Question

The DTFT of a sequence \(x\left[ n \right]\) is given by \(X\left( {{e^{iω }}} \right)\). Since \(X\left( {{e^{iω }}} \right)\) is period function of ω, it can be expressed classical Fourier series as,

\(X\left( {{e^{iω }}} \right) = \sum\limits_{n = - \infty }^\infty {{C_n}{e^{in{ω _0}ω }}} \)

Where ω₀ is a fundamental frequency. Which of the following statement is correct?

1. \({\omega _0} = \pi\), \({C_n} = - x\left[ n \right]\)
2. \({\omega _0} = \pi\), \({C_n} = x\left[ -n \right]\)
3. ω₀ = 1, \({C_n} = x\left[ -n \right]\)
4. ω0 = 1, \({C_n} = - x\left[ -n \right]\)

Answer 1

Correct Answer - Option 3 : ω₀ = 1, \({C_n} = x\left[ -n \right]\)

Concept:

If f(n) is a signal then its DTFT will be defined As.

\(F\left( {{e^{j\omega }}} \right) \leftrightarrow \;\mathop \sum \nolimits_{n = - \infty }^\infty f\left( n \right){e^{ - j\omega n}}\)

Analysis:

x(n) ↔ X(e^jω)

\(X\left( {{e^{j\omega }}} \right) = \mathop \sum \nolimits_{n = - \infty }^\infty x\left( n \right){e^{ - j\omega n}} \ ..(i)\)         

It is given that:

 \(X\left( {{e^{j\omega }}} \right) = \mathop \sum \nolimits_{n = - \infty }^\infty {C_n}{e^{jn{\omega _0}\omega }} \ ..(ii)\)        

On Replacing (n) by (-n) in equation (i)

\(X\left( {{e^{j\omega }}} \right) = \mathop \sum \nolimits_{n = - \infty }^\infty x\left( { - n} \right){e^{j\omega n}} \ ..(iii)\)         

On comparing equation (ii) and (iii)

C_n = x(-n)

ω₀ = 1