Is my solution correct?

Question

$\textbf{Question:}$

$y_a(t)$ is a rectangular waveform defined as:
$$\ y_a(t) = \begin{cases} 2 &t \in [0,1/25)s\\ 0 &t \in [1/25,1/15)s \end{cases}$$ and $y_a(t)$ is periodic with a period of 1/15 seconds. Clearly identify the complex sinusoidal components of $y_r(t)$ that are not aliased and that are aliased. Can we increase the sampling rate (from $T_s = \frac{1}{100}$) to prevent aliasing in this case? Why?

$\textbf{Answer:}$

First find Fourier Series Expansion of $y_a(t)$:

$$a_k = \frac {1}{\pi} \left(\frac{1-e^{-j \frac{30}{25}\pi k}}{jk}\right)$$

$$\tilde{y}_a(t) = \sum_{k=-\infty}^{\infty} a_k \cdot e^{j(30\pi)kt}$$

Look at this expansion other than the DC and the first harmonic ($k= -1,0,1$) all other components will be aliased right (Harmonics at frequencies $>30\pi$ are all bigger than $f_s=100$) ?

And we cannot increase the sampling rate to get rid of the aliasing since we would have to increase it to somewhere in infinity which is impossible. In essence, $y_a(t)$ is not bandlimited.

Is this solution sufficient?

Also I don't really say a way of writing my $a_k$ as a real valued sinusoidal. Can anyone help with that as well?

$\textbf{Edit:}$ Actually more than just the DC and the first harmonic will be recovered unaliased since the $f_s = 100$ but $\omega_s=200\pi$ which will allow $k=-3,-2,-1,0,1,2,3$ to pass since at $k=|3|$ the $\omega_a = 90\pi$ and $2\omega_a < \omega_s$ so the Nyquist rate is still satisfied.

Answer 1

You basically have a rectangular wave with frequency $f = 15$, amplitude $A = 2$ and duty cycle $D = 15/25$.

Let me start from the end. To write the series as real-valued trigonometric functions, you can simply apply the definition:

$y_a(t) = \frac{a_0}{2} + \sum_{n = 1}^{\infty} [a_n \cos(w n t) + b_n \sin(w n t)]$

where $ w = 2 \pi f = 2 \pi 15 = 30 \pi$.

$a_0 = \frac{2}{T} \int_{0}^{T} y_a(t) dt$

$a_n = \frac{2}{T} \int_{0}^{T} y_a(t) \cos(w n t) dt$

$b_n = \frac{2}{T} \int_{0}^{T} y_a (t) \sin(w n t) dt$

The definite integrals are quite easy to compute: just write them as the sum of two integrals:

$\int_{0}^{T} y_a(t) g(t) dt = \int_{0}^{D T} 2 g(t) dt + \int_{D T}^{T} 0 \cdot g(t) dt = \int_{0}^{D T} 2 g(t) dt$.

with $g(t) = 1$ for $a_0$, $g(t) = \cos(w n t) $ for $a_n$ and $g(t) = \sin(w n t) $ for $b_n$.

If you do it, you will see that the results are the following:

$a_0 = 2 A D$

$a_n = \frac{2 A}{n \pi} \sin(n \pi D) $

$b_n = 0 \;\; \forall n \in \mathbb{N}^+$

As you can see, the signal is not band-limited: there are infinite harmonics. No matter how big $n$ is, $a_n \neq 0$. Which means that no matter how big the sampling rate is, you will never recover all the harmonics: there will always be some aliasing.

With your current sampling rate, $f_s = 100$. The harmonics that are not aliased are those with frequencies between 0 and Nyquist, i.e between $0$ and $f_s/2 = 100/2 = 50 Hz$.

Since $f = 15$, you will have harmonics at $0$ (DC component), $1 \cdot 15 $ (fundamental harmonic), $2 \cdot 15 = 30$ ... The last harmonic bellow Nyquist is at frequency $3 \cdot 15 = 45$. So you will only have three sinusoidal components of yr(t) that are not aliased with the current sampling rate. From your question, I see that you have already found their expression (it would be replacing $k = 0, 1, 2, 3$ in your $a_k$).

Hope this helped.