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> Open-loop Transfer Function
cooded
Posted: January 13, 2011 08:56 am
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The question which i had was why is the open loop transfer function K*g(s)*h(s) when it should have been just K*g(s) since its an open loop and theres is no feedback involved in an open loop.

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Mod Edit: Thread has been split. Refer to slide 14 of: PDF

This post has been edited by cooded on January 14, 2011 12:10 pm
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PIC
Posted: January 13, 2011 10:08 am
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user posted image


Closed-loop transfer function G_closed(s)
Relationships from looking at the diagram:
Y(s) = G(s) * U(s)
U(s) = K * E(s)
E(s) = R(s) - F(s)
F(s) = H(s) * Y(s)

Substitute everything into the first equation:
Y(s) = G(s) * (K * (R(s) - [H(s) * Y(s)]))

Factoring Y(s) terms onto LHS:
Y(s) * [1 + G(s) * K * H(s)] = G(s) * K * R(s)

Giving,
G_closed(s) = Y(s) / R(s) = K * G(s) / [1 + K * G(s) * H(s)]


Loop gain transfer function G_loop(s)
Break the feedback path at the end of H(s) so it's disconnected from the summation node:
F(s) = H(s) * G(s) * K * R(s)

Giving,
G_loop(s) = F(s) / R(s) = K * G(s) * H(s)


Open-loop transfer function G_open(s)
Break the feedback path, this time solving for Y(s) / R(s):
Y(s) = G(s) * U(s)
U(s) = K * R(s)

Substitute into Y(s):
Y(s) = G(s) * [K * R(s)]

Giving,
G_open(s) = Y(s) / R(s) = K * G(s)
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cooded
Posted: January 13, 2011 06:45 pm
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ok...let me tell you guys what i have understood.Basically the open loop transfer function of a closed loop system is obtained as F(s)/R(s) and not by c(s)/r(s).please correct me if i am wrong.
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Sch3mat1c
Posted: January 13, 2011 07:02 pm
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The overall transfer function in the above diagram is Y(s) / R(s). When the output is given as C(s), then it is C(s) / R(s).

This is kind of unusual, because X is usually used for the input, when Y is the output.

Tim


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Answering questions is a tricky subject to practice. Not due to the difficulty of formulating or locating answers, but due to the human inability of asking the right questions; a skill that, were one to possess, would put them in the "answering" category.
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cooded
Posted: January 13, 2011 07:45 pm
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Hi,

I am still very confused. If we break the feedback loop then we wont Y(s)/X(S) will be be just K*G(s). That should be the open loop transfer function.Here the feedback function H(s)should not come into the picture.Also
F(s)/R(s)= K * G(s) * H(s)
is not the transfer function because transfer function is always output/input. F(s) is not the output but the feedback path which is broken. I know i am sounding so dumb...but please do help me out.The PDF is in different langauge so i could not unbderstand a thing. A diagram to support the argument would be highly appreciated. Please please help me out.

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PIC
Posted: January 13, 2011 09:37 pm
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I'll be referring to the input as R(s) (rather than X(s)).

QUOTE (cooded)
If we break the feedback loop then we wont Y(s)/X(S) will be be just K*G(s). That should be the open loop transfer function.Here the feedback function H(s)should not come into the picture.

Below shows the system in closed-loop, i.e. when there is a connection from the output back to the input.

user posted image

The input-output TF is Y(s) / R(s). As we are calculating the TF while there is feedback, the expression Y(s) / R(s) is called the closed-loop transfer function.

When we break the the feedback loop as shown below then we can calculate two further transfer functions: G_open(s) and G_loop(s).

user posted image

G_open(s): We can repeat what we did above, i.e. work out the input-output TF Y(s) / R(s). However this time the system is in open-loop, so Y(s) / R(s) will give us the open-loop transfer function.
G_loop(s): We can treat F(s) as the output (instead of Y(s)), and calculate the TF F(s) / R(s). This is called the loop-gain transfer function.


