Exercise 1a transfer functions - solutions

EE4107
-‐
Cybernetics
Advanced

Faculty of Technology, Postboks 203, Kjølnes ring 56, N-3901 Porsgrunn, Norway. Tel: +47 35 57 50 00 Fax: +47 35 57 54 01

Exercise
1a:
Transfer
functions
(Solutions)

Transfer
functions

Transfer
functions
are
a
model
form
based
on
the
Laplace
transform.
Transfer
functions
are
very

useful
in
analysis
and
design
of
linear
dynamic
systems.

A
general
Transfer
function
is
on
the
form:

𝐻 𝑆 =
𝑦(𝑠)
𝑢(𝑠)

Where
𝑦
is
the
output
and
𝑢
is
the
input.

A
general
transfer
function
can
be
written
on
the
following
general
form:

𝐻 𝑠 =
𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(𝑠)
𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑠)
=
𝑏! 𝑠! + 𝑏!!! 𝑠!!! + ⋯ + 𝑏! 𝑠 + 𝑏!
𝑎! 𝑠! + 𝑎!!! 𝑠!!! + ⋯ + 𝑎! 𝑠 + 𝑎!

The
Numerators
of
transfer
function
models
describe
the
locations
of
the
zeros
of
the
system,
while

the
Denominators
of
transfer
function
models
describe
the
locations
of
the
poles
of
the
system.

Differential
Equations

While
the
transfer
function
gives
an
external
in-‐out
representation
of
a
system,
will
the
differential

equations
of
a
system
give
an
internal
representation
of
a
system.

We
can
find
the
transfer
function
from
the
differential
equation
by
using
Laplace
and
Laplace

transformation
pairs.
Likewise,
we
can
find
the
differential
equation
from
the
transfer
function

using
inverse
Laplace.

The
following
transformation
pair
is
much
used:

Differentiation:

1.order
systems:

𝑥 ⟺ 𝑠𝑥(𝑠)

For
higher
order
systems:

𝑥(!)
⟺ 𝑠!
𝑥(𝑠)

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Integration:

𝑥 ⟺
1
𝑠
𝑥(𝑠)

Time-‐delay:

𝑢 𝑡 − 𝜏 ⟺ 𝑢(𝑠)𝑒!!"

Static
Time-‐response

In
some
cases
we
want
to
find
the
constant
value
𝑦!
of
the
time
response
when
the
time
𝑡 → ∞.

We
can
then
use
the
final
value
theorem
(sluttverditeoremet):

𝑦! = lim
!→!
𝑦 𝑡 = lim
!→!
𝑠 ∙ 𝑦(𝑠)

MathScript

MathScript
has
several
functions
for
creating
transfer
functions:

Function
Description
Example

tf
Creates
system
model
in
transfer
function
form.
You
also
can

use
this
function
to
state-‐space
models
to
transfer
function

form.

>num=[1];
>den=[1, 1, 1];
>H = tf(num, den)
Sys_order1
Constructs
the
components
of
a
first-‐order
system
model
based

on
a
gain,
time
constant,
and
delay
that
you
specify.
You
can
use

this
function
to
create
either
a
state-‐space
model
or
a
transfer

function
model,
depending
on
the
output
parameters
you

specify.

>K = 1;
>tau = 1;
>H = sys_order1(K, tau)
Sys_order2
Constructs
the
components
of
a
second-‐order
system
model

based
on
a
damping
ratio
and
natural
frequency
you
specify.
You

can
use
this
function
to
create
either
a
state-‐space
model
or
a

transfer
function
model,
depending
on
the
output
parameters

you
specify.

>dr = 0.5
>wn = 20
>[num, den] = sys_order2(wn, dr)
>SysTF = tf(num, den)
step
Creates
a
step
response
plot
of
the
system
model.
You
also
can

use
this
function
to
return
the
step
response
of
the
model

outputs.
If
the
model
is
in
state-‐space
form,
you
also
can
use
this

function
to
return
the
step
response
of
the
model
states.
This

function
assumes
the
initial
model
states
are
zero.
If
you
do
not

specify
an
output,
this
function
creates
a
plot.

