General expression for the DTFT of a periodic impulse

Given the DTFT pair 1 $\overset{\mathcal{F}}{\longleftrightarrow }2πδ_{2π}(Ω)$ , use the time-scaling property to find a general expression for the DTFT of $δ_{N_0}[n]$ .

Question

General expression for the DTFT of a periodic impulse

Given the DTFT pair 1 [latex]\overset{\mathcal{F}}{\longleftrightarrow }2πδ_{2π}(Ω)[/latex], use the time-scaling property to find a general expression for the DTFT of [latex]δ_{N_0}[n][/latex].

Accepted Answer

The constant 1 can be expressed as [latex]δ_1[n][/latex] . The periodic impulse [latex]δ_{N_0}[n][/latex] is a time-scaled version of [latex]δ_1[n][/latex]scaled by the integer [latex]N_0[/latex] . That is

[latex]\delta_{N_0}[n]= \begin{cases}\delta_1\left[n / N_0 ight], & n / N_0 ext { an integer } \ 0, & ext { otherwise }\end{cases}[/latex]

Therefore, from (7.20)

[latex] ext { If } \mathrm{z}[n]=\left\{\begin{array}{ll} \mathrm{x}[n / m], & n / m ext { an integer } \ 0, & ext { otherwise } \end{array} ight\} ext { then }\left\{\begin{array}{l} \mathrm{z}[n] \stackrel{\mathcal{F}}{\longleftrightarrow} \mathrm{X}(m F) \ \mathrm{z}[n] \stackrel{\mathcal{F}}{\longleftrightarrow} \mathrm{X}\left(e^{j m \Omega} ight) \end{array} ight\}[/latex] (7.20)

[latex]δ_{N_0}[n]\overset{\mathcal{F}}{\longleftrightarrow }2πδ_{2π}(N_0Ω) = (2π/N_0)δ_{2π/N_0}(Ω)[/latex].

Question 7.4: General expression for the DTFT of a periodic impulse Given ......

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