# Time domain and frequency domain analysis equivalence

Let us imagine an LTI system with physically realizable input (ruling out fancy mathematical functions and the concomitant complexities and paradoxes) completely known from -$\infty$ to $\infty$. We want to calculate the output. We can analyse it in time domain using the linear constant coefficient differential equation and in the frequency domain using the Fourier representation of the signal and frequency response of the system.

The outputs calculated from both the methods should be identical. In time domain, we get (transients + steady state) response of the system, In frequency domain, the sinusoids are eternal and hence transients are generally neglected. But since we are talking about the same system with the same input, the outputs calculated from both the methods should be the same.

Will they be identical in spite of neglecting the transients in the frequency domain?

• It is not true that in the frequency domain "transients are generally neglected". Where from did you get this idea? Time and frequency domain analyses must and will give the same result. – Matt L. Jun 5 '16 at 8:13
• @ Matt L:I got the idea from this. Consider finding the output of a filter to a given input signal. We say that the input signal has these, these components some of which are removed by the filter and we are left with this particular frequency etc. So here, we are here not considering the transient response of the filter isn't it? We are only considering the steady-state behaviour of the filter, I guess. – Seetha Rama Raju Sanapala Jun 5 '16 at 12:51
• This is too simplistic a view. It sounds like you only consider signals with discrete frequency components. If your input signal is a sinusoid that is switched on at a certain time, then there will be transients in the output signal, no matter if you compute it in the time domain or in the frequency domain. – Matt L. Jun 5 '16 at 14:17
• @ Matt L:I understand that as I stated in my question itself - we should get the same result from both the methods. I am considering a general problem not just discrete frequency components. I stated in my comment the generally given explanation in frequency domain, Is there a reference where a simple problem is solved in time and frequency domains and shown to be identical?. – Seetha Rama Raju Sanapala Jun 5 '16 at 14:35
• Also have a look at this answer of mine to a related question. – Matt L. Jun 5 '16 at 14:35

now, even a steady-state sinusoid can be thought of, in the limit, as a sum of weighted pulses, each with a beginning and some kinda end. how do all of these "transient" signals add up to a steady-state sinusoid? but they do:

\begin{align} x(t) &= \lim_{T \to 0} \sum_{n=-\infty}^{+\infty} x(nT) \cdot \operatorname{rect}\left( \frac{t-nT}{T} \right) \\ &= \lim_{T \to 0} \sum_{n=-\infty}^{+\infty} x(nT) \cdot \frac{1}{T}\operatorname{rect}\left( \frac{t-nT}{T} \right) \cdot T \\ &= \int\limits_{-\infty}^{\infty} x(u) \cdot \delta(t-u) \cdot du \end{align}

where $\operatorname{rect}(u) \triangleq \begin{cases} 1 \quad |u|<\tfrac12 \\ 0 \quad |u| > \tfrac12 \end{cases}$

The Fourier integral models the entire time function as an infinitely dense and infinitely large set of sinusoids. in a similar manner, those steady-state functions can add to something with a transient.

$$x(t) = \int\limits_{-\infty}^{+\infty} X(f) \, e^{j2\pi ft} \ df$$

• In your answer, your first equation seems to be right but I get a doubt, should rect be replaced by delta function that is generally used in text books to represent discrete signals? – Seetha Rama Raju Sanapala Jun 5 '16 at 3:37
• it becomes something that looks like a delta function in the limit. i will modify it a little to show. – robert bristow-johnson Jun 5 '16 at 3:46

Your premise is false. The frequency domain does not neglect transients.

However finite transients in one domain become infinite in extent in the other domain. e.g. A finite width rect in one domain becomes an infinite width Sinc in the other FT domain to correctly and completely represent the rect.

Because of this, one can be fooled by the illusion of neglect (of transients) by not looking at enough of the frequency response (or enough sinusoids).