A “Beatnote” signal has the following expression: With f 1 = 50 Hz and f 2 = 60

Question

A “Beatnote” signal has the following expression:

With f1 = 50 Hz and f2 = 60 Hz.

By examining the expression of x(t), what are the frequencies exist in x(t)?

Numerically sample x(t) with a sampling frequency fs³ 1000Hz, obtain a sampled sequence {x[n]} with sufficient number of data points.

Conduct FFT on the sampled sequence {x[n]}, and generate the magnitude spectrum chart.

How many dominant frequencies in the spectrum charts, what are the values of the dominant frequencies? How these values related to your answer in step (1)?

A "Beatnote" signal has the following expression: x(t) = cos (2·f1 . t) . cos (2·fy t) Withf1 = 50 Hz and. = 60 Hz. 1. By examining the expression of x(f), what are the frequencies exist in x(t)? 2. Numerically sample x(t) with a sampling frequency fs2 1000Hz, obtain a sampled sequence {x[n]; with sufficient number of data points. Conduct FFT on the sampled sequence x[n]j, and generate the magnitude spectrum chart. How many dominant frequencies in the spectrum charts, what are the values of the dominant frequencies? How these values related to your answer in step (1)? 3. 4.

Explanation / Answer

x(t)= cos (2 *pi *50* t) cos (2 *pi* 60*t) = 1/2 cos (a+b)+ cos (a-b)= 1/2 [ cos (2* pi 110 t ) + cos (-10 t)]

=1/2 cos (2*pi*110 t) - cos2*pi* 10 t)

w1=2 *pi *f1=2* pi* 50

w2=2*pi*f2=2*pi*60

samplinf fequecy= 1000 hz

sampling time T= 1/F= 1/1000

1. the siganal x(t ) is the differnce of two signals cos (2*pi*110t)and cos (2*pi* 10 t)

2. when sampling a contionus signal x(t) we get a discrete signal x[n]= x{nT} whereT is sampling time.

cos (2*pi* 110 t ) becomes cos (2*pi* 110 *1*n /1000) where n= [ ...,-1,0,1,2,....]

cos (2* pi*10 t)becomes cos (2*pi* 10.n*1/1000 where n=[ .....-1,0,1,2....]

this can be solved by plotting athe signal ..

3.the fft will express the finite lenght discrete signal as sum of sinusodal components equally spaced in[0 2pi]

let the lenght of FFt is M then they are spaced at 2pi/M radians

the kth coefficient is related to discrete frequency as Wk=2pi/M radians for k=[0,1,..M-1]

the calculations can be found using matlab or by graphical method

matlab programmig

// To draw x(t)
// To draw cosine wave1
F1=input('enter frequency')
t=0:0.001:1
subplot(4,1,1)
cos_wave1=cos(2*pi*f1*t)
plot(t,cos_wave1)
//To draw cosine wave2

F2=input('enter frequency')
t=0:0.001:1
subplot(4,1,1)
cos_wave2=cos(2*pi*f2*t)
plot(t,cos_wave2)
t=0:0.001:1
x(t)= cos wave1+ cos wave 2
plot(t,x(t

//To find FFt aand magnitude response
x= input(‘enter the sequence’) // Print the content
M= input(‘enter length’) // Print the content
y=fft(x,M) // Discrete Fourier Transform
subplot(3,2,1) // Plotting first section
stem(y) // Plot the discrete sequence data
subplot(3,2,3) // Plotting the third section  
n=0:length(y)-1 // Length of the spectrum
stem(n,abs(y)) // Plot discrete sequence data
subplot(3,2,4) // Plotting the 4th section
stem(n,angle(y)) // Plot discrete sequence data

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