# I'm really confused about this bit in my assignment. Please include as much detail as possible. Thank youThis is part of an assignment. I was just wondering how a beat frequency is produced when...

I'm really confused about this bit in my assignment. Please include as much detail as possible. Thank you

This is part of an assignment. I was just wondering how a beat frequency is produced when two high frequency AC currents combine? Also why is the beat frequency equal to the difference between the two high frequencies?

As i said, please include as much detail as possible. Thanks heaps

### 1 Answer | Add Yours

Suppose for simplicity that the amplitudes of the two AC currents (signals) are identical.

`y_1(t) = A*cos(omega_1*t)`

`y_2(t) = A*cos(omega_2*t)`

We know the basic trigonometric identities (sum and difference of angles)

`cos(a+b) = cos(a)*cos(b) -sin(a)*sin(b)`

`cos(a-b) = cos(a)*cos(b) +sin(a)*sin(b)`

Summing together we get

`cos(a+b) + cos(a-b) = 2*cos(a)*cos(b)`

By taking `a = (omega_1*t + omega_2*t)/2`and `b = (omega_1*t -omega_2*t)/2` we obtain

`cos(omega_1*t) +cos(omega_2*t) = 2*cos((omega_1*t +omega_2*t)/2)*cos((omega_1*t -omega_2*t)/2)`

or

`cos(omega_1*t) +cos(omega_2*t) = 2*cos[2*pi*((F_1 +F_2)/2)*t]*cos[2*pi*((F_1 -F2_2)/2*t])`

This means the frequency of the beats is half the frequency of the sum or difference of the frequencies original signals.

In the case the signals are not electrical but audible sounds, the ear perceives a sound with the frequency `(F1+F2)/2` being modulated in amplitude by `cos[2*pi*((F1-F2)/2)*t]` . But because the ear can feel only the envelope of the sound (in other words only the absolute value of the `cos[...]` ) it feels like the frequency of the beats is equal to the difference of original frequencies.