>num=[1,1];
>den=[1,-1,3];
>H=tf(num,den);
>t=[0:0.01:10];
>step(H,t);
Example:

Given
the
following
transfer
function:

𝐻 𝑠 =
1
𝑠 + 1

In
MathScript
we
will
use
the
following
code:

% Define Transfer function

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Advanced

num = [1];
den = [1, 1];
H = tf(num, den)
% Step Response
step(H)
This
gives
the
following
step
response:

A
general
transfer
function
can
be
written
on
the
following
general
form:

𝐻 𝑠 =
𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(𝑠)
𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑠)
=
𝑏! 𝑠! + 𝑏!!! 𝑠!!! + ⋯ + 𝑏! 𝑠 + 𝑏!
𝑎! 𝑠! + 𝑎!!! 𝑠!!! + ⋯ + 𝑎! 𝑠 + 𝑎!

The
Numerators
of
transfer
function
models
describe
the
locations
of
the
zeros
of
the
system,
while

the
Denominators
of
transfer
function
models
describe
the
locations
of
the
poles
of
the
system.

In
MathScript
we
can
define
such
a
transfer
function
using
the
built-‐in
tf
function
as
follows:

num = [bm, bm_1, bm_2, … , b1, b0];
den = [an, an_1, an_2, … , a1, a0];
H = tf(num, den)
Task
1:
Differential
equations
to
Transfer
functions

Task
1.1

Given
the
following
differential
equation:

𝑥 = −0.5𝑥 + 2𝑢

Find
the
following
transfer
function:

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𝐻 𝑠 =
𝑥(𝑠)
𝑢(𝑠)

Solution:

Laplace
gives:

𝑠𝑥(𝑠) = −0.5𝑥(𝑠) + 2𝑢(𝑠)

Further:

𝑠𝑥 𝑠 + 0.5𝑥(𝑠) = 2𝑢(𝑠)

Further:

𝑥 𝑠 (𝑠 + 0.5) = 2𝑢(𝑠)

Further:

𝑥 𝑠
𝑢(𝑠)
=
2
𝑠 + 0.5
=
4
2𝑠 + 1

This
gives:

𝐻 𝑠 =
𝑥 𝑠
𝑢(𝑠)
=
4
2𝑠 + 1

Task
1.2

Given
the
following
2.order
differential
equation:

𝑦 + 𝑦 + 5𝑦 = 5𝑥

Find
the
following
transfer
function:

𝐻 𝑠 =
𝑦(𝑠)
𝑥(𝑠)

Solution:

We
get:

𝑠!
𝑦 𝑠 + 𝑠𝑦 𝑠 + 5𝑦 𝑠 = 5𝑠𝑥(𝑠)

Further:

𝑦 𝑠 [𝑠!
+ 𝑠 + 5] = 5𝑠𝑥(𝑠)

This
gives
the
following
transfer
function:

𝑦 𝑠
𝑥(𝑠)
=
5𝑠
𝑠! + 𝑠 + 5

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Task
2:
Transfer
functions
to
differential
equatons

Given
the
following
system:

𝐻 𝑠 =
𝑥(𝑠)
𝑢(𝑠)
=
3
0.5𝑠 + 1

Task
2.1

Find
the
differential
equation
from
the
transfer
function
above.

Solution:

We
get:

𝑥(𝑠) 0.5𝑠 + 1 = 3𝑢(𝑠)

Further:
0.5𝑠𝑥 𝑠 + 𝑥(𝑠) = 3𝑢(𝑠)

Inverse
Laplace
gives:

0.5𝑥 + 𝑥 = 3𝑢

This
gives
the
following
differential
equation:

𝑥 = −2𝑥 + 6𝑢

Task
2.2

Draw
a
block
diagram
of
the
system.

Solution:

We
can
draw
the
following
block
diagram:

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Note!
Even
when
the
system
is
in
the
time
plane
we
normally
use
the
symbol
.
Other
symbols
that

are
commonly
used
for
the
integrator
are:

or

Task
3:
2.order
system

Given
the
following
transfer
function:

𝐻 𝑠 =
𝑦(𝑠)
𝑢(𝑠)
=
2𝑠 + 3
𝑠! + 4𝑠 + 3

Task
3.1

Find
the
differential
equation
for
the
system.

