Question

1) Show that cos A * cos B = 0.5 [cos(A+B) + cos(A-B)] . Some English is essential in
addition to the algebra. Note that cosine is an even function so that cos(A-B) = cos(BA).
2) A mixer has a local oscillator at 1157 kHz and a bandpass filter centred at 455 kHz.
What are the centre frequencies of signals which would cause output from the filter?
How is it that a radio receiver accepts only one channel of input signal?

Answer 1

Answer on Question #45204, Engineering, Other

Task: 1) Show that $\cos A \cdot \cos B = 0.5(\cos(A + B) + \cos(A + B))$. Some English is essential in addition to the algebra. Note that cosine is an even function so that $\cos(A - B) = \cos(BA)$.

2) A mixer has a local oscillator at $1157\,\mathrm{kHz}$ and a bandpass filter centered at $455\,\mathrm{kHz}$. What are the center frequencies of signals which would cause output from the filter? How is it that a radio receiver accepts only one channel of input signal?

Solution:

1)

$$\begin{array}{l} \cos (A + B) = \cos A \cos B - \sin A \sin B \\ \cos (A - B) = \cos A \cos B + \sin A \sin B \Rightarrow \\ 0.5(\cos(A + B) + \cos(A + B)) = \cos A \cos B - \sin A \sin B + \cos A \cos B + \sin A \sin B = \\ = 0.5 \cdot 2 \cos A \cos B = \cos A \cos B \Rightarrow \\ \cos A \cos B = \cos A \cos B \end{array}$$

So, we show that $\cos A \cdot \cos B = 0.5(\cos(A + B) + \cos(A + B))$

2) What are the center frequencies of signals which would cause output from the filter?

Answer: $f_{cf} = f_{l} - f_{c} = 1157 - 455 = 702\,\mathrm{kHz}$

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