# Derive an expression for the closed loop transfer function

Derive an expression for the closed loop transfer function:

h(s) = Y(s)

----

R(s) for a gain k = 10.

Work i have done so far:

Ea(s) = R(s) - B(s) = R(s) - H(s)Y(s)

Because the output is related to the actuating signal by G(s), we have

Y(s) = G(s) Ea(s)

Therefore:

Y(s) = G(s) [R(s) - H(s)Y(s)]

Now, solving for Y(s), we obtain:

Y(s) [ 1 + G(s)H(s) ] = G(s)R(s)

Therefore the transfer function relating the output Y(s) to the input R(s) is:

Y(s) = G(s)

---- -------------------------------

R(s) 1 + G(s) H(s)

Now, as the gain is equal to 10, I'm not sure if i have to just substitute that value into the last equation or not?

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#### Solution Summary

The solution derives an expression for the closed loop transfer function (see attachment).