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

Is it an FIR or IIR system?

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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 $|a|\le1$.

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$$\begin{align*} H(\omega) &= \frac{1-a^4\exp(-4j\omega)}{1-a^4\exp(-j\omega)}\\ &= (1-a^4\exp(-4j\omega))\sum_{n=0}^\infty (a^4\exp(-j\omega))^n\\ &= \sum_{n=0}^\infty a^{4n}\exp(-nj\omega)-\sum_{n=0}^\infty a^{4n+4}\exp(-(n+4)j\omega)\\ &= \sum_{n=0}^3 a^{4n}\exp(-nj\omega) + \sum_{n=4}^\infty [a^{4n} - a^{4n-12}]\exp(-nj\omega)\\ h[n] &= \begin{cases}0, &n < 0,\\ a^{4n}, &n = 0, 1, 2, 3,\\ a^{4n} - a^{4n-12}, &n \geq 4.\end{cases} \end{align*} $$ 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 $4$).

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  • $\begingroup$ Nice alternate derivation. I fixed my claim on stability in my answer also. $\endgroup$ – Jason R Feb 4 '12 at 2:26

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