Answer :
Answer:
To find the power of the signal \( w(t) = \frac{1}{2} Z(2t) \), where \( Z(t) \) is a periodic signal defined as:
\[ Z(t) = \begin{cases}
y(t) & \text{for } 0 \leq t \leq 6 \\
y(t + 6) & \text{for all } t
\end{cases} \]
and \( y(t) \) is given by:
\[ y(t) = \begin{cases}
-t + 2 & \text{for } 0 \leq t \leq 1 \\
t - 1 & \text{for } 1 \leq t \leq 2 \\
-\frac{t}{3} + \frac{5}{3} & \text{for } 2 \leq t \leq 5 \\
0 & \text{otherwise}
\end{cases} \]
We will follow these steps:
1. **Find the Power of \( Z(t) \)**:
Since \( Z(t) \) is periodic with period 6, its power is given by the energy over one period divided by the period. The energy of \( y(t) \) is \( E = 3 \) J. Thus, the energy of \( Z(t) \) over one period (0 to 6) is:
\[
E_Z = 3 \text{ J} \times \frac{6}{6} = 3 \text{ J}
\]
The power of \( Z(t) \) is:
\[
P_Z = \frac{E_Z}{\text{Period}} = \frac{3 \text{ J}}{6 \text{ s}} = 0.5 \text{ W}
\]
2. **Find the Power of \( Z(2t) \)**:
The signal \( Z(2t) \) scales the time axis by a factor of \( \frac{1}{2} \). When the time scale changes, the power of the signal is scaled by the reciprocal of the time scaling factor squared. Therefore:
\[
P_{Z(2t)} = \frac{P_Z}{4} = \frac{0.5 \text{ W}}{4} = 0.125 \text{ W}
\]
3. **Find the Power of \( \frac{1}{2} Z(2t) \)**:
The power of a signal scaled by a factor \( \alpha \) is scaled by \( \alpha^2 \). Here, \( \alpha = \frac{1}{2} \):
\[
P_{w(t)} = \left(\frac{1}{2}\right)^2 \times P_{Z(2t)} = \frac{1}{4} \times 0.125 \text{ W} = 0.03125 \text{ W}
\]
Thus, the power of \( w(t) \) is \( 0.03125 \) watts.
