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While trying to determine the unit step and impulse response for a discrete LTI system I had several problems:

$$ \begin{aligned} (Hx)[n]&=(Hx)[n-1]+\frac{1}{N}(x[n]-x[n-N]) \\x[n]&=(Hx)[n]=0\;\forall\;n\lt0 \end{aligned} $$

Starting with the Impulse response:

$$h[n]=(H\delta)[n]=(H\delta)[n-1]+\frac{1}{N}(\delta[n]-\delta[n-N])$$

Can I argue that because of the condition $x[n]=(Hx)[n]=0\;\forall\;n\lt0$ the impulse response only exists where $n\gt 1\;\&\;n\gt N?$ But what about the other cases?

Concerning the Step response I am quite lost and am not even sure if this is how it is calculated:

$$a[n]=(H\sigma)[n]\sum_{k=-\infty}^{\infty}h[k]\sigma[n-k]$$

Help as to how to proceed and solutions are greatly appreciated

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This is a homework type problem, so I'll give you a few hints to help you solve it yourself (and learn something while doing so).

From what is given, we can assume that the given difference equation describes a causal discrete-time system. Let $h[n]$ denote the impulse response. It must satisfy

$$h[n]=h[n-1]+\frac{1}{N}\big(\delta[n]-\delta[n-N]\big),\qquad n\ge 0\tag{1}$$

with $h[-1]=0$. Note that the second term on the right-hand side of $(1)$ only contributes to $h[n]$ for $n=0$ and $n=N$. Now start at $n=0$ and continue up to $n=N$ to see what happens. It will turn out that the impulse response has a finite length, i.e., the given system is an FIR filter.

The step response is the convolution of the impulse response with the unit step sequence, which simplifies to

$$a[n]=\sum_{k=0}^{\infty}h[k]u[n-k]=u[n]\sum_{k=0}^nh[k]\tag{2}$$

So the step response $a[n]$ is just the cumulative sum of the impulse response $h[n]$.

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  • $\begingroup$ The recursive part of the response would be carried over to the next response? Therefore, starting from 0 for the 1st I would have 1/N up until n=N? $\endgroup$ – dsisko Jan 4 at 16:20
  • $\begingroup$ @dsisko: That's it. At $n=N$ that value gets subtracted so $h[n]$ equals zero for all $n\ge N$. $\endgroup$ – Matt L. Jan 4 at 16:52
  • $\begingroup$ @Matt L : In this particular case, is the step response just a moving average in the time domain ? $\endgroup$ – mark leeds Jan 6 at 2:24
  • $\begingroup$ @markleeds: The filter itself is a moving average. The step response increases linearly up to its final value which is just the sum of all impulse response coefficients. $\endgroup$ – Matt L. Jan 6 at 6:28

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