The continuous time signal $x(t)$ is real, periodic with period $T$ and satisfies the Dirichlet conditions.
The Fourier series representation of $x(t) = \sum_{-\infty}^\infty a_n e^{j\left(\frac{2\pi nt}{T}\right)}$ and $x(t)$ satisfies the following: $x\left(t - \frac{T}{2}\right) = -x(t)$.
For any integer $m$, which of the following options is correct?
Solution
The problem concerns a real, periodic continuous-time signal $x(t)$ with period $T$ that satisfies Dirichlet conditions and a specific symmetry property: $x\left(t - \frac{T}{2}\right) = -x(t)$. This property is known as half-wave symmetry. We need to determine the condition for its Fourier series coefficients $a_n$.
Fourier Series Representation
The signal $x(t)$ is given by its complex exponential Fourier series:
$x(t) = \sum_{n=-\infty}^\infty a_n e^{j n \omega_0 t}$where $\omega_0 = \frac{2\pi}{T}$ is the fundamental angular frequency.
Applying the Half-Wave Symmetry
First, let's express $x\left(t - \frac{T}{2}\right)$ using the Fourier series:
$x\left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^\infty a_n e^{j n \omega_0 (t - T/2)}$Using $\omega_0 T/2 = \pi$, we get:
$x\left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^\infty a_n e^{j n \omega_0 t} e^{-j n \pi}$Recall that $e^{-j n \pi} = (-1)^n$. Therefore:
$x\left(t - \frac{T}{2}\right) = \sum_{n=-\infty}^\infty a_n (-1)^n e^{j n \omega_0 t}$Deriving the Condition on Coefficients
Now, substitute this back into the given symmetry property $x\left(t - \frac{T}{2}\right) = -x(t)$:
$\sum_{n=-\infty}^\infty a_n (-1)^n e^{j n \omega_0 t} = -\sum_{n=-\infty}^\infty a_n e^{j n \omega_0 t}$Combine the terms:
$\sum_{n=-\infty}^\infty a_n (-1)^n e^{j n \omega_0 t} + \sum_{n=-\infty}^\infty a_n e^{j n \omega_0 t} = 0$ $\sum_{n=-\infty}^\infty a_n [(-1)^n + 1] e^{j n \omega_0 t} = 0$Due to the linear independence of the complex exponential functions $e^{j n \omega_0 t}$, the coefficients must satisfy:
$a_n [(-1)^n + 1] = 0 \quad \text{for all integers } n$Analyzing Even Index Coefficients
We are interested in the coefficients where the index $n$ is an even integer. Let $n = 2m$, where $m$ is any integer.
For even $n$, $(-1)^n = (-1)^{2m} = 1$. Substituting this into the derived condition:
$a_{2m} [1 + 1] = 0$ $a_{2m} [2] = 0$ $2a_{2m} = 0$This implies that:
$a_{2m} = 0$Note: For odd indices ($n = 2m+1$), the condition becomes $a_{2m+1}[(-1)^{2m+1} + 1] = a_{2m+1}[-1 + 1] = 0$, which simplifies to $0 = 0$. This provides no constraint on the odd coefficients.
Conclusion
The half-wave symmetry property $x\left(t - \frac{T}{2}\right) = -x(t)$ directly leads to the conclusion that all Fourier series coefficients corresponding to even indices must be zero. Thus, $a_{2m} = 0$ for any integer $m$.
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:
