Go back
Transfer Functions

Transfer Functions

Science

R
Standard memberRemoved

Joined
10 Dec 06
Moves
8528
Clock
23 Mar 16
Vote Up
Vote Down

I'm looking for some layman explanation on how transfer functions work within the framework of classical mechanics. If it could be related specifically to the example below that would be even better.

A system consists of a gasoline engine (prime mover) with a Torque vs Speed curve: T(w), and Gearbox and a Drum (load) that is to be rotated.

The engine is initially in operation at a fixed speed uncoupled from the load. At time "t" equal to zero the load is engaged (via clutch). If the load has a high mass inertia the engine responds to the step change in load by decreasing it speed abruptly; followed by in increase in speed; followed by an overshoot and sinusoidal ring out to steady state. Typical of the 2nd Order ODE transfer function.

Assuming no system losses, the mechanics of the system would typically be modeled as:

T(w) = I_drum*(GR)*d(w)/dt

I suppose I have to solve the these two ODE's simultaneously to find w(t)?

D
Losing the Thread

Quarantined World

Joined
27 Oct 04
Moves
87415
Clock
23 Mar 16
Vote Up
Vote Down

Originally posted by joe shmo
I'm looking for some layman explanation on how transfer functions work within the framework of classical mechanics. If it could be related specifically to the example below that would be even better.

A system consists of a gasoline engine (prime mover) with a Torque vs Speed curve: T(w), and Gearbox and a Drum (load) that is to be rotated.

The engine ...[text shortened]... _drum*(GR)*d(w)/dt

I suppose I have to solve the these two ODE's simultaneously to find w(t)?
The only transfer functions I remember from my physics days are from statistical physics, and are used to calculate partition functions for things like the Ising model. Can you explain what you mean by a transfer function - saying what's being transferred will help!

R
Standard memberRemoved

Joined
10 Dec 06
Moves
8528
Clock
23 Mar 16
Vote Up
Vote Down

Originally posted by DeepThought
The only transfer functions I remember from my physics days are from statistical physics, and are used to calculate partition functions for things like the Ising model. Can you explain what you mean by a transfer function - saying what's being transferred will help!
I'm not very familiar with the concept myself. As far as I know a transfer function is defines how a system output frequency responds to a input frequency. Here is a paper explaining it.

http://www.cds.caltech.edu/~murray/books/AM05/pdf/am06-xferfcns_16Sep06.pdf

In this case Power is being transferred from the rotating motor shaft to the initially static drum.

The second order linear DE for a step input is described in this paper.

https://www.facstaff.bucknell.edu/mastascu/eControlHTML/SysDyn/SysDyn2.html#TheSystem

I'm afraid that is about the best I can do for you. Hope it helps?

K

Germany

Joined
27 Oct 08
Moves
3118
Clock
23 Mar 16
Vote Up
Vote Down

It's a fairly long time ago since I used transfer functions, but as I recall they can be used to solve ODE's through Laplace transforms. That's generally not the only way to solve ODE's though. What problem specifically are you trying to solve?

R
Standard memberRemoved

Joined
10 Dec 06
Moves
8528
Clock
23 Mar 16
1 edit
Vote Up
Vote Down

Originally posted by KazetNagorra
It's a fairly long time ago since I used transfer functions, but as I recall they can be used to solve ODE's through Laplace transforms. That's generally not the only way to solve ODE's though. What problem specifically are you trying to solve?
In general I just want to learn a bit about them. My interest arose when I was trying to analyze a problem similar to the example I posted in the OP.

The basic premise: An engine is idling along at a certain rpm under no load. The system is a Drum and a gearbox. At time "t" = 0 the load is coupled to the engine (the Drum is initially at rest). For a brief moment at the instant of coupling, the engine see's a high impulse torque from trying to instantaneously accelerate the load which causes it to change its rotational rate (if the inertial load is too much, the engine will stall). This impulse and the response of the motor speed that follows cannot be modeled strictly from mechanics...as far as I can tell. (Maybe it can and I'm missing it? This is again something I would like to discuss.)

