2
$\begingroup$

It sounds like a very elementary question on system theory but I got quite confused about it, so hopefully you guys can enlighten me.

I'm considering an ideal analog integrator, i.e., a system with transfer function $G(s) = \frac 1s$. I'm aware of the fact that it is marginally stable and cannot be built like that, it's more of a theoretical question really.

The impulse response of the system is a unit-step function $g(t) = \sigma(t)$ (considering one-sided Laplace transform here).

Now, I was wondering what happens if I excite this signal with a cosine function of non-zero frequency $f_0$. First gut feeling: I would expect no blow-up as the cosine oscillates and hence the integrator should give us again a harmonic of the same frequency. The system is linear after all. Also, its transfer function does not have a singularity for any nonzero frequency, so again, no blow-up expected, things should work nicely. However, how to compute it really?

Options I see:

  • Time domain. Let's forget the Laplace for a second and just focus on the fact that the impulse response of our ideal integrator is a step function, so all we need to do is convolve with it: $$ \begin{align} y(t) & = \int_{-\infty}^\infty \sigma(\tau) \cos(2\pi f_0 (t-\tau)) {\rm d}\tau \\ & = \int_0^\infty \cos(2\pi f_0 (t-\tau)) {\rm d}\tau \\ & = \left[\frac{-1}{2\pi f_0}\sin(2\pi f_0(t-\tau))\right]_0^\infty \\ & = \frac{1}{2\pi f_0}\left[\sin(2\pi f_0 t) - \lim_{\tau\rightarrow \infty}\sin(2\pi f_0(t-\tau))\right], \end{align}$$ where the first part looks fine but the second part clearly shows that the limit does not exist. So this integral cannot be computed.
  • Laplace domain: This one clearly shows why it didn't work: The region of convergence (RoC) for the unit-step function is ${\rm Re}(s)>0$, hence the Laplace transform does not exist on the imaginary axis. Hence our input signal is outside the RoC and we cannot expect a stable output signal. Makes sense theoretically, a little surprising, since one may still expect an integrator to convert a cosine into a sine. I was ready to stop here (indeed I did at first), but then I thought further.
  • Fourier domain: in the engineering literature, you'll read that a system with an impuse response $\sigma(t)$ does have a Fourier transform, and it is given by $G(f) = \frac 12 \delta(f) + \frac{1}{\jmath 2\pi f}$. Surprising, since substituting $s = \jmath 2\pi f$ is not allowed here, as the RoC excludes ${\rm Re}(s)=0$. I expect this is where the $\delta(f)$ comes in: it's not a real Fourier transform but a "generalized" one, using distributions (which engineers usually treat in a notoriously sloppy way). Let's believe for a sec that's true, then we can compute our output signal in the Fourier domain: $$Y(f) = \left[ \frac 12 \delta(f) + \frac{1}{\jmath 2\pi f} \right] \cdot \left[ \frac 12 \delta(f-f_0) + \frac 12 \delta(f+f_0)\right] = \frac{1}{\jmath 2\pi f_0} \left[\frac 12 \delta(f+f_0) - \frac 12 \delta(f+f_0)\right]$$ which transforms into $\frac{1}{2\pi f_0} \sin(2\pi f_0(t))$. Whoa! Just like our time-domain approach but now without the diverging limit! What's going on? I'm assuming this one is wrong as it disagrees with the time-domain result and I'm assuming the reason is we're ouside the RoC, but I'm not sure.

The real bummer came when I thought even further. Let's now think of a windowed cosine, i.e., one multiplied with a rectangular window from $0$ to $T$. Now I can clearly compute the integral in time domain (it'll be the windowed sine) and it will clearly agree with what I get via Fourier. Now let $T$ grow: for every finite $T$, this will be true. Now, this drives me nuts: If I have a very long cosine, the integrator spits out a very long sine. But if I have an infinitely long cosine, I cannot even compute any value, even the result for very small $t$ is undefined. What?

I mean, the integrator is causal, it shouldn't be influenced by the signal's content in the very far future, right?

Is all this the result of considering an ideal integrator that cannot be built anyways along with an infinitely long cosine, that doesn't exist anyways? Still, the theory should be consistent, no?

*edit: It seems the problem disappears if I make sure my input signal is causal as well, restricting it to be zero for $t<0$. I'm kind of surprised this is necessary as I always thought that causal systems would take care of this more or less automatically by "not responding" to anything before $t<0$. Is this where I'm wrong (which would mean my time-domain computation is wrong and the system does indeed procude a sine wave as expected)?

