Question

Knowing that the signal $x(t)$ has the Fourier transform $W = 65$ $X(\omega) = \begin{cases} 1 - \frac{|\omega|}{W}, & |\omega| < W \\ 0, & \text{elsewhere} \end{cases}$ determine the total energy of the signal, the bandwidth that includes 90% of this energy, the dc component of the signal in time domain $x(0)$, the integral $\int_{-\infty}^{\infty} x(t)dt$ as well as the energy of the signal $y(t) = \frac{d^2x(t)}{dt^2}$ (10p).

          Knowing that the signal $x(t)$ has the Fourier transform
$W = 65$
$X(\omega) = \begin{cases} 1 - \frac{|\omega|}{W}, & |\omega| < W \\ 0, & \text{elsewhere} \end{cases}$
determine the total energy of the signal, the bandwidth that includes 90% of this energy, the dc component of
the signal in time domain $x(0)$, the integral $\int_{-\infty}^{\infty} x(t)dt$ as well as the energy of the signal $y(t) = \frac{d^2x(t)}{dt^2}$ (10p).

Knowing that the signal x(t) has the Fourier transform
W = 65
X(ω) = 1 - (|ω|)/(W), |ω| < W
0, elsewhere
determine the total energy of the signal, the bandwidth that includes 90% of this energy, the dc component of
the signal in time domain x(0), the integral ∫-∞^∞ x(t)dt as well as the energy of the signal y(t) = (d^2x(t))/(dt^2) (10p).

Added by Jose Luis S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Stylianos Gregoriou

Step 1

Since the Fourier transform of the signal is given as W=65 elsewhere, we can square it to get the squared magnitude. |W|^2 = 65^2 The total energy is then given by integrating this squared magnitude over all frequencies: Total Energy = ∫ |W|^2 df Since the Show more…

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Knowing that the signal (t)has the Fourier transform W=65 elsewhere determine the total energy of the signal, the bandwidth that includes 90% of this energy, the de component of