K

Germany

Joined
27 Oct 08
Moves
3118
Clock
24 Mar 16
Vote Up
Vote Down

Originally posted by joe shmo
In general I just want to learn a bit about them. My interest arose when I was trying to analyze a problem similar to the example I posted in the OP.

The basic premise: An engine is idling along at a certain rpm under no load. The system is a Drum and a gearbox. At time "t" = 0 the load is coupled to the engine (the Drum is initially at rest). For a ...[text shortened]... I can tell. (Maybe it can and I'm missing it? This is again something I would like to discuss.)
How much do you know about solving differential equations and linear algebra? I wouldn't recommend starting to learn about transfer functions and Laplace transforms without knowing a bit more basic stuff. For instance, what is the general solution to this differential equation?

y" = y

sdp
troll taunter

wherever I am needed

Joined
13 Dec 12
Moves
40201
Clock
24 Mar 16
Vote Up
Vote Down

2nd differential of a function is equal to the function?

Clearly e^x is involved and, as 2 differentials (hence minus x minus) e^-x

So I would guess a general answer, A and B variables to be determined, would be

y = A*e^x +B*e^-x ?

With respect, feel that is a long way from the original problem

K

Germany

Joined
27 Oct 08
Moves
3118
Clock
24 Mar 16
Vote Up
Vote Down

Originally posted by st dominics preview
2nd differential of a function is equal to the function?

Clearly e^x is involved and, as 2 differentials (hence minus x minus) e^-x

So I would guess a general answer, A and B variables to be determined, would be

y = A*e^x +B*e^-x ?

With respect, feel that is a long way from the original problem
Very good (although it would be more common to denote A and B as boundary or initial conditions rather than variables)! It's totally unrelated to the problem but it is a problem one should be able to solve within about 10 seconds before one starts to think about using Laplace transforms to solve ODE's.

sdp
troll taunter

wherever I am needed

Joined
13 Dec 12
Moves
40201
Clock
24 Mar 16
Vote Up
Vote Down

Originally posted by KazetNagorra
Very good (although it would be more common to denote A and B as boundary or initial conditions rather than variables)! It's totally unrelated to the problem but it is a problem one should be able to solve within about 10 seconds before one starts to think about using Laplace transforms to solve ODE's.
fair enough

took me at least 2 minutes, so i am clearly totally unworthy!

R
Standard memberRemoved

Joined
10 Dec 06
Moves
8528
Clock
25 Mar 16
2 edits
Vote Up
Vote Down

Originally posted by KazetNagorra
How much do you know about solving differential equations and linear algebra? I wouldn't recommend starting to learn about transfer functions and Laplace transforms without knowing a bit more basic stuff. For instance, what is the general solution to this differential equation?

y" = y
How much do you know about solving differential equations and linear algebra?


On a scale of 1 to 10, 1 being completely inept and 10 being able to solve problems in these fields of study in my sleep I would say I'm about a 3. Which means I can muddle through the solution if I refresh from my coursework in the subjects.

The method that is easiest to apply:

As st domonick said: we are looking for a solution of the form y = e^(m*x)

y" - y = 0 ... homogeneous equation

m^2*e^(m*x) - e^(m*x) = 0

e^(m*x)*(m^2 - 1) = 0

m^2 - 1 = 0 ...Auxiliary Equation

Two real and distinct roots: m = =+/- 1

y(x) = C1*e^(x) + C2*e^(-x)

Now to satisfy the initial conditions and solve for the coefficients C1 and C2:

y(0) = C1 + C2 ...condition 1

y'(x) = C1*e^(x) - C2*e^(-x)

y'(0) = C1 - C2 condition 2

From condition 1 and 2:

