.7
rad/s is
?𝝅
??.?
= ?. 𝟓?? 𝒔
. At 10 m/s, the vehicle will travel 10*0.587 = 5.87 m.
So, the
wavelength is 5.87 m (distance between peaks).
(4.e)
(10%) Assume at a vehicle speed of 10 m/s, the input r(t) is a sinewave with amplitude 0.2
m and a frequency of 2 hz.
Use the dynamic gain shown on the frequency response to determine
the steady state amplitude of
y(t). Don’t forget, the frequency response is shown for rad/s
frequencies.
2 hz = 12.5 rad/s.
The dynamic gain for 12.5 is observed to be 4 dB. The phase
shift is observed to be 47.6 deg (0.83 rad). The output will be a 12.5 rad/s sinewave with an
amplitude of 0.2*10
4/20
= 0.2*1.58 = 0.316 m.
That is, for r(t) = 0.2sin(12.5t), the output will
be y(t) = 0.316sin(12.5t + 0.83)
5.
(10%) Use ‘lsim’ to confirm your answer in (4.e). If the input has a period of 0.502 seconds,
then to get at least 100 points per cycle, the time increment needs to be about 0.005 seconds.
Since the largest time constant is 0.298 seconds, to get to steady state will take 5*0.298 = 1.5
seconds. So, we will run the simulation at least 2.5 seconds to see about 2 steady state cycles.
As shown on the figure below, the steady state amplitude of y(t) is 0.318 m which agrees with
the product of the dynamic gain and the input amplitude in (4.e).
G=tf([1200
0
0],[1
48
1776
14400
172800]);
t=0:0.005:2.5;
r=0.2*sin(12.5*t);
y=lsim(G,r,t);
plot(t,y,
'r'
,t,r,
'k--'
,
'LineWidth'
,3)
xlabel(
'time, s'
,
'FontSize'
,18)
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4
ylabel(
'y(t) and r(t), m'
,
'FontSize'
,18)
legend(
'change in gap size, y(t) m'
,
'input irregularities, r(t) m'
,
'Location'
,
'best'
)
grid
6.
(10%) Use the inverse Laplace transform to confirm your answer in (4.d).
The Laplace transform of 0.2sin(12.5t) is
0.2
12.5
?
2
+12.5
2
=
2.5
?
2
+156.25
.
So, Y(s) becomes
𝑌(?) = [
1200?
2
?
4
+ 48?
3
+ 1776?
2
+ 14400? + 172800
] [
2.5
?
2
+ 12. 5
2
]
=
3000
[(? + 3.35)
2
+ 10. 7
2
][(? + 20.6)
2
+ 30.8
2
][(? + 0)
2
+ 12. 5
2
]
?(?) =
?
−3.35?
10.7
?
1
sin(10.7? + ∅
1
) +
?
−20.6?
30.8
?
2
sin(30.8? + ∅
2
) +
?
−0?
12.5
?
3
sin (12.5? + ∅
3
)
Where
>> s=0+12.5j;
>> M3=abs(3000*s^2/(s^4+48*s^3+1776*s^2+14400*s+172800))
M3 =
3.9780
>> Phi3=angle(3000*s^2/(s^4+48*s^3+1776*s^2+14400*s+172800))
Phi3 =
0.8212
?
3
= |
3000?
2
?
4
+ 48?
3
+ 1776?
2
+ 14400? + 172800
|
?=0+𝑗12.5
= 3.978
5
∅
3
= (
3000?
2
?
4
+ 48?
3
+ 1776?
2
+ 14400? + 172800
)
?=0+𝑗12.5
= 0.8212
After about 1.5 seconds, the first two exponential terms have decreased to almost zero; so,
?(?) = 0 + 0 +
?
−0?
12.5
3.978 sin(12.5? + 0.8212)
= 3.18sin (12.5 + 0.8212)
This amplitude of 3.18 and phase of 0.82 matches the results in (4.e) and part 5.
7.
