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Suppose you begin with the following model of a discrete impulse response:

$$h[n] = \begin{cases} \displaystyle\sum_{k=1}^K \alpha_k x_k^n & n \ge 0 \\ 0 & n < 0 \end{cases}$$

i.e. a linear superposition of geometric series, where $\alpha_k$'s and $x_k$'s are free parameters.

Am I correct in understanding that

• This is the impulse response of an IIR filter of order $K$.
• The filter's poles are the $x_k$ values.

My reasoning follows from taking the Z-transform on both sides:

$$ H(z) = \sum_{k=1}^K \alpha_k \frac{1}{1 - x_k z^{-1}} $$

Combining all into a single fraction (where $\beta_k$'s represent appropriately computed coefficients):

$$ H(z) = \frac{\displaystyle\sum_{k=1}^{K} \beta_k z^{k}}{\displaystyle\prod_{k=1}^K (z - x_k)}$$

From the denominator one can infer the poles are $x_k$, right? And the order of both numerator and denominator polynomials is $K$, hence that's also the order of the filter?

Also:

• Will this hold for unstable versions where $|x_k| < 1$ is not satisfied?
• Do I have to account for the Z-transform's ROC in this case, and if so, how?
• Is there a simple rule for $\alpha_k$'s to make this an all-pole filter?

Suppose you begin with the following model of a discrete impulse response:

$$h[n] = \begin{cases} \displaystyle\sum_{k=1}^K \alpha_k p_k^n & n \ge 0 \\ 0 & n < 0 \end{cases}$$

i.e. a linear superposition of geometric series, where $\alpha_k$'s and $p_k$'s are free parameters.

Am I correct in understanding that

• This is the impulse response of an IIR filter of order $K$.
• The filter's poles are the $p_k$ values.

My reasoning follows from taking the Z-transform on both sides:

$$ H(z) = \sum_{k=1}^K \alpha_k \frac{1}{1 - p_k z^{-1}} $$

Combining all into a single fraction (where $\beta_k$'s represent appropriately computed coefficients):

$$ H(z) = \frac{\displaystyle\sum_{k=1}^{K} \beta_k z^{k}}{\displaystyle\prod_{k=1}^K (z - p_k)}$$

From the denominator one can infer the poles are $p_k$, right? And the order of both numerator and denominator polynomials is $K$, hence that's also the order of the filter?

Also:

• Will this hold for unstable versions where $|p_k| < 1$ is not satisfied?
• Do I have to account for the Z-transform's ROC in this case, and if so, how?
• Is there a simple rule for $\alpha_k$'s to make this an all-pole filter?

Suppose you begin with the following model of a discrete impulse response:

$$h[n] = \begin{cases} \sum_{k=1}^K \alpha_k x_k^n & n \ge 0 \\ 0 & n < 0 \end{cases}$$

i.e. a linear superposition of geometric series, where $\alpha_k$'s and $x_k$'s are free parameters.

Am I correct in understanding that

• This is the impulse response of an IIR filter of order $K$.
• The filter's poles are the $x_k$ values.

My reasoning follows from taking the Z-transform on both sides:

$$ H(z) = \sum_{k=1}^K \alpha_k \frac{1}{1 - x_k z^{-1}} $$

Combining all into a single fraction (where $\beta_k$'s represent appropriately computed coefficients):

$$ H(z) = \frac{\sum_{k=1}^{K} \beta_k z^{k}}{\prod_{k=1}^K (z - x_k)}$$

From the denominator one can infer the poles are $x_k$, right? And the order of both numerator and denominator polynomials is $K$, hence that's also the order of the filter?

Also:

• Will this hold for unstable versions where $|x_k| < 1$ is not satisfied?
• Do I have to account for the Z-transform's ROC in this case, and if so, how?
• Is there a simple rule for $\alpha_k$'s to make this an all-pole filter?

Suppose you begin with the following model of a discrete impulse response:

$$h[n] = \begin{cases} \displaystyle\sum_{k=1}^K \alpha_k x_k^n & n \ge 0 \\ 0 & n < 0 \end{cases}$$

i.e. a linear superposition of geometric series, where $\alpha_k$'s and $x_k$'s are free parameters.

Am I correct in understanding that

• This is the impulse response of an IIR filter of order $K$.
• The filter's poles are the $x_k$ values.

My reasoning follows from taking the Z-transform on both sides:

$$ H(z) = \sum_{k=1}^K \alpha_k \frac{1}{1 - x_k z^{-1}} $$

Combining all into a single fraction (where $\beta_k$'s represent appropriately computed coefficients):

$$ H(z) = \frac{\displaystyle\sum_{k=1}^{K} \beta_k z^{k}}{\displaystyle\prod_{k=1}^K (z - x_k)}$$

From the denominator one can infer the poles are $x_k$, right? And the order of both numerator and denominator polynomials is $K$, hence that's also the order of the filter?

Also:

• Will this hold for unstable versions where $|x_k| < 1$ is not satisfied?
• Do I have to account for the Z-transform's ROC in this case, and if so, how?
• Is there a simple rule for $\alpha_k$'s to make this an all-pole filter?

Suppose you begin with the following model of a discrete impulse response:

$$h[n] = \begin{cases} \sum_{k=1}^K \alpha_k x_k^n & n \ge 0 \\ 0 & n < 0 \end{cases}$$

i.e. a linear superposition of geometric series, where $\alpha_k$'s and $x_k$'s are free parameters.

Am I correct in understanding that

• This is the impulse response of an IIR filter of order $K$.
• The filter's poles are the $x_k$ values.

My reasoning follows from taking the Z-transform on both sides:

$$ H(z) = \sum_{k=1}^K \alpha_k \frac{1}{1 - x_k z^{-1}} $$

Combining all into a single fraction (where $\beta_k$'s are appropriately computed):

$$ H(z) = \frac{\sum_{k=1}^{K} \beta_k z^{k}}{\prod_{k=1}^K (z - x_k)}$$

From the denominator one can infer the poles are $x_k$, right? And the order of both numerator and denominator polynomials is $K$, hence that's also the order of the filter?

Also:

• Will this hold for unstable versions where $|x_k| < 0$ is not satisfied?
• Do I have to account for the Z-transform's ROC in this case, and if so, how?
• Is there a simple rule for $\alpha_k$'s to make this an all-pole filter?

Suppose you begin with the following model of a discrete impulse response:

$$h[n] = \begin{cases} \sum_{k=1}^K \alpha_k x_k^n & n \ge 0 \\ 0 & n < 0 \end{cases}$$

i.e. a linear superposition of geometric series, where $\alpha_k$'s and $x_k$'s are free parameters.

Am I correct in understanding that

• This is the impulse response of an IIR filter of order $K$.
• The filter's poles are the $x_k$ values.

My reasoning follows from taking the Z-transform on both sides:

$$ H(z) = \sum_{k=1}^K \alpha_k \frac{1}{1 - x_k z^{-1}} $$

Combining all into a single fraction (where $\beta_k$'s represent appropriately computed coefficients):

$$ H(z) = \frac{\sum_{k=1}^{K} \beta_k z^{k}}{\prod_{k=1}^K (z - x_k)}$$

From the denominator one can infer the poles are $x_k$, right? And the order of both numerator and denominator polynomials is $K$, hence that's also the order of the filter?

Also:

• Will this hold for unstable versions where $|x_k| < 1$ is not satisfied?
• Do I have to account for the Z-transform's ROC in this case, and if so, how?
• Is there a simple rule for $\alpha_k$'s to make this an all-pole filter?