Question

1. Consider a signal x(t) defined over the interval t =(-1,1]as shown below. X(t) ЛИ (a) Is x(t) an energy signal? Find its energy. (b) Find the Fourier transform X (jo)of x(t).

Answer 1

(a) A finite duration signal is always an energy signal and a periodic signal is always a power signal. This can be verified by seeing the formula for energy and power. The power of a signal, P is, P= lim 1 *(t)dt + T JO The energy of a signal, E is, E = * x*(t)dt -00 Since, the signal is non periodic, the time period, T, does not exist. Thus, its power will appear as on, which also signifies that it's not a power signal. Let's calculate the energy of the signal. The given function, x(t) can be written as, - -1<t<0 x(t) = 0<t<1 otherwise The energy, E, is given by, E=L,(-1)º dt +f*(t)? dt Thus, the energy of the signal, x(t) is - Joules. al (b) As per the definition, the Fourier transform is given by, X(jo) = x(t)e jet dt Substitute the function, x(t).

x(jw) = L^, (-1)e-jødt +L• (t)e=jordt The top - e - - n 1 [: j?=-1] H-*(-jeve * -050)+(-29) Solve it further as, X(jo)= ={ -ül-jorelo telo) + 3 (sexe* + e*)} - ha az (el® +e+®) + 3% (0-3° -elo) enote-jo -= COS O 2 ejo 2 e jo -= sino Solve it further as, X (jo) = + 1,sine OL 0 0 Therefore, the Fourier transform of signal, x(t) is 21 cos o - -- + +sino I tried to answer in as detail as possible. Hope, it helps you.