# Calculate (10)/(1+F(s)G(s)) (Y(s))/(R(s))=(F(s)G(s))/(1+F(s)G(s))

We have a system with output given by
$\frac{Y\left(s\right)}{R\left(s\right)}=\frac{F\left(s\right)G\left(s\right)}{1+F\left(s\right)G\left(s\right)}$
where $F\left(s\right)G\left(s\right)=K\frac{s+1}{{s}^{2}+s+1}$
Let K=4 and R(s)=10/s. Using the final value theorem, calculate the steady state error of the system.
Now, I thought the error would be given by $\underset{s\to 0}{lim}\frac{10\cdot F\left(s\right)G\left(s\right)}{1+F\left(s\right)G\left(s\right)}$, but the solution says to calculate $\frac{10}{1+F\left(s\right)G\left(s\right)}$. Can anyone explain why?
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Sauppypefpg
We have to find the steady state error
$\underset{t\to \mathrm{\infty }}{lim}\left(r\left(t\right)-y\left(t\right)\right)$
According to the final value theorem
$\underset{t\to \mathrm{\infty }}{lim}\left(r\left(t\right)-y\left(t\right)\right)=\underset{s\to 0}{lim}\left[s\left(R\left(s\right)-Y\left(s\right)\right)\right]$
The Laplace transform of the error is given by
$R\left(s\right)-Y\left(s\right)=R\left(s\right)-\frac{F\left(s\right)G\left(s\right)}{1+F\left(s\right)G\left(s\right)}R\left(s\right)=\frac{1}{1+F\left(s\right)G\left(s\right)}R\left(s\right)$
Hence, for a step reference with Laplace transform $R\left(s\right)=10/s$ the steady state error value can be derived by
$\underset{t\to \mathrm{\infty }}{lim}\left(r\left(t\right)-y\left(t\right)\right)=\underset{s\to 0}{lim}\left[s\frac{1}{1+F\left(s\right)G\left(s\right)}\frac{10}{s}\right]=\underset{s\to 0}{lim}\left[\frac{10}{1+F\left(s\right)G\left(s\right)}\right]$