My prof gave a question

\( x= \delta(t-1) + 5Cos(2t) \ \ \ \longrightarrow \ \ \ \ \ \fbox{H( \omega)= \frac {1}{j\omega +1}} \ \ \ \longrightarrow \ \ \ y=find \ using \ super \ position\)

Now for the solution , he treats each of the input functions separately with the boxed function.

he writes

\( x_1 = \delta(t-1) \longrightarrow \fbox{h(t)=e^{-t} u(t)} \longrightarrow h(t-1)=e^{-(t-1)}u(t-1) = y_1\)

Then,

\( x_2 = 5Cos(2t) \longrightarrow \fbox {H(\omega)=\frac{1}{j\omega+1}} \longrightarrow 5\ \mid H(\omega)\mid \ Cos(2t + \theta) \)

\( where \mid H(\omega) \mid=( \frac{1}{(2)^2+1})^{\frac{1}{2}}and \theta =- tan^{-1} ( \frac{2}{1})=-63^o \)

So using super sposition , we get

\( y(t)=\ e^{-(t-1)} u(t-1)\ \ \ + \ \ \ \frac{5}{\sqrt{5}} Cos(2t-63^o) \)

Now my question is that,

1 - Why did he used time domain for the impulse function and frequency domain for cos(2t).

2 - How did he get expression for \(y_2\) , the one with magnitude of H(w).

thnx