As drawn, it isn't true. yp does not equal y.

I think the real plant (top P(s) should also contains an additional a delay of tn. Then it might be useful. The bottom one is the plant without the delay.

thanks for reply

yes you are right,there is the delay equal tn in the top p(s).is the same idea of smith predictor ,so by this situation is there any theorem to prove yp=y?

best regard

If the model of the plant without delay P(s) at the bottom is identical to the real plant without delay, P(s) at the top, and the model of the delay (bottom) is identical to the actual delay in the plant (top), then y will be equal to the output of the plant without the delay of the plant. If your model of the plant is a bit wrong, or the delay is a bit wrong, you will get some error. Try drawing it with the delay in the top plant shown properly. Trace through the signals and you will find that y = y_p exp(+s tau_n), i.e. the output of the plant predicted ahead by time tau_n, assuming your model of the plant and delay are exact..

thanks @Paul you are right ,under assumption that the model of observer is equal the model of real plant as well as delays are equal yp=y ,but how can i prove that mathematically ,logically it seem equal but any assumption needs to prove ,i hope you get my point

Under your assumptions, yp does not equal y.

so when yp=y ?in which assumption ?

y is a output of the plant including dead-time and yp is a output without delay. So they are not equal. However this yp is used to design a controller for a plant with delay. Refer book "chemical process control", by Stephanopoulos for dead-time compensator or smith predictor. They also given a proof using block diagram reduction.

Hi, from your diagram you have: yp = Pu + y - Pu*exp(-tn*s). From the definition of y --> y = Pu.

yp = 2y - y*exp(-tn*s). yp --> y as tn --> 0. Hope this helps.

The earlier comment about Smith Predictors is helpful. I suggest you google it and find a wealth of information on the internet.

Regards