I'm a bit confused with some details about frequency domain or studying systems with Laplace and Fourier analysis.

Let's say we have a system with a transfer function $H(s)$ with no poles and zero's at origin. We know that the transfer function is a ratio between output and input. Also, an input of $X(s)=1$ represents a single impulse (Dirac) in time domain, which can be viewed as a signal containing all frequencies in frequency domain. Also, applying this input to the system will lead to an output $Y(s)$ which is equal to $H(s)$ which is like the system's unique characteristic (its transfer function) and also can be viewed as a response to all frequencies. This signal $H(s)$ can be transformed back to time domain.

Let's say we have a system whose impulse response is something like this:

Impulse response

When I studied Fourier Transform we used to study system's response and Bode plot by analysing variations in $\omega$. So, for example, making $\omega=0$ we could find the system's DC gain. Making $\omega$ low values we can observe the system response to low frequencies and the same for high frequencies with high values in $\omega$. It was very useful to study filters, for example.

But note that all of this is only using the transfer function. And the transfer function may have an appearance like this above. And this does not make sense to me. If we think of the time->frequency transform, we should think that in frequency domain we are finding all frequencies which compose our signal in time domain, correct? This explains the Dirac containing all frequencies, for example. So, when making s=0 we should find H(s)=0 since this signal above does not have a DC level.

When we think about the transfer function as a system response to $X(s)=1$ it seems ok. But if we think about the transfer function as the same representation of this signal above, it does not make any sense. Also, if we apply a step signal, our system may respond like this:

Step response

We saw that for this system (with no poles at origin), we can find something like $H(0)=K$ which represents the system's DC gain. Applying a step response will lead us to a signal $\frac{H(s)}{s}$. So now making now s=0 will lead us to $+\infty$. Again, if we think of this signal being the one above but represented in frequency domain, making s=0 should left us a DC finit level since the time domain representation show us that the signal contain a limited DC level.

I know the final and initial value theorems but still I want to understand these relations between these domains.

Note that if we want to study a filter response, we should apply some signal to the input and see what is the output. Mathematically we should convolute these signals in time domain or multiply in frequency domain and see what is the resulting signal. In time domain it all makes sense since we just convolute and see the system's response. In frequency domain it also makes sense just like the case where you use $X(s)=1$ which represents all frequencies and multiply by the system and see the resulting signal. But when thinking about the same signal being represented in both domains, I think I'm messing things up.

I just want a help to understand relations between 's' variable and time domain. Making s=0 shouldn't lead me to the DC component of the signal in time domain? Generically, each value of 's' shouldn't lead me to 'how much' of this frequency is present in this signal?

in short:

A signal like the first one does not have any DC component so it goes to zero when $t=\infty$. So why its representation in frequency domain have a DC component when we put s=0? Also, the second signal does have a DC level when $t=\infty$. But in frequency domain the signal goes to $\infty$ when s=0 (which I thought it should represent DC level). It happens with Heaviside function too. Its representation in frequency domain is $\frac{1}{s}$ altough when s=0 it goes to $\infty$. But it has a DC limited component. How do we explain all of that?

  • $\begingroup$ Your question needs to be more specific. Maybe something along the lines of "How do I get the transfer function from this impulse response?", since the main problem seems to be that you don't understand how to get a transfer function from the impulse response. $\endgroup$
    – Jim Clay
    Commented Dec 27, 2013 at 16:23
  • $\begingroup$ I do not want to know any answer. I know how to calculate these things. I just want to understand those meanings that are specified in the question. $\endgroup$ Commented Dec 27, 2013 at 18:11
  • $\begingroup$ I've just edited my question and added some extra info. Thanks for commenting anyway. I hope it helps. $\endgroup$ Commented Dec 27, 2013 at 18:22

2 Answers 2


The impulse response and the transfer function of an LTI system are different, sometimes extremely different (except for a specific shape of Gaussian).

The DC component of any impulse response is it's average value. Your example time-domain squiggle seems to have a non-zero average value (integrate to determine the area under the curve), thus is has a non-zero DC component in the frequency domain.

  • $\begingroup$ Isn't the transfer function always a ratio between output and input? The transform of impulse isn't equal to one in frequency domain? So, why impulse response is not always equal to the transfer function? $\endgroup$ Commented Dec 27, 2013 at 18:14
  • $\begingroup$ The Fourier transform of an impulse is not an impulse (quite the opposite). A time-domain impulse contains all frequencies, therefore it is non-zero from -inf to +inf in the other domain. And vice versa. $\endgroup$
    – hotpaw2
    Commented Dec 27, 2013 at 18:45
  • $\begingroup$ I think you misunderstood me. I said that the impulse (dirac) in time domain can also be represented in the frequency domain as a function F(s)=1. Isn't it? So, if the transfer function is always a ratio between Y(s) and X(s) (output and input), why the transfer function is not always equal to the impulse response (X(s)=1)? $\endgroup$ Commented Dec 27, 2013 at 19:04

There appears to be some confusion between Laplace and Fourier ,,frequencies'' - Actually, only the bases of the Fourier domain are periodic. The Laplace domain is only formally similar. Actually, in both cases, a parameter of $0$ will display ,,the DC part of the signal'', which makes it even harder to understand. Additionally, you seem to confuse $t$ and $s$. With $s=0$, you're trying to express a step at time $0$, which the Laplace transform just doesn't do well. You could take a limit, however.


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