Solution:

We
do
as
follows:

𝑦 𝑠 𝑠!
+ 4𝑠 + 3 = 𝑢 𝑠 [2𝑠 + 3]

This
gives:

𝑠!
𝑦 𝑠 + 4𝑠𝑦 𝑠 + 3𝑦 𝑠 = 2𝑠𝑢 𝑠 + 3𝑢(𝑠)

This
gives
the
following
differential
equation:

𝑦 + 4𝑦 + 3𝑦 = 2𝑢 + 3𝑢

Note!
We
have
used
the
rule:

𝑥(!)
⟺ 𝑠!
𝑥(𝑠)

Task
4:
Static
Time-‐response

Task
4.1

Given
the
following
system:

𝐻(𝑠) =
𝑦(𝑠)
𝑢(𝑠)
=
3
2𝑠 + 1

Find
the
static
time-‐response.

We
will
use
a
step
for
the
control
signal
( 𝑢 𝑡 = 1).

Note!
The
Laplace
Transformation
pair
for
a
step
is
as
follows:

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1
𝑠
⇔ 1

Solution:

We
have:

𝐻(𝑠) =
𝑦(𝑠)
𝑢(𝑠)
=
3
2𝑠 + 1

Meaning
that:

𝑦(𝑠) =
3
2𝑠 + 1
𝑢(𝑠)

where
𝑢 𝑠 =
!
!

This
means:

𝑦 𝑠 =
3
2𝑠 + 1
∙
1
𝑠
=
3
2𝑠 + 1 𝑠

Then
we
use
the
final
value
theorem
(sluttverditeoremet):

𝑦! = lim
!→!
𝑦 𝑡 = lim
!→!
𝑠 ∙ 𝑦 𝑠 = lim
!→!
𝑠
3
2𝑠 + 1 𝑠
= lim
!→!

3
2𝑠 + 1
=
3
2 ∙ 0 + 1
= 3

Task
4.2

Given
the
following
system:

𝐻(𝑠) =
𝑦(𝑠)
𝑢(𝑠)
=
6(𝑠 + 1)
9𝑠 + 0.25

Find
the
static
time-‐response.

We
will
use
a
step
for
the
control
signal
( 𝑢 𝑡 = 1).

Solution:

We
get:

𝑦(𝑠) =
6(𝑠 + 1)
9𝑠 + 0.25
𝑢(𝑠)

where
𝑢 𝑠 =
!
!

This
means:

𝑦 𝑠 =
6(𝑠 + 1)
9𝑠 + 0.25 𝑠

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Then
we
get
using
the
final
value
theorem:

𝑦! = lim
!→!
𝑦 𝑡 = lim
!→!
𝑠 ∙ 𝑦 𝑠 = lim
!→!
𝑠
6 𝑠 + 1
9𝑠 + 0.25 𝑠
= lim
!→!

6(𝑠 + 1)
9𝑠 + 0.25
=
6(0 + 1)
9 ∙ 0 + 0.25
=
6
0.25
= 24

Task
5:
1.order
transfer
functions

Given
the
following
system:

𝐻(𝑠) =
𝑦(𝑠)
𝑢(𝑠)
=
2
4𝑠 + 1

Task
5.1

What
are
the
values
for
the
gain
𝐾
and
the
time
constant
𝑇
for
this
system?

Sketch
the
step
response
for
the
system
using
“pen
and
paper”.

Find
the
step
response
using
MathScript
and
compare
the
result
with
your
sketch.

Solutions:

Gain
𝐾
and
the
time-‐constant
𝑇:

𝐾 = 2

𝑇 = 4

Step
response
for
a
1.order
system:

MathScript:

clear

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clc
K=2;
T=4;
num=[K];
den=[T, 1];
H = tf(num, den);
step(H)
This
gives
the
following
plot:

For
a
1.order
system
with
time-‐delay
we
have:

Task
5.2

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EE4107
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Find
the
differential
equation
from
the
transfer
function
above
and
draw
a
block
diagram
of
the

system
(“pen
and
paper”).