$\endgroup$
2
$\begingroup$

Let's go through a few ways to solve this:

  1. Fourier transform: an ideal integrator is an LTI system, so its response to a sinusoidal input signal is a sinusoid with the amplitude and phase changed according to the frequency response evaluated at the input frequency (if it exists). For the ideal integrator we have $$H(\omega)=\pi\delta(\omega)+\frac{1}{j\omega}\tag{1}$$ so for $\omega_0> 0$, the response to $x(t)=\cos(\omega_0t)$ is $$\begin{align}y(t)&=|H(\omega_0)|\cos(\omega_0t+\arg\{H(\omega_0)\})\\&=\frac{1}{\omega_0}\cos(\omega_0t-\pi/2)\\&=\frac{1}{\omega_0}\sin(\omega_0t) \end{align}\tag{2}$$ This result can of course be verified by computing the inverse transform of the product of the frequency response and the Fourier transform of the input signal, as you showed in your question.

  2. Convolution: since we're dealing with a marginally stable system we must be prepared to deal with integrals that do not converge in the conventional sense, and we must resort to distribution theory and to generalized limits. The convolution integral can be written as $$\begin{align}y(t)&=(x\star h)(t)\\&=\int_{-\infty}^{\infty}\cos(\omega_0\tau)u(t-\tau)d\tau\\&=\int_{-\infty}^{t}\cos(\omega_0\tau)d\tau\\&=\frac{1}{\omega_0}\left[\sin(\omega_0t)-\lim_{\tau\to -\infty}\sin(\omega_0\tau)\right]\tag{3}\end{align}$$ As you've already pointed out, the limit in $(3)$ doesn't exist as an ordinary limit. However, since we're dealing with distributions, we need to consider generalized limits. As a generalized limit it does exist and it evaluates to zero as a consequence of the Riemann-Lebesgue lemma. This shows that the results obtained via Fourier transform and via convolution agree.

  3. Laplace transform: this is quickly dealt with because we simply cannot use the Laplace transform here. The Laplace transform of the input signal doesn't exist. That's a good example for people who claim that the Laplace transform is more general than the Fourier transform, because in this case the Fourier transform leads to a correct result, whereas the Laplace transform is not applicable.

EDIT: I'll add a justification as to why the limit in $(3)$ tends to zero if interpreted as a generalized limit. This is not a mathematical proof, but it shows that if we accept $(1)$ as the frequency response of an ideal integrator, then this implies that the limit in $(3)$ approaches zero.

The convolution integral can be written as

$$\begin{align}y(t)&=\int_{-\infty}^{t}\cos(\omega_0\tau)d\tau\\&=\int_{0}^t\cos(\omega_0\tau)d\tau+\int_{-\infty}^0\cos(\omega_0\tau)d\tau\\&=\frac{1}{\omega_0}\sin(\omega_0t)+\int_{0}^{\infty}\cos(\omega_0t)dt\tag{4}\end{align}$$

We need to show that the improper integral on the right-hand side of $(4)$ equals zero (as a generalized limit). Note that that integral equals the real part of the frequency response of an ideal integrator, and from $(1)$ we have

$$\text{Re}\big\{H(\omega_0)\big\}=\int_{0}^{\infty}\cos(\omega_0t)dt=\pi\delta(\omega_0)=0,\quad\omega_0>0\tag{5}$$

| improve this answer | |
$\endgroup$
  • $\begingroup$ Wow, thanks, this seems to be very helpful! Can you elaborate a little more on the convolution part? I haven't heard of generalized limits before and I don't understand how they work together with the integral we're trying to solve here. The Riemann-Lebesgue Lemma (that I didn't know either) seems to be about L1 functions, but the cosine is not L1 integrable. Hence I'm not able to fully make the connection yet. $\endgroup$ – Florian Jan 13 at 12:19
  • $\begingroup$ @Florian: You're welcome, I'll add some stuff to my answer later on. $\endgroup$ – Matt L. Jan 13 at 12:45
  • $\begingroup$ @Florian: I've added a justification for the given interpretation of the limit. $\endgroup$ – Matt L. Jan 14 at 12:59
  • $\begingroup$ Thanks, much appreciated. I still cannot connect all the dots, as I don't see the connection to the Riemann-Lebesque lemma yet but I guess my homework for now is to read up on the generalized limits. That could help. For now, I'm willing to accept what I got. :) $\endgroup$ – Florian Jan 14 at 16:39
  • 1
    $\begingroup$ @Florian: It's a bit more involved than that, but basically you have $\lim_{t\to\infty}\cos(\omega t)=\lim_{t\to\infty}\sin(\omega t)=0$ as generalized limits as a consequence of the Riemann-Lebesgue lemma. I recommend "The Fourier Integral and Its Applications" by A. Papoulis (Appendix I). $\endgroup$ – Matt L. Jan 14 at 17:41
1
$\begingroup$

An ideal integrator is ambiguous with regard to an integration constant. The preconditions of the Laplace transform (domain of convergence a half-plane) require more than exponential decay for negative time values and fix that problem in that manner. Usually this is assured in Laplace transforms by taking functions to be zero below some fixed value, but some functions (like Gaussian impulses) converge even while being nonzero over the entire domain.

But your everlasting cosine wave does not have a Laplace transform. The best you can work with is a Fourier transform, and it has Dirac pulses in its frequency domain making things tricky.

So basically you are correct: the DC offset of your result is indeterminate.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.