Control problem. Transfer function of an electrical system

[rowardHoark]

[URL]http://img28.mediafire.com/bacfa47633147eefb0c3433511d3f1415g.jpg[/URL]

1. Electrical system given. Find a transfer function. Correct answer UR(s)/U(s)=1/(s+2)

2. My attempt
Use Kirchhoff's voltage law u(t)-i(t)*R1-UR(t)=0; u(t)=i(t)R1+uR(t); apply Laplace Transform (L.T.) U(s)=I(s)R1+UR(s)

i(t)=i1(t)+i2(t)=1/L*\intuR(t) dt+uR(t)/R2; take a L.T. assuming zero initial conditions I(s)=1/(L*s)*UR(s)+UR(s)/R2=UR(s)[1/(L*s)+1/R2]; since L=R1=R2=1 I(s)=UR(s)*(1/s+1); UR(s)=I(s)/(1/s+1)

H(s)=HR(s)/U(s)=[I(s)/(1/s+1)]/[I(s)R1+UR(s)]=[I(s)/(1/s+1)]/[I(s)+I(s)/(1/s+1)]=[1/(1/s+1)]/[1+1/(1/s+1)]=s/(1+2s)

Where am I making a mistake?

 

[xcvxcvvc]

I'm not going to look through your nonlatex work but I'll post a set up that should be correct.

\frac{u_R(t)-u(t)}{R_1}+\frac{u_R(t)}{sL}+\frac{u_R(t)}{R_2}=0

 

[rowardHoark]

I'm not going to look through your nonlatex work but I'll post a set up that should be correct.

\frac{u_R(t)-u(t)}{R_1}+\frac{u_R(t)}{sL}+\frac{u_R(t)}{R_2}=0

I am assuming you are using the Kirchhoff's current law?

I agree with everything, except the frac \frac{u_R(t)}{sL}.Can you, please explain, how does this results in current? How can you have a time domain function in the numerator and a frequency variable in the denominator?