First Order System's?

[ozi-rock]

Ok my question is about first order system's. The equation is a0y + a1 (dy/dt) = b0x which can be written as y + τDy = (b0/a0)x where τ is the time constant and = (a1/a0) and D = (d/dt).

 

This means that we can rewrite the equation as y(1 + τD) = mx.

 

What I want to know is if I have a time constant of 50 what does the D do, will it change my 50 to a zero?

 

I'm very confused because if that is what happens then unless your time constant had an x in it you' always just have y(1) = mx.

 

Can anyone clear this up for me?

[lawcat]

I think, when you talk about first order systems, you are talking about oscillatory systems (harmonics). All oscillatory systems have steady state and transient state (or transient response). Your d/dt is pertinent to transient response. But, if you are taking a1/ao as constant, that still does not affect the possibility that d/dt will affect y.

 

y(1 + a1/a0*D) = mx; y + Dy = mx; (notice that a1/a0 bears no relationship to d/dt; while a1/a0 does not change, y changes with espect to t))

 

y represents the previous position, Dy represents how previous y affects new y. If change in y with respect to time is constant, then Dy, the slope, is zero (previous y no longer affects new y); position only depends on m constant This happens in steady state. However, in transient state, Dy affects y. Notice that Dy diminishes y in time, towards the steady state; which means that transient state always precedes the steady state. If it does not, then you have exponentially growing y due to Dy, and the system is unstable.

 

You can also look at it this way:

 

Your tau is the period, and the inverse of it is the frequency. The general solution will contain [y = e^(-1/tau)*t] plus some constant of the function mx. You can see that in time, exponential part--the transient part--will go to zero due to negative sign in the power of e; and it only depends on the period and time--not on x. Then, you will have left the [mx] part, the steady state part. See, http://en.wikipedia.org/wiki/Linear_differential_equation (see bottom of the page).

[ozi-rock]

Ok I get that I think but I taught I'd be able to answer a question knowing that but I'm useless, did I mention I hate this subject :shrug:

 

The question is:

 

An unmanned submarine is equipped with temperature and depth measuring instruments and has radio equipment which can transmit the output readings of these instruments to the surface. The submarine is initially floating on the surface of the sea with the instrument output readings in steady state. The depth measuring instrument is approximately zero order and the temperature transducer first order with a time constant of 50 seconds. The water temperature on the sea surface T0 is 20°C and the temperature Tx at a depth of x metres is given by the relation

 

Tx = T0 –0.01x

 

a) If the submarine starts diving at time zero, and thereafter goes down at a velocity of 0.5m/s construct a table showing the temperature and depth measurements reported at intervals of 100 seconds over the first 500 seconds of travel. Show also in the table the error in each temperature reading.

 

 

Ok so the system has an input, two processes and an output. We're given an equation to find the actual temperature so we know that and it is the input.

 

It goes through the dept measuring instrument and remains the same on the other side i.e. 1 goes in 1 goes out.

 

It then proceeds to the first order system which involves the 50 seconds and I still don't know what happens here. How does this change the input?

 

Am I right with the first bit?

[lawcat]

I believe that the specified time constant 50, represents the delay. In general, first, you have to construct the table with: (1) 5 x 100s intervals represented horizontally (100s, 200s, 300s, 400s, 500s); these will represent measurements of time; and (2) five columns representing Position, Temperature, Measured Position, Measure Temperature, and error (difference over real temperature);

 

I may be wrong on my differential equations, and if anyone wnat s to chip in they are welcomed.

 

First, your depth measurement at intervals 100, 200, 300 ,400, and 500s will be accurate and equal to actual depth.

 

Second, your Measured Temperature will not be equal to Actual Temperature. Actual temperature is given by Ta = To - 0.01x; Measure Temperature is given by first order diff: Ta - 50dT/dx = Tmeasured.

dT/dx can be obtained from Ta --> d(To - 0.01x)/dx = -0.01

Therefore, yourmeasure temperature is

 

Tmeasured = Ta - 50 d(Ta)/dx ---> Tmeasured = (To - 0.01x) - (50*(-0.01));

Tmeasured = T0 - 0.01x + 0.5

 

Depth, and its reading, will be as follows:

 

100s -> 0.5m/s x 100s = 50m = x1

200s -> 0.5m/s x 200s = 100m =x2

300s -> 150m =x3

400s -> 200m = x4

500s -> 250m = x5

 

Actual Temperatire, Ta = To - 0.01x:

 

100s = 20 - 0.01 x 50m = 19.5

200s - > 19

300s -> 18.5

400s -> 18

500s -> 17.5

 

Tmeasured = To - 0.01x + 0.5:

 

100s -> 20 - 0.01x 50m + 0.5 = 20

200s -> 19.5

300s -> 19

400s -> 18.5

500s -> 18

 

Error:

at 100s -> 20 -19.5/ 19.5 -> 2.6%; etc.

 

___

**EDIT**

If your instructor is anal, you should include m in all depth readings, as some constant may influence the depth readings as compared to actual depth. Thus Xreading = m*Xactual. Thus, at 100s, Xraeading = (m*100s*0.5m/s); or a constant, asi in Xreading = 100s*0.5m/s + C.

[ozi-rock]

Your close, but not there yet. The answer given for a depth of 50 is 19.716.

 

I was thinking it might be something like

 

Tmeasured = To - 0.01(x) + (-0.01(x))/(1 + 0.5)

 

But that didnt work, maybe I've done something simple that you can see?

 

Thanks for all your help

[lawcat]

Is the next one 19.216, and then 18.716 and then 18.216, 17.716?

[ozi-rock]

You are very close, the answers go: 19.716, 19.245, 18.749, 18.25, 17.75.

 

I still maintain this lecturer pull numbers out of no where

[lawcat]

Why not post the full problem?

[ozi-rock]

Well I posted the entire question, it was just the answers I did'nt post and I posted them in the last message, thanks again for all your help.