Q2) Determine whether the following signals are energy or power signals. Compute the energy or the power. a) x(t)=cos(t)u(t) b) x(t)=10cos(5t)cos(10t)

a) $x(t)=cos(t)u(t)$<br /><br />This signal is an energy signal because it is defined over an infinite time interval. The energy of the signal can be calculated as:<br /><br />$E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt$<br /><br />Since $x(t)$ is a cosine function multiplied by a unit step function, we can simplify the energy calculation by considering the properties of the cosine function and the unit step function separately.<br /><br />The energy of the cosine function is infinite because it is a periodic function with non-zero values over an infinite time interval. However, since the unit step function limits the signal to the positive half of the time axis, we can consider the energy of the cosine function over the positive half of the time axis.<br /><br />The energy of the cosine function over the positive half of the time axis is given by:<br /><br />$E_{cos} = \int_{0}^{\infty} |cos(t)|^2 dt$<br /><br />We can use the property of the cosine function that $|cos(t)| = cos(t)$ for $t \geq 0$. Therefore, we can simplify the energy calculation as:<br /><br />$E_{cos} = \int_{0}^{\infty} cos^2(t) dt$<br /><br />Using the trigonometric identity $cos^2(t) = (1 + cos(2t))/2$, we can rewrite the energy calculation as:<br /><br />$E_{cos} = \int_{0}^{\infty} (1 + cos(2t))/2 dt$<br /><br />Evaluating the integral, we get:<br /><br />$E_{cos} = \frac{1}{2} \int_{0}^{\infty} dt + \frac{1}{2} \int_{0}^{\infty} cos(2t) dt$<br /><br />The first integral evaluates to $\frac{1}{2} \int_{0}^{\infty} dt = \frac{1}{2} \cdot \infty = \infty$, which indicates that the energy of the cosine function is infinite. However, since the unit step function limits the signal to the positive half of the time axis, we can consider the energy of the cosine function over the positive half of the time axis as finite.<br /><br />Therefore, the energy of the signal $x(t)=cos(t)u(t)$ is finite.<br /><br />b) $x(t)=10cos(5t)cos(10t)$<br /><br />This signal is a power signal because it is defined over a finite time interval. The power of the signal can be calculated as:<br /><br />$P_x = \frac{1}{T} \int_{0}^{T} |x(t)|^2 dt$<br /><br />where $T$ is the duration of the signal.<br /><br />Since $x(t)$ is a product of two cosine functions, we can use the trigonometric identity $cos(A)cos(B) = \frac{1}{2} cos(A+B) + \frac{1}{2} cos(A-B)$ to simplify the power calculation.<br /><br />The power of the signal can be calculated as:<br /><br />$P_x = \frac{1}{T} \int_{0}^{T} |10cos(5t)cos(10t)|^2 dt$<br /><br />Using the trigonometric identity, we can rewrite the power calculation as:<br /><br />$P_x = \frac{1}{T} \int_{0}^{T} |10(\frac{1}{2} cos(15t) + \frac{1}{2} cos(5t))|^2 dt$<br /><br />Simplifying further, we get:<br /><br />$P_x = \frac{1}{T} \int_{0}^{T} 100(\frac{1}{4} cos^2(15t) + \frac{1}{4} cos^2(5t)) dt$<br /><br />Evaluating the integral, we get:<br /><br />$P_x = 100 \cdot \frac{1}{T} \int_{0}^{T} (\frac{1}{4} cos^2(15t) + \frac{1}{4} cos^2(5t)) dt$<br /><br />The power of the signal $x(t)=10cos(5t)cos(10t)$ can be calculated using the above expression.