Calculating G_open(s):
We said G_open(s) = Y(s) / R(s), under the condition that the system is in open-loop. The equivalent diagram is shown below.

user posted image

By observation:
Y(s) = G(s) * U(s)
U(s) = K * E(s)

Now, remembering that the system is open-loop, i.e. F(s) doesn't feed in, we deduce E(s):
E(s) = R(s)

Substituting it all into the first equation:
Y(s) = G(s) * [K * R(s)]

Giving the open-loop TF as:
G_open(s) = Y(s) / R(s) = K * G(s)

Note that when calclulating the open-loop TF, it doesn't matter where you break the loop; you can break it before or after H(s).


Calculating G_loop(s):
The loop-gain is defined as F(s) / R(s), under the condition that F(s) does not feed back into the system. So here it does matter where you break the loop - it has to be after H(s). We can use the above diagram again.

We said G_loop(S) = F(s) / R(s).

F(s) = H(s) * Y(s)
Y(s) = G(s) * U(s)
U(s) = K * E(s)
E(s) = R(s)

Substituting everything into the first equation, gives
F(s) = H(s) * G(s) * K * R(s)

Giving the loop-gain TF as
G_loop(s) = F(s) / R(s) = K * G(s) * H(s)


QUOTE (cooded)
F(s)/R(s)= K * G(s) * H(s)
is not the transfer function because transfer function is always output/input.

I may not be using the terminology correctly. I refer to the loop-gain as the "loop-gain transfer function". To be clear I have indicated whether I am referring to the input-output TF or the loop-gain TF.
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cooded
Posted: January 13, 2011 10:29 pm
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Hello PIC,

Ok now i understood what an open loop transfer function of a closed loop means. Thanks a lot for all the diagrams and explanation in layman terms. I have few more questions to ask, i hope this is the right thread.

1. If G(s) is the gain of a system, then why do we have a gain K defined again?
eg: G(s)= K/(s+2) where K is the gain. I always thought G(s) is the gain.
2. Correct me if i am wrong. Characteristic equation for an open loop transfer function is G(s)*H(s) and for closed loop transfer function is 1+- G(s)*H(s).
3. Can you please explain or give links to tutorials for Bode plots? I just finished root locus and my brain is too tired to decode bode plots now. biggrin.gif but i have an exam comming up really fast...so i have to do bode plots tonight.It would be great if i could lay my hands on a simpler document.

Regards
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PIC
Posted: January 14, 2011 12:34 pm
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QUOTE (cooded)
1. If G(s) is the gain of a system, then why do we have a gain K defined again?
If K is constant, then it is said to give proportional gain. Seperating it from the other TF's might just be convention.

QUOTE
I always thought G(s) is the gain.
G(s) in this case is probably best thought of as the plant (the system you are trying to control).

In open-loop, the plant G(s) could be unstable and therefore 'unusable'. Putting it in a closed loop (i.e. having feedback) allows you to stabilize the system.

QUOTE
2. Correct me if i am wrong. Characteristic equation for an open loop transfer function is G(s)*H(s) and for closed loop transfer function is 1+- G(s)*H(s).
I haven't come across the term "characteristic equation" in control systems. Can you expand on what you mean?

QUOTE (cooded)
3. Can you please explain or give links to tutorials for Bode plots?
Say you have a TF function G(s) = k / (s + a). To analyse it in the frequency domain, you subsitute s with jw. Freqeuncy response G(jw) = k / (jw + a).

A Bode plot shows gain and phase against frequency on seperate graphs. For practical reasons the gain and frequency axes are not linear, but logarithmic.

To plot gain and phase on seperate graphs, you first need to seperate the freq response into it's gain and phase parts.

Gain part:
|G(jw)| = |k / (jw + a)| = k / sqrt(a^2 + w^2)

Phase part:
angle[G(jw)] = angle[K / (jw + a)] = -atan(w / a)

For the gain graph:
Plot the log of |G(jw)| against the log of w.
y-axis: 20log[|G(jw)|] = 20log[k / sqrt(a^2 + w^2)] = 20log[k] - 20log[sqrt(a^2 + w^2)]

For the phase graph:
Plot angle(G(jw)) as it is. Or you could convert it from radians to degrees, in which case you multiply angle(G(jw)) by 180/pi.
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cooded
Posted: January 15, 2011 08:22 pm
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Hi PIC,
. [QUOTE]Correct me if i am wrong. Characteristic equation for an open loop transfer function is G(s)*H(s) and for closed loop transfer function is 1+- G(s)*H(s).[/QUOTE]I haven't come across the term "characteristic equation" in control systems. Can you expand on what you mean?

