Question

\(x[n]\) is a real and odd signal. Which one of the following statements is true about its Discrete Time Fourier Transform (DTFT) \(X(e^{j\omega})\)?

• A. \(X(e^{j\omega})\) is real and even
• B. \(X(e^{j\omega})\) is imaginary and even
• C. \(X(e^{j\omega})\) is real and odd
• D. \(X(e^{j\omega})\) is imaginary and odd

Hint:

Real + even signal gives real + even DTFT; real + odd signal gives purely imaginary + odd DTFT.

Answer 1

The Correct Option is D

Rather than expanding the DTFT sum from scratch, we can use a standard result every signals-and-systems student should know by heart: the symmetry of $x[n]$ (real, even, odd) fixes the symmetry of its DTFT $X(e^{j\omega})$. Let's check each option using that rule.

1. real and even: This is the pairing for a real AND even signal, for example a cosine-shaped sequence symmetric about $n=0$. Our signal is odd, not even, so this option does not apply here.
2. imaginary and even: An "even" DTFT ($X(-\omega) = X(\omega)$) always goes together with an even input signal. Since $x[n]$ here is odd, its transform cannot be even. Wrong.
3. real and odd: A transform that is purely real can only arise from an input sequence that is real AND even. Our $x[n]$ is odd, so its transform is never purely real, ruling this option out too.
4. imaginary and odd: The standard real-sequence symmetry rule states: real + even signal gives a real + even transform; real + odd signal gives an imaginary + odd transform. Since $x[n]$ is given as real and odd, its DTFT must come out purely imaginary and odd in $\omega$. This matches.

A quick way to see why this rule holds: writing $X(e^{j\omega}) = \sum_n x[n]\cos(\omega n) - j\sum_n x[n]\sin(\omega n)$, an odd $x[n]$ makes every term of the cosine sum cancel with its mirror term at $-n$ (cosine is even, so the odd sequence values at $n$ and $-n$ are opposite and multiply by equal cosine factors, giving a perfect cancellation), while the sine sum's terms reinforce each other instead of cancelling (sine is odd, so two sign flips cancel out, leaving the terms equal). The real part vanishes and only the odd imaginary part survives.

Let's summarize:

• Real + even signal gives a real + even DTFT.
• Real + odd signal gives a purely imaginary + odd DTFT.
• Here $x[n]$ is real and odd, so $X(e^{j\omega})$ must be imaginary and odd.

So the correct choice is that $X(e^{j\omega})$ is imaginary and odd, option (D).