- #1
jackonelli
- 2
- 0
Hi there.
I've been struggling with this problem for days now (4 days, no joke) and I feel like I have a mental block and really cannot get any further.
I have a system that's described by
[tex] f(t) = g''(t) + 15g'(t) + 1600g(t) [/tex] Where the input is [tex]g(t)[/tex]
The problem is to, with this information, determine the coefficients in another system, where the input is[tex]f(t)[/tex]and the output is given by [tex]u(t)[/tex], so that [tex]u(t) = g(t)[/tex]
This other system is given by
[tex]c_2u''(t)+c_1u'(t)+c_0u(t) = b_2f''(t) + b_1f'(t) + b_0f(t)[/tex]
I think this is supposed to be simple and I think I make it more difficult in my head than it is. I first substituted f(t) in the second differential equation with the left hand side of the first equation:
[tex]c_2u''(t)+c_1u'(t)+c_0u(t) = b_2f''(t) + b_1f'(t) + b_0(g''(t) + 15g'(t) + 1600g(t))[/tex]
[tex]=> c_2u''(t)+c_1u'(t)+c_0u(t) = b_2f''(t) + b_1f'(t) + b_0g''(t) + b_015g'(t) + b_01600g(t)[/tex]I´m supposed to get numerical values for all coefficients but I really can’t figure out what coefficients makes u(x) = g(t). I don't really know how to proceed. I've tried a lot of other ways too, for example solve for u(t) and g(t) explicitly:[tex]u(t) = \frac{b_2}{c_0}f''(t) + \frac{b_1}{c_0}f'(t) + \frac{b_0}{c_0}f(t)-\frac{c_2}{c_0}u''(t)-\frac{c_1}{c_0}u'(t)[/tex] and [tex] g(t) = \frac{1}{1600}g''(t) + \frac{15}{1600}g'(t) -\frac{1}{1600}f(t)[/tex]
And then putting their right hand sides equal each other, but this didn't really get me anywhere.
Just before posting this question I’ve been staring and trying for another 4 hours, and I feel I’m that my self-confidence is at an all-time low and I'm starting to ask myself if I really should be doing math at all (Yes, it's really a first world issue, I know).
REALLY grateful for any help!
I've been struggling with this problem for days now (4 days, no joke) and I feel like I have a mental block and really cannot get any further.
I have a system that's described by
[tex] f(t) = g''(t) + 15g'(t) + 1600g(t) [/tex] Where the input is [tex]g(t)[/tex]
The problem is to, with this information, determine the coefficients in another system, where the input is[tex]f(t)[/tex]and the output is given by [tex]u(t)[/tex], so that [tex]u(t) = g(t)[/tex]
This other system is given by
[tex]c_2u''(t)+c_1u'(t)+c_0u(t) = b_2f''(t) + b_1f'(t) + b_0f(t)[/tex]
I think this is supposed to be simple and I think I make it more difficult in my head than it is. I first substituted f(t) in the second differential equation with the left hand side of the first equation:
[tex]c_2u''(t)+c_1u'(t)+c_0u(t) = b_2f''(t) + b_1f'(t) + b_0(g''(t) + 15g'(t) + 1600g(t))[/tex]
[tex]=> c_2u''(t)+c_1u'(t)+c_0u(t) = b_2f''(t) + b_1f'(t) + b_0g''(t) + b_015g'(t) + b_01600g(t)[/tex]I´m supposed to get numerical values for all coefficients but I really can’t figure out what coefficients makes u(x) = g(t). I don't really know how to proceed. I've tried a lot of other ways too, for example solve for u(t) and g(t) explicitly:[tex]u(t) = \frac{b_2}{c_0}f''(t) + \frac{b_1}{c_0}f'(t) + \frac{b_0}{c_0}f(t)-\frac{c_2}{c_0}u''(t)-\frac{c_1}{c_0}u'(t)[/tex] and [tex] g(t) = \frac{1}{1600}g''(t) + \frac{15}{1600}g'(t) -\frac{1}{1600}f(t)[/tex]
And then putting their right hand sides equal each other, but this didn't really get me anywhere.
Just before posting this question I’ve been staring and trying for another 4 hours, and I feel I’m that my self-confidence is at an all-time low and I'm starting to ask myself if I really should be doing math at all (Yes, it's really a first world issue, I know).
REALLY grateful for any help!
This was only a sub-problem, so maybe I'll be back later...