The modeling equation gathered from
the free body diagram is in the time domain.
We will now learn how to convert it into the frequency domain by
performing the Laplace Transform and determining the transfer function of this
first order system.
In this example we will once again
consider a truck pulling a trailer where the friction and wind resistance acts as a velocity dependant
force, with a coefficient b.
HOW TO FIND THE TRANSFER FUNCTION
From the free body diagram we were
able to extract the following governing equation:
f(t) -
m v' -
b v = 0
The notation of the Laplace Transform
is L{ }
L{ v(t) } = V(s)
L{ v’(t) } = sV(s) – v(0)
L{ f(t) } = F(s)
When
finding the transfer function, ‘zero’ initial conditions must be assumed,
so v(0) = 0.
Substituting
into the governing equation we get:
m[sV(s)]
+ b[V(s)] – F(s) = 0
Rearranging we get:
[ms + b]V(s) = F(s)
The transfer function is defined as
the output Laplace function over the input Laplace function, so the transfer
function of this first order system is:
V(s)/F(s)
= 1/[ms+b]
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 variable.
m = 20,000kg
b = 500kg/s
>>
m =
20000;
>>
b =
500;
In order to enter the transfer
function into MATLAB, you must separate the numerator and denominator, which
in this example are denoted by ‘num’ and ‘den’ respectively.
The format for either matrix is to enter the coefficients of sn
in descending order.
>>
num = [
1 ];
>>
den = [
m b ];
>>
first_tf
= tf(num, den)
Transfer function:
1
-------------
20000 s + 500
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. In this case
we will plot the unit step response, taking the input to be 1 unit. This
models what would happen if you applied 1N of force pushing this truck.
>>
u = 1;
>>
step(u
* first_tf)
The MATLAB output will be the following plot
of the step response:
Obviously this truck is able to exert
far more than 1N of force, which is why the truck moves so little. More
realistically this truck should be able to exert around 20,000N of force.
>>
u =
20,000;
>>
step(u
* first_tf)
This is far more like it. Given
the truck is able to exert 20,000N of force, this truck, on level ground, has
a top speed of 40m/s and takes 250s to reach that speed.
IMPULSE RESPONSE USING THE TRANSFER
FUNCTION
MATLAB can also plot the impulse
response of a transfer function.
>>
u = 1;
>> impulse(u * first_tf)
The MATLAB output will be the following plot
of the unit impulse response:
Now given an input impulse of 20,000N:
>> u = 20000;
>> impulse(u *
first_tf)
This shows that given an impulse of
20,000N, the speed of the truck drops from 1m/s and takes approximately 250s
to come to rest.
BODE PLOT USING THE TRANSFER
FUNCTION
MATLAB’s bode command plots the
frequency response of a system as a bode plot.
>>
bode(first_tf)
The MATLAB output will be the following bode
plot of the frequency response:
While trucks are rarely tested in this
way, this plot shows the response of this truck with a sinusoidal application
of the throttle.
STATE SPACE FROM TRANSFER FUNCTION
In the next section of this example
problem we will learn how to determine the state-space representation of this
first order system. To
find a state space representation of the system from the transfer function in the form
x' =
Ax +
Bu
y =
Cx
+ Du
use
MATLAB's tf2ss command:
>>
[A, B,
C, D] = tf2ss(num,den)
The MATLAB output will be:
A =
-0.0250
B =
1
C =
5.0000e-005
D =
0
In order to find the entire state
space system instead of the separate matrices from the transfer function, use
the following command:
>> first_ss = ss(first_tf)
MATLAB will assign the state space
system under the name first_ss,
and output the following:
a =
x1
x1 -0.025
b =
u1
x1 0.007813
c =
x1
y1 0.0064
d =
u1
y1
0
Continuous-time model.
