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Is this a linear system?

  • Thread starter magnifik
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  • #1
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y(t) = x(t) + [tex]\int[/tex] (t - [tex]\tau[/tex])x([tex]\tau[/tex])d[tex]\tau[/tex]

for it to be linear, T[kx(t)] = kT[x(t)] so i have
T[kx(t)] = kx(t) + [tex]\int[/tex] (t - [tex]\tau[/tex])x([tex]\tau[/tex])d[tex]\tau[/tex]
and
kT[x(t)] = k[x(t) + [tex]\int[/tex] (t - [tex]\tau[/tex])x([tex]\tau[/tex])d[tex]\tau[/tex]] = kx(t) + k[tex]\int[/tex] (t - [tex]\tau[/tex])x([tex]\tau[/tex])d[tex]\tau[/tex]
so they aren't equal and aren't linear. however, i'm not sure about this answer because for the first part, T[kx(t)], i'm not sure if x([tex]\tau[/tex]) should also be multiplied by k, making it a linear system

any help would be appreciated. thx.
 

Answers and Replies

  • #2
352
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You should, in fact, be multiplying [tex]x(\tau)[/tex] by [tex]k[/tex] inside the integral -- simple substitution is how you see this. If [tex](Tx)(t) = x(t) + \int (t - \tau) x(\tau) \,d\tau[/tex] and [tex]x_1(t) = kx(t)[/tex] then [tex](Tx_1)(t) = x_1(t) + \int (t - \tau) x_1(\tau) \,d\tau = kx(t) + \int (t - \tau) kx(\tau) \,d\tau[/tex].
 

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