Sea la señal x[n] = 5 + sen(1/4πn) + cos(4πn+1/9π). Determine los coeficientes para la serie de Fourier a_k de los siguientes valores de k.

Steps to Solve

1. Identify the components of the signal

The signal $x[n]$ is given by: $x[n] = 5 + \sin(\frac{1}{4}\pi n) + \cos(4\pi n + \frac{1}{9}\pi)$ This signal consists of three components: a DC component (5), a sine wave, and a cosine wave.

2. Express sine and cosine in complex exponentials

We can use Euler's formula to express the sine and cosine functions as complex exponentials:

$\sin(\theta) = \frac{e^{j\theta} - e^{-j\theta}}{2j}$ $\cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2}$

Thus:

$\sin(\frac{1}{4}\pi n) = \frac{e^{j\frac{1}{4}\pi n} - e^{-j\frac{1}{4}\pi n}}{2j}$ $\cos(4\pi n + \frac{1}{9}\pi) = \frac{e^{j(4\pi n + \frac{1}{9}\pi)} + e^{-j(4\pi n + \frac{1}{9}\pi)}}{2}$

3. Rewrite the signal x[n] using complex exponentials

Substitute the complex exponential forms into the expression for $x[n]$:

$x[n] = 5 + \frac{e^{j\frac{1}{4}\pi n} - e^{-j\frac{1}{4}\pi n}}{2j} + \frac{e^{j(4\pi n + \frac{1}{9}\pi)} + e^{-j(4\pi n + \frac{1}{9}\pi)}}{2}$

4. Identify the Fourier series coefficients

The general form of the Fourier series is: $x[n] = \sum_{k} a_k e^{j \frac{2\pi}{N}kn}$

Comparing this with our expression for $x[n]$, we want to identify the coefficients $a_k$ for different values of $k$.

For the DC component, $5 = a_0 e^{j0n}$, so $a_0 = 5$. Actually here we consider $N=1$.

For the sine component $\sin(\frac{1}{4}\pi n) = \frac{e^{j\frac{1}{4}\pi n} - e^{-j\frac{1}{4}\pi n}}{2j}$: $\frac{1}{4}\pi = \frac{2\pi}{N}k$, with $N=8$, then $\frac{1}{4}\pi = \frac{2\pi}{8} k = \frac{1}{4}\pi k$ so $k=1$. Therefore $a_1 = \frac{1}{2j} = -\frac{j}{2}$. For $k=-1$, $a_{-1} = -\frac{1}{2j} = \frac{j}{2}$.

For the cosine component $\cos(4\pi n + \frac{1}{9}\pi) = \frac{e^{j(4\pi n + \frac{1}{9}\pi)} + e^{-j(4\pi n + \frac{1}{9}\pi)}}{2}$:

$4\pi = \frac{2\pi}{N}k$, with $N=1/2$ then $k=1$. Therefore $a_1 = \frac{e^{j\frac{1}{9}\pi}}{2}$ For $k=-1$, $a_{-1} = \frac{e^{-j\frac{1}{9}\pi}}{2}$

When $N = 1$, $x[n] = 5 + \frac{e^{j\frac{1}{4}\pi n} - e^{-j\frac{1}{4}\pi n}}{2j} + \frac{e^{j\frac{1}{9}\pi}e^{j4\pi n} + e^{-j\frac{1}{9}\pi}e^{-j4\pi n}}{2}$ The signal is discrete but not periodic

If we consider the signal is periodic in $N=8$, we can write: $x[n] = 5 + \frac{e^{j\frac{1}{4}\pi n} - e^{-j\frac{1}{4}\pi n}}{2j} + \frac{e^{j\frac{1}{9}\pi}e^{j\frac{2\pi}{8}16 n} + e^{-j\frac{1}{9}\pi}e^{-j\frac{2\pi}{8}16 n}}{2}$ so, $a_0 = 5$, $a_1 = \frac{1}{2j} = -\frac{j}{2}$, $a_{-1} = -\frac{1}{2j} = \frac{j}{2}$, $a_{16} = \frac{e^{j\frac{1}{9}\pi}}{2}$ y $a_{-16} = \frac{e^{-j\frac{1}{9}\pi}}{2}$