Question

If x(n)=A, -M<n<M,; x(n)=0, elsewhere. Then what is the Fourier transform of the signal?

(a) A\(\frac{sin⁡(M-\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}\)

(b) A^2\(\frac{sin⁡(M+\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}\)

(c) A\(\frac{sin⁡(M+\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}\)

(d) \(\frac{sin⁡(M-\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}\)

This question was posed to me by my school teacher while I was bunking the class.

My doubt stems from Properties of Fourier Transform  for Discrete Time Signals topic in section Frequency Analysis of Signals and Systems of Digital Signal Processing

Answer 1

Right option is (c) A\(\frac{sin⁡(M+\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}\)

For explanation: Clearly, x(n)=x(-n). Thus the signal x(n) is real and even signal. So, we know that

\(X(ω)=X_R(ω)=A(1+2∑_{n=1}^∞ cos⁡ωn)\)

On simplifying the above equation, we obtain

X(ω)=A\(\frac{sin⁡(M+\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}\).