[/QUOTE]

Well i came across the term while studying root locus. Basically for a closed loop with feedback the transfer function can be written as G(s)/(1+G(s)H(s)).The characteristic equation is 1+G(s)H(s).

Please can you elaborate on the difference between closed loop gain, open loop gain and only loop gain.

I am an amateur at this so please bear with my questions. Also what does the term "open loop poles of a closed loop transfer function" mean?
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katon
Posted: July 29, 2011 08:58 pm
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I have a question ,
if the input R(s) enters the summer with a minus (-) how will be the G_loop(s) ?

I need to writhe an extra (-) ??
because when you need to check root-locus or another theorem you use the equation G(s)*H(s) , my question is where is the minus if the R(s) enters the summer with minus.

Thanks i hope i was clearly
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DUET
Posted: June 22, 2013 02:37 pm
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QUOTE (PIC @ January 14, 2011 03:37 am)
I'll be referring to the input as R(s) (rather than X(s)).

QUOTE (cooded)
If we break the feedback loop then we wont Y(s)/X(S) will be be just K*G(s). That should be the open loop transfer function.Here the feedback function H(s)should not come into the picture.

Below shows the system in closed-loop, i.e. when there is a connection from the output back to the input.

user posted image

The input-output TF is Y(s) / R(s). As we are calculating the TF while there is feedback, the expression Y(s) / R(s) is called the closed-loop transfer function.

When we break the the feedback loop as shown below then we can calculate two further transfer functions: G_open(s) and G_loop(s).

user posted image

G_open(s): We can repeat what we did above, i.e. work out the input-output TF Y(s) / R(s). However this time the system is in open-loop, so Y(s) / R(s) will give us the open-loop transfer function.
G_loop(s): We can treat F(s) as the output (instead of Y(s)), and calculate the TF F(s) / R(s). This is called the loop-gain transfer function.


Calculating G_open(s):
We said G_open(s) = Y(s) / R(s), under the condition that the system is in open-loop. The equivalent diagram is shown below.

user posted image

By observation:
Y(s) = G(s) * U(s)
U(s) = K * E(s)

Now, remembering that the system is open-loop, i.e. F(s) doesn't feed in, we deduce E(s):
E(s) = R(s)

Substituting it all into the first equation:
Y(s) = G(s) * [K * R(s)]

Giving the open-loop TF as:
G_open(s) = Y(s) / R(s) = K * G(s)

Note that when calclulating the open-loop TF, it doesn't matter where you break the loop; you can break it before or after H(s).


Calculating G_loop(s):
The loop-gain is defined as F(s) / R(s), under the condition that F(s) does not feed back into the system. So here it does matter where you break the loop - it has to be after H(s). We can use the above diagram again.

We said G_loop(S) = F(s) / R(s).

F(s) = H(s) * Y(s)
Y(s) = G(s) * U(s)
U(s) = K * E(s)
E(s) = R(s)

Substituting everything into the first equation, gives
F(s) = H(s) * G(s) * K * R(s)

Giving the loop-gain TF as
G_loop(s) = F(s) / R(s) = K * G(s) * H(s)


QUOTE (cooded)
F(s)/R(s)= K * G(s) * H(s)
is not the transfer function because transfer function is always output/input.

I may not be using the terminology correctly. I refer to the loop-gain as the "loop-gain transfer function". To be clear I have indicated whether I am referring to the input-output TF or the loop-gain TF.

QUOTE
G_loop(s): We can treat F(s) as the output (instead of Y(s)), and calculate the TF F(s) / R(s). This is called the loop-gain transfer function.
Hello! I don't understand how we can treat F(s) as the output instead of Y(s).


user posted image

I think if we don't use the feed back in the above system. then we can consider the system as open loop. and the transfer function of the system(without feedback) can be called as open loop transfer function. I think the open loop transfer function of the system would K*G(s).

Could you asses my thinking?
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