Draw Signal graph from a frequency domain representation

How can i Draw the signal graph from this frequency domain over interval [0; 50] ms

Answer 1

[Star Strider]

That is not possible with only the information shown in the plot image.

Inverting the Fourier transform of the signal requires that both the real and imaginary parts be present. These can be recovered from the amplitude and phase spectrum data if both are provided (that is essentially trivial), however without the phase data, half the necessary data are lost, and so inverting the Fourier transform is not possible.

[Md Naeemur Rahman]

Actually this is the question i received with that figure and I was asked to draw signal graph-

[Star Strider]

O.K., then forget about using the phase information and doing an inverse Fourier transform. Just add sine curves with the appropriate frequencies and amplitudes. Choose an appropriate sampling frequency (two times the Nyquist frequency) and create a time vector using it that goes from 0 to 50 ms.

[Md Naeemur Rahman]

I was trying to figure out the equation, I suppose this would be like this-

1.5+2*sin(2*pi*100t)+sin(2*pi*300t)+0.5*sin(2*pi*400*t) where t 0 to 50 ms

However i'm bit confused, how can i implement this in Matlab code?

[Md Naeemur Rahman]

i=1:1:5;
f=[0 100 200 300 400];
A=[1.5 2 0 1 0.5];
a=A.^0.5;
t=0:0.001:1000;
for i=1:5
for j=1:length(t)
harm(i,j)=a(i)*(sin(((2*pi/f(i))*t(j))));
end
end
signl=nansum(harm);
plot(t,signl)

This is my final approach , Is this okay ?

[Star Strider]

I just now ran it and it runs without error.

I suggest that you reconsider ‘t’ since it does not appear to satisfy the requirements.

[Md Naeemur Rahman]

I'm stuck here! Any suggestion?

[Md Naeemur Rahman]

t=0: 1e-3: 50e-3;
y1 = 1.5+2*sin(2*pi*100*t)+sin(2*pi*300*t)+0.5*sin(2*pi*400*t);
p = plot(t,y1);

I also tried this way, I'm not sure which approach would be more accurate

[Star Strider]

Yes!

Choose an appropriate sampling frequency that meets the requrements for the 50 ms limit and still produces the appropriate result. It may be necessary to do a fft on the result to check it.

An example of that would be —

t = ...;                                                % Time Vector
s = ...;                                                % Signal Vector
L = numel(t);                                           % Signal Length
Ts = mean(diff(t));                                     % Sampling Interval
Fs = 1/Ts;                                              % Sampling Frequency
Fn = Fs/2;                                              % Nyquist Frequency
nfft = 2^nextpow2(L);                                   % FFT Length
FTs = fft(s,nfft)/L;                                    % Fourier Transform
Fv = linspace(0, 1, nfft/2+1)*Fn;                       % Frequency Vector
Iv = 1:numel(Fv);                                       % Index Vector

figure
stem(Fv, abs(FTs(Iv))*2)
grid

With —

t = linspace(0,10,500);
s = sin(2*pi*10*t);

(that is simply for demonstration purposes and has no relation to your assignment) that code produces:

L = numel(t);                                           % Signal Length
Ts = mean(diff(t));                                     % Sampling Interval
Fs = 1/Ts;                                              % Sampling Frequency
Fn = Fs/2;                                              % Nyquist Frequency
nfft = 2^nextpow2(L);                                   % FFT Length
FTs = fft(s,nfft)/L;                                    % Fourier Transform
Fv = linspace(0, 1, nfft/2+1)*Fn;                       % Frequency Vector
Iv = 1:numel(Fv);                                       % Index Vector

figure
stem(Fv, abs(FTs(Iv))*2)
grid

So you can use it to check the result of your code. It should reproduce the plot you were initially given, and that satisfy the other requrements.