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For a sine wave, $A \sin(\omega_0 t)$, with an amplitude $A$, after FFT and calculating the magnitude spectrum we can get a single peak with a height also equal to $A$. This requires proper scaling.

Let us say we have an exponentially decaying sine wave $x(t)$, then, the "recovered" peak intensity in the frequency domain is less than $A$ because of dampening the original signal.

$$x(t) = A \sin(\omega_0 t) e^{-\alpha t} \, u(t)$$

where $u(t)$ is the Heaviside unit step function.

I am trying to find an analytical expression to predict the peak amplitude of the exponentially decaying sinuoid in the frequency domain for discrete Fourier transform. Visually, the peak amplitude after FFT is the average amplitude of the damped sine wave in the time domain.

In the continuous version of FT:

$$X(f) = \int_{0}^{\infty} A \sin(\omega_0 t) e^{-\alpha t} e^{-j 2 \pi f t} \, dt$$

The solution is $$ X(f)=\frac{A\omega_0}{(\alpha+j 2 \pi f)^2+\omega_0^2}$$

How can we connect this continuous result with the observation in the discrete Fourier transform to predict the peak maximum value for a given $\alpha$, $A$ and $\omega_0$?

This query is related to spectroscopy because damped sinusoids are used in nuclear magnetic resonance.

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  • $\begingroup$ Y'all need to be a little careful when mixing "$f(t)$" with "$X(f)$" in the same problem. They be different $f$'s. $\endgroup$ Mar 12 at 18:17
  • $\begingroup$ A z transform/CZT analysis might be useful for your problem. $\endgroup$
    – Ash
    Mar 15 at 3:07

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I am trying to find an analytical expression to predict the peak amplitude of the exponentially decaying sinuoid in the frequency domain for discrete Fourier transform.

You'd need to do that in the time domain.

It's simply the roots of the time derivative of $A\sin(\ldots)e^{-\alpha t}$, i.e.

\begin{align} \frac{\mathrm d} {\mathrm dt} x(t) &= \frac{\mathrm d} {\mathrm dt} \left(A\sin(\omega_0 t) e^{-\alpha t}\right) \\ &= A \left[ \left( \frac{\mathrm d}{\mathrm dt} \sin(\omega_0 t) \right) e^{-\alpha t} + \sin(\omega_0 t) \left( \frac{\mathrm d}{\mathrm dt} e^{-\alpha t} \right) \right] \\ &= A \left[ \omega_0\cos(\omega_0t) e^{-\alpha t} + -\alpha\sin(\omega_0t)e^{-\alpha t} \right] \\ &\overset!=0 \\ &\iff \\ \omega_0\cos(\omega_0t) e^{-\alpha t} &= \alpha\sin(\omega_0t)e^{-\alpha t} &\|e^{\ldots} \ne 0 \\&\iff\\ \\ \omega_0\cos(\omega_0t) &= \alpha\sin(\omega_0t) \\ \frac{\omega_0}{\alpha}&= \tan(\omega_0t) \\&\iff\\ \omega_0 t &= \arctan\left(\frac{\omega_0}{\alpha} \right) + n\pi, \quad n\in \mathbb N_0\\\ t &= \frac{\arctan\left(\frac{\omega_0}{\alpha} \right)}{\omega_0} + n\frac{\pi}{\omega_0} \end{align}

Note that these are the times of the extrema, not only of the maxima. However, since the dampened sine still alternates between maxima and minima, the set of maximum times is $$T_\max= \left\{ \frac{\arctan\left(\frac{\omega_0}{\alpha} \right)}{\omega_0} + (2n+1)\frac{\pi}{\omega_0} \middle| n\in\mathbb N \right\} $$

Insert these $t$ into your $x(t)$ and you get the values of the local maxima.

How can we connect this continuous result with the observation in the discrete Fourier transform to predict the peak maximum value for a given $\alpha$, $A$ and $\omega_0$?

You do the inverse discrete Fourier transform, and then work in time.

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    $\begingroup$ Thank you Marcus. I got your point in finding the time and corresponding amplitudes of the exponentially damped sinusoid at the extrema. We will have a set of maximum values in $T_{max}$ in the time domain. The key question is when we do FFT of $f(t)$, there will be a single peak at $\omega_0$. What will be the theoretical amplitude of this single peak in the frequency domain? $\endgroup$
    – AChem
    Mar 12 at 17:50
  • $\begingroup$ you should see two peaks, not one; this signal is real, as you can see by the denominator of your continuous-time FT. And due to leakage, that peak will also not lie on a single bin, and the amplitude would hence depend not only on $\omega/\alpha$, but also on the relative position of $\omega$ and your sampling rate and your FFT length. $\endgroup$ Mar 12 at 17:58
  • $\begingroup$ note that the inherent windowing due to the finite length of the DFT introduces another convolution with a sinc function; hence, be a bit careful when going the other way around, i.e., when estimating the parameters from position and value of the peaks in the DFT. $\endgroup$ Mar 12 at 18:04
  • $\begingroup$ Right, I was talking about one-sided spectrum which is used for chemical spectroscopy. This part has been pretty confusing as to how we can predict the "theoretical" amplitude of the damped sinusoid in the frequency domain. As you said that it depends on $\omega$ and $\alpha$, but what will be the actual relations. This is why I was trying the CT FT. Let me mention that "observationally" the amplitude is the average value of the extrema of the sampled data. Hard to discover the mathematical basis of this "average." $\endgroup$
    – AChem
    Mar 12 at 18:09
  • $\begingroup$ @AChem it's not too surprising that it's at least proportional to that; you multiply a sine with an exponential in the time domain, so in the frequency domain, you get the convolution of the Fourier transform of both – so you convolve the Fourier transform of a sine, which is two Dirac deltas, each of amplitude $\frac A{2i}$ with the FT of the exponential decay. Convolution with a delta is just a shift, so all you get at the position of the Dirac delta is $\frac A{2i}$ times the zeroth frequency coefficient of that FT of the decay, plus the alias from the higher-frequency components of… $\endgroup$ Mar 12 at 18:51

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