Question

A periodic signal \(x(t)\) of period \(T_0\) is given by \( x(t) = \begin{cases} 1, & |t|<T_1 \\ 0, & T_1<|t|<\frac{T_0}{2} \end{cases} \). The dc component of \(x(t)\) is

• A. \( \frac{T_1}{T_0} \)
• B. \( \frac{T_1}{2T_0} \)
• C. \( \frac{2T_1}{T_0} \)
• D. \( \frac{T_0}{T_1} \)

Hint:

The DC component of a periodic signal is simply its average value. For a rectangular pulse of height A, width W, and period T, the DC component is \( A \times \frac{W}{T} \). In this case, A=1, W=2\(T_1\), and T=\(T_0\).

Answer 1

The Correct Option is C

Step 1: Define the DC component. The DC component of a periodic signal is its average value over one period. For a signal \(x(t)\) with period \(T_0\), the DC component, often denoted as \(a_0\), is calculated by the formula: \[ a_0 = \frac{1}{T_0} \int_{T_0} x(t) \,dt \] where the integral is taken over any interval of length \(T_0\). We can choose the interval \( [-\frac{T_0}{2}, \frac{T_0}{2}] \).
Step 2: Set up the integral based on the signal definition. The signal \(x(t)\) is 1 for \(|t|<T_1\) (which means \(-T_1<t<T_1\)) and 0 otherwise within the period. So, the integral is non-zero only in the range from \(-T_1\) to \(T_1\). \[ a_0 = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \,dt = \frac{1}{T_0} \int_{-T_1}^{T_1} (1) \,dt \]
Step 3: Evaluate the integral. \[ a_0 = \frac{1}{T_0} \left[ t \right]_{-T_1}^{T_1} \] \[ a_0 = \frac{1}{T_0} (T_1 - (-T_1)) = \frac{1}{T_0} (2T_1) = \frac{2T_1}{T_0} \] This matches option (C).