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)?

The correct answer is 2.

Step-by-step Explanation:

1. Understanding the Constant Term of a Fourier Series:
The Fourier series representation of a periodic function $F\left(t\right)$ with period $T$ decomposes the function into a sum of sine and cosine waves, along with a constant (DC) term. This constant term represents the average value of the function over one complete period $T$, and is calculated as:

$$a_0$$

2. Analyzing the Waveform from the Image:
By examining the provided plot, we can observe the following:

• The vertical axis is labeled $F\left(t\right)$ and has a peak value of $4$.
• The horizontal axis is labeled $t\text{(seconds)}$ and marked with integer seconds: $-3,-2,-1,0,1,2,3,4$.
• From $t=0$ to $t=1$ second, the function value is $F\left(t\right)=4$.
• From $t=1$ to $t=2$ seconds, the function value is $F\left(t\right)=0$.
• This cycle repeats periodically, meaning the fundamental period of the waveform is $T=2$ seconds.

3. Calculating the Constant Term (Average Value):
We integrate the function over one full period from $t=0$ to $t=2$:

$$a_0=\frac{1}{2}\left({\int }_0^{1}4dt+{\int }_1^{2}0dt\right)$$

Since the integral of $0$ is $0$, this simplifies to:

$$a_0=\frac{1}{2}{\left[4t\right]}_0^1=\frac{1}{2}\times \left(4-0\right)=2$$
Thus, the constant term in the Fourier series expansion of $F\left(t\right)$ is indeed $2$.