System with a 2nd input

This was how i tried to formulate the answer but ive been told its incorrect.

$$R(t)=u(t)$$ $$D(t)=u(t)=1/s$$ $$K=10$$ $$H=1$$

$$G=\frac{K}{s(s+1)(s+2)} =\frac{K}{s^3+3s^2+2s} $$

If someone could help me understand how i tackle this system that would be appreciated.

  • 1
    \$\begingroup\$ Well first of all, it’s uncertain what operation you’re doing at the summing blocks (i.e. additional or subtraction, it doesn’t show). Secondly, you don’t multiply equations at a summing block like you did with r(t) and d(t). Thirdly, you did not take into account that there is a feedback loop. \$\endgroup\$
    – user103380
    May 31, 2020 at 14:39
  • \$\begingroup\$ I believe we assume its addition. hmm... so if i were to add 1/s and then form the open loop function i should get the answer? \$\endgroup\$
    – dorudrakky
    May 31, 2020 at 14:59
  • \$\begingroup\$ Yeah sorry but im not that good at this, im not sure how to do all that \$\endgroup\$
    – dorudrakky
    May 31, 2020 at 15:18

1 Answer 1


Considering your block diagram with the signs showed as in the figure below:

Block diagram - Dirceu

The principle of superposition can be applied in this linear system in order to investigate the effects that two inputs (regulation \$r(t)\$ and disturbance \$d(t)\$) on output. In this way, we have the two transfer functions in \$s\$ domain:

Doing \$d(t) = 0\$: $$ G_1(s) = \frac{Y(s)}{R(s)}$$ and doing \$r(t)=0\$: $$ G_2(s) = \frac{Y(s)}{D(s)} $$:

Then, using the standard feedback formula: $$ G_1(s) = \frac{K}{s^2+3s+2+K} $$ With \$K=10\$: $$ G_1(s) = \frac{10}{s^2+3s+12} $$

For \$G_2(s)\$ determination, note the negative sign on input of block \$K\$: $$ Y(s) = -\frac{K}{(s+1)(s+2)}Y(s) + D(s) $$ Or: $$ G_2(s) = \frac{s^2+3s+2}{s^2+3s+2+K} $$

With \$K=10\$: $$ G_2(s) = \frac{s^2+3s+2}{s^2+3s+12} $$ The expression for output is: $$ Y(s) = G_1(s)R(s) + G_2(s)D(s) $$

Finally, replace both \$R(s)\$ and \$D(s)\$ by \$1/s\$ (the Laplace transform of unit step input \$u(t)\$).

$$Y(s) = \frac{1}{s}$$ An simulation graph: Simu_Dirceu

  • 1
    \$\begingroup\$ Hey, all i got to say is wow. Im thankful for everyones help and this answer is just so well thought out. The time you took out of your day to help me is invaluable. Looking at this im starting to understand how you came to that solutions and the methods required for it. Again really thank you and i hope you have a good day. \$\endgroup\$
    – dorudrakky
    May 31, 2020 at 16:00
  • \$\begingroup\$ just to confirm my final answer came out as 1/s will this be what is used to form a step response graph? If so the graph shows x=y so how would i analyse it? \$\endgroup\$
    – dorudrakky
    May 31, 2020 at 17:05
  • \$\begingroup\$ \$Y(s) = 1/s\$. The time graph is that one (inverse Laplace transform for \$1/s\$). \$\endgroup\$ May 31, 2020 at 17:36
  • \$\begingroup\$ so should my step response graph come out as a straight line with a gradient of zero? \$\endgroup\$
    – dorudrakky
    May 31, 2020 at 18:17
  • \$\begingroup\$ See the above simulation. \$\endgroup\$ May 31, 2020 at 19:39

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