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 . 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 | kHz f 1 + f 4 = 10 . 5 kHz | f 4 f 1 | =9 . 5 kHz f 2 + f 3 . 4 kHz | f 2 f 3 | =2 . 6 kHz f 2 + f 4 = 10 . 9 kHz | f 2 f 4 | . 1 kHz The components that pass through the Flter are 4 kHz, 4.4 kHz, 9.1 kHz, and 9.5 kHz. Note that when di±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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hwset1solution - 1-1 Solution Signal 1 contains components...

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