- #1
magnifik
- 360
- 0
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.
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.