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?
Answer
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
$$H(\omega)=A(\omega)+jB(\omega)$$
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:
$$x(t)=u(t)\sin(\omega_0t)$$
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
$$\Im\{Y(\omega)\}=-\frac{\pi}{2}[A(\omega_0)\delta(\omega-\omega_0)-A(-\omega_0)\delta(\omega+\omega_0)]-\frac{\omega_0B(\omega)}{\omega^2-\omega_0^2}\tag{1}$$
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.
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