where A is the amplitude and R the phasor. A 1 and 1 are respectively the amplit
ID: 1833626 • Letter: W
Question
where A is the amplitude and R the phasor.
A1 and 1 are respectively the amplitude and phase of the signal received over path1
A2 and 2 are the amplitude and phase of the signal received over path2
t1 is the propagation delay corresponding to path1
t2 is the propagation delay corresponding to path2
f is the frequency
Therefore, the result shows that amplitude A of the received signal depends on the phase difference 2-1 , on the delay difference t2-t1, and , most interestingly, on the frequency f.
where f0 is the frequency for which 0=d2-d1, the path difference, and =2-1.
A. what is the peak-to-peak(or null-to-null) frequency difference?
B. verify that the expression for the peaks predicts the values observable in the following Figure:
Explanation / Answer
We have the term cos(-2pi*f(t2-t1) + (phi2-phi1)) which can also be represented as cos(2*pi*f*(t2-t1) - (phi2-phi1)) as cos(-theta) = cos(theta) t2- t1 is essentially the time lag between the receiving of the two signals. which is due to difference between the distance. This is represented as t2-t1 = (d2-d1)/velocity of the wave This determines whether the waves construct or destruct. as, for constructive intereference d2-d1 = lambda velocity of wave = lambda * f0 t2-t1 = 1/f0 putting the value back in the cosine function maxima of cosine occurs at cos(2*n*pi) so, 2*n*pi = 2*pi*f/f0 - ?f f = n*f0 + ?f*f0/2pi for destructive interference cos((2n+1)pi/2) should be used (2n+1)*pi/2 = 2*pi*f/f0 - ?f f = (2n+1)f0/2 + ?f*f0/(2pi) f = n*f0 + f0/2 + ?f*f0/(2pi) b) frequency difference is delta f = (n+1) f0 + ?f*f0/2pi - nf0 - ?f*f0/2pi delta f = f0 c) to do the third part i need the value of f0 so that i can compare! I hope this solves your problem.
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