# Feedback system

1. Sep 2, 2006

### LM741

hey guys.
attached is a negative feedback system. This is used to control the system and maintain stability. But something doesn't make sense to me.
If you look at the diagram, when the output is y(t)=4sint (as required) then the feedback signal becomes f(t)=2sint, but that means that the signal put back into the system will be zero (because i take my feedback signal and subtract it from my original signal) that means i will get zero output every now and then...doesn't seem very practical...can someone please shed some light ...
thanks

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2. Sep 2, 2006

### Dirty D

Was this a problem you had? can you post the rest of it and not just the answer you came up with? Otherwise, yea, what you got here is a failure to communicate. signal goes in, nothing comes out.

3. Sep 2, 2006

### LM741

no its just something that bothers me about a negative feedback system - it just doesn't make sense because when the output is what we want (4sint) then like i said - then we'll get nothing coming out - just doesn't seem right... say out output rises to 6sint fro example, then the negative feedback system is mean to counteract that - but dont' see how it does that coz then all that happens is our output becmoes (si -sf = 2sint - 3sint = -sint) -sint....
hope you can understand my dillema...

4. Sep 2, 2006

### SGT

Your reasoning is wrong. The amplifier gain (2) acts on the error signal se and not on the input signal.
y = 2*se = 2*(si - sf)
so
sf = 0.5*y = 0.5*2*(si - sf) = si - sf
so
2*sf = si
or sf = si/2
se = si/2
y = si

The gain of your feedback amplifier is 1

5. Sep 2, 2006

### Corneo

You could have figured this out by know that L = ab = 2*.5 = 1.

6. Sep 3, 2006

### LM741

corneo>>you missing the point

7. Sep 3, 2006

### leright

y=2*se
sf = 0.5*y
y=2*si - 2*sf
y=2*si - y
y=si

gain = 1

8. Sep 3, 2006

### leright

Also, note that you can generalize your your amplifier gain by substituting your gain values for general gain values to yield gain = K1/(1 + K1K2), where K1 is the gain of the block on the top, and K2 is the gain of the block on the bottom.

9. Sep 3, 2006

### Gokul43201

Staff Emeritus
For completeness, I'm going to derive the general result of leright's post#8 using the generalization of the calculation done by SGT in post#4 and leright in post#7.

Terminology and definitions:
Vi = input voltage
Vo = output voltage
Vf = feedback voltage
Ve = error voltage = Vi - Vf (for negative feedback)
A = open loop gain or open loop transfer function = Vo/Ve (= Vo/Vi, when Vf=0)
B = feedback ratio = Vf/Vo
L = closed loop gain or overall transfer function = Vo/Vi

Proof:
Vo = A*Ve {defn} = A*(Vi - Vf) {defn} = A*(Vi - B*Vo) {defn} = A*Vi - A*B*Vo
=> Vo*(1+AB) = AVi
=> L = Vo/Vi = A/(1+AB)

~~~~~~~~~~~~~~~~~~~~~

To address the question in the OP, the input function Vi is not 2sint; that is the error function, Ve.

10. Sep 14, 2006

### LM741

sup guys

apparently my frame of thought was wrong. I asked the lecturer. YOu can't look at the system like that. The signal is sort of instantaneous..or something like that..

just to let you guys know...
thanks for any feedback though.
john