Question Number 212816 by Akayx last updated on 24/Oct/24
Answered by mahdipoor last updated on 24/Oct/24
![f(t)=Σ_(k=0) ^∞ 2(t−4k)(u_(4k) −u_(4k+2) )= 2Σ_(k=0) ^∞ [(t−4k)u_(4k) −(t−(4k+2))u_(4k+2) +2u_(4k+2) ]= L(f)= 2Σ_(k=0) ^∞ [(e^(−4ks) /s^2 )−(e^(−(4k+2)s) /s^2 )+((2e^(−(4k+2)s) )/s) ]= 2[(1/(s^2 (1−e^(−4s) )))−(e^(−2s) /(s^2 (1−e^(−4s) )))+((2e^(−2s) )/(s(1−e^(−4s) ))) ]= ((2(1−e^(−2s) +2se^(−2s) ))/(s^2 (1−e^(−4s) )))](https://www.tinkutara.com/question/Q212819.png)
$${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}\left({t}−\mathrm{4}{k}\right)\left({u}_{\mathrm{4}{k}} −{u}_{\mathrm{4}{k}+\mathrm{2}} \right)= \\ $$$$\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left({t}−\mathrm{4}{k}\right){u}_{\mathrm{4}{k}} −\left({t}−\left(\mathrm{4}{k}+\mathrm{2}\right)\right){u}_{\mathrm{4}{k}+\mathrm{2}} +\mathrm{2}{u}_{\mathrm{4}{k}+\mathrm{2}} \right]= \\ $$$${L}\left({f}\right)= \\ $$$$\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{{e}^{−\mathrm{4}{ks}} }{{s}^{\mathrm{2}} }−\frac{{e}^{−\left(\mathrm{4}{k}+\mathrm{2}\right){s}} }{{s}^{\mathrm{2}} }+\frac{\mathrm{2}{e}^{−\left(\mathrm{4}{k}+\mathrm{2}\right){s}} }{{s}}\:\right]= \\ $$$$\mathrm{2}\left[\frac{\mathrm{1}}{{s}^{\mathrm{2}} \left(\mathrm{1}−{e}^{−\mathrm{4}{s}} \right)}−\frac{{e}^{−\mathrm{2}{s}} }{{s}^{\mathrm{2}} \left(\mathrm{1}−{e}^{−\mathrm{4}{s}} \right)}+\frac{\mathrm{2}{e}^{−\mathrm{2}{s}} }{{s}\left(\mathrm{1}−{e}^{−\mathrm{4}{s}} \right)}\:\right]= \\ $$$$\frac{\mathrm{2}\left(\mathrm{1}−{e}^{−\mathrm{2}{s}} +\mathrm{2}{se}^{−\mathrm{2}{s}} \right)}{{s}^{\mathrm{2}} \left(\mathrm{1}−{e}^{−\mathrm{4}{s}} \right)} \\ $$