# Multi-loop systems block reduction techniques

Do I start by multiplying 18/s+3 * 1/s+3 = 18/(s+3)^(2)?

Then is it (18/(s+3)^(2))/(1+(18/(s+3)^(2)*4)) = 18 / ((s + 3)^2 * (72 / (s + 3)^2 + 1))?

• You can reformat in latex to make it easier to read. For instance 18/(s+3)^(2) becomes $\dfrac{18}{(s+3)^2}$ by using this: $\dfrac{18}{(s+3)^2}$ <-- you ought to show your steps too because of the poor/difficult format of your final formula and, the lack of steps make it tricky to follow what you did. Commented Apr 8 at 11:08

The most left part in the 2nd line is still correct - however, the most right part is wrong. After multiplying both - numerator and denominator - with (s+3)² the whole expression reduces to

18/[(s+3)² + 72].

But note that this expression is the transfer function of the small inner loop only.

• Do I then multiply that by the Gain of 50? Commented Apr 8 at 13:55
• Is $\dfrac{900}{s^2+6s+81}$ the correct answer? Commented Apr 8 at 14:06
• @ NonComposMentis No - I dont think so (the last numer (81) is not correct.)
– LvW
Commented Apr 8 at 14:48
• @NonComposMentis No. It's $\frac{900}{s^2\,+\,6s\,+\,981}$: ratsimp(solve(Eq(((vin-vout)*50-4*vout)*18/(s+3)*1/(s+3),vout),vout)[0]/vin)` Commented Apr 8 at 14:49
• @ periblepsis Instead of 972 I arrive at (972+9).....ahh I see, you have corrected the constant term.
– LvW
Commented Apr 8 at 14:51