Let a continuous-time signal be $x(t)=e^{j9t}+e^{j5t}$, where $j=\sqrt{-1}$ and $t$ is in seconds. The fundamental period of magnitude of $x(t)$, in seconds, is

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.