Match List - I with List - II. Choose the correct answer from the options given below :List - I (Sequence x[n]) List - II (Fourier Transform X(Ω)) (A) \(e^{j\Omega_{0}n}x[n]\) (I) \(\left(1-e^{-j\Omega}\right)X(\Omega)\) (B) \(n\,x[n]\) (II) \(X(\Omega-\Omega_{0})\) (C) \(x[n]-x[n-1]\) (III) \(e^{-j\Omega n_{0}}\) (D) \(\delta[n-n_{0}]\) (IV) \(j\dfrac{dX(\Omega)}{d\Omega}\)
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Each entry is one standard property of the discrete-time Fourier transform: (A)-(II), (B)-(IV), (C)-(I), (D)-(III) — option 2.
(A) — frequency shifting. Multiplying a sequence by a complex exponential translates its spectrum:
\(e^{j\Omega_{0}n}x[n]\ \longleftrightarrow\ X\left(\Omega-\Omega_{0}\right)\)
It is the dual of time shifting, and it is exactly what modulation does — multiplying by a carrier moves the baseband spectrum up to the carrier frequency.
(B) — differentiation in frequency. Differentiating the transform brings down a factor of n:
\(n\,x[n]\ \longleftrightarrow\ j\dfrac{dX(\Omega)}{d\Omega}\)
This can be seen directly from the definition \(X(\Omega)=\sum x[n]e^{-j\Omega n}\): differentiating with respect to \(\Omega\) brings down \(-jn\) from the exponent, and multiplying by j corrects the sign.
(C) — the first difference. Time shifting by one sample multiplies the transform by \(e^{-j\Omega}\), so by linearity
\(x[n]-x[n-1]\ \longleftrightarrow\ X(\Omega)-e^{-j\Omega}X(\Omega)=\left(1-e^{-j\Omega}\right)X(\Omega)\)
The factor \(\left(1-e^{-j\Omega}\right)\) vanishes at \(\Omega=0\), which is the discrete-time counterpart of differentiation removing a constant — a first difference is a high-pass operation.
(D) — the shifted impulse. The sum collapses to a single term:
\(\sum_{n}\delta\left[n-n_{0}\right]e^{-j\Omega n}=e^{-j\Omega n_{0}}\)
Its magnitude is 1 at every frequency — an impulse contains all frequencies equally — and the shift appears entirely as a linear phase. This is the origin of the rule that a pure delay is a linear-phase system.
A quick way to check the code. Only (D) has a transform that is a bare function of \(\Omega\) with no \(X\) in it, since \(\delta[n-n_{0}]\) is the only entry in List I that is a specific sequence rather than a general one. Pairing (D) with (III) alone eliminates options 1 and 3, and (B) with (IV) settles the rest.
Hence, the correct code is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).
Similar Questions
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Important Questions from Properties of Fourier Transform
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