(5%) Considering your answer in (4.e), what is the minimum design value for the suspension
stroke to prevent bottoming out for these road and speed conditions? The stroke needs to be
some number greater than 0.32 m;
maybe a value of 0.35 or 0.4.
8. Assume an input corresponding to a smooth highway and with a vehicle speed of 10 m/s.
?
̂
??
(?) =
?
̂
??
(𝜅)
𝜐
=
𝐴
𝜅
𝑁
𝜐
=
𝐴𝜐
𝑁−1
?
𝑁
No.
Description
N
A
1
Smooth runway
3.8
4.3 x 10
-11
3
Smooth highway
2.1
4.8 x 10
-7
4
Highway with gravel
2.1
4.4 x 10
-6
5
Pasture
1.6
3.0 x 10
-4
6
Plowed field
1.6
6.5 x 10
-4
For a smooth highway,
?
̂
??
(?) =
?
̂
𝑦𝑦
(𝜅)
𝜐
=
𝐴
𝜅
𝑁
𝜐
=
𝐴𝜐
𝑁−1
𝑓
𝑁
=
(4.8?10
−7
)10
1.1
𝑓
2.1
=
6?10
−6
𝑓
2.1
8.a (5%) Plot the input PSD for
0.5
≤
f
≤
18; use a frequency increment of 0.001 hz. Note, this
is approximately the band of input frequencies based on the -20 dB criteria in (4.b).
>> f=0.5:0.001:18;
>> Grr=(6e-6)./f.^2.1;
% ./ and .^ are required since f is an array of values
>> plot(f,Grr,'r','linewidth',3)
>> ylabel('PSD of input, m^2/hz','fontsize',18)
>> xlabel('input frequency, hz','fontsize',18)
>> grid
6
8.b (13%) Use ‘StochInput’ to generate values for r(t) corresponding to a smooth highway and
then use ‘lsim’ with the input values for r(t) to get corresponding values for y(t). You will first
need to determine values for the number of points N and the time increment H. Also, use ‘mean’
and ‘std’ to get values for the mean and standard deviation of r(t) and y(t). Considering the
results of 8.a, the results in 4.b, and the guidance on page 8-28 in the book, base your values for
N and H on a frequency increment of
∆? ≤ 0.001
hz out to a frequency of at least 20 hz.
Also,
take into consideration the magnitude of the eigenvalues and the time constants.
Based on the largest eigenvalue magnitude of 37.1,
? ≤
1
10(37.1)
= 0.0027
.
Considering that the
highest frequency generated will be
?
𝑚𝑎?
=
1
2?
≥ 20 ℎ?
, this gives
? ≤
1
40
= 0.025
. So, we will
use H=0.0027 (actually rounded to 0.0025).
To get a value for N, in order to get enough points
for an accurate representation of the PSD,
∆? =
1
𝑁?
≤ 0.001
.
So, for H=0.0025,
? ≥
1
0.001(0.0025)
= 400,000.
But N must be a power of 2; N = 2
19
= 524,288.
H: 0.0025
N: 524,288
𝜇
?
: − 6.9?10
−18
≈ 0 𝜎
?
: 0.1592 𝜇
?
: − 2?10
−7
≈ 0 𝜎
?
: 0.0025
Comment on the effectiveness of the suspension to filter out the effects of the road irregularities
r(t). The suspension significantly filtered the input so that the gap variations were almost 0.
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7
function
Homework_6_part_8b
G=tf([1200
0
0],[1
48
1776
14400
172800]);
N=2^19;H=0.0025;
[r,t]=StochInput(N,H);
y=lsim(G,r,t);
rmean=mean(r)
rStandardDev=std(r)
ymean=mean(y)
yStandardDev=std(y)
figure(1)
plot(t,r,
'r'
,t,y,
'k--'
,
'LineWidth'
,3)
xlabel(
'time, s'
,
'fontsize'
,18)
ylabel(
'displacement, m'
,
'FontSize'
,18)
legend(
'input r(t)'
,
'change in gap y(t)'
,
'location'
,
'best'
)
grid
[Grr,f]=comppsd(r,N,H);
figure(2)
plot(f,Grr,
'r'
,
'LineWidth'
,3)
xlabel(
'frequency, hz'
,
'FontSize'
,18)
xlim([0.5
20])
ylabel(
'PSD of r(t), m^2/hz'
,
'FontSize'
,18)
% function [Y,t]=StochInput(N,H)
% Generates N points in time Y(t) at a time interval of H with zero mean.
