# Thread: Differential Equation system Problem

1. ## Differential Equation system Problem

Can anybody help?

A system is modelled by Tx' + x = Ky

given the following results deduce the parameters T and K

. When the input y(t) of 5 units was applied the output x(t) ultimately settled with a value of 20 units

. when a sinusoidal input was applied at an input frequency of 10 rad/s 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 ( T & K deduced previously)

2. Originally Posted by al2308
Can anybody help?

A system is modelled by Tx' + x = Ky

given the following results deduce the parameters T and K

. When the input y(t) of 5 units was applied the output x(t) ultimately settled with a value of 20 units

. when a sinusoidal input was applied at an input frequency of 10 rad/s 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 ( T & K deduced previously)
$T \frac{dx}{dt} + x = 5K$ has the solution $x = 20$.

$T \frac{dx}{dt} + x = K \sin (10t)$ has the solution $x = A \sin \left( 10\left( t - \frac{\pi}{4} \right) \right)$.

Use the above to get the value of T and K.

3. I have no idea on where to even start this one. Has anybody got any pointers? I assume you end up with a simultaneous equation of some form. But i'm not sure how to get there.

4. Originally Posted by al2308
I have no idea on where to even start this one. Has anybody got any pointers? I assume you end up with a simultaneous equation of some form. But i'm not sure how to get there.
You substitute the solutions given into the DE's and solve for what you don't know:

Sub x = 20 into the first:

20 = 5K => K = 4.

Sub K = 4 into the second:

$T \frac{dx}{dt} + x = 4 \sin (10t)$

Now sub $x = A \sin \left( 10\left( t - \frac{\pi}{4} \right) \right) =$(expand using compound angle formula) into the DE. Equate coefficients of sin and cos to get two simultaneous equations in A and T. Solve those equations to get the value of A and T.

5. My knowledge of the compound angle formula is a bit rusty. So i've cheated and used my calculator. This has returned a very long expansion.

A(sqrt2/2 sin(t) - sqrt2/2 Cos(t)) x {512(sqrt2/2 cos(t) + sqrt2/2 sin(t))^9 - 1024(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^7 + 672(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^5 - 160(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^3 + 10(sqrt2/2 Cos(t) + sqrt2/2 sin(t))}

this is not the smallest expresion to put into a DE. is my calculator right? i suspect it is! as i've tested it out on some smaller equations, or is there an easier method of solving the original problem.

6. Originally Posted by al2308
My knowledge of the compound angle formula is a bit rusty. So i've cheated and used my calculator. This has returned a very long expansion.

A(sqrt2/2 sin(t) - sqrt2/2 Cos(t)) x {512(sqrt2/2 cos(t) + sqrt2/2 sin(t))^9 - 1024(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^7 + 672(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^5 - 160(sqrt2/2 Cos(t) + sqrt2/2 sin(t))^3 + 10(sqrt2/2 Cos(t) + sqrt2/2 sin(t))}

this is not the smallest expresion to put into a DE. is my calculator right? i suspect it is! as i've tested it out on some smaller equations, or is there an easier method of solving the original problem.
Actually I was anticipating you'd do something like the following:

$\sin \left( 10\left( t - \frac{\pi}{4} \right) \right)$

$= \sin \left( 10 t - \frac{10 \pi}{4} \right)$

$= \sin \left( 10 t - \frac{5 \pi}{2} \right)$

$= \sin \left( 10 t - \frac{\pi}{2} \right)$

$= - \cos (10 t)$.

You could also use symmetry and the complementary angle formula to get the last line.

Substituting a solution where the argument of the trig function is 10t is pretty essential since the term on the right hand side is a a trig with an argument of 10t. Hard to justify equating coefficients of sin and cos on each side if they are sining and cosing different arguments ......

7. I wasn't far off then!!!

ok so X=-Acos(10t)

X'=10Asin(10t) (hopefully)

putting these into the DE

T(10ASin(10t) + Acos(10t) = 4Sin(10t)

I'm not confident this is correct or where i'm headed as trig tends to do stange things!

8. Originally Posted by al2308
I wasn't far off then!!!

ok so X=-Acos(10t)

X'=10Asin(10t) (hopefully)

putting these into the DE

T(10ASin(10t) + Acos(10t) = 4Sin(10t)

I'm not confident this is correct or where i'm headed as trig tends to do stange things!
Well, there's a problem now because you require:

10AT = 4 AND A = 0 .......

I'm not sure how you'll resolve this - perhaps get clarification on the original question .....?