Question 1. Determine the Fourier transform of the following signals: (a) f(t)=\t-et- (c) h(t) = 2* [u(t) –u(t –3)] (d) i(t) = 2*13+j2)* u(t – 5) (b) ht)= sin(Tt):e 12xt tiết)

(20)(a)f(t)=|t| e^{-|t|}f(t)=t e^{-t} \cdot u(t)-t e^{t} u(-t)we know that e^{-a t} u(t) \longleftrightarrow \frac{1}{a+j w}e^{a t} u(-t) \longleftrightarrow \frac{1}{a-j \omega}\begin{array}{l}F(\omega)=j x(t) \longrightarrow j \frac{d}{d \omega} x(\omega) \\=j\left[\frac{-1}{(1+j \omega)^{2}} \cdot j-\frac{-1}{(1-j \omega)^{2}} \cdot-j\right] \\\end{array}\begin{array}{l}=j\left[\frac{-j}{(1+j \omega)^{2}}+\frac{-j}{(1-j \omega)^{2}}\right] \\ =\frac{1}{(1+j \omega)^{2}}+\frac{1}{(1-j \omega)^{2}} \\ =(1-j \omega)^{2}+(1+j \omega)^{2}=\frac{1-\omega^{2}-2 j \omega+1-\omega^{2}+2 j \omega}{\left.1+\omega^{2}\right)^{2}} \\ {[(1+j \omega)(1-j \omega)]^{2}} \\ F(\omega)=\frac{2\left(1-\omega^{2}\right)}{\left(1+\omega^{2}\right)^{2}}=\frac{2\left(1-\omega^{2}\right)}{\left(1+\omega^{2}\right)^{2}} \\ \text { (b) } \quad h(t)=\sin (\pi r) e^{-j 2} e_{\omega} \omega_{0} \cdot u(t) \\ h(t)=\sin (\omega t) e^{-j \omega_{0} t} u(t) \\ x(t) \longleftrightarrow x / \omega) \\ e^{j \omega_{0} t} x(t) \longleftrightarrow x\left(\omega-\omega_{0}\right) \\ x(t) \longleftrightarrow x(\omega)=\text { ? } \\ \begin{aligned} x(t)=\sin \pi t & \frac{e^{j \pi t}-e^{-j \pi t}}{2 j} \\ & =\frac{1}{2 j}\left[e^{j \pi t-j \pi t}-e^{-j}\right]\end{aligned} \\ \leftrightarrow \frac{1}{2 j}[2 \pi \delta(\omega-\pi)-2 \pi \delta(\omega+\pi)] \\\end{array}jiot\text { e } x(t) \longleftrightarrow x\left(\omega+\omega_{0}\right)intt e^{-j 2 \pi t} \longleftrightarrow \frac{\pi}{j}[\delta(\omega-\pi+2 \pi)-\delta(\omega+\pi+2 \pi)]sintt.\begin{array}{l}\leftrightarrow \frac{\pi}{j}[\delta(\omega+\pi)-\delta(\omega+3 \pi)] \\H(\omega)=\frac{\pi}{2}[\delta(\omega+\pi)-\delta(\omega+3 \pi)] \\H(f)=\frac{1}{2 j}[\delta(f+1 / 2)-\delta(f+3 / 2)] \\\rightarrow h(t)=e^{-2 t}[u(t)-u(t-3)] \\\begin{aligned}& e^{-2 t} u(1-3) \\& =e^{-2 t} u(t)-e^{-2(t-3+3)} u(t-3) \\h(t) & =e^{-2 t} u(t)-e^{(t-3)} e^{-6} u(t-3)\end{aligned} \\H(\omega)=?_{0} \\e^{2 t} u(\ell) \longleftrightarrow \frac{1}{2+j \omega} \\x(t) \longleftrightarrow x(w) \\x\left(t-t_{0}\right) \longleftrightarrow e^{-j \omega t_{0}} x(\omega) \\e^{-6} \cdot e^{-2(r-3)} \cdot u(t-3) \longleftrightarrow \frac{1}{2+j \omega} \cdot e^{-j \cdot \omega \cdot 3} \cdot e^{-6} \\\end{array}\begin{aligned}H(\omega) & =\frac{1}{2+j \omega}+e^{-6} \cdot \frac{1}{2+j \omega} \cdot e^{-3 j \omega} \\H(\omega) & =\frac{1}{2+j \omega}\left[1+e^{-3 j \omega-6}\right] \\\therefore H(\omega) & =\frac{1}{2+j \omega}\left[1+e^{-(6+3 j \omega)}\right]\end{aligned}(a)\begin{array}{l}x(t)=e^{-3 t} u(t-5)=e^{-3(t-\sqrt{t} 5)} \cdot u(t-5) \\=e^{-3(t-5)} \\x(\omega)=\frac{1}{3+j \omega} \cdot e^{-j \omega \cdot 5} \cdot e^{-15} \quad u(t-5) \cdot e^{-15} \\\therefore I(\omega)=e^{-15} \frac{1}{3+j(\omega+2)} \\\therefore J(\omega)=\frac{1}{3+j(\omega+2)} \\-15 \quad-5 j(\omega+2) \\I(\omega)=\frac{1}{3+} \cdot e^{-(15+5 j \omega+10 j)} \\\end{array} ...