C1 = 1/2*( y(0) + y'(0) ) and C2 = 1/2*( y(0) - y'(0) )

y = 1/2*( y(0) + y'(0) )*e^(x) + 1/2*( y(0) - y'(0) )*e^(-x)

Alternative: Laplace Transform

L{ y" } = Y(s)
L{ y" } = s^2*Y(s) - s*y(0) - y'(0)

L{ y" } = L{ y }

s^2*Y(s) - s*y(0) - y'(0) = Y(s)

Y(s) = ( s*y(0) + y'(0) )/(s^2 - 1)

Y(s) = s*y(0)/( ( s - 1 )*( s + 1 ) ) + y'(0)/( ( s - 1 )*( s + 1 ) )

Decompose denominator into partial fractions

Y(s) = 1/2*y(0)*( 1/(s-1) + 1/(s+1) ) + 1/2*y'(0)*1/(s-1) - 1/2*y'(0)*1/(s+1) )

Y(s) = 1/2*( y(0) - y'(0) )*1/( s+1 ) + 1/2*( y(0) + y'(0) )*1/(s -1)

L{Y(s)}^(-1) = 1/2*( y(0) - y'(0) )* L{1/( s+1 )}^(-1) + 1/2*( y(0) + y'(0) )* L{1/( s - 1 )}^(-1)

y(x) = 1/2* ( y(0) - y'(0) )*e^(-x) + 1/2*( y(0) + y'(0) )*e^(x)

EDIT: Found error and corrected it.

K

Germany

Joined
27 Oct 08
Moves
3118
Clock
25 Mar 16
Vote Up
Vote Down

Great! A transfer function is nothing but solving an ODE using a Laplace transform but typically one does not write it explicitly. Engineers love this method because they can take their computer, compute the transfer function, and ta-daa, the equation is solved. It then allows you to mix and match different inputs and outputs by simply multiplying the transfer functions.

R
Standard memberRemoved

Joined
10 Dec 06
Moves
8528
Clock
25 Mar 16
Vote Up
Vote Down

Originally posted by KazetNagorra
Great! A transfer function is nothing but solving an ODE using a Laplace transform but typically one does not write it explicitly. Engineers love this method because they can take their computer, compute the transfer function, and ta-daa, the equation is solved. It then allows you to mix and match different inputs and outputs by simply multiplying the transfer functions.
After doing a bit more reading it must not be "Transfer Functions" that I am actually concerned with. I thought it was because the motor - rotating drum system "behaves" like a 2nd order linear transfer function, but they are unrelated to solving the initial problem I posed.

So...how is the problem I posed tackled in classical mechanics? The response of the motors rotational dynamics when it is coupled to the drum is clearly "higher order" than the mechanics I have available to describe it.

What I mean; if I were trying to determine the speed of the motor as a function of time "w(t)" from the instant of coupling to steady state value from observation it would be of the form:

d²/dt²( w(t) ) + A*d/dt ( w(t) ) + B*w(t) = C*u(t)

However, the mechanics I have to describe the relationship are confined to a first order ODE in motor rpm " w(t)"

T( w(t) ) = I_drum* d/dt ( w(t) )

So, do I have to use "higher order mechanics" to solve this problem?

K

Germany

Joined
27 Oct 08
Moves
3118
Clock
25 Mar 16
Vote Up
Vote Down

Originally posted by joe shmo
After doing a bit more reading it must not be "Transfer Functions" that I am actually concerned with. I thought it was because the motor - rotating drum system "behaves" like a 2nd order linear transfer function, but they are unrelated to solving the initial problem I posed.

So...how is the problem I posed tackled in classical mechanics? The response o ...[text shortened]... = I_drum* d/dt ( w(t) )

So, do I have to use "higher order mechanics" to solve this problem?
I'm not confident in saying whether your set of differential equations describes the problem or not, but typically one resorts to numerical techniques to solve differential equations of order 3 or higher - unless there is again some trick here involving the multiplication of transfer functions, again, I'm not really sure.

Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More.