# Math Help - differential equation problem - help please!

1. ## differential equation problem - help please!

I'm stuck on this problem and am unsure how to tackle it.

It is known that an industrial process system is accurately modelled by the standard first order differential equation:

Tx' + x = Ky

i) Given the following results of experimental measurement, deduce the parameters T and K,

When an input Y(t) of 5 units was applied, it was found that the system output x(t) ultimately settled with a value of 20 units.

When a sinusoidal input was applied, it was found that, at an input frequency of 10 radians, the output lagged behind the input by exactly -45 degrees.

ii) If y(t) was a step change drive, at what time would x(t) reach half of its final steady state value (using the values of K and T already deduced)?

I've got as far as:

Tx' + x = K5 when x=20

=> 20 = K5

K = 4

Can anyone show me where to go from here?

I've found a passage in my notes which says "Another useful response of the first order system is the one we observe when the input is a sinusoid. As we saw earlier, all we need do is replace the derivative with jw (I'm using 'w' as omega as I can't find how to change it on my PC) as follows:

Tjwx + x = Ky

How does it work using this formula?

2. Originally Posted by boab
I'm stuck on this problem and am unsure how to tackle it.

It is known that an industrial process system is accurately modelled by the standard first order differential equation:

Tx' + x = Ky

i) Given the following results of experimental measurement, deduce the parameters T and K,

When an input Y(t) of 5 units was applied, it was found that the system output x(t) ultimately settled with a value of 20 units.

When a sinusoidal input was applied, it was found that, at an input frequency of 10 radians, the output lagged behind the input by exactly -45 degrees.

ii) If y(t) was a step change drive, at what time would x(t) reach half of its final steady state value (using the values of K and T already deduced)?

I've got as far as:

Tx' + x = K5 when x=20

=> 20 = K5

K = 4

Can anyone show me where to go from here?

I've found a passage in my notes which says "Another useful response of the first order system is the one we observe when the input is a sinusoid. As we saw earlier, all we need do is replace the derivative with jw (I'm using 'w' as omega as I can't find how to change it on my PC) as follows:

Tjwx + x = Ky

How does it work using this formula?
First lets get the general solution to this equation:

Tx'+x=K y(t)

where y(t) is the forcing function.

Divide through by T and identify the integrating factor:

$x'+x/T=(K/T) y(t)$

$\frac{d}{dt} x(t) e^{t/T}=(K/T) y(t)e^{t/T}$

so the general solution is:

$x(t) e^{t/T}=(K/T) \int_0^t y(\tau) e^{\tau/T} ~d\tau+C$

Take it from there.

RonL

3. Capt. Black, thanks for taking the time to help. I'm obviously missing something as this hasn't really cleared anything up. I'm not familiar with the use of tau.

If anyone else can shed any light it would be appreciated