# Is my solution correct?

$$\textbf{Question:}$$

$$y_a(t)$$ 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.

• What's $y_r(t)$ ? The question also feels out of order. $y_a(t)$ is a continuous wave form so there is no aliasing unless you want to sample it. Sampling only comes up AFTER the questions asks about aliasing. Dec 22, 2022 at 12:31
• If you want a real formulation for $\tilde{y}_a(t)$ you can use Euler's formula Dec 22, 2022 at 12:34
• $y_r(t)$ is the recovered signal after the sampling procedure. @Hilmar how so?
– user64710
Dec 22, 2022 at 12:42

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.

• Thanks a lot. I should have used that definition for FSE...
– user64710
Dec 22, 2022 at 15:22
• A few typos, but I got the gist.
– user64710
Dec 22, 2022 at 15:51