The continuous time signal $x\left(t\right)$ is real, periodic with period $T$ and satisfies the Dirichlet conditions. The Fourier series representation of $x\left(t\right)=\sum _{n=-\infty }^{\infty }{a}_n{e}^{j2\pi nt/T}$ and $x\left(t\right)\text{ satisfies the following:}x(t-\frac{T}{2})=-x\left(t\right)$. For any integer $m$, which of the following options is correct?

The correct option is:
$a_{2m}=0$

Step-by-Step Explanation:

1. Understand the Given Condition:
We are given a real, periodic continuous-time signal $x\left(t\right)$ with period $T$. The signal exhibits half-wave symmetry, which is defined mathematically as:
$x(t-\frac{T}{2})=-x\left(t\right)$

2. Fourier Series Representation:
The Fourier series of $x\left(t\right)$ is given by:
$x\left(t\right)=\sum _{n=-\infty }^{\infty }{a}_n{e}^{j2\pi nt/T}$

3. Apply the Time-Shift Property of Fourier Series:
If a periodic signal $x\left(t\right)$ has Fourier coefficients $a_n$, then the shifted signal $x(t-t_0)$ has Fourier coefficients $a_n{e}^{-jn{\omega }_0{t}_0}$, where ${\omega }_0=\frac{2\pi }{T}$ is the fundamental frequency.
Setting the time shift $t_0=\frac{T}{2}$, we obtain the Fourier coefficients for $x(t-\frac{T}{2})$:
$a_n{e}^{-jn\left(\frac{2\pi }{T}\right)\left(\frac{T}{2}\right)}=a_n{e}^{-jn\pi }$

4. Simplify the Exponential Term:
Using Euler's identity, we know that for any integer $n$:
$e^{-jn\pi }=(-1{)}^n$
Thus, the Fourier series coefficients of $x(t-\frac{T}{2})$ are:
$a_n(-1{)}^n$

5. Relate the Coefficients Using the Given Symmetry:
From the equation $x(t-\frac{T}{2})=-x\left(t\right)$, we can equate their respective Fourier series coefficients:
$a_n(-1{)}^n=-a_n$
Rearranging this equation gives:
$a_n[1+(-1{)}^n]=0$

6. Analyze the Case for Even and Odd Harmonics:
• If $n$ is even (i.e., $n=2m$ for any integer $m$):
$1+(-1{)}^{2m}=1+1=2$
Substituting this back into the relation:
$2a_{2m}=0\Rightarrow a_{2m}=0$
• If $n$ is odd (i.e., $n=2m+1$):
$1+(-1{)}^{2m+1}=1-1=0$
This results in $0\cdot a_{2m+1}=0$, which allows $a_{2m+1}$ to take any non-zero value.

Hence, half-wave symmetry guarantees that all even-indexed Fourier coefficients are zero:
$a_{2m}=0$