I am interested in characterising the time-dependent response of a system to a sinusoidal input.

The steady state response of the system is also sinusoidal, however I wonder if there is a standard way to represent the information in a manner that is more detailed than the Frequency Response Function to capture the transient information too (as a function of frequency input)?

E.g. The system where input: $s(t) = \sin(\omega t), 0 \le t < \infty $ gives output: $y(t) = (1+e^{-t}) \sin(\omega t), 0 \le t < \infty$.

The frequency response would be unity, and the impulse response could not be used to capture the transient behaviour?


3 Answers 3


As already pointed out in Hilmar's answer, a linear time-invariant (LTI) system is completely described by its impulse response, or - equivalently - by its frequency response, which is the Fourier transform of its impulse response. I think that there is a misunderstanding concerning the Fourier transform, namely that it can only be used to describe steady-state responses to infinitely long sinusoidal input signals, but this is definitely not the case. I think it's illustrative to look at an example:

Let's assume we have a causal real-valued LTI system with frequency response


where $A(\omega)$ and $B(\omega)$ are the real and imaginary parts of $H(\omega)$, respectively. Now let's define an input signal to this system:


Its Fourier transform is

$$X(\omega)=\frac{\pi}{2j}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)] +\frac{\omega_0}{\omega_0^2-\omega^2}$$

Using $A(\omega)$ and $B(\omega)$, the imaginary part of the Fourier transform of the output signal $y(t)$ can be written as


Since $y(t)=0$ for $t<0$ (because this is true for $x(t)$ and because the system is causal), it is fully described by the imaginary part of its Fourier transform:

$$y(t)=-\frac{2}{\pi}\int_{0}^{\infty}\Im\{Y(\omega)\}\sin\omega t\; d\omega,\quad t>0\tag{2}$$

From (1) and (2) we get

$$y(t) = A(\omega_0)\sin\omega_0t+\frac{2\omega_0}{\pi}\int_{0}^{\infty}\frac{B(\omega)}{\omega^2-\omega_0^2}\sin\omega t\;d\omega,\quad t>0\tag{3}$$

Equation (3) fully describes the response of the LTI system to the causal input signal $x(t)=u(t)\sin\omega_0 t$, including transients. This example is supposed to show that the frequency response (or the impulse response) is sufficient for a complete description of the system's response to arbitrary input signals.

EDIT: Derivation of Equation (2):

Let $Y(\omega)=Y_R(\omega)+jY_I(\omega)$ be the Fourier transform of $y(t)$. We need the following basic properties of the Fourier transform (see here):

$$y(t)\textrm{ real-valued}\Longrightarrow Y(\omega)=Y^*(-\omega)$$ from which follows $$Y_R(\omega)=Y_R(-\omega)\textrm{ and }Y_I(\omega)=-Y_I(-\omega)$$

If $y_e(t)=0.5[y(t)+y(-t)]$ is the even part of $y(t)$ and $y_o(t)=0.5[y(t)-y(-t)]$ is the odd part of $y(t)$ then

$$y_e(t)\Longleftrightarrow Y_R(\omega)\\ y_o(t)\Longleftrightarrow jY_I(\omega)$$ These relation can be found in the link above.

If $y(t)$ is real-valued the inverse transform can be written as

$$y(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}Y(\omega)e^{j\omega t}d\omega= \frac{1}{2\pi}\int_{-\infty}^{\infty}\Re\{Y(\omega)e^{j\omega t}\}d\omega=\\ =\frac{1}{2\pi}\int_{-\infty}^{\infty}[Y_R(\omega)\cos(\omega t)-Y_I(\omega)\sin(\omega t)]d\omega=\\ =\frac{1}{\pi}\int_{0}^{\infty}[Y_R(\omega)\cos(\omega t)-Y_I(\omega)\sin(\omega t)]d\omega$$

where the last equality follows from the fact that the integrand is an even function of $\omega$. Consequently we have

$$y_e(t)=\frac{1}{\pi}\int_{0}^{\infty}Y_R(\omega)\cos(\omega t)d\omega\\ y_o(t)=-\frac{1}{\pi}\int_{0}^{\infty}Y_I(\omega)\sin(\omega t)d\omega$$

With these preliminaries we can finally state the result using the fact that for causal $y(t)$ the following holds for $t>0$ (because $y(-t)=0$ for $t>0$): $$y(t)=2y_e(t)=2y_o(t),\quad t>0$$

This means

$$y(t)=\frac{2}{\pi}\int_{0}^{\infty}Y_R(\omega)\cos(\omega t)d\omega=\\ =-\frac{2}{\pi}\int_{0}^{\infty}Y_I(\omega)\sin(\omega t)d\omega,\quad t>0$$

The last equality is the same as Equation (2) above.

