Skip to main content
Signal Processing

Return to Answer

added 145 characters in body
Source Link
Matt L.
Matt L.
  • 92.4k
  • 10
  • 81
  • 184

It's important to specify at which frequency you want unity gain. But assuming you mean DC ($\omega=0$), because that filter has a low pass characteristic, the DC gain of an IIR filter is given by

$$G_{DC}=\frac{\sum_kb[k]}{\sum_ka[k]}\tag{1}$$

It's also common to normalize the denominator coefficients such that $a[0]=1$. In your example that would give

a = [1.00000 -0.29240 0.10849]

and

b = [0.00000 131163547.59967]

Finally, normalizing by the DC gain $(1)$ will give you a filter with unity gain at DC. This leaves the denominator coefficients unchanged, and the new normalized numerator coefficients are given by

b = [0.00000 0.81609]

It's important to specify at which frequency you want unity gain. But assuming you mean DC ($\omega=0$), because that filter has a low pass characteristic, the DC gain of an IIR filter is given by

$$G_{DC}=\frac{\sum_kb[k]}{\sum_ka[k]}\tag{1}$$

It's common to normalize the denominator coefficients such that $a[0]=1$. In your example that would give

a = [1.00000 -0.29240 0.10849]

and

b = [0.00000 131163547.59967]

Finally, normalizing by the DC gain $(1)$ will give you a filter with unity gain at DC.

It's important to specify at which frequency you want unity gain. But assuming you mean DC ($\omega=0$), because that filter has a low pass characteristic, the DC gain of an IIR filter is given by

$$G_{DC}=\frac{\sum_kb[k]}{\sum_ka[k]}\tag{1}$$

It's also common to normalize the denominator coefficients such that $a[0]=1$. In your example that would give

a = [1.00000 -0.29240 0.10849]

and

b = [0.00000 131163547.59967]

Finally, normalizing by the DC gain $(1)$ will give you a filter with unity gain at DC. This leaves the denominator coefficients unchanged, and the new normalized numerator coefficients are given by

b = [0.00000 0.81609]

added 11 characters in body
Source Link
Matt L.
Matt L.
  • 92.4k
  • 10
  • 81
  • 184

It's important to specify at which frequency you want unity gain. But assuming you mean DC ($\omega=0$), because that filter has a low pass characteristic, the DC gain of an IIR filter is given by

$$G_{DC}=\frac{\sum_kb[k]}{\sum_ka[k]}\tag{1}$$

It's common to normalize the denominator coefficients such that $a[0]=1$. In your example that would give

a = [1.00000 -0.29240 0.10849]

and

b = [0.00000 1209000000]131163547.59967]

Finally, normalizing by the DC gain $(1)$ will give you a filter with unity gain at DC.

It's important to specify at which frequency you want unity gain. But assuming you mean DC ($\omega=0$) because that filter has a low pass characteristic, the DC gain of an IIR filter is given by

$$G_{DC}=\frac{\sum_kb[k]}{\sum_ka[k]}\tag{1}$$

It's common to normalize the denominator coefficients such that $a[0]=1$. In your example that would give

a = [1.00000 -0.29240 0.10849]

and

b = [0 1209000000]

It's important to specify at which frequency you want unity gain. But assuming you mean DC ($\omega=0$), because that filter has a low pass characteristic, the DC gain of an IIR filter is given by

$$G_{DC}=\frac{\sum_kb[k]}{\sum_ka[k]}\tag{1}$$

It's common to normalize the denominator coefficients such that $a[0]=1$. In your example that would give

a = [1.00000 -0.29240 0.10849]

and

b = [0.00000 131163547.59967]

Finally, normalizing by the DC gain $(1)$ will give you a filter with unity gain at DC.

Source Link
Matt L.
Matt L.
  • 92.4k
  • 10
  • 81
  • 184

It's important to specify at which frequency you want unity gain. But assuming you mean DC ($\omega=0$) because that filter has a low pass characteristic, the DC gain of an IIR filter is given by

$$G_{DC}=\frac{\sum_kb[k]}{\sum_ka[k]}\tag{1}$$

It's common to normalize the denominator coefficients such that $a[0]=1$. In your example that would give

a = [1.00000 -0.29240 0.10849]

and

b = [0 1209000000]