sol6 - -108 MHz. Recall that the ourier Transform of cos(2...

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ECE 351, Systems I Mar. 9, 2009 OSU, Winter 2009 Solutions - Problem Set 6 Problem 1 Let us denote y c ( t ) = x 2 ( t ) + 3 x ( t ). Then, the multiplication x ( t ) · x ( t ) is a convolutionin the frequency domain: Y c ( ) = 1 2 π { X ( ) * X ( ) } + 3 X ( ) . The convolution of X ( ) with itself will result in a signal whose f max is 16 kHz. The f max of 3 X ( ) is 8 kHz.Hence, f max of Y c ( ) becomes 16 kHz, and Y ( e jωT ) will be equal to Y c ( ) (up to a scale factor) if 1 T > 32 kHz. As an alternative viewpoint, consider: cos 2 ( ωt ) = 1 2 (1 + cos(2 ωt )). Thus, the squaring of the input increases the maximum frequency by a factor of 2. Problem 2 We want to Fnd what frequency bands from 250 MHz to 1000 MHz will alias into the 88 MHz to 108 MHz ±M band when sampled at f s = 500 MHz. In other words we want to determine what frequencies from 250 MHz to 1000 MHz “appear” to be in the 88 MHz to 108 MHz band after sampling. We are interested in those f 0 [250 MHz , 1000 MHz] that when x c ( t ) = cos(2 πf 0 t ) is sampled at f s = 500 MHz, the resulting signal has a nonzero frequency component inside 88
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Unformatted text preview: -108 MHz. Recall that the ourier Transform of cos(2 f t ) yields two impulses, one at frequency-f Hz (or equivalently at-2 f rad/sec), and the other at frequency f Hz. When x c ( t ) is sampled, it creates replicas of X c ( f ) with a separation of f s . Thus, we will have impulses at-f + kf s and at f + kf s , for each integer (both negative and positive) k . So, we are interested in identifying those f for which either-f + kf s or f + kf s is inside 88-108 MHz for some integer k . Then, it can be seen that there are 3 bands in the 250 MHz to 1000 MHz region that will alias into the 88 MHz to 108 MHz M band: (a) 392-412 MHz - or 2 392-2 412 rad/sec (b) 588-608 MHz - or 2 588-2 608 rad/sec (c) 892-912 MHz - or 2 892-2 912 rad/sec 1 Problem 3 2...
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sol6 - -108 MHz. Recall that the ourier Transform of cos(2...

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