# exam2soln - BME 311 1[15 Fourier series and transforms...

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BME 311 EXAM #2 SOLUTIONS April 19, 2006 1. [15] Fourier series and transforms: (a) This is a periodic CT signal, so we use the discrete Fourier series to decompose it: c k = X (e jkω 0 ) = 1 T 0 Z T 0 x ( t )e - jkω 0 t d t = 1 2 Z 1 0 e - t e - jkπt d t = 1 2 Z 1 0 e - (1+ jkπ ) t d t = 1 2 - 1 1 + jkπ e - (1+ jkπ ) t ± ± ± 1 0 = 1 2(1 + jkπ ) h 1 - e - (1+ jkπ ) i = 1 2(1 + jkπ ) ² 1 - 1 e ( - 1) k ³ . (b) This is an aperiodic CT signal, so we use the Fourier transform: Y ( ) = Z -∞ y ( t )e - jωt d t = Z 1 0 e - t e - jωt d t = Z 1 0 e - (1+ ) t d t = - 1 1 + e - (1+ ) t ± ± ± 1 0 = 1 1 + ² 1 - e - e ³ . (c) This is a periodic DT signal, so we use the discrete Fourier transform. Although I intended the fundamental period to be N = 10, I accepted other values that were reasonable given the graph of x [ n ]. c k = Z (e jkω 0 ) = 1 N X n = h N i z [ n ]e - jkω 0 n = 1 10 - 1 X n = - 4 z [ n ]e - jkω 0 n = 1 10 ² 1 4 e j 4 0 + 1 2 e j 3 0 + 3 4 e j 2 0 + e jkω 0 ³ . 2. [20] Taking the Fourier transform of the input-output relation, we obtain: d 2 y ( t ) d t 2 + 2 d y ( t ) d t + 4 y ( t ) = x ( t ) + d x ( t ) d t l F - ω 2 Y ( ) + 2 jωY ( ) + 4 Y ( ) = X ( ) + jωX ( ) ´ ( ) 2 + 2 + 4 µ Y ( ) = [1 + ] X ( ) . (a) The frequency response of this receptor is:

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exam2soln - BME 311 1[15 Fourier series and transforms...

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