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hwset1solution - 1-1 Solution Signal 1 contains components...

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1-1 Solution Signal 1 contains components at f 1 = 0 . 5 kHz and f 2 = 0 . 9 kHz and signal 2 contains components at f 3 = 3 . 5 kHz and f 4 = 10 kHz. The output of the multiplier is [cos ω 1 t + cos ω 2 t ][cos ω 3 t + cos ω 4 t ] . When expanded, the product includes 4 terms: cos ω i t cos ω j t , where ( i, j ) { (1 , 3) , (1 , 4) , (2 , 3) , (2 , 4) } . Using the identify cos ω i t cos ω j t = 1 2 cos( ω i + ω j ) t + 1 2 cos( ω i ω j ) t it is clear that the output of the multiplier contains the following frequencies: f 1 + f 3 = 4 kHz | f 1 f 3 | = 3 kHz f 1 + f 4 = 10 . 5 kHz | f 4 f 1 | = 9 . 5 kHz f 2 + f 3 = 4 . 4 kHz | f 2 f 3 | = 2 . 6 kHz f 2 + f 4 = 10 . 9 kHz | f 2 f 4 | = 9 . 1 kHz The components that pass through the filter are 4 kHz, 4.4 kHz, 9.1 kHz, and 9.5 kHz. Note that when di ff erence frequencies are being calculated the absolute value is used to ensure that the result is positive. In other words, since we are discussing real-valued signals, every negative-frequency component is accompanied by a positive-frequency counterpart. When asked for “the frequency” of a sinusoidal signal, we report the positive number.
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