Transfer Function Matrices

Discussion in 'Homework Help' started by SteveP72, Nov 9, 2010.

1. SteveP72 Thread Starter New Member

Nov 9, 2010
5
0
I have got to the following stage of finding a transfer function. Can any one help me to complete the next step?

Ei = 3 4 = Eo
Ii 1.25 2 Io

Where Ei & Ii are inputs and Eo & Io are outputs, with the matrix being the transfer function.

Tanks SteveP72

2. SteveP72 Thread Starter New Member

Nov 9, 2010
5
0
Hi I posted this morning, but I don't think I was as clear as I could have been.

I have been working on the transfer function for a simple electrical circuit with three resistors. So far I have correctly worked out what the transfer matrix is (this is confirmed using Matlab). I'm a little stuck as to what I need to do with the matrix.

(Ei) (1 2) = (Eo)
(Ii ) (3 4) = (Io)

Where Ei & Ii = Inputs, Eo & Io = Outputs, and [1 2; 3 4] = the transfer function matrix. The figures in the matrix are not the actual ones I'm working with.

3. The Electrician AAC Fanatic!

Oct 9, 2007
2,347
347
You should always post a schematic when you ask for help involving a circuit.

You say you need to determine the transfer function. There is more than one transfer function for a network, such as the voltage transfer ratio (sometimes called voltage gain), current transfer ratio (AKA current gain), transfer admittance, etc. Which one are you trying to find?

4. SteveP72 Thread Starter New Member

Nov 9, 2010
5
0
Hi

Thanks for the reply. Actually I'm a mechanical engineer and though the question given is electrical. It was given to us because it can be used in mechanical situations such as silencers and pumping situations.

I'll need to look at my notes to get the diagram for you, that will be tommorow now.

5. Georacer Moderator

Nov 25, 2009
5,151
1,266
Do you want to find all of the transfer functions? That is the dependances of every output variable from every input variable?

In general the problem is solved as follows:
$\left[ \begin{array}{c} y1 \\ y2 \end{array} \right]= \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{2} \end{array} \right] \cdot \left[ \begin{array}{c} x1 \\ x2 \end{array} \right] \\
\Leftrightarrow \left{ \begin{array}{l} y1=a_{11} \cdot x1 + a_{12} \cdot x2 \\ y2=a_{21} \cdot x1 + a_{22} \cdot x2 \end{array} \right\
$

As this is a two input, two output network, we get two transfer function with two variables each.

Nov 25, 2009
5,151
1,266

7. hgmjr Moderator

Jan 28, 2005
9,030
214
I think we are going to need to see the circuit?

hgmjr

8. hgmjr Moderator

Jan 28, 2005
9,030
214
I have merged both of the threads into one.

hgmjr

9. SteveP72 Thread Starter New Member

Nov 9, 2010
5
0
Sorry for not getting back sooner but I've been tied up.

Below is the complete task and as far as I have got with it. I know my answer so far is correct because as part of the task I have to check it using Matlab. This is enough to pass this part of the assignment, though I can get better marks if I can make sense of it, and thats what I'm tring to do. In other words for an input of say 6 and an output of say 10, what would the transfer function be.

ie 6 * A = 10 Would be 10 / 6 = A Therefore A would be 1.666 rec.

I've tried to cut and paste the circuit diagram to the end of this. It seems to have put somthing but its showing nothing. The circuit is very simple. Think of 2 parrallel lines, the top one having a resistor of a value of 4. Each side of this resistor the parrallel lines are connected with the first having a resistor of a value of 4, and the one down stream having a value of 2.

I've been told by a fellow student that all I need to do next is find the determinant of the inverse of the matrix to give me the transfer function I need.

Thanks for your help and replies.

Paragraph 1: Manual Calculations

Divide the diagram below into three blocks and derive the relationship between I1, E1 and I2, E2 in matrix form

From The Circuit We Get: Below

(E1) = ( 1 0) (1 R2) (1 0) (E2)
(I1 ) (1/R1 1) (0 1) (1/R3 1) (I2)

From Above We Get Matrices Below

(1 0) (1 4) (1 0)
(¼ 1) (0 1) (½ 1)

Multiply Above Matrices To Get The Final Matirx Below

(1 0) (1 4) = (1 4) Then (1 4) (1 0) = (3 4)
(¼ 1) (0 1) (¼ 2) (¼ 2) (½ 1) (1¼ 2)

Thus

(E1) = (3 4) (E2)
(I1 ) (2) (I2 )

Paragraph 2: Matlab Program

a=[1 0; 0.25 1]
b=[1 4; 0 1]
c=[1 0; 0.5 1]
d=a*b
e=d*c

Paragraph 3: Matlab Results

a =
1.0000 0
0.2500 1.0000
b =
1 4
0 1
c =
1.0000 0
0.5000 1.0000

d =
1.0000 4.0000
0.2500 2.0000

e =
3.0000 4.0000
1.2500 2.0000

Hopfuly you should see the circuit below

10. The Electrician AAC Fanatic!

Oct 9, 2007
2,347
347
I don't see the circuit.

When you post, scroll down to the "Additional options" window and use the "manage attachments" function.

11. SteveP72 Thread Starter New Member

Nov 9, 2010
5
0
Thanks for everyones input. My fellow student was wrong if I follow his advise I get a TF value of 1, which would be useless as a transfer function.

Having spent some time scouring my notes I have found a function that Georacer suggests, and using it I get 2 useful TF values. So it's task completed.

Its my first time using a forum site so forgive my if I didn't follow the rule correctly. I take your comments on board for nextime.

Regards SteveP72.