SIGNALS & SYSTEMS

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If you could kindly use this textbook (B. P. Lathi, 2nd ed. New York: Oxford University

Press., 2010, p. xvi, 975., ISBN: 9780195392562)

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Section 1: Mathematical Analysis of System

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Solution 1:

The following free-body diagram shows all the equation on the mass m shown in Figure 2:

Solution 2:

On balancing the forces along the vertical direction, we can write:

ma=C ( ∂ x ( t )∂ t

−∂ y (t )∂ t )+k (x ( t )− y ( t ) )+F ,where F=0

Or,

m ∂2 y ( t )∂ t 2 =C ( ∂x (t )

∂ t−∂ y (t )∂ t )+k (x (t )− y ( t ) )

Or,

∂2 y ( t )∂t 2

+Cm∂ y ( t )∂t

+ kmy ( t )=C

m∂ x ( t )∂ t

+ kmx (t )

Solution 3:

We can assume that the ground motion is given by a harmonic function:

x (t )=A ei st

and that the response of the mass is given by:

y (t )=Be i st

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After differentiating, substituting, and solving, we obtain the solution for the steady-state

displacement of the mass:

y (t )= k+isCk−s2m+ isC

Ae i st

We can now define the transfer function between the displacement of the mass and the input

ground displacement:

H (s )= y (t )x (t )

H (s )= k+ isCk−s2m+isC

Solution 4:

By comparing equations (1) and (2), we can write:

ωn=√ kmζ= C

2√km

Solution 5:

Based on equation (2), the characteristic equation can be written as:

s2+2ωn ζs+ωn2=0

The eigen values can written as:

s=−ωn (ζ ±√ (ζ 2−1 ))

Solution 6:

∂2 y (t )∂t 2

+2ωnζ∂ y ( t )∂ t

+ωn2 y (t )=0

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a) Undamped Free Vibration

∂2 y (t )∂t 2

+ωn2 y (t )=0

On Solving the above equation of motion:

y ( t )=√ y (0 )2+( y (0 )ωn )cos(ωn t−tan−1 y ( 0 )

y (0 )ωn )b) Sub Critical Damping

y ( t )=e−ζ ωn t√ y (0 )2+( y (0 )ωn )cos(ωn t √ (1−ζ 2)−tan−1 y (0 )

y (0 )ωn )c) Critical Damping

∂2 y (t )∂t2

+2ωn∂ y (t )∂t

+ωn2 y ( t )=0

d) Super Critical Damping

∂2 y (t )∂t 2

+2ωnζ∂ y ( t )∂ t

+ωn2 y (t )=0

Solution 7:

ωn=√ 21000340

=7.85905 rads

f n=ωn2π

=7.859052π

=1.2581Hz

Solution 8:

ζ= C2√km

When,

ζ=1 ,

C s=2√km=2√340×21000=5344.155686

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Section 2: System analysis using Matlab

In this section, the system responses should be analysed using Matlab. Refer to the document

“A Brief MATLAB Guide” in order to understand how to represent LTI systems in Matlab,

and hence how to determine impulse response, step response and frequency response of

systems.

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Solution 9:

Solution 10:

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Solution 11:

Solution 12:

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The above plot shows to response to excitation when tires were also considered in the

analysis, Figure 1.

Solution 13:

Section 3: Including the wheel & tyre response in the analysis

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Solution 15:

ωn=√ kmSolution 16:

Impulse Response:

In order to find the unit step response, we multiply H(s) by 1:

Y (s )=H (s )= k+isCk−s2m+isC

= 192000+100 is192000−45 s2+100is

Step Response:

In order to find the unit step response, we multiply H(s) by 1/s:

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Y (s )=1sH (s )=1

sk+isC

k−s2m+isC

Frequency Response:

From equation (2), we have:

∂2 y ( t )∂t 2

+2ωnζ∂ y ( t )∂ t

+ωn2 y (t )=F ( t )

[ (iω )2+2iωωnζ+ωn2 ]Y (ω )=F(ω)

The frequency response can be written as:

H ( iω )=Y (ω)F (ω)

= 1( iω )2+2iωωn ζ+ωn

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Solution 17:

The responses are usually considered to be harmonic in nature and the results of both the

responses are oscillatory (sines and cosines) which seems pretty close to actual responses

from vehicle wheel.

Solution 18:

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Solution 19:

Solution 20:

The difference in frequency response on ignoring the tires spring and damper (Figure 2) and

one considering them (Figure 1) can be clearly evident by the plots produced by the

frequency plots, magnitude and phase plots, from the MATLAB code.

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