Question:
$F(t)$ is a periodic square wave function as shown. It takes only two values, 4 and 0, and stays at each of these values for 1 second before changing. What is the constant term in the Fourier series expansion of $F(t)$?
$F(t)$ is a periodic square wave function as shown. It takes only two values, 4 and 0, and stays at each of these values for 1 second before changing. What is the constant term in the Fourier series expansion of $F(t)$?
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The constant term of a Fourier series is always the average value of the function over one full period.
Updated On: Dec 1, 2025
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The Correct Option is B
Solution and Explanation
We are given that $F(t)$ is a periodic square wave that alternates between the values 4 and 0, each for 1 second. This means the total period is $T = 2$ seconds.
Step 1: Identify the average value.
The constant term in a Fourier series is the average value of the function over one period. The formula for the average (or DC term) is: \[ a_0 = \frac{1}{T} \int_0^T F(t)\, dt . \] Step 2: Compute the integral over one period.
In one period (2 seconds), the function is 4 for 1 second and 0 for 1 second. Therefore: \[ \int_0^T F(t)\, dt = 4(1) + 0(1) = 4. \] Step 3: Compute the average.
\[ a_0 = \frac{4}{2} = 2. \] Final Answer: 2
Step 1: Identify the average value.
The constant term in a Fourier series is the average value of the function over one period. The formula for the average (or DC term) is: \[ a_0 = \frac{1}{T} \int_0^T F(t)\, dt . \] Step 2: Compute the integral over one period.
In one period (2 seconds), the function is 4 for 1 second and 0 for 1 second. Therefore: \[ \int_0^T F(t)\, dt = 4(1) + 0(1) = 4. \] Step 3: Compute the average.
\[ a_0 = \frac{4}{2} = 2. \] Final Answer: 2
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