Understanding the Continuous-Time Signal

We are given a continuous-time signal defined as:

$x(t) = e^{j9t} + e^{j5t}$

Here, $j$ represents the imaginary unit ($\sqrt{-1}$) and $t$ is time in seconds. The question asks for the fundamental period of the magnitude of this signal, denoted as $|x(t)|$.

Calculating the Magnitude of x(t)

To find the magnitude, we first express $x(t)$ in terms of its real and imaginary parts using Euler's formula, $e^{j\theta} = \cos(\theta) + j\sin(\theta)$:

$x(t) = (\cos(9t) + j\sin(9t)) + (\cos(5t) + j\sin(5t))$

Group the real and imaginary parts:

$x(t) = (\cos(9t) + \cos(5t)) + j(\sin(9t) + \sin(5t))$

The square of the magnitude of a complex number $z = a + jb$ is given by $|z|^2 = a^2 + b^2$. Applying this to $x(t)$:

$|x(t)|^2 = (\cos(9t) + \cos(5t))^2 + (\sin(9t) + \sin(5t))^2$

Expand the terms:

$|x(t)|^2 = (\cos^2(9t) + 2\cos(9t)\cos(5t) + \cos^2(5t)) + (\sin^2(9t) + 2\sin(9t)\sin(5t) + \sin^2(5t))$

Rearrange and use the trigonometric identity $\cos^2(\theta) + \sin^2(\theta) = 1$:

$|x(t)|^2 = (\cos^2(9t) + \sin^2(9t)) + (\cos^2(5t) + \sin^2(5t)) + 2(\cos(9t)\cos(5t) + \sin(9t)\sin(5t))$

$|x(t)|^2 = 1 + 1 + 2(\cos(9t)\cos(5t) + \sin(9t)\sin(5t))$

Now, use the cosine angle subtraction identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$:

$|x(t)|^2 = 2 + 2\cos(9t - 5t)$

$|x(t)|^2 = 2 + 2\cos(4t)$

Therefore, the magnitude is:

$|x(t)| = \sqrt{2 + 2\cos(4t)}$

Determining the Fundamental Period

The signal whose period we need to find is $|x(t)| = \sqrt{2 + 2\cos(4t)}$.

The fundamental period of a signal is the smallest positive time duration $T$ after which the signal repeats itself, meaning $|x(t+T)| = |x(t)|$ for all $t$.

We need to find the period of the function $|x(t)|$. The periodicity is determined by the term $\cos(4t)$.

The general form of a cosine function is $\cos(\omega t)$, where $\omega$ is the angular frequency. The fundamental period ($T_0$) of $\cos(\omega t)$ is given by the formula:

$T_0 = \frac{2\pi}{|\omega|}$

In our expression, the term inside the cosine function is $4t$. Thus, the angular frequency $\omega = 4$ rad/s.

The fundamental period of $\cos(4t)$ is:

$T_{\cos(4t)} = \frac{2\pi}{|4|} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ seconds}$

Since the magnitude $|x(t)| = \sqrt{2 + 2\cos(4t)}$ depends directly on $\cos(4t)$, its fundamental period will be the same as the fundamental period of $\cos(4t)$. We can confirm this by checking the condition for periodicity:

We require the smallest $T > 0$ such that $|x(t+T)| = |x(t)|$.

$\sqrt{2 + 2\cos(4(t+T))} = \sqrt{2 + 2\cos(4t)}$

Squaring both sides:

$2 + 2\cos(4t + 4T) = 2 + 2\cos(4t)$

$2\cos(4t + 4T) = 2\cos(4t)$

$\cos(4t + 4T) = \cos(4t)$

This equality holds if $4t + 4T = 4t + 2k\pi$ for some integer $k$.

$4T = 2k\pi$

$T = \frac{2k\pi}{4} = \frac{k\pi}{2}$

The smallest positive value for $T$ occurs when $k=1$.

Therefore, the fundamental period of $|x(t)|$ is:

$T = \frac{\pi}{2} \text{ seconds}$

Conclusion

The calculation shows that the fundamental period of the magnitude of the signal $x(t)$ is determined by the period of the $\cos(4t)$ term, which is $\frac{\pi}{2}$ seconds.