How to determine time-domain amplitude from frequency domain

I'm a bit stuck here, so any help is appreciated.

So I'm trying to derive the amplitude (time-domain) of a damped sine from the frequency domain. As a test case I'll be generating my damped sine so I know what the amplitude should look like. For my I application I will not be able to do so as I will be using real signals that I dont have the parameters for (I am actually deriving them!)

The only parameter I'm struggeling with is the amplitude, I got my frequency, phase and damping with less than 1% error. My original signal looks like the snippet below and note that I only have a few samples (100!).

fs = 2e6;
t = 0:(100-1);

b = -zeta*2*pi*freq/fs;     
c = 2*pi*freq/fs;
x = ampl*(exp(b.*t)).*sin(c*t+phi/180*pi);

I apply a Hamming window to my signal and perform an fft. Run it through my algorithm and I get accurate parameters for Freq, Zeta and Phi (frequency, damping and phase). So you may assume that those are known. I do compute a magnitude (i.e. amplitude in frequency domain, I'll refer to it as magnitude to prevent ambiguity) however I'm unable to correlate it to the initial (time-domain) amplitude.

Given that I can reconstruct the original signal exactly with an ifft I would imagine that I could derive the time-domain amplitude, below I've provided a working example that reconstructs the input signal.

%% parameters to construct signal
fs = 2e6;
freq = 180e3;
ampl = 44;
phi  = 0;
zeta = 0.12;

n_sample = 100;
t = 0:(n_sample-1);

%% construct signal
b = -zeta*2*pi*freq/fs;     
c = 2*pi*freq/fs;
x = ampl*(exp(b.*t)).*sin(c*t+phi/180*pi);
x=x';

%% Apply window, zeropad and FFT
w = hamming(length(x));
y = [w.*x; zeros(28,1)];
xFFT = fft(y)/length(y);

%% Inverse FFT
w = [w; ones(28,1)];
z = ifft(xFFT.*length(y))./w;

%% Plot
figure()
hold on
plot(x)
plot(y)
plot(z)
legend('original', 'windowed', 'reverse')

Any ideas on how I can compute the initial amplitude? So in the example above, how I can find 44?

Answer 1

[David Goodmanson]

Hi pimovietc,

I'm not sure what is going on with the use of fs in some of your expressions so I will use a different but similar example. If I understand this correctly, you have a signal

x = A*exp(b.*t).*sin(w*t+phi*pi/180);

and through some process in the frequency domain you have determined b,w and phi. Now you need to find A. It's not even necessary to have phi. Using the same t array as before,

t = t(:);
x = x(:);
u = [exp(b*t).*sin(w*t) exp(b*t).*cos(w*t)];    % two basis vectors
d = u\x;                      % determine basis vector coefficients  
A = norm(d)
phi = atan2d(d(2),d(1))                         % atan2d for degrees

or, if you are confident in your calculation of phi,

t = t(:);
x = x(:);
u = exp(b.*t).*sin(w*t+phi*pi/180);
A = u\x

.

[pimovietc]

Thank you very much for your assistance.

I could no longer see the forest for the trees at this one. In case someone stumbles upon this thread in the future and is also confused by my usage of fs.

Fs is my sampling frequency and its presence within my expressions is to make sure that I'm still working with the correct "relative units" without using the actual units. Just a different method of formulation.

Once again thank you for your assistance!