In this example we will once again
consider the motion of a truck seat equipped with suspension mounted inside a
truck which has a suspension of its own.
The dynamic equations derived from the
free body diagram are in the time domain. We
will now learn how to convert the model into the frequency domain by using the
Laplace Transform and determining the transfer function from the input
yg
to the output
z2.
HOW TO FIND THE LAPLACE TRANSFORM
GOVERNING EQUATION
From the free body diagrams the time
domain differential equations are:
0 = -m1z1"
+ b1(yg(t)' -
z1') + k1(yg(t)
- z1) - b2(z1' -
z2') - k2(z1
- z2)
0 = - m2z2"
+ b2(z1' -
z2') + k2(z1 -
z2)
The notation for the Laplace Transform
operation is L{ }.
The Laplace Transform relations for a variable and its first and
second derivatives are the following:
L{ z(t)
} = Z(s)
L{ z’(t)
} = sZ(s) – z(0)
L{ z"(t)
} = s2Z(s) – sz(0) –
z'(0)
To determine the transfer function,
the initial conditions for this
system are assumed to be zero. z1(0) =
z2(0) = 0,
z1'(0) =
z2'(0) = 0.
Taking the Laplace Transform of the
governing equation and rearranging yields the following:
(m1s2 + (b1
+ b2)s + (k1
+ k2))Z1(s)
- (b2s + k2)Z2(s)
= (b1s + k1)Yg(s)
-(b2s + k2)Z1(s)
+ (m2s2 +
b2s + k2)Z2(s) = 0
To solve for the variable Z2(s)
in terms of Yg(s) it is useful to arrange the equations into
vector-matrix form and apply linear algebra:
The
next step is to multiply both sides of the above equation
by the inverse of the 2-by-2 matrix. The formula for the inverse of a 2-by-2 matrix is:
where:
In this example:
After
substituting in q, and multiplying both sides of the equation by the inverse,
the result is:
Continuing
to reduce the right hand side gives:
Finally
the expression for Z2(s)
is:
Substituting in for g21,
q, and 
gives:
The transfer function is the ratio of
the output Laplace transform to the input Laplace transform. So the transfer
function of this fourth order system is:
HOW TO INPUT THE TRANSFER FUNCTION
INTO MATLAB
We will now enter this transfer
function into MATLAB. Because
MATLAB cannot manipulate symbolic variables, we will now assign numerical
values to each parameter.
m1 = 10000 kg
m2 = 150 kg
b1 = 300,000 Ns/m
b2 = 1200 Ns/m
k1 = 100,000 N/m
k2 = 11000 N/m
>> m1 = 10000; m2 = 150; b1 = 300000; b2 =1200; k1 = 100000; k2 = 11000;
MATLAB's tf
function allows the variable s
to be used symbolically in the frequency domain. To convert s
to the Laplace variable, type:
>> s = tf('s')
Transfer function:
s
Once s
has been assigned as the Laplace variable, transfer functions can be specified
directly as rational expressions in s.
To assign our example transfer function to the system named higher_tf,
type the following:
>> higher_tf = (b2*s+k2)*(b1*s+k1)/((m1*s^2 + (b1+b2)*s + (k1+k2))*(m2*s^2+b2*s+k2)-(b2*s+k2)^2)
Transfer function:
3.6e008 s^2 + 3.42e009 s + 1.1e009
------------------------------------------------------------------
1.5e006 s^4 + 5.718e007 s^3 + 4.867e008 s^2 + 3.42e009 s + 1.1e009
STEP RESPONSE USING THE TRANSFER
FUNCTION
Now that we have the transfer function
entered into MATLAB we can use MATLAB to do many useful calculations including
determining the step response. Because
the transfer function is in the form of output over input and we are interested
in finding the output, the transfer function must be multiplied by the input
function. The step response models how the
truck seat would respond to a sudden change in the level, yg = 0.1m, of the pavement like an
uneven transition between an older road and a newly paved section.
>> yg = 0.1;
>> step(yg * higher_tf)
BODE PLOT USING THE TRANSFER FUNCTION
MATLAB’s bode command plots the
frequency response of a system as a bode plot. This helps to model how the
truck will respond over rutty pavement of various frequencies.
>> bode(higher_tf)
