3. b) for the system in the figure, sketch a(w), b(w), c(w), y(w), given: (8 pt)\n$x(\\omega)=tri(\\frac{\\om…

Answer:

To sketch (A(\omega), B(\omega), C(\omega), Y(\omega)), we need to use the properties of Fourier - transforms and modulation.

1. First, recall the modulation property of the Fourier - transform: If (x(t)) has a Fourier - transform (X(\omega)), then (x(t)\cos(\omega_0t)) has a Fourier - transform (\frac{1}{2}[X(\omega-\omega_0)+X(\omega + \omega_0)]).
   • Given (X(\omega)=\text{tri}(\frac{\omega}{10})), and (A(t)=X(t)\cos(50t)).
     • By the modulation property, (A(\omega)=\frac{1}{2}[\text{tri}(\frac{\omega - 50}{10})+\text{tri}(\frac{\omega + 50}{10})]).
2. Then, to find (B(\omega)), we use the frequency - response of the system (H(\omega)).
   • Since (B(\omega)=A(\omega)H(\omega)), and (H(\omega)=\begin{cases}1, &- 50\leq\omega\leq - 40,40\leq\omega\leq50\0, &\text{otherwise}\end{cases}).
     • (B(\omega)=\frac{1}{2}\begin{cases}\text{tri}(\frac{\omega - 50}{10}), &40\leq\omega\leq50\\text{tri}(\frac{\omega + 50}{10}), &- 50\leq\omega\leq - 40\0, &\text{otherwise}\end{cases})
3. Next, to find (C(\omega)), we use the modulation property again. Since (C(t)=B(t)\cos(100t)).
   • By the modulation property, (C(\omega)=\frac{1}{2}[B(\omega - 100)+B(\omega + 100)]).
     • Substitute the expression of (B(\omega)) into the above formula.
     • For (B(\omega - 100)):
       • When (40\leq\omega - 100\leq50) (i.e., (140\leq\omega\leq150)), (B(\omega - 100)=\frac{1}{2}\text{tri}(\frac{\omega - 150}{10})); when (-50\leq\omega - 100\leq - 40) (i.e., (50\leq\omega\leq60)), (B(\omega - 100)=\frac{1}{2}\text{tri}(\frac{\omega - 50}{10})).
     • For (B(\omega + 100)):
       • When (40\leq\omega + 100\leq50) (i.e., (-60\leq\omega\leq - 50)), (B(\omega + 100)=\frac{1}{2}\text{tri}(\frac{\omega + 50}{10})); when (-50\leq\omega + 100\leq - 40) (i.e., (-150\leq\omega\leq - 140)), (B(\omega + 100)=\frac{1}{2}\text{tri}(\frac{\omega+150}{10})).
4. Finally, to find (Y(\omega)), we use the frequency - response of the system (G(\omega)).
   • Since (Y(\omega)=C(\omega)G(\omega)) and (G(\omega)=\begin{cases}1, &-100\leq\omega\leq100\0, &\text{otherwise}\end{cases}).
     • We need to consider the intersection of the support of (C(\omega)) and the support of (G(\omega)).
     • (Y(\omega)=\frac{1}{2}\begin{cases}\text{tri}(\frac{\omega - 50}{10}), &50\leq\omega\leq60\\text{tri}(\frac{\omega + 50}{10}), &-60\leq\omega\leq - 50\0, &\text{otherwise}\end{cases})

To sketch these functions:

• Sketch of (A(\omega)):
  • The triangular function (\text{tri}(\frac{\omega}{10})) has a base from (- 10) to (10) and a peak at (\omega = 0) with height (1). After modulation by (\cos(50t)), (A(\omega)) consists of two triangular - shaped functions centered at (\omega=-50) and (\omega = 50) with height (\frac{1}{2}) and base width (20).
• Sketch of (B(\omega)):
  • We take the product of (A(\omega)) and (H(\omega)). So, (B(\omega)) has non - zero values only in the intervals ([-50,-40]) and ([40,50]). In ([40,50]), it is half of the right - shifted triangular function (\text{tri}(\frac{\omega - 50}{10})), and in ([-50,-40]), it is half of the left - shifted triangular function (\text{tri}(\frac{\omega + 50}{10})).
• Sketch of (C(\omega)):
  • After the second modulation by (\cos(100t)), (C(\omega)) has components shifted by (\pm100) from (B(\omega)). But we need to consider the non - zero parts of (B(\omega)) first.
• Sketch of (Y(\omega)):
  • We take the product of (C(\omega)) and (G(\omega)). So, (Y(\omega)) has non - zero values only in the intervals ([-60,-50]) and ([50,60]) based on the support of (G(\omega)) and the non - zero parts of (C(\omega)).

Explanation:

Step1: Apply modulation property for (A(\omega))

If (x(t)) has Fourier - transform (X(\omega)), then (x(t)\cos(\omega_0t)) has Fourier - transform (\frac{1}{2}[X(\omega-\omega_0)+X(\omega+\omega_0)]). Here (x(t)) has (X(\omega)=\text{tri}(\frac{\omega}{10})) and (\omega_0 = 50), so (A(\omega)=\frac{1}{2}[\text{tri}(\frac{\omega - 50}{10})+\text{tri}(\frac{\omega + 50}{10})]).

Step2: Find (B(\omega)) using (H(\omega))

Since (B(\omega)=A(\omega)H(\omega)) and (H(\omega)) is a rectangular function, we multiply (A(\omega)) by (H(\omega)) to get (B(\omega)) which is non - zero only in ([-50,-40]) and ([40,50]).

Step3: Apply modulation property for (C(\omega))

Apply the modulation property again for (C(t)=B(t)\cos(100t)). So (C(\omega)=\frac{1}{2}[B(\omega - 100)+B(\omega + 100)]).

Step4: Find (Y(\omega)) using (G(\omega))

Since (Y(\omega)=C(\omega)G(\omega)) and (G(\omega)) is a rectangular function, we multiply (C(\omega)) by (G(\omega)) to get (Y(\omega)) which has non - zero values only in the intervals where (G(\omega)) is non - zero and (C(\omega)) is non - zero.