A continuous time periodic signal $x(t)$ is
$x(t) = 1 + 2 \cos 2\pi t + 2 \cos 4\pi t + 2 \cos 6\pi t$
If $T$ is the period of $x(t)$, then $\frac{1}{T} \int_{0}^{T}|x(t)|^2 dt =$ ___________ (round off to the nearest integer).
Signal Components and Period Determination
The given continuous-time signal is: $x(t) = 1 + 2 \cos(2\pi t) + 2 \cos(4\pi t) + 2 \cos(6\pi t)$
This signal consists of a DC component and three cosine terms.
- DC component: $1$
- First cosine term: $2 \cos(2\pi t)$ (Frequency $f_1 = 1$ Hz, Period $T_1 = 1$ s)
- Second cosine term: $2 \cos(4\pi t)$ (Frequency $f_2 = 2$ Hz, Period $T_2 = 1/2$ s)
- Third cosine term: $2 \cos(6\pi t)$ (Frequency $f_3 = 3$ Hz, Period $T_3 = 1/3$ s)
The fundamental period $T$ of the sum of periodic signals is the least common multiple (LCM) of their individual periods.
Therefore, $T = \text{LCM}(T_1, T_2, T_3) = \text{LCM}(1, 1/2, 1/3) = 1$ second.
Average Power Calculation
The average power $P_{avg}$ is defined as:
$P_{avg} = \frac{1}{T} \int_{0}^{T}|x(t)|^2 dt$Because the cosine terms are orthogonal over the fundamental period $T$, the average power of the sum is the sum of the average powers of its components.
- The average power of the DC component $A$ is $A^2$. For $A=1$, the power is $1^2 = 1$.
- The average power of a cosine term $A \cos(\omega t)$ is $A^2/2$.
- For $2 \cos(2\pi t)$, the average power is $\frac{2^2}{2} = 2$.
- For $2 \cos(4\pi t)$, the average power is $\frac{2^2}{2} = 2$.
- For $2 \cos(6\pi t)$, the average power is $\frac{2^2}{2} = 2$.
Summing the average powers:
$P_{avg} = 1 + 2 + 2 + 2 = 7$Final Result
The calculated average power is $7$.
Rounding $7$ to the nearest integer results in $7$.
Important Questions from Fourier Series
The Fourier series representation of a square wave is shown in the figure below. The fluctuations seen near $x = \pm 1$ are named after which one of the following scientists?
The Fourier series expansion of the function $f(x) = |\cos(x)|$ in the interval $(-\pi, \pi)$ is $\alpha + \beta \left[ \frac{1}{3}\cos(2x) - \frac{1}{15}\cos(4x) + \dots \right]$.
The values of $\alpha$ and $\beta$ respectively are ______.
Periodic function $f(x)$ is given below. $$f(x) = \begin{cases} -1, & \text{when } -\pi < x < 0 \\ 1, & \text{when } 0 < x < \pi \end{cases} ; f(x + 2\pi) = f(x)$$ The CORRECT option representing the Fourier series expansion of $f(x)$ is:
A time-limited waveform $g(x)$ is specified as follows: $$g(x) = \begin{cases} -k, & -\pi < x \le 0 \\ +k, & 0 < x \le \pi \\ 0, & \text{otherwise} \end{cases}$$ A new waveform $f(x)$ is constructed from $g(x)$ as follows: $$f(x) = \sum_{m=-\infty}^{\infty} g(x + 2\pi m), \quad \text{for all } x \in \mathbb{R}$$ The sum of the coefficients of the third harmonics of the sine and cosine terms in the trigonometric Fourier series expansion of $f(x)$ is $\frac{2}{3\pi}$.
What is the value of $k$?
The effective value of the current for the Sawtooth waveform as shown is:
