Question:medium
The continuous time signal \(x(t)\) is real, periodic with period \(T\), and satisfies the Dirichlet conditions.
The Fourier series representation of \(x(t)\) is
\[
x(t)=\sum_{n=-\infty}^{\infty}a_ne^{j\left(\frac{2\pi nt}{T}\right)}
\]
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?
The continuous time signal \(x(t)\) is real, periodic with period \(T\), and satisfies the Dirichlet conditions.
The Fourier series representation of \(x(t)\) is
\[ x(t)=\sum_{n=-\infty}^{\infty}a_ne^{j\left(\frac{2\pi nt}{T}\right)} \]
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?
The Fourier series representation of \(x(t)\) is
\[ x(t)=\sum_{n=-\infty}^{\infty}a_ne^{j\left(\frac{2\pi nt}{T}\right)} \]
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?
Show Hint
Substitute t - T/2 into the Fourier series and compare coefficients with -x(t); only the even-indexed coefficients get forced to zero.
Updated On: Jul 20, 2026
- \(a_{2m}=0\)
- \(a_{2m}=1\)
- \(a_{2m}=a_{2m+1}\)
- \(a_{2m}=-1\)
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Name the symmetry.
The rule $x(t-T/2)=-x(t)$ says the wave repeats itself upside down after half a period. This is called half-wave symmetry, and it is well known to kill off all even harmonics in a Fourier series, including the average value $a_0$.
Step 2: Prove it directly from the series.
Shift the exponential series by $T/2$:
\[ x\left(t-\frac{T}{2}\right)=\sum_n a_n e^{j\frac{2\pi n t}{T}}\,e^{-j\pi n}=\sum_n a_n(-1)^n e^{j\frac{2\pi n t}{T}} \]
because $e^{-j\pi n}=(-1)^n$ for integer $n$.
Step 3: Match this to $-x(t)$.
Since the two expansions must agree coefficient by coefficient,
\[ a_n(-1)^n=-a_n\quad\text{for every }n \]
Step 4: Split by parity of n.
If $n$ is even, say $n=2m$, then $(-1)^n=1$ and the equation becomes $a_{2m}=-a_{2m}$, forcing $a_{2m}=0$. If $n$ is odd, $(-1)^n=-1$ and the equation becomes $-a_n=-a_n$, which holds automatically and tells us nothing new about odd coefficients.
Step 5: Read off the answer.
So every even-indexed coefficient must vanish, while odd-indexed coefficients are free.
\[ \boxed{a_{2m}=0} \]
The rule $x(t-T/2)=-x(t)$ says the wave repeats itself upside down after half a period. This is called half-wave symmetry, and it is well known to kill off all even harmonics in a Fourier series, including the average value $a_0$.
Step 2: Prove it directly from the series.
Shift the exponential series by $T/2$:
\[ x\left(t-\frac{T}{2}\right)=\sum_n a_n e^{j\frac{2\pi n t}{T}}\,e^{-j\pi n}=\sum_n a_n(-1)^n e^{j\frac{2\pi n t}{T}} \]
because $e^{-j\pi n}=(-1)^n$ for integer $n$.
Step 3: Match this to $-x(t)$.
Since the two expansions must agree coefficient by coefficient,
\[ a_n(-1)^n=-a_n\quad\text{for every }n \]
Step 4: Split by parity of n.
If $n$ is even, say $n=2m$, then $(-1)^n=1$ and the equation becomes $a_{2m}=-a_{2m}$, forcing $a_{2m}=0$. If $n$ is odd, $(-1)^n=-1$ and the equation becomes $-a_n=-a_n$, which holds automatically and tells us nothing new about odd coefficients.
Step 5: Read off the answer.
So every even-indexed coefficient must vanish, while odd-indexed coefficients are free.
\[ \boxed{a_{2m}=0} \]
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