% N must be a power of 2! The N points will be generated
% so as to have a desired one-sided PSD defined in the function
% dpsd(f) at the end of this file; you will need to modify dpsd to give
% your desired PSD; f is the frequency in Hertz.
% The frequency resolution will be 1/NH.
% The smallest frequency will be 0 Hz.
% The largest frequency will be 1/2H Hz. A plot of the PSD out to 1/2H may
% not give you enough resolution on the lower frequencies; so, you may
% want to use the axis command to limit the frequency axis.
% Note, the frequency 1/2H is called the folding or Nyquist frequency.
%
% If you don't want this m-file to automatically plot y(t) and the
% desired PSD, you need to add "%" to all plot commands in this m-file.
%
% The time span for y(t) will be from 0 to NH.
function
[Y,t]=StochInput(N,H)
no2=N/2;
% Step 1: Generate frequencies and PSD values at each frequency
f=0:1/(N*H):1/(2*H);
%Note, the first frequency is zero but we don't
%compute GY at f=0 to avoid potential computational problems.
GY at f=0
%should be zero for a zero mean process.
GY(1)=0;
% The desired PSD is defined in function dpsd at the end of this file
for
m1=2:no2+1;
GY(m1)=dpsd(f(m1));
end
% Step 1: Generate N/2 random phase angles with uniform denisty
%
between 0 and 2pi
TH=random(
'unif'
,0,2*pi,no2,1);
% Step 2: Generate the appropriate amplitude at each frequency using
% the random phase angles.
for
m=2:no2;
C=sqrt(GY(m)*N*H/2);
8
CTH=cos(TH(m-1))*C;
STH=sin(TH(m-1))*C;
Y(m)=CTH+j*STH;
k=N+2-m;
Y(k)=CTH-j*STH;
end
% Step 3:
Make sure the N/2+1 value is real that the negative frequency
% values will be the mirror image of the positive frequency values.
no2p1=no2+1;
C=sqrt(GY(no2p1)*N*H/2);
CTH=cos(TH(no2))*C;
Y(no2p1)=CTH+j*0.e0;
% Make sure the zero frequency value is zero to achieve
% a zero mean time series.
Y(1)=0.e0+j*0.e0;
% Generate Y(t), the inverse transform of the N Fourier Transform
% values Y(f).
[Y]=ifft(Y);
% Rescale to get the correct time series corresponding to this desired PSD.
for
m=1:N;
Y(m)=real(Y(m)/H);
end
% Plot the PSD and time series
t=0:H:(N-1)*H;t=t';
Y=Y';
%-------------------------------------
% function[PSD,f]=comppsd(Y,N,H)
% This function generates N/2 unique values for the
% PSD of Y(t) at a frequency resolution of 1/NH. The
% zero frequency value represents the mean value squared times 2NH.
% Note, the PSD is plotted for all values of frequency.
% After generating the plot, for better resolution, you can change
% the scales of the plot using the command axis([XMIN XMAX YMIN YMAX]).
function
[PSD,f]=comppsd(Y,N,H)
[Y]=fft(Y);
NO2=N/2;
for
m=1:NO2
PSD(m)=2*H*abs(Y(m))^2/N;
end
f=0:1/(N*H):(N-2)/(2*N*H);
%plot(f,PSD)
%xlabel('Frequency (Hertz)')
%ylabel('PSD (units of y^2)/(cycles/sec)')
end
%-------------------------------------
function [GY]=dpsd(f)% This function will be called for each value of f
% You need equations or logic for GY describing the desired PSD.
GY=4.8e-7*10^1.1/f^2.1;
end
end
end