Using the DTFT property, find h[n] of a system where:enter image description here

Is it an FIR or IIR system?


2 Answers 2


While this is by your admission homework (and fairly basic), I'll bite. Recall the definition of the DTFT:

$$ X(\omega) = \sum_{n=0}^{\infty}x[n] e^{-j\omega n} $$

And recall the definition of the frequency response $H(\omega)$:

$$ H(\omega) = \frac{Y(\omega)}{X(\omega)} $$

where $x[n]$ is the input to the system and $y[n]$ is its output. Combine these two equations:

$$ \begin{eqnarray*} H(\omega)X(\omega) &=& Y(\omega) \\ \frac{1 - a^4 e^{-j 4 \omega}}{1 - a^4 e^{-j \omega}}X(\omega) &=& Y(\omega) \\ (1 - a^4 e^{-j 4 \omega})X(\omega) &=& (1 - a^4 e^{-j \omega})Y(\omega) \\ X(\omega) - a^4 e^{-j 4 \omega} X(\omega) &=& Y(\omega) - a^4 e^{-j \omega} Y(\omega) \end{eqnarray*} $$

Now, perform the inverse DTFT on both sides of the equation. By definition, $X(\omega)$ and $x[n]$ are a transform pair; likewise for $Y(\omega)$ and $y[n]$. For the other two terms, recall the time-shifting property of the DTFT:

$$ x[n-k] \leftrightarrow e^{-jk\omega} X(\omega) $$

which can be shown easily from the definition of the DFT. Using this property, the equation inverse transforms to the difference equation specification for the system:

$$ x[n] - a^4 x[n-4] = y[n] - a^4 y[n-1] $$

$$ y[n] = x[n] - a^4 x[n-4] + a^4 y[n-1] $$

This is the definition of a recursive filter, which are usually IIR; that is the case for this one. Finding the impulse response is easy; let $x[n] = \delta[n]$ and find that the system output is:

$$ y[n] = a^{4n}u[n] - a^{4(n-4)+4}u[n-4] $$

$$ y[n] = a^{4n}u[n] - a^{4n-12}u[n-4] $$

enter image description here

The above is plotted for $a=0.99$. It should be noted that the system is only stable for $|a|\le1$.

  • $\begingroup$ I've tried to calculate the impulse response but got tangled. Could you show how it's done? thank you. $\endgroup$ Jun 23, 2020 at 14:14

$$\begin{align*} H(\omega) &= \frac{1-a^4\exp(-4j\omega)}{1-a^4\exp(-j\omega)}\\ &= (1-a^4\exp(-4j\omega))\sum_{n=0}^\infty (a^4\exp(-j\omega))^n\\ &= \sum_{n=0}^\infty a^{4n}\exp(-nj\omega)-\sum_{n=0}^\infty a^{4n+4}\exp(-(n+4)j\omega)\\ &= \sum_{n=0}^3 a^{4n}\exp(-nj\omega) + \sum_{n=4}^\infty [a^{4n} - a^{4n-12}]\exp(-nj\omega)\\ h[n] &= \begin{cases}0, &n < 0,\\ a^{4n}, &n = 0, 1, 2, 3,\\ a^{4n} - a^{4n-12}, &n \geq 4.\end{cases} \end{align*} $$ Since the impulse response extends to $\infty$, this is an IIR filter. JasonR states in his answer that the filter is stable only if $|a| < 1$. In fact, the filter is stable when $|a| \leq 1$, and is unstable only for $|a| > 1$. However, when $|a| = 1$, from the geometric series formula $1 + r + r^2 + r^3 = \frac{1-r^4}{1-r}$, we get that $$H(\omega) = \frac{1-\exp(-4j\omega)}{1-\exp(-j\omega)} = 1 + \exp(-j\omega)+\exp(-2j\omega)+\exp(-3j\omega)$$ is the transfer function of a (stable) FIR filter that can be described as a short-term integrator or short-term averager (with gain $4$).

  • $\begingroup$ Nice alternate derivation. I fixed my claim on stability in my answer also. $\endgroup$
    – Jason R
    Feb 4, 2012 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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