  • $\begingroup$ So is there a function $B(\omega)$ such that the integral in (3) results in something like $$-A(\omega_0)\sin(\omega_0 t) + (1+e^{-t})\sin(\omega_0 t)$$ so that cancellation makes the right side of (3) work out to be exactly $(1+e^{-t})\sin(\omega_0 t)$ as the OP wants it to be? $\endgroup$ Commented Mar 25, 2014 at 22:35
  • $\begingroup$ and where does the $\frac{\pi}{2j}[\delta(\omega - \omega_0) - \delta(\omega + \omega_0)]$ term come from? I've always seen it omitted in Laplace transform tables of $u(t) \sin(\omega_0 t)$ $\endgroup$
    – xyz
    Commented Mar 26, 2014 at 4:11
  • $\begingroup$ @DilipSarwate Of course not, as we both know. My interpretation of the OP's made up example is that he wants to see a combination of steady-state and transient response, and that's also what he gets. Just not in the form he expects it. $\endgroup$
    – Matt L.
    Commented Mar 26, 2014 at 7:32
  • $\begingroup$ @James You won't find it in Laplace transform tables, because the Fourier transform is different from the Laplace transform. If you know the basic Fourier transform pair $e^{j\omega_0t}\Longleftrightarrow 2\pi\delta(\omega-\omega_0)$, then you can probably see how such a term comes about when considering the transform of a sinusoid. $\endgroup$
    – Matt L.
    Commented Mar 26, 2014 at 8:03
  • $\begingroup$ @MattL.: Are the weakly singular infinite integrals such as the one in the example above best tackled with numerical quadrature directly or is there some kind of trick using contour integration that is typically used for this case? $\endgroup$
    – xyz
    Commented Apr 29, 2014 at 20:50

The first question to be decided here if the system whose behavior is described by the input-output relationship:

$$u(t)\sin(\omega t) \to u(t)(1+e^{-t})\sin(\omega t)\tag{1}$$ is indeed a LTI system. The unstated assumption here is that $(1)$ holds for all $\omega$. Now, if the system is indeed a causal LTI system with impulse response $h(t)$, then its response to the signal $u(t)\sin(\omega t)$ is given by $$y(t) = \int_{-\infty}^{\infty} h(\tau)u(t-\tau)\sin(\omega (t-\tau))\,\mathrm d\tau = \int_{0}^t h(\tau)\sin(\omega(t-\tau))\,\mathrm d\tau, \quad t > 0$$ where (as usual) we have changed the limits on the integral by taking advantage of the fact that $u(t-\tau)$ makes a transition from $1$ to $0$ as soon as $\tau$ sweeps past $t$ as $\tau$ increases from $-\infty$ to $\infty$, and second by using our knowledge that $h(\tau) =0$ for all $\tau < 0$. Now, this integral should work out to be of the form $$a(\omega, t)\sin(\omega t) + b(\omega, t)\cos(\omega t) \tag{2}$$ because the antiderivative of $\sin(\omega(t-\tau))$ is $\frac{1}{\omega}\cos(\omega(t-\tau))$ and so $\omega$ should show up somewhere in the coefficients of $\sin(\omega t)$ and $\cos(\omega t)$. But it doesn't; we are told that $(2)$ works out to be $(1+e^{-t})$, that is, $b(\omega, t) = 0$ and $a(\omega, t) = 1+e^{-t}$ is a function of $t$ only. So, there is something fishy here.


Every Linear Time Invariant system is fully described by either it's frequency response or it's impulse response. One can turned into the other via the Fourier Transform.

Either one can be used to model the behavior of the system to anything you want to throw at it. For example if you are interested in the "transient behavior at different frequencies", you could create a sine wave burst (i.e. a short signal with a few cycles of a sine wave embedded in zeros) and convolve this with the impulse response. The result would show the transient response to a "narrow band" signal.

  • $\begingroup$ But take, for example, the system where input $\sin(\omega t)$ gives output: $(1+e^{-t}) \sin(\omega t)$ (is this time invariant?), the frequency response would be unity, and the impulse response could not be used to capture the transient behaviour? $\endgroup$
    – xyz
    Commented Mar 22, 2014 at 10:29

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