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[SOLVED] Amplitude and Phase of a sinusoidal signal

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Sonia1234

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Hello friends,

I have a silly question (I admit it's silly but I'm not getting it :cry:).
I have this signal x[n]=Acos(ω n+φ) where I have the value of n and it's corresponding x[n] value. Also I have the value of ω which is 2Π/15. Now all I need is to find the value of amplitude A and φ which I soo not getting it :cry:. Any guidance would be wholeheartedly appreciated :).

Waiting for your inputs!!
 

I assumed you expected one single combination of x[n] and n.

My first guess is that there are infinite combinations of A en phi that fullfill your question. This is a case of one equation with two unknowns. This means you need two combinations of x[n] and n to solve this problem (so it becomes two equations with two unknowns).
 

The phase is not an absolute quantity, it has to be referred to another signal or to a fixed time vector, that mean you know where the time t=0 is

However you can find the amplitede since it hasn't affected by the phase: if x[n] is a time vector, we can write it as xn(t), the you can find the maximum of x[n], this will be A ==> A=max{xn(t)}

In case you have the time vector, simply take note of the time position (in seconds) for which you found the maximum (as said the cos have to start from 0 seconds in you vector). Let's suppose it is t1, since we know the frequency:

Period=1/frequency
1 Period = 2•Π rad

then

phase(rad) = 2•Π•t1/Period

In you case:
frequency=2•Π/15 ==> Period = 15/(2•Π)

phase(rad) = 4•Π²•t1/15
 
Thanks but I think maybe I am forgetting some parameters and confusing you all. Okay so to be clear here's the image of the question.
Part a is fine. When computing part be I get β=2cosω. Then after computing by method cosω=x[n+1]+x[n-1] / 2x[n], I get ω=0.4190=2Π/15.
After this, I would have frequency (f). Okay so now I need to compute A and φ which I am so not getting.
 

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OK, we can take two points. Let's say:

x[m] = A*cos(m*w+fi)
x[n] = A*cos(n*w+fi)

expanding the trigonometry (for sake of simplicity I will omit the [] and I'll call the points xm and xn):

xm = A*[cos(m*w)*cos(fi)-sin(m*w)*sin(fi)]
xn = A*[cos(n*w)*cos(fi)-sin(n*w)*sin(fi)]

solving the first with respect to A:

A = xm/[cos(m*w)*cos(fi)-sin(m*w)*sin(fi)]

substituting this result into the second equation and rearranging:

[xn/xm*cos(m*w)-cos(n*w)]*cos(fi) = [xn/xm*sin(m*w)-sin(n*w)]*sin(fi)

from which:

sin(fi)/cos(fi) = tg(fi) = [xn/xm*cos(m*w)-cos(n*w)]*cos(fi)]/[xn/xm*sin(m*w)-sin(n*w)]

now "fi" is know and you can substitute it into the expression for A.

I found: A=5, fi = Π/5
 
@albbg: This shows that you need at least two combinations of n and x[n] to solve Sonia1234's problem. I was too lazy to do the math, Thanks.
 

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