Question

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?

• A. \(a_{2m}=0\)
• B. \(a_{2m}=1\)
• C. \(a_{2m}=a_{2m+1}\)
• D. \(a_{2m}=-1\)

Hint:

Substitute t - T/2 into the Fourier series and compare coefficients with -x(t); only the even-indexed coefficients get forced to zero.

Answer 1

The Correct Option is A

Step 1: Write the shifted signal in terms of the Fourier coefficients.
Starting from
\[ x(t)=\sum_{n=-\infty}^{\infty}a_ne^{j\frac{2\pi nt}{T}}, \]
replace \(t\) by \(t-\frac{T}{2}\):
\[ x\left(t-\frac{T}{2}\right)=\sum_{n=-\infty}^{\infty}a_ne^{j\frac{2\pi n}{T}\left(t-\frac{T}{2}\right)}=\sum_{n=-\infty}^{\infty}a_ne^{j\frac{2\pi nt}{T}}e^{-j\pi n} \]

Step 2: Simplify the extra exponential factor.
Since \(n\) is an integer,
\[ e^{-j\pi n}=(-1)^n \]
So,
\[ x\left(t-\frac{T}{2}\right)=\sum_{n=-\infty}^{\infty}a_n(-1)^ne^{j\frac{2\pi nt}{T}} \]

Step 3: Apply the given condition.
We are told
\[ x\left(t-\frac{T}{2}\right)=-x(t)=-\sum_{n=-\infty}^{\infty}a_ne^{j\frac{2\pi nt}{T}} \]
Comparing the two Fourier series term by term (Fourier coefficients are unique), for every \(n\):
\[ a_n(-1)^n=-a_n \]

Step 4: Solve this equation for even and odd n separately.
For even \(n\), write \(n=2m\), so \((-1)^n=1\), giving
\[ a_{2m}=-a_{2m}\ \Rightarrow\ 2a_{2m}=0\ \Rightarrow\ a_{2m}=0 \]
For odd \(n\), \((-1)^n=-1\), giving
\[ -a_n=-a_n \]
which is always true and places no restriction on odd-indexed coefficients.

Step 5: Interpret the result.
The condition \(x\left(t-\frac{T}{2}\right)=-x(t)\) is the half-wave symmetry condition. It forces every even-indexed Fourier coefficient, including the DC term \(a_0\), to be zero, so only odd harmonics can be present in the signal.

Step 6: Analyze the options.

(A) a_2m = 0: Matches the result derived above. Correct.

(B) a_2m = 1: Contradicts the derivation, which gives zero, not one. Incorrect.

(C) a_2m = a_2m+1: There is no general equality forced between an even-indexed coefficient and the next odd-indexed one; the derivation only forces the even one to vanish. Incorrect.

(D) a_2m = -1: Contradicts the derivation. Incorrect.

Step 7: Final conclusion.
Therefore, for any integer \(m\), \[ \boxed{a_{2m}=0} \]