Understanding LTI System Response to Sinusoidal Inputs

This problem involves finding the output signal \(y(t)\) when an input signal \(x(t)\) is passed through a Linear Time-Invariant (LTI) system with a given impulse response \(h(t)\). The output of an LTI system is the convolution of the input signal and the impulse response: \(y(t) = x(t) * h(t)\). However, for signals composed of sums of sinusoids, we can analyze the system's response in the frequency domain using its frequency response, \(H(\omega)\), which is the Fourier Transform of the impulse response \(h(t)\).

Analyzing the Input Signal \(x(t)\)

The input signal is given by:

\(x(t) = 2 \sin(10\pi t) + 5 \cos(15\pi t) + 7 \sin(42\pi t) + 4 \cos(45\pi t)\)

This signal is composed of four sinusoidal components with the following angular frequencies (\(\omega\)):

• Component 1: \(2 \sin(10\pi t)\) has \(\omega_1 = 10\pi\) rad/s.
• Component 2: \(5 \cos(15\pi t)\) has \(\omega_2 = 15\pi\) rad/s.
• Component 3: \(7 \sin(42\pi t)\) has \(\omega_3 = 42\pi\) rad/s.
• Component 4: \(4 \cos(45\pi t)\) has \(\omega_4 = 45\pi\) rad/s.

Calculating the Frequency Response \(H(\omega)\)

The impulse response is given by:

\(h(t) = 2 \left( \frac{\sin (10 \pi t)}{\pi t} \right) \cos (40 \pi t)\)

To find the frequency response \(H(\omega)\), we take the Fourier Transform of \(h(t)\). We use the following Fourier Transform pairs and properties:

• The Fourier Transform of \( \frac{\sin(at)}{\pi t} \) is \( \text{rect}\left(\frac{\omega}{2a}\right) \). For \(a = 10\pi\), the transform of \( \frac{\sin(10\pi t)}{\pi t} \) is \( \text{rect}\left(\frac{\omega}{20\pi}\right) \). This function is \(1\) for \(|\omega| \le 10\pi\) and \(0\) otherwise.
• The Fourier Transform of \( \cos(\omega_0 t) \) is \( \pi [\delta(\omega - \omega_0) + \delta(\omega + \omega_0)] \).
• The modulation property states that if \(f(t) \leftrightarrow F(\omega)\), then \(f(t)\cos(\omega_0 t) \leftrightarrow \frac{1}{2}[F(\omega - \omega_0) + F(\omega + \omega_0)]\).

Let \(g(t) = \frac{\sin(10\pi t)}{\pi t}\). Then \(G(\omega) = \text{rect}\left(\frac{\omega}{20\pi}\right)\).

Now, \(h(t) = 2 g(t) \cos(40\pi t)\). Applying the modulation property:

\(H(\omega) = \mathcal{F}\{2 g(t) \cos(40\pi t)\}\)

\(H(\omega) = 2 \cdot \frac{1}{2} [G(\omega - 40\pi) + G(\omega + 40\pi)]\)

\(H(\omega) = G(\omega - 40\pi) + G(\omega + 40\pi)\)

Since \(G(\omega) = 1\) for \(|\omega| \le 10\pi\):

• \(G(\omega - 40\pi) = 1\) when \(| \omega - 40\pi | \le 10\pi\), which means \(30\pi \le \omega \le 50\pi\).
• \(G(\omega + 40\pi) = 1\) when \(| \omega + 40\pi | \le 10\pi\), which means \(-50\pi \le \omega \le -30\pi\).

Therefore, the frequency response \(H(\omega)\) is:

\(H(\omega) = 1\) for \( \omega \in [-50\pi, -30\pi] \cup [30\pi, 50\pi] \)

\(H(\omega) = 0\) otherwise.

This indicates that the system acts as a bandpass filter, allowing frequencies between \(30\pi\) and \(50\pi\) (and their negative counterparts) to pass through, while attenuating others.

Determining the Output Components

We now check which components of the input signal \(x(t)\) fall within the passband of the system \(|\omega| \in [30\pi, 50\pi]\).

• Component 1 (\(\omega_1 = 10\pi\)): \(10\pi\) is not in the range \([30\pi, 50\pi]\). This component is filtered out, so its output contribution is \(0\).
• Component 2 (\(\omega_2 = 15\pi\)): \(15\pi\) is not in the range \([30\pi, 50\pi]\). This component is also filtered out, contributing \(0\) to the output.
• Component 3 (\(\omega_3 = 42\pi\)): \(42\pi\) is within the passband \([30\pi, 50\pi]\). Here, \(H(42\pi) = 1\). The input is \(7 \sin(42\pi t)\). The output contribution is \(7 \cdot H(42\pi) \cdot \sin(42\pi t) = 7 \cdot 1 \cdot \sin(42\pi t) = 7 \sin(42\pi t)\).
• Component 4 (\(\omega_4 = 45\pi\)): \(45\pi\) is within the passband \([30\pi, 50\pi]\). Here, \(H(45\pi) = 1\). The input is \(4 \cos(45\pi t)\). The output contribution is \(4 \cdot H(45\pi) \cdot \cos(45\pi t) = 4 \cdot 1 \cdot \cos(45\pi t) = 4 \cos(45\pi t)\).

Final Output Signal

The total output signal \(y(t)\) is the sum of the contributions from each component:

\(y(t) = 0 + 0 + 7 \sin(42\pi t) + 4 \cos(45\pi t)\)

\(y(t) = 7 \sin(42\pi t) + 4 \cos(45\pi t)\)