0
$\begingroup$

I am dealing with a physics problem which is related to signal processing. The problem requires me to calculate the instantaneous force acting on a body which depends on some physical parameter $x$. Assume that $x(t)$ is periodic in time for the moment. Since $x(t)$ is periodic, then it can be expanded as a Fourier series with different frequency components (and it doesn't really matter if $x(t)$ is causal). The calculation for the instantaneous force involves adding a complex phase shift (which may depend on the frequency) to each of the frequency component. To do that, I can use the convolution theorem and take the convolution of $x(t)$ with some kernel $\kappa(t)$ whose Fourier transform gives me the required phase shifts, i.e. $\tilde{\kappa}(\omega) \propto e^{i\delta(\omega)}$ where $\delta(\omega)$ is the phase shift.

Now if in reality $x(t)$ is not periodic and is causal since I only know its values in the past, can I still apply the same kernel to get the instantaneous force? I have been told that I should use Laplace transform instead of Fourier transform. I see the point of it being bilateral by definition, but I am not sure how it is actually different to Fourier transform. Does applying the convolution theorem to a causal signal still give me the desired phase shifts?

$\endgroup$
  • $\begingroup$ if $x(t)$ is periodic, it cannot be causal (so it does really matter) unless, i s'pose, if $x(t)=0$, that $x(t)$ is both periodic and causal. but i think that special case is not very useful. $\endgroup$ – robert bristow-johnson Apr 13 '17 at 1:06
  • $\begingroup$ now, if what you want is an FIR filter that will convert the phase of each sinusoid in your periodic $x(t)$ from the phase that it is in the input to a specified phase in the output $y(t)$, that can be done, in a rather straight-forward manner. is that what you want? $\endgroup$ – robert bristow-johnson Apr 13 '17 at 1:18
  • $\begingroup$ @robertbristow-johnson I believe I have already done that for the periodic case. For example if I want the phase shift to be proportional to the frequency, i.e. $\delta(\omega) = \omega\tau$ where $\tau$ is a constant, then I can just take the convolution of the input with a delta function $\delta(t - \tau)$. My actual question is, does this still apply when I have a causal signal? $\endgroup$ – David Young Apr 13 '17 at 8:40
  • $\begingroup$ @DavidYoung: As I said in my answer, what the system does (e.g., delay the input by $\tau$, as in your example) is independent of the input signal. The input-output relationship of an LTI system is described by a convolution, so the properties of the input signal do not matter. $\endgroup$ – Matt L. Apr 13 '17 at 13:05
0
$\begingroup$

What you want is an all-pass filter with frequency response

$$H(\omega)=e^{j\phi(\omega)}\tag{1}$$

where $\phi(\omega)$ is the desired phase shift (and $j$ is how we denote the imaginary unit over here). This system is called an all-pass filter because clearly $|H(\omega)|=1$ holds.

The type of input signal is irrelevant, it can be periodic, non-periodic, causal, or non-causal; if you filter it with a linear time-invariant (LTI) filter with a frequency response given by $(1)$ then the desired phase shift will be achieved.

Your problem is the (causal and stable) realization of such a filter. In general, for a given phase shift $\phi(\omega)$ the frequency response given by $(1)$ cannot be implemented exactly; it can only be approximated.

$\endgroup$
  • $\begingroup$ Is there a way to show that this is indeed true? For example, can i consider a simple sine wave as input, and truncate it by multiplying it by the Heaviside function. Now if I apply an all-pass filter, then $FT(output) = e^{i\phi(\omega)} \cdot FT(H(x)\cdot input)$ where $FT(H(x)\cdot input)$ can be computed by taking the convolution between FT(H(x)) and FT(input). $\endgroup$ – David Young Apr 23 '17 at 0:26
  • $\begingroup$ @DavidYoung: The output signal of an LTI system can always be computed by convolving the input signal with the system's impulse response. Not sure what it exactly is that you're doubting. $\endgroup$ – Matt L. Apr 23 '17 at 18:11
  • $\begingroup$ I'm dealing with a physics problem that requires me to apply an all-pass filter to some causal input signal to get the instantaneous "force", and for some reason the output doesn't have to correct frequency response which is why I'm asking these questions. $\endgroup$ – David Young Apr 23 '17 at 21:11
  • $\begingroup$ One thing I am worried about is that I obviously have to truncate the convolution integral at some limit and I am wondering what effects that might have. To look into that, I consider a simple sine wave input signal, and multiply it by a top hat function: $H(t) = 1$ for $0 < t < T$ and zero otherwise, that way I have my "causal" input signal. The Fourier transform of the top hat function is $\tilde{H}(\omega) \propto e^{-i\omega T/2}\frac{sin(\omega T/2)}{\omega}$. And since the Fourier transform of the sine wave is just a delta function, then the convolution of them gives me an extra phase. $\endgroup$ – David Young Apr 23 '17 at 21:18

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.