Let's break down the problem in steps to calculate the output $y(n)$ based on the provided information.

Problem Breakdown:

1. Input Signal $x(n)$:

The input signal $x(n)$ is given as: $x(n) = 12 \cos\left(\frac{\pi n}{2}\right) + \frac{3}{2} \cos\left(\frac{2\pi n}{3}\right)$ This is a sum of two cosine signals with different frequencies.

• The first term $12 \cos\left(\frac{\pi n}{2}\right)$ has a frequency of $\frac{1}{4}$ (since $\cos(2\pi f n)$, so $f_1 = \frac{1}{4}$).
• The second term $\frac{3}{2} \cos\left(\frac{2\pi n}{3}\right)$ has a frequency of $\frac{1}{3}$ (so $f_2 = \frac{1}{3}$).

2. Channel's Amplitude Response $S(f)$:

The channel is modeled as a linear time-invariant (LTI) system with an amplitude response given by: $S(f) = \frac{2f}{f + 2}$

This response will affect each frequency component of the input signal $x(n)$.

3. Effect of Channel on Each Frequency Component:

For each frequency component of $x(n)$, the channel will scale it based on its amplitude response $S(f)$.

• For the first term $12 \cos\left(\frac{\pi n}{2}\right)$, the frequency is $f_1 = \frac{1}{4}$. So, the amplitude response will be: $S\left(\frac{1}{4}\right) = \frac{2 \times \frac{1}{4}}{\frac{1}{4} + 2} = \frac{\frac{1}{2}}{\frac{9}{4}} = \frac{1}{2} \times \frac{4}{9} = \frac{2}{9}$ The new amplitude for this term will be: $12 \times \frac{2}{9} = \frac{24}{9} = \frac{8}{3}$ So the output for this term will be $\frac{8}{3} \cos\left(\frac{\pi n}{2}\right)$.

• For the second term $\frac{3}{2} \cos\left(\frac{2\pi n}{3}\right)$, the frequency is $f_2 = \frac{1}{3}$. So, the amplitude response will be: $S\left(\frac{1}{3}\right) = \frac{2 \times \frac{1}{3}}{\frac{1}{3} + 2} = \frac{\frac{2}{3}}{\frac{7}{3}} = \frac{2}{7}$ The new amplitude for this term will be: $\frac{3}{2} \times \frac{2}{7} = \frac{6}{14} = \frac{3}{7}$ So the output for this term will be $\frac{3}{7} \cos\left(\frac{2\pi n}{3}\right)$.

4. Final Output $y(n)$:

The final output $y(n)$ will be the sum of the two terms after being affected by the channel: $y(n) = \frac{8}{3} \cos\left(\frac{\pi n}{2}\right) + \frac{3}{7} \cos\left(\frac{2\pi n}{3}\right)$

This is the output signal after passing through the LTI system.

Do you need more details on any step or further clarifications?

Related Questions:

1. How do you find the frequency of a cosine signal from its equation?
2. What is the role of an amplitude response in an LTI system?
3. How do you apply a system's amplitude response to a given signal?
4. Can you explain why different frequencies are affected differently by the channel?
5. How would the output change if the channel response $S(f)$ was different?

Tip: Always identify the frequencies of the input signal's components before applying the system's response to each.