Modeling a Control System using Transfer Functions

[yaro99]

Homework Statement

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Homework Equations

Listed under 2.1 in the image above.
This is the only relevant equation that I'm aware of, but I'm almost sure that there is something else I need to know before I can solve the problem.

The Attempt at a Solution

I tried solving for the preamplifier first, then I realized I was just running around in circles and not actually solving anything.In order to get V_e(s), I need to first find the output of the first potentiometer and subtract it from that of the second one.

output for first potentiometer: K_pot*θ_i(s)
output for second potentiometer: K_pot*θ_a(s)

V_e(s) = K_pot(θ_i(s)-θ_a(s))

Now, V_p(s) = K*V_e(s) = K*K_pot(θ_i(s)-θ_a(s))

Transfer function of preamplifier: [K*K_pot(θ_i(s)-θ_a(s))]/[K_pot(θ_i(s)-θ_a(s))] = K

Clearly the transfer function is not the same thing as the function for the physical block, so I must be doing something wrong here. I just don't see how I can fill out the table using just the equation I was given.

 

[donpacino]

it looks like you have a fundamental misunderstanding of what transfer function you are solving for.

K is typically a fixed value that you choose (using a technique like root locus for example)

When you are told you need to sole for a transfer function, typically you solve for the output over the input (azimuth angle over desired azimuth angle). so your transfer function for the preamp would simply be K

there are many ways to do that. two simple ways are the mason gain formula and simple block diagram algebra.
https://en.wikipedia.org/wiki/Mason's_gain_formula you can follow this if you want to do masons gain

block diagram algebra for a problem like this is simple:

V_e=V_i-theta_o*K_pot_2 ...as you already mentioned

simply multiply that by the forward chain until you get to your output. then manipulate the equation so you have your output over your input

give it a shot

 

[yaro99]

simply multiply that by the forward chain until you get to your output. then manipulate the equation so you have your output over your input

I don't understand this part, forgive my ignorance. Why am I multiplying, and what is the forward chain? We are talking about just the preamp here right?

 

[LvW]

I must admit that I do not completely understand the problem:
1.) All the transfer functions of the variuous blocks are given in the Block diagram
2.) Where is Block 6 (integrator)?
3.) As far as I understand, you are required (according to 2.1) to derive the "open-loop transfer function".. That is the transfer function between input and output WITHOUT any feedback. Hence, it is simply the product of the shown 5 Blocks in the forward path.

So - what is the problem?

 

[donpacino]

I don't understand this part, forgive my ignorance. Why am I multiplying, and what is the forward chain? We are talking about just the preamp here right?

just multiply the gains of the the blocks.

but as LvW pointed out, I may have a misunderstanding as well. I assumed you need to take the closed loop transfer function, but you might need to take the open loop transfer function. with open loop it becomes much easier

 

[yaro99]

but as LvW pointed out, I may have a misunderstanding as well. I assumed you need to take the closed loop transfer function, but you might need to take the open loop transfer function. with open loop it becomes much easier

How would I do this for each particular block? I am supposed to fill out the table for each one.
Also, in the table, what would be the difference between the Block and the Transfer Function?

 

[donpacino]

How would I do this for each particular block? I am supposed to fill out the table for each one.
Also, in the table, what would be the difference between the Block and the Transfer Function?

how much experience do you have with block diagrams?

looking at V_e and V_p, by definition V_p=V_e*preamp which is V_p=V_e*K

the gain of the block is the transfer function of the block by definition

 

[yaro99]

how much experience do you have with block diagrams?

Not much. My professor gave us one (bad) lecture on them and then gave us this HW. We are supposed to look up transfer functions on our own but my book was unhelpful.

the gain of the block is the transfer function of the block by definition

That's what confused me. If that's true then I'm wondering why the table asks for both.

 

[donpacino]

Not much. My professor gave us one (bad) lecture on them and then gave us this HW. We are supposed to look up transfer functions on our own but my book was unhelpful.
That's what confused me. If that's true then I'm wondering why the table asks for both.

so for one line of the table

physical black=preamp
tranfer function=k
input=v_e
output=v_p

i would interpret block as the name, not the transfer function