bon said:Homework Statement
ok so I am given that I(t) = A cos (w1 t)cos(w2 t)
where w2<<w1
then I am asked to express I as the sum of two cosine functions with angular frequences P and Q which I have:
I = A/2 (cosPt + cosQt) where P = w1+w2 and Q=w1-w2
Im then asked to evaluate the bandwith in terms of w1 and w2
Is this just |P-Q| in which case it would be 2w2? I am confused :S thanks
The formula for finding the sum of two cosine functions with angular frequencies is:
f(x) = A1cos(ω1x + φ1) + A2cos(ω2x + φ2), where A1 and A2 are the amplitudes, ω1 and ω2 are the angular frequencies, and φ1 and φ2 are the phase shifts of the two cosine functions.
Yes, the sum of two cosine functions with different angular frequencies can be simplified using the trigonometric identity:
cos(x+y) = cos(x)cos(y) - sin(x)sin(y)
The amplitudes and phases of the two cosine functions affect the sum by determining the overall shape and position of the resulting waveform. The amplitudes determine the maximum and minimum values of the sum, while the phases determine the starting point and direction of the waveform.
Yes, the sum of two cosine functions can have a negative amplitude if the two individual cosine functions have opposite amplitudes and are out of phase by π radians (180 degrees).
The angular frequency affects the period of the sum of two cosine functions by determining how many cycles occur within a given interval. The higher the angular frequency, the more cycles will occur and thus the shorter the period will be.