Return to Answer

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle.

Even within the case of the FIR filters - which are unconditionally stable- we can end up with an unstable adaptive FIR filter if the loop gain at certain frequencies is large enough.

With FIR filters we are essentially solving iteratively a second-order convex optimization problem. This problem has no local minimums and the Hessian which plays a crucial role in analyzing the convergence of the filter is constant. The error term is linearly related to filter weight. This makes the problem both

1. Well behaved=> So that you can easily converge

2. Easy to analyze => So that you can find a scaling factor that leads to the fastest convergence.

With the adaptive IIR filters, the problem is not convex and is nonlinear. If you look at the following block diagram you may think at first look that it is linear with respect to filter coefficients. However, you can inspect that the input to $B(z)$ block contains the coefficients from $A(z)$ and previous iterations of $B(z)$. Compared to the adaptive FIR case we will have a system which is:

1. Not Well behaved=> The surface can have local minimums and the slope can change erratically compared to an FIR filter cost function.

2. Hard to analyze => To find a good scaling factor we need to analyze the error surface. In the adaptive IIR case, it is hard if not impossible. Also, the Hessian which plays a crucial role in the analysis of the convergence is even harder to calculate for an adaptive IIR filter.

3. Dependency on the previous samples (make it hard to pass over a spike in one of the stages or other instabilities)

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle.

Even for the FIR filters - which are unconditionally stable- we can end up with an unstable adaptive FIR filter if the loop gain at certain frequencies is large enough.

With FIR filters we are essentially solving iteratively a second-order convex optimization problem. This problem has no local minimums and the Hessian which plays a crucial role in analyzing the convergence of the filter is constant. The error term is linearly related to filter weight. This makes the problem both

1. Well behaved=> So that you can easily converge

2. Easy to analyze => So that you can find a scaling factor that leads to the fastest convergence.

With the adaptive IIR filters, the problem is not convex and is nonlinear. If you look at the following block diagram you may think at first look that it is linear with respect to filter coefficients. However, you can inspect that the input to $B(z)$ block contains the coefficients from $A(z)$ and previous iterations of $B(z)$. Compared to the adaptive FIR case we will have a system that is:

1. Not Well behaved=> The surface can have local minimums and the slope can change erratically compared to an FIR filter cost function.

2. Hard to analyze => To find a good scaling factor we need to analyze the error surface. In the adaptive IIR case, it is hard if not impossible. Also, the Hessian which plays a crucial role in the analysis of the convergence is even harder to calculate for an adaptive IIR filter.

3. Dependency on the previous samples (make it hard to pass over a spike in one of the stages or other instabilities)

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle.

Even within the case of the FIR filters - which are unconditionally stable, we can end up with an unstable adaptive FIR filter if the loop gain at certain frequencies is large enough.

With FIR filters we are essentially solving iteratively a convex optimization problem without local minimums. The error term is linearly related to filter weight, this makes the problem both

1. Well behaved=> So that you can easily converge

2. Easy to analyze => So that you can find a scaling factor that leads to the fastest convergence.

With the IIR filters, the problem is not convex and is nonlinear. If you look at the following block diagram you may think at first look that it is linear with respect to filter coefficients. However, you can inspect that the input to $B(z)$ block contains the coefficients from $A(z)$ and previous iterations of $B(z)$. Compared to the FIR case we will have

1. Not Well behaved=> The surface can have local minimums and the slope can change erratically compared to an FIR filter cost function.

2. Hard to analyze => To find a good scaling factor we need to analyze the error surface. In the IIR case, it is hard if not impossible. Also, the Hessian which plays a crucial role in the analysis of the convergence is even very harder to calculate for an IIR adaptive filter. But for an FIR filter is a constant number.

3. Dependency on the previous samples (make it hard to pass over a spike in one of the stages or other instabilities)

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle.

Even within the case of the FIR filters - which are unconditionally stable- we can end up with an unstable adaptive FIR filter if the loop gain at certain frequencies is large enough.

With FIR filters we are essentially solving iteratively a second-order convex optimization problem. This problem has no local minimums and the Hessian which plays a crucial role in analyzing the convergence of the filter is constant. The error term is linearly related to filter weight. This makes the problem both

1. Well behaved=> So that you can easily converge

2. Easy to analyze => So that you can find a scaling factor that leads to the fastest convergence.

With the adaptive IIR filters, the problem is not convex and is nonlinear. If you look at the following block diagram you may think at first look that it is linear with respect to filter coefficients. However, you can inspect that the input to $B(z)$ block contains the coefficients from $A(z)$ and previous iterations of $B(z)$. Compared to the adaptive FIR case we will have a system which is:

1. Not Well behaved=> The surface can have local minimums and the slope can change erratically compared to an FIR filter cost function.

2. Hard to analyze => To find a good scaling factor we need to analyze the error surface. In the adaptive IIR case, it is hard if not impossible. Also, the Hessian which plays a crucial role in the analysis of the convergence is even harder to calculate for an adaptive IIR filter.

3. Dependency on the previous samples (make it hard to pass over a spike in one of the stages or other instabilities)

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle. Even with an unconditionally stable IIR filter, we can end up with an unstable loop if the loop gain at certain frequency is large enough. With FIR filters we are essentially solving iteratively a convex optimization problem without local minimums. The error term is linearly related to filter weight, this makes the problem both 1-Well behaved=> So that you can easily converge 2-Easy to analyze => So that you can find a scaling factor that leads to the fastest convergence.

With the IIR filters, the problem is not convex and is nonlinear. If you look at the following block diagram you may think that it is linear with respect to filter coefficients. However, you can inspect that the input to $B(z)$ block contains the coefficients from $A(z)$ and previous iterations of $B(z)$.

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle.

Even within the case of the FIR filters - which are unconditionally stable, we can end up with an unstable adaptive FIR filter if the loop gain at certain frequencies is large enough.

With FIR filters we are essentially solving iteratively a convex optimization problem without local minimums. The error term is linearly related to filter weight, this makes the problem both

1. Well behaved=> So that you can easily converge

2. Easy to analyze => So that you can find a scaling factor that leads to the fastest convergence.

With the IIR filters, the problem is not convex and is nonlinear. If you look at the following block diagram you may think at first look that it is linear with respect to filter coefficients. However, you can inspect that the input to $B(z)$ block contains the coefficients from $A(z)$ and previous iterations of $B(z)$. Compared to the FIR case we will have

1. Not Well behaved=> The surface can have local minimums and the slope can change erratically compared to an FIR filter cost function.

2. Hard to analyze => To find a good scaling factor we need to analyze the error surface. In the IIR case, it is hard if not impossible. Also, the Hessian which plays a crucial role in the analysis of the convergence is even very harder to calculate for an IIR adaptive filter. But for an FIR filter is a constant number.

3. Dependency on the previous samples (make it hard to pass over a spike in one of the stages or other instabilities)