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Given a disturbance reduction system

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Create a system that will reduce $U(s)$ noise to $100$ times its value. Find the A and G gain value to satisfy the requirement

My attempt: I've analyzed the system and here's what I got

$$Y(s) = \frac{A\cdot H(s) + U(s)}{1+(A\cdot H(s) + U(s) )G(s)} X(s)$$

I don't know what to do next. Any clue? Thanks in advance

Update

Here's what I got after digesting Hilmar's comment: $$Y(s) = \frac{A\cdot H(s)}{1+A\cdot H(s)G(s)}X(s) + \frac{1}{1+A\cdot H(s)G(s)}U(s)$$

Since we want to reduce the noise without sacrificing the $X(s)$ signal $$\frac{A\cdot H(s)}{1+A\cdot H(s)G(s)}=1...(1)$$ $$\frac{1}{1+A\cdot H(s)G(s)}=\frac{1}{100}...(2)$$ Then $$1+A\cdot H(s)G(s) = 100 $$ Subtituting to the first equation $$A\cdot H(s)=100\\ A = \frac{100}{H(s)}$$ Then $$G(s) = 99$$ How can I get the A value?

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    $\begingroup$ Hint: $U(s)$ is not a transfer function, it's a signal. Once you are done this should look like $Y(s)= P(s)\cdot X(s) + Q(s) \cdot U(s)$ You have a term in there that's $U(s) \cdot X(s) and that doesn't make sense: signals don't multiply. $\endgroup$
    – Hilmar
    Jun 5 at 12:41
  • $\begingroup$ @Hilmar is it $Y(s) = \frac{AH(s)}{1+AG(s)H(s)}X(s) + \frac{1}{1 + AG(s)H(s)}U(s)$? $\endgroup$
    – love you
    Jun 5 at 16:10
  • $\begingroup$ You lost your dependency on s somewhere. I don't think that G(s) should be constant but the phrasing of the question is not very clear. I'm guessing you are supposed express G(s) as a function of H(s). $\endgroup$
    – Hilmar
    Jun 5 at 16:51
  • $\begingroup$ It's also possible that if $Y(s) = P(s) \cdot X(s)+Q(s) \cdot U(s)$ they simply want you to solve for $Q(s) = .01$ . That reduces the noise by a factor of 100 compared to the non-feedback case and doesn't care about what happens to $X(s)$. But as I said, the phrasing of the question is unclear (at least to me). $\endgroup$
    – Hilmar
    Jun 5 at 16:57
  • $\begingroup$ @Hilmar it wants us to reduce the noise by a factor of 100 without affecting the $X(s)$ signal. $\endgroup$
    – love you
    Jun 5 at 17:34
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The same system block is given below.

The system's block diagram.

I’ve assumed that the output of the block that has the transfer function $H(s)$ and the disturbance’s Laplace transform $U(s)$ are summed up just before the output $Y(s)$. And $E(s)$ is the steady-state error of the output. And, of course, the system is thought to be an LTI system.

So, now that these knowledge are given, we can start to analyse the system. The output $Y(s)$ can be formulated as;

$$Y(s) = C(s) + U(s)$$ $$Y(s) = E(s)A(s)H(s) + U(s)$$ $$Y(s) = [X(s)-Y(s)G(s)]A(s)H(s) + U(s)$$ $$Y(s) = X(s)A(s)H(s) + Y(s)G(s)A(s)H(s) + U(s)$$ $$Y(s) = {X(s)\frac{A(s)H(s)}{1 + G(s)A(s)H(s)}} + {U(s)\frac{1}{1 + G(s)A(s)H(s)}}$$ According to the comments, at the output stage, the input $X(s)$ is wanted to be neither attenuated nor amplified i.e. the transfer function for $X(s)$ is going to be $1$.

On the other hand, which is the main point of the question, $U(s)$ should be attenuated to $0.01$ times of its magnitude.

For accomplishing these expectations;

$$\frac{1}{1 + G(s)A(s)H(s)} = 0.01$$ $$1 + G(s)A(s)H(s) = 100$$ $$G(s)A(s)H(s) = 99$$

and;

$$\frac{A(s)H(s)}{1 + G(s)A(s)H(s)} = 1$$ $$A(s)H(s) = 1 + G(s)A(s)H(s)$$

By using the result of the penultimate calculation lines in the last equation;

$$A(s)H(s) = 100$$

and;

$$G(s) = 0.99$$

Consequently, in order to attenuate the disturbance to the desired level, the requirements are;

$$A(s) = \frac{100}{H(s)}$$

and;

$$G(s) = 0.99$$

Lastly, at this point, figuring out a precise value for $A(s)$ is ill-advised as it depends on the transfer function $H(s)$ and other features of the overall/closed-loop system e.g. amount of steady-state error.

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  • $\begingroup$ It the original question $A$ appears to be just a constant gain and not a function of $s$ $\endgroup$
    – Hilmar
    Jun 6 at 23:33
  • $\begingroup$ @Hilmar Thank you for your attention, but I've wanted to generalise the analysis. And I suppose doing so doesn't trigger something serious as in mathematics, we write down constant functions in this way e.g. $f(x) = 3$. $\endgroup$ Jun 7 at 8:08
  • $\begingroup$ @Hilmar Additionally, for the transfer function $A(s)$, being constant depends upon the transfer function $H(s)$. As long as $H(s)$ is a constant, so is $A(s)$. Otherwise, in the case of $H(s)$ is being a first, second, etc. order transfer function e.g. $H(s) = \frac{1}{s+2}$, according to the results, $A(s)$ should be another order of transfer function. $\endgroup$ Jun 7 at 9:17
  • $\begingroup$ I don't think that's the case. This looks like a pretty standard problem where you want the output to be $Y(s) = H(s) \cdot X(s)$ and use a feedback loop to remove additive noise. In this case $H(s)$ is the desired transfer function and $A$ and $G(s)$ create the feedback loop. Then A would indeed be a constant and H(s) and G(s) be functions. as written But I agree that the question is poorly worded, so multiple interpretations are possible here. $\endgroup$
    – Hilmar
    Jun 8 at 11:32

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