# Calculating "simple" transfer function help

Discussion in 'Homework Help' started by AzureX, Dec 12, 2009.

1. ### AzureX Thread Starter New Member

Dec 12, 2009
5
0
Hi guys

Can anyone help me calculating the transferfunction for this lowpass filter? I've been told that it should be of fifth order.

Any help with really be appreciated. I think it should be done with the node technique. If you figure it out, it would be really nice if you explain shortly the few steps you've taken.

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2. ### beenthere Retired Moderator

Apr 20, 2004
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Why are the components drawn as rectangles, instead of the recognized schematic symbols? What are the components? What are the units "pC" and "uL"?

3. ### AzureX Thread Starter New Member

Dec 12, 2009
5
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Sorry the drawing may be a bit confusing.

Its because they are all seen as laplace components hence resistors. pF is pico farad and uL is micro Henry. So components with a L next to them is inductors and C means capacitors.

Regards Mikael

4. ### AzureX Thread Starter New Member

Dec 12, 2009
5
0
Is noone able to solve this task?

I have a big assignment for next friday, so it is quite urgent.

Regards Mikael

5. ### steveb Senior Member

Jul 3, 2008
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469
It's difficult to answer without knowing what theoretical tools you have been taught and are expected to use in your class.

The usual approach to solve this is to do the problem in the Laplace complex frequency domain. This is done by replacing all capacitors (you have three of them) with equivalent capacitive reactance 1/(sC), and replacing all inductors (you have two of them) with equivalent inductive reactance sL. Note that s is the complex frequency parameter used for Laplace Transforms. If you haven't been taught Laplace Transforms yet, then this is not going to make sense to you right now.

The problem is now reduced to a simple "resistive" type of network and various techniques can be used to solve for Vo/Vi, which is the transfer function. The transfer function is in terms of "s" the complex frequency. If you want to find the magnitude and phase response in terms of real frequency, you substitute in s=jw, where j is the square root of -1, and w is the angular frequency. Note that w=2*pi*f, where f is the usual cyclic frequency in Hertz. The new transfer function, in terms of frequency "f", can be separated into magnitude and phase response by considering that T=Vo/Vi will be a complex number R+j*I where R is the real part and I is the imaginary part.

Magnitude response then becomes sqrt(R^2+I^2) and phase response becomes arctan(I/R)

6. ### AzureX Thread Starter New Member

Dec 12, 2009
5
0

Im fully aware of how laplace works, and how to make bodeplots, impulse response etc.
The only problem is in fact to solve the transfer function. I usually use the node technique.

If anyone is able to do that i would deeply appreciate your help.

Regards Mikael

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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Why don't you use the node technique for this problem?

8. ### AzureX Thread Starter New Member

Dec 12, 2009
5
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As i said, i am aware that i should use the nodal technique. Im just not able to solve the task with this technique because i lack intelligence for it im afraid

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
In post #5, you said:

I took that to mean that you have done it in the past, so I naturally wondered why you don't do it again.

Your circuit is a ladder network with 2 inductors in the series path, and 3 capacitors in shunt.

For a solution, lets designate the inductors as L1 and L2, left to right. Designate the capacitors C1, C2 and C3, left to right.

Number the 3 nodes along the top edge as V1, V2 and V3, left to right. The bottom edge is the reference node (ground).

So, V1 is the same as Vi, and V3 is the same as Vo.

Using Kirchoff's current law (KCL), you can set up 3 equations, 1 for each of the nodes.

The way you do that is to sum expressions for the current through each component connected to a particular node; the sum is set equal to zero.

For example, at node 1, there are only two components connected there, L1 and C1.

The current through L1 is equal to the voltage across L1 (which is V1-V2), divided by the impedance of L1: (V1 - V2)/(s*L1). The current through C1 is (V1 - 0)/(1/(s*C1)). So, the first equation is:

(V1 - V2)/(s*L1) + (V1 - 0)/(1/(s*C1)) = 0

The second node has 3 components connected to it, so there will be 3 currents to sum to zero. The second equation will be:

(V2 - V1)/(s*L1) + (V2 - 0)/(1/(s*C1)) + (V2 - V3)/(s*L2) = 0

It is conventional to take the currents leaving a node to be positive, so to get the signs of the currents right in the expressions for the currents at a node, the numerator of each expression should have the voltage at that node as positive. For example at node 1, the numerator for the voltage across L1 should be (V1-V2), but when calculating the current in L1 when you are summing at node 2, the numerator should be (V2-V1).

I'll leave it up to you to set up the equation at node 3. Then you will have 3 equations in 3 unknowns.

You don't have a load shown in your circuit, but if you add one you will just have another term in your third equation.

Now, you said you want the transfer function. A network like this can have more than one transfer function, but I suspect that you want what is called the voltage transfer ratio, Vo/Vi.

You can get that from your 3 equations. Give it a try and report back if you have a problem.