0
$\begingroup$

Convolution is used to find the output when the input and the impulse response is known. I am curious if there is a similar operator, like convolution, or a method that can determine the output of a system. I am asking this question, right of the top of my head, so I don't have any concrete example to apply it to.

If I am not wrong, one can use deconvolution to find the impulse response and go from there, but this is not what I am searching for.

$\endgroup$
  • 1
    $\begingroup$ In addition to Matt's accurate answer below: You can also find the impulse response from the step response by differentiation. That means an alternative way to calculate the output is convolving with the derivative of the step response. ( The deeper reason for this is that LTI systems commute and that the unit impulse is the derivative of the unit step. ) $\endgroup$ – Jazzmaniac Jun 25 '14 at 10:43
1
$\begingroup$

This can indeed be done. Write the input function $x(t)$ as

$$x(t)=x(-\infty)+\int_{-\infty}^t x^{\prime}(\tau)d\tau= x(-\infty)+\int_{-\infty}^{\infty} x^{\prime}(\tau)u(t-\tau)d\tau\tag{1}=\\ =x(-\infty)+(x^{\prime}*u)(t)$$

where $x^{\prime}(t)$ is the derivative of $x(t)$, and $u(t)$ is the unit step function. From (1), the system's response $y(t)$ to the input $x(t)$ can be written in terms of the step response $a(t)$:

$$y(t)=a(\infty)x(-\infty)+\int_{-\infty}^{\infty} x^{\prime}(\tau)a(t-\tau)d\tau\tag{2}$$

So, apart from the first constant term, the output signal $y(t)$ is given by the convolution of the derivative of the input signal with the system's step response.

Equation (2) is valid because the system's response to the constant $x(-\infty)$ is $H(0)x(-\infty)$, where $H(0)$ is the system's frequency response at DC:

$$H(0)=\int_{-\infty}^{\infty}h(t)dt=a(\infty)$$

The integral on the right-hand side of (2) is the convolution

$$((x^{\prime}*u)*h)(t)=(x^{\prime}*(u*h))(t)=(x^{\prime}*a)(t)$$

EDIT: The discrete-time case is completely analogous. Write the input sequence as

$$x[n]=x[-\infty]+\sum_{k=-\infty}^n(x[k]-x[k-1])=x[-\infty]+\sum_{k=-\infty}^{\infty}(x[k]-x[k-1])u[n-k]$$

This results in the following expression for the output sequence in terms of the step response:

$$y[n]=x[-\infty]a[\infty]+\sum_{k=-\infty}^{\infty}(x[k]-x[k-1])a[n-k]$$

Again, apart from the constant first term, the output sequence is obtained by convolving the first order difference of the input sequence with the system's step response.

$\endgroup$
  • $\begingroup$ If the system was in discrete time, how would the derivative be used. $\endgroup$ – user29568 Jun 25 '14 at 9:28
  • $\begingroup$ @user29568: I edited my answer to include the discrete-time case. $\endgroup$ – Matt L. Jun 25 '14 at 10:11
  • $\begingroup$ Of course, we can make use of this in Fourier domain as well? For example, if the input function was periodic it would translate to $$Y(e^{j\Omega})= (1-e^{j\Omega})X(e^j\Omega)H(e^j\Omega)$$ where $H(e^j\Omega)$ is the DT fourier transform of the unit sample response. $\endgroup$ – user29568 Jul 12 '14 at 8:46
  • $\begingroup$ Is it possible to do this? $\endgroup$ – user29568 Jul 12 '14 at 8:52
  • 1
    $\begingroup$ @user29568: The frequency domain equation is $$Y(e^{j\Omega})=x[-\infty]a[\infty]2\pi\delta(\omega)+(1-e^{-j\Omega})X(e^{j\Omega})A(e^{j\Omega}‌)$$ (I forgot the $\delta$ impulse in my previous comment). This equation holds whether or not the input is periodic. The problem is that $x[-\infty]$ sometimes does not exist. If $a[\infty]=0$ that's no problem, otherwise the equation does not give you the correct value at $\Omega=0$. So $\Omega=0$ is the only problem, for all other values of $\Omega$ you can just use $$Y(e^{j\Omega})=(1-e^{-j\Omega})X(e^{j\Omega})A(e^{j\Omega}‌​)$$ $\endgroup$ – Matt L. Jul 12 '14 at 13:50

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.