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

Step 1: Write down the DTFT definition.
The DTFT of a discrete sequence \(x[n]\) is:
\[ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n]\,e^{-j\omega n} \]
Using Euler's formula, \(e^{-j\omega n} = \cos(\omega n) - j\sin(\omega n)\), so we can split the sum into a real and an imaginary part:
\[ X(e^{j\omega}) = \sum_{n} x[n]\cos(\omega n) \;-\; j\sum_{n} x[n]\sin(\omega n) \]

Step 2: Use the fact that \(x[n]\) is real.
Since \(x[n]\) is real, both sums above are real numbers, there is no leftover imaginary contribution hiding inside them; the split above is already the true real and imaginary parts of \(X(e^{j\omega})\).

Step 3: Use the fact that \(x[n]\) is odd.
An odd sequence satisfies \(x[-n] = -x[n]\), and \(x[0] = 0\). Now look at each sum term by term, pairing \(n\) with \(-n\):
For the cosine (real) sum: \(\cos(\omega n)\) is an even function of \(n\), so the term at \(n\) is \(x[n]\cos(\omega n)\) and the term at \(-n\) is \(x[-n]\cos(-\omega n) = -x[n]\cos(\omega n)\). These two cancel exactly, so:
\[ \sum_n x[n]\cos(\omega n) = 0 \]
This means the real part of \(X(e^{j\omega})\) is zero for every \(\omega\), so \(X(e^{j\omega})\) is purely imaginary.
For the sine (imaginary) sum: \(\sin(\omega n)\) is an odd function of \(n\), so the term at \(n\) is \(x[n]\sin(\omega n)\) and the term at \(-n\) is \(x[-n]\sin(-\omega n) = (-x[n])(-\sin(\omega n)) = x[n]\sin(\omega n)\). These two are equal, they add rather than cancel, so the sine sum survives and is, in general, nonzero.

Step 4: Check the symmetry of \(X(e^{j\omega})\) in \(\omega\).
Replacing \(\omega\) by \(-\omega\) flips the sign of \(\sin(\omega n)\) inside the surviving sum, so:
\[ X(e^{-j\omega}) = -j\sum_n x[n]\sin(-\omega n) = -X(e^{j\omega}) \]
A transform that flips sign when its frequency variable is negated is, by definition, odd in \(\omega\).

Step 5: Final Answer.
Combining both results, \(X(e^{j\omega})\) has no real part (purely imaginary) and is odd in \(\omega\). \[ \boxed{X(e^{j\omega}) \text{ is imaginary and odd}} \] This corresponds to option (D).