Calculating Closed Loop Gain in a Noisy System

AI Thread Summary
Calculating the closed loop gain in a noisy system involves incorporating noise into the gain formula. The initial equations provided suggest a method to express the output voltage in terms of input voltage and noise. The discussion emphasizes the need for clarity in defining whether the gain is closed loop or not, as well as specifying the input-output relationship. A proposed adjustment to the gain calculation introduces a modified variable to account for noise, leading to a new expression for the closed loop gain. The conversation highlights the complexity of accurately determining gain in the presence of noise.
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Hi,

Is it possible to calculate the gain of a closed loop system with Noise? Here's a diagram describing what i mean:

http://img206.imageshack.us/img206/8955/cltfmh8.png

I've got the following so far:

Vout = \frac{A1A2Vin}{1 + A1A2B} + \frac{AVn}{1 + A1A2B}

But the question is asking for the closed loop gain. Would that just be:

Vout / Vin = \frac{A1A2}{1 + A1A2B} + \frac{AVn}{Vin(1 + A1A2B)}


Thanks.
 
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name said:
Is it possible to calculate the gain of a closed loop system with Noise? Here's a diagram describing what i mean:

http://img206.imageshack.us/img206/8955/cltfmh8.png

I've got the following so far:

V_{out} = \frac{A_1 A_2}{1 + A_1 A_2 B} V_{in} + \frac{A_2}{1 + A_1 A_2 B} V_n

But the question is asking for the closed loop gain.

if the question is well posed, it needs to still specify the gain (closed loop or not) from some specified input to a specified output. your first equation (i prettied it up a little) actually shows two different transfer functions.
 
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What if i do this:

A' = A2 (A1 + Vn)

and then...

Vout / Vin = \frac{A'}{1 + A'B}

Is this logically correct?
 
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