The problem is asking for the exponential Fourier series coefficients $D_n$ for the signal $x_p(t) = 2 \, \text{tri}\left(\frac{t}{2}\right) \ast \delta_8(t)$, where $\text{tri}(t)$ refers to the triangular function, and $\ast$ denotes convolution. Let’s break this down step-by-step:

1. Understand the Functions Involved:

• Triangular Function $\text{tri}(t/2)$: The triangular function $\text{tri}(t)$ is typically defined as a piecewise linear function that is 1 at $t = 0$, and decays linearly to 0 at $t = \pm 1$. So $\text{tri}(t/2)$ will have a similar shape but scaled in time by a factor of 2.

• Dirac Comb $\delta_8(t)$: The $\delta_8(t)$ represents an impulse train (Dirac comb) with period 8. Mathematically: $\delta_8(t) = \sum_{k=-\infty}^{\infty} \delta(t - 8k)$

• Convolution with $\delta_8(t)$: Convolution with $\delta_8(t)$ means the signal will be repeated periodically every 8 units of time.

2. Fourier Series of Periodic Signals:

Since the signal is periodic with period 8 (due to the convolution with $\delta_8(t)$), we can express the signal as a Fourier series: $x_p(t) = \sum_{n=-\infty}^{\infty} D_n e^{j n \omega_0 t}$ where $\omega_0 = \frac{2\pi}{T} = \frac{\pi}{4}$ is the fundamental angular frequency (since $T = 8$).

3. Convolution in Time Domain and Multiplication in Frequency Domain:

In the time domain, $x_p(t) = 2 \, \text{tri}(t/2) \ast \delta_8(t)$ involves convolution with an impulse train. In the frequency domain, this corresponds to multiplication of the Fourier coefficients of the triangular function with the Fourier coefficients of the impulse train.

• The Fourier transform of $\delta_8(t)$ is another impulse train in the frequency domain, with impulses located at multiples of $\omega_0 = \frac{\pi}{4}$.

4. Find Fourier Series Coefficients for $2 \, \text{tri}(t/2)$:

The triangular function $2 \, \text{tri}(t/2)$ has a known Fourier transform. The Fourier series coefficients $D_n$ of a triangular function can be found using the Fourier transform of the $\text{tri}$-function, which decays as $\text{sinc}^2$-type functions. The coefficients for $\text{tri}(t/2)$ scaled by 2 are proportional to $\text{sinc}^2$-like terms, given by:

\frac{4}{n^2 \pi^2}, & n \neq 0 \\ 1, & n = 0 \end{cases}$$ for the triangular function. ### 5. **Final Fourier Series Coefficients**: After convolution with $$\delta_8(t)$$, the Fourier series coefficients for $$x_p(t)$$ will be nonzero only at multiples of $$n \omega_0 = \frac{n \pi}{4}$$, corresponding to the periodic impulses introduced by the Dirac comb. Thus, the exponential Fourier series coefficients $$D_n$$ will be the sampled values of the Fourier transform of the triangular function at these frequencies. Specifically: $$D_n = \frac{4}{(n \pi/4)^2} \quad \text{for} \ n \neq 0$$ Would you like more detailed clarification or explanations on any step? ### Further questions you could explore: 1. How do you derive the Fourier transform of the triangular function? 2. What properties of convolution are used to simplify this problem? 3. How does the Dirac comb affect the signal in the time domain and frequency domain? 4. Can you explain why only specific Fourier coefficients remain after convolution with $$\delta_8(t)$$? 5. How does the scaling of the triangular function affect the Fourier coefficients? ### Tip: Convolution in the time domain corresponds to multiplication in the frequency domain, which simplifies analysis for periodic signals.