Solutions:

For
a
1.order
system
in
general
we
have:

𝐻(𝑠) =
𝑦(𝑠)
𝑢(𝑠)
=
𝐾
𝑇𝑠 + 1

or:

𝑦 𝑠 =
𝐾
𝑇𝑠 + 1
𝑢(𝑠)

Which
gives:

𝑦 𝑠 (𝑇𝑠 + 1) = 𝐾𝑢(𝑠)

𝑇𝑠𝑦 𝑠 + 𝑦(𝑠) = 𝐾𝑢(𝑠)

In
the
time
domain
we
get
the
following
differential
equation
(using
Inverse
Laplace):

𝑦 =
1
𝑇
(−𝑦 + 𝐾𝑢)

We
can
draw
the
following
block
diagram
of
the
system:

Where
𝐾 = 2
and
𝑇 = 4
for
our
system:

Note!
Even
when
the
system
is
in
the
time
plane
we
normally
use
the
symbol
.
Other
symbols
that

are
commonly
used
for
the
integrator
are:

or
.

Task
5.3

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From
the
block
diagram
in
Task
1.2,
find
the
transfer
function
𝐻(𝑠) =
!(!)
!(!)

(The
answer
shall
of
course
be
𝐻(𝑠) =
!(!)
!(!)
=
!
!!!!
)

Solutions:

From
the
block
diagram
in
the
previous
task
we
get
the
following
transfer
function:

𝐻 𝑠 =
𝑦 𝑠
𝑢 𝑠
= 2 ∙
0.25 ∙
1
𝑠
1 + 0.25 ∙
1
𝑠
=
0.5
𝑠 + 0.25
=
2
4𝑠 + 1

As
expected,
the
result
is
the
same
as
the
transfer
function
given
in
Task
1.1.

Note!
We
have
used
both
the
serial
and
feedback
rules
that
yield
for
block
diagram
reduction.

Task
5.4

Find
the
solution
for
the
differential
equation
and
plot
it
(“pen
and
paper”).

We
will
use
a
step
for
the
control
signal
( 𝑢 𝑡 = 1).

Note!
The
Laplace
Transformation
pair
for
a
step
is
as
follows:

1
𝑠
⇔ 1

Tip!
You
also
need
to
use
the
following
Laplace
transform
pair:

𝐾
𝑇𝑠 + 1 𝑠
⇔ 𝐾(1 − 𝑒!!/!
)

Compare
to
the
results
from
Task
1.1.

Solutions:

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For
a
1.order
system
in
general
we
have:

𝐻(𝑠) =
𝑦(𝑠)
𝑢(𝑠)
=
𝐾
𝑇𝑠 + 1

Here
we
will
find
the
mathematical
expression
for
the
step
response
( 𝑦(𝑡)):

𝑦 𝑠 = 𝐻 𝑠 𝑢(𝑠)

Where

𝑢 𝑠 =
𝑈
𝑠

We
use
inverse
Laplace
and
find
the
corresponding
transformation
pair
in
order
to
find
𝑦(𝑡)).

𝑦 𝑠 =
𝐾
𝑇𝑠 + 1
∙
𝑈
𝑠

We
use
the
following
Laplace
transform
pair:

𝐾
𝑇𝑠 + 1 𝑠
⇔ 𝐾(1 − 𝑒!!/!
)

This
gives:

𝑦 𝑡 = 𝐾𝑈(1 − 𝑒!!/!
)

Setting
𝐾 = 2,
𝑇 = 4
and
𝑈 = 1
gives:

𝑦 𝑡 = 2(1 − 𝑒
!
!
! !
)

We
can
plot
this
in
MathScript:

clear
clc
K=2;
T=4;
U=1;
t=0:0.1:20;
% Method 1 - Transfer Function
num=[K];
den=[T, 1];
H = tf(num, den);
figure(1)
step(H, t)

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% Method 2 - Plot the solution of the differential equation
y = K*U*(1-exp(-t/T));
figure(2)
plot(t,y)
We
get
the
same
results
(of
course).

