# Sine wave multiplication

Discussion in 'Homework Help' started by geft, Dec 15, 2011.

1. ### geft Thread Starter New Member

Dec 8, 2011
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0
A radio receiver is designed using a band-pass filter with frequency centered on 455kHz. It can be tuned to different stations by multiplying the incoming signal by a sine wave of correctly chosen frequency. The incoming signal is a 600kHz sine wave. What must be the frequency of the multiplying sine wave to make the signal pass through the filter?

All I have is a trigonometric identity. I'm confused. How do I even multiply sine waves?

2. ### MrChips Moderator

Oct 2, 2009
12,436
3,360
You have two choices, 600-455 kHz or 600+455 kHz, take your pick.
In mathematics x = A1*sin(w1*t) * A2*sin(w2*t)
In electronics v = v1*v2 using a transconductance amplifier or something similar.

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3. ### geft Thread Starter New Member

Dec 8, 2011
19
0
Thanks, but could you please elaborate on the concept? How does adding or subtracting a frequency make the sine wave pass through the filter?

4. ### MrChips Moderator

Oct 2, 2009
12,436
3,360
When you multiply two sine waves, you end up with the sum and difference frequencies.

sin(at)*cos(bt) = [sin(a-b)t + sin(a+b)t]/2

sin(at)*sin(bt) = [cos(a-b)t - cos(a+b)t]/2

So if the input frequency is 600KHz and the local oscillator frequency is 1055kHz, you end up with 455kHz and 1655kHz. Your IF filter is tuned to 455kHz and so you reject the 1655kHz signal. (IF stands for Intermediate Frequency).

Last edited: Dec 15, 2011
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5. ### lopoditi New Member

Jul 24, 2011
6
1
the multiplying sine wave should be 145 kHz

6. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
@lopoditi

Here in Homework Help forum, the main goal is to educate the inquirer, not give him a plain result as an answer. If you feel you can contribute, please explain the reasoning behind your answers, and if possible don't give the OP the answer before he tries a bit himself, after he is being given a hint or two.

Thank you.

7. ### Papabravo Expert

Feb 24, 2006
10,138
1,789
It could be but it doesn't have to be because the solutions to the problem are NOT unique. When you use the lower frequency (145 kHz), the sum and the difference are closer together requiring a filter with steeper skirts. When you use the higher frequency (1055 kHz), the sum and the difference are further apart which makes the filtering easier.