Consider the LTI system: $$x\left( t \right) \to LTI\, system \to y\left( t \right) $$
The graph of the $x(t) $ signal is the following:
Remark: The graph when $t>0$ is a quarter of a circle.
Q1: Find $$ \int_{-\infty }^{\infty} X(\omega )d\omega $$ Q2: Find the energy spectral density of the output signal of the system, if the unit impulse response is $$ h(t)=e^{7-t} u\left( t-7 \right) $$ and the new input signal is $x_{2}(t)=x(t)[u(t+7)-u(t)] $ . The expression $u(t)$ denotes the unit step function.
Q3: Find the energy of this last output signal.
Remark: In Q1, $X(w) $ denotes either Fourier transform or Laplace tranform, I am not sure. I think it most probably is Fourier transform.
Edit: Thanks to the help of @Stelios , I was able to solve Q1 (I think). The inverse Fourier transform is given by: $$ f(x)=\int _{-\infty }^{\infty }{\hat {f}}(\xi )\ e^{2\pi ix\xi }\,d\xi $$
However, in this case, we should use the alternative version of the inverse Fourier transform: $$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i \omega t} d\omega $$ $$ \Rightarrow x( 0 )= \frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{0} d\omega $$
But, by looking at the graph, we know that $x(0)=16 $ . $$ \Rightarrow \int_{-\infty }^{\infty} X(w)dw =2\pi \, x(0)=32\pi $$
And that would be the answer to Q1.
Now, for Q2, I think I can find the output signal $y(t) $ by using the convolution integral.
$$ y(t)=(h*x_2)(t) =(x_2*h)(t) =\int_{-\infty}^{\infty} x(\tau )[u(\tau +7)-u(\tau ) ]
e^{7-t+\tau } u(t-\tau -7) d\tau $$
But the function $u_2(\tau ):= u(\tau +7)-u(\tau ) $ is equal to $1$ only when $-7< \tau <0 $ . Therefore we will have: $$ y(t)= \int_{-7}^{0} x(\tau ) ) e^{7-t+\tau } u(t-\tau -7) d\tau $$
Then, for the case when $7<t $, we will have: $$ y(t)= \int_{-7}^{0} x(\tau ) e^{7-t+\tau } d\tau $$
And for the case when $0<t<7 $, we will have: $$ y(t)= \int_{-7}^{t-7} x(\tau ) e^{7-t+\tau } d\tau $$
And when $t<0 $, we will have $y(t)=0 $. Also, $x(\tau )=16 $ in the interval $[-7,0]$, then:
$$ y(t)= \begin{cases} 0 \quad & ; t<0 \\ 16 \left( 1-e^{-t} \right) & ; 0<t<7 \\ 16\, e^{-t} \left( e^7 -1 \right) & ; 7<t \\ \end{cases} $$
To find the energy: $$ E_y =\int_{-\infty}^{\infty} \mid y(t) \mid ^2 dt = \int_{0}^{\infty} \mid y(t) \mid ^2 dt $$
which should not be a complicated integral. Now, for the energy spectral density: $$ S_{yy}= \mid \hat{y} (f) \mid ^2 $$
where $\hat{y} (f) $ denotes the Fourier transform of $y(t)$ , and $f$ denotes the frequency. I then get: $$ \hat{y}(f)= \int_0^{7} 16(1-e^{-t} )e^{2\pi ift} dt + \int_{7}^{\infty} 16(e^7 -1)e^{-2\pi ift} dt $$
Are all the steps above correct?