Pset5 Solution - 1. (4.3) This function has the symmetry...

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1. (4.3) This function has the symmetry property: ) ( ) ( t y t y = or sine-like symmetry. Therefore 0 = n A . ∫∫ + = = = 2 / 02 / 0 2 / 2 / ] ) sin( ) ( ) sin( ) ( [ 2 ) sin( ) ( 2 ) sin( ) ( 2 TT T T T T n dt t n t y dt t n t y T dt t n t y T dt t n t y T B ω ] | ) cos( | ) cos( )[ ( 2 ] ) sin( ) ( ) sin( [ 2 2 / 2 / 0 2 / / T T T T t n t n n F T dt t n F dt t n F T + = + = Recall that π 2 = T . ] ) 1 ( 1 [ 2 )] cos( 1 [ 2 n n n F n n F B = = . e.g. F B 4 1 = , 0 2 = B , 3 4 3 F B = , . .. ...) 5 sin 5 1 3 sin 3 1 (sin 4 ) ( + + + = t t t F t y where lb F 1000 = , sec / 1000 002 . 0 2 2 rad T = = = 2. (5.13) f=50Hz, 95 . 0 ) 2 ( 1 1 2 = + τ f Solve the above equation to get 00105 . 0 = sec. 3. (5/19) 30 . 0 = ξ , 950 = f Hz, 1800 = n f Hz. Error is given by %. 27 27 . 0 1 ) / 2 ( ] ) / ( 1 [ 1 1 2 2 2 = = + = n n s d f f f f P P 439 . 0 ) / ( 1 / 2 tan 2 = = n n f f f f φ , o 7 . 23 = 4. (a) The minimum sampling rate is 4000Hz as dictated by the highest frequency present in the signal.
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