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: Determine the DC component.The DC component, \(a_0\), of a periodic signal \(x(t)\) with period \(T_0\) represents its average value over one period. It is calculated as:\[a_0 = \frac{1}{T_0} \int_{T_0} x(t) \,dt\]This integral is evaluated over any interval of length \(T_0\), such as \( [-\frac{T_0}{2}, \frac{T_0}{2}] \).
Step 2: Define the integral based on the signal.The signal \(x(t)\) equals 1 for \(|t|<T_1\) (i.e., \(-T_1<t<T_1\)) and 0 elsewhere within the period. Consequently, the integral is only non-zero between \(-T_1\) and \(T_1\).\[a_0 = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \,dt = \frac{1}{T_0} \int_{-T_1}^{T_1} (1) \,dt\]
Step 3: Calculate 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 result corresponds to option (C).