Task
6:
Transfer
functions
in
MathScript

Define
the
following
transfer
functions
in
MathScript.

Task
6.1

Given
the
following
transfer
function:

𝐻(𝑠) =
2𝑠! + 3𝑠 + 4
5𝑠 + 9

Solutions:

MathScript
Code:

num = [2, 3, 4];
den = [5, 9];
H = tf(num, den)
Task
6.2

Given
the
following
transfer
function:

𝐻(𝑠) =
4𝑠! + 3𝑠 + 4
5𝑠! + 9

Solutions:

MathScript
Code:

num = [4, 0, 0, 3, 4];
den = [5, 0, 9];
H = tf(num, den)
Note!
If
some
of
the
orders
are
missing,
we
just
put
in
zeros.
The
transfer
function
above
can
be

rewritten
as:

𝐻(𝑠) =
4𝑠! + 0 ∙ 𝑠! + 0 ∙ 𝑠! + 3𝑠 + 4
5𝑠! + 0 ∙ 𝑠 + 9

Task
6.3

Given
the
following
transfer
function:

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𝐻(𝑠) =
7 + 3𝑠 + 2𝑠!
5𝑠 + 6𝑠!

Solutions:

We
need
to
rewrite
the
transfer
function
to
get
it
in
correct
orders:

𝐻(𝑠) =
2𝑠! + 3𝑠 + 7
6𝑠! + 5𝑠

MathScript
Code:

num = [2, 3, 7];
den = [6, 5, 0];
H = tf(num, den)
Task
7:
Differential
equations
to
Transfer
functions

Task
7.1

Given
the
following
differential
equation:

𝑥 = −0.5𝑥 + 2𝑢

Find
the
following
transfer
function:

𝐻 𝑠 =
𝑥(𝑠)
𝑢(𝑠)

Solution:

Laplace
gives:

𝑠𝑥(𝑠) = −0.5𝑥(𝑠) + 2𝑢(𝑠)

Further:

𝑠𝑥 𝑠 + 0.5𝑥(𝑠) = 2𝑢(𝑠)

Further:

𝑥 𝑠 (𝑠 + 0.5) = 2𝑢(𝑠)

Further:

𝑥 𝑠
𝑢(𝑠)
=
2
𝑠 + 0.5
=
4
2𝑠 + 1

This
gives:

𝐻 𝑠 =
𝑥 𝑠
𝑢(𝑠)
=
4
2𝑠 + 1

15

EE4107
-‐
Cybernetics
Advanced

Task
8:
PI
Controller

A
PI
controller
is
defined
as:

𝑢(𝑡) = 𝐾! 𝑒 +
𝐾!
𝑇!
𝑒𝑑𝜏
!
!

Where
u
is
the
controller
output
and
𝑒 is
the
control
error:

𝑒 𝑡 = 𝑟 𝑡 − 𝑦(𝑡)

Task
8.1

Find
the
transfer
function
for
the
PI
Controller:

𝐻! 𝑠 =
𝑢(𝑠)
𝑒(𝑠)

Solutions:

Using
Laplace
gives:

𝑢 𝑠 = 𝐾! 𝑒 𝑠 +
𝐾!
𝑇!
∙
1
𝑠
𝑒 𝑠

Then
we
get:

𝐻! 𝑠 =
𝑢(𝑠)
𝑒(𝑠)
= 𝐾! +
𝐾!
𝑇! 𝑠
=
𝐾! 𝑇! 𝑠
𝑇! 𝑠
+
𝐾!
𝑇! 𝑠
=
𝐾! 𝑇! 𝑠 + 𝐾!
𝑇! 𝑠
=
𝐾!(𝑇! 𝑠 + 1)
𝑇! 𝑠

This
gives
the
following
transfer
function
for
the
PI
controller:

𝐻! 𝑠 =
𝑢(𝑠)
𝑒(𝑠)
=
𝐾!(𝑇! 𝑠 + 1)
𝑇! 𝑠

Additional
Resources

• http://home.hit.no/~hansha/?lab=mathscript
Here you will find tutorials, additional exercises, etc.

Exercise 1a transfer functions - solutions

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