Sine wave multiplication

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

  1. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    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.
     
    geft likes this.
  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).

    Check this link:

    http://en.wikipedia.org/wiki/Superheterodyne_receiver
     
    Last edited: Dec 15, 2011
    geft likes this.
  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.
     
Loading...