Question

An acoustic signal is composed of the first three harmonics of a wave of fundamental frequency 425 Hz. If these harmonics are described, in order, by cosine waves with amplitudes of 0.630, 0.900, and 0.650, what is the total amplitude of the signal at time 0.825 seconds? Assume the waves have phase angles θ_n = 0.

Answer 1

Fundamental frequency,f1 =425 Hz

So, Ist Harmonic Frequency = 425 Hz

2nd Harmonic Frequency,f2= 425*2=850 Hz

3rd Harmonic Frequency,f3 = 1275 Hz

Ist Harmonic is detemined by following wave :-

y1(t) = 0.630 cos(2*3.14 *f1*t +θn )

2nd Harmonic is detemined by following wave :-

y2(t) = 0.900 cos(2*3.14 *f2*t +θn )

3rd Harmonic is detemined by following wave :-

y3(t) = 0.650 cos(2*3.14 *f3*t +θn )                                   ,Here θn = 0.

Total amplitude of the signal at time t = y1(t) + y2(t) + y3(t)                    ,   (Waves superposition principle)

So, after time t = 0.825 sec

Total amplitude = 0.630 cos(2*3.14 *425*0.825 ) + 0.900 cos(2*3.14 *850*0.825) + 0.650 cos(2*3.14 *1275*0.825 )    

                        =-0.595+ 0.709 - 0.351

                         = -0.237

Total amplitude of the signal at time 0.825 seconds = -0.237