InterviewSolution
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InterviewSolution
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What is the Fourier transform of x(t)?(a) X (F) = \(\frac{1}{2} [X_l (F-F_c)+X_l^* (F-F_c)]\)(b) X (F) = \(\frac{1}{2} [X_l (F-F_c)+X_l^* (F+F_c)]\)(c) X (F) = \(\frac{1}{2} [X_l (F+F_c)+X_l^* (F-F_c)]\)(d) X (F) = \(\frac{1}{2} [X_l (F-F_c)+X_l^* (-F-F_c)]\)This question was posed to me in unit test.I want to ask this question from Sampling of Band Pass Signals topic in portion Sampling and Reconstruction of Signals of Digital Signal Processing |
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Answer» CORRECT choice is (d) X (F) = \(\FRAC{1}{2} [X_l (F-F_c)+X_l^* (-F-F_c)]\) For EXPLANATION: X (F) = \(\int_{-\infty}^∞ x(t)e^{-j2πFt} dt\) =\(\int_{-\infty}^∞ \{Re[x_l (t) e^{j2πF_c t}]\}e^{-j2πFt} dt\) Using the identity, Re(ε)=1/2(ε+ε^*) X (F) = \(\int_{-\infty}^∞ [x_l (t) e^{j2πF_c t}+x_l^* (t)e^{-j2πF_c t}] e^{-j2πFt} dt\) =\(\frac{1}{2}[X_l (F-F_c)+X_l^* (-F-F_c)]\). |
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