DISP 2003Lecture 6 – Part 2

Digital Filters 4

Coefficient quantizationZero input limit cycleHow about using float?

Philippe Baudrenghien, AB-RF

27 March 2003 P Baudrenghien AB/RF 2

Quantization of coefficients

• The coefficients ak and br must be quantized into our two’s complement fractional

• This creates a distortion in the achieved frequency response

• We measure the sensitivity to coefficient quantization by comparing the poles/zeros of the infinite precision and finite precision realizations

∑

∑N

0=k

k-k

M

0=r

r-r

z a

z b

=X(z)

Y(z)=H(z)

27 March 2003 P Baudrenghien AB/RF 3

Displacement of the poles (1)• Our transfer function has N poles p0, p1, …, pN-1 defined by

the zeros of the denominator (Nth order polynomial with coefficients a0, a1, …, aN)

• Let us introduce an error am on the coefficient am

• This displaces all poles: p0 -> p0 + p0 , p1 -> p1 + p1 , …

• The displacement pn of the nth pole due to the error am is given by

∏N

n≠r 1,=rrn

m-Nn

mn

)p-(p

p aΔ - =pΔ

27 March 2003 P Baudrenghien AB/RF 4

Displacement of the poles (2)

• High sensitivity if the filter has many poles. High order some poles are close together

• Potentially unstable if one pole is displaced outside the unit circle

• Conclusions use cascade of biquads (only 2 poles per transfer function) beware of poles close to the unit circle

∏N

n≠r 1,=rrn

m-Nn

mn

)p-(p

p aΔ - =pΔ

27 March 2003 P Baudrenghien AB/RF 5

Elliptic IIR. Direct Form (1)• Consider the fourth order

Elliptic IIR LPF (Example 4, slide 11 in lecture 5, Part II)

• Direct Form II implementation (slide 15 in lecture 5, Part II)

• Quantize its coefficients with 8 bits …

• … and the filter becomes unstable. Two poles are moved outside the unit circle

Pole/zero plot of the infinite precision (blue) and finite precision (red) Elliptic IIR. Direct Form II, 8 bits coefficients.

27 March 2003 P Baudrenghien AB/RF 6

Elliptic IIR. Direct Form (2)• … so we increase the

number of bits: 9 bits coefficients …

• … and it is stable but the frequency response shows severe distortion

Pole/zero plot of the infinite precision (blue) and finite precision (red) Elliptic IIR.Comparison of the achieved frequency response (green) with the reference (blue). Direct Form II, 9 bits coefficients.

27 March 2003 P Baudrenghien AB/RF 7

Elliptic IIR. Direct Form (3)

• …we have to use 12 bits coefficients to obtain a reasonable result

• … but there still is a big ripple at the end of the Pass-band

Comparison of the achieved frequency response (green) with the reference (blue). Direct Form II, 12 bits coefficients

27 March 2003 P Baudrenghien AB/RF 8

Elliptic IIR. Cascade of biquads • We now implement the IIR

with a cascade of 2 biquads (slide 16, lecture 5, Part II)

• The realization is now stable with only 8 bits coefficients

• The frequency response is a reasonable match to the reference

Comparison of the achieved frequency response (green) with the reference (blue). Cascade of two biquads, 8 bits coefficients

Lesson: Always use a cascade of biquads!

27 March 2003 P Baudrenghien AB/RF 9

Equiripple FIR (1)• Coefficient quantization

displaces the zeros as well and the FIR response will also suffer distortion

• Consider the 31 coefficients Equiripple FIR (slide 17, lecture 5, Part I)

• First use 8 bits coefficients. The result is quite good

Pole/zero plot of the infinite precision (blue) and finite precision (red) Equiripple FIR.Comparison of the achieved frequency response (green) with the reference (blue). 8 bits coefficients.

27 March 2003 P Baudrenghien AB/RF 10

Equiripple FIR (2)• So we decrease the

resolution further: 5 bits coefficients…

• … the impulse response is much distorted …

• … the zeros are not displaced much …

Impulse response of infinite precision (blue dots) and finite precision (green squares). Pole/zero plot of the infinite precision (blue) and finite precision (red). Equiripple FIR.5 bits coefficients

27 March 2003 P Baudrenghien AB/RF 11

Equiripple FIR (3)• … and the

frequency response is still acceptable. It is actually almost within specs …

Comparison of the achieved frequency response (green) with the reference (blue). Equiripple FIR. 5 bits coefficients.

Lesson: FIR are much less sensitive to coefficient quantization

27 March 2003 P Baudrenghien AB/RF 12

Zero Input Limit Cycle

• Back to the quantization of data

• So far (Part I) we have considered that the quantization noise sources (e1(n),e2(n),…) were white and independent of the input data x(n). This does not hold if the data covers a very small portion of the dynamic range

27 March 2003 P Baudrenghien AB/RF 13

Zero Input Limit Cycle. Rounding• First order system

• We have

IC: y(0)=1 switch input off: x(n) = 0 precision = 1 decimal digit a = -0.9

• Rounding (downwards) to the first fractional digit

a

y(n)

z-1

++

X

Q[.]quantizer

X

1-a

x(n)Q[.]

Precision Finite x(n)]a)-Q[(1 + 1)]-y(n Q[a = y(n)

Precision Infinite x(n)a)-(1 + 1)-y(n a = y(n)

= y(10)

0.4- = Q[-0.36] = 0.4] x Q[-0.9 = y(9)

0.4 = Q[0.36] = 0.4]- x Q[-0.9 = y(8)

0.4- = Q[-0.36] = 0.4] x Q[-0.9 = y(7)

0.4 = Q[0.45] = 0.5]- x Q[-0.9 = y(6)

0.5- = Q[-0.54] = 0.6] x Q[-0.9 = y(5)

0.6 = Q[0.63] = 0.7]- x Q[-0.9 = y(4)

0.7- = Q[-0.72] = 0.8] x Q[-0.9 = y(3)

0.8 = Q[0.81] = 0.9]- x Q[-0.9 = y(2)

0.9- = Q[-0.9] = 1] x Q[-0.9 = y(1)

1=y(0)

Limit cycle of frequency ½ and amplitude 0.4

27 March 2003 P Baudrenghien AB/RF 14

Zero Input Limit Cycle. Truncation• Truncation• With zero input, an oscillation

can be present at the output of a first order filter if rounding is used. For second and higher order filters a zero input limit cycle can appear with both rounding and truncation.

• The frequency of the limit cycle is close to the frequency where the filter has maximum response

No limit cycle 0 = Q[0] = 0] x Q[-0.9 = y(19)

0 = Q[0.09] = 0.1]- x Q[-0.9 = y(18)

0.1- = Q[-0.09] = 0.1] x Q[-0.9 = y(17)

0.1 = Q[0.18] = 0.2]- x Q[-0.9 = y(16)

0.2- = Q[-0.18] = 0.2] x Q[-0.9 = y(15)

0.2 = Q[0.27] = 0.3]- x Q[-0.9 = y(14)

0.3- = Q[-0.27] = 0.3] x Q[-0.9 = y(13)

0.3 = Q[0.36] = 0.4]- x Q[-0.9 = y(12)

0.4- = Q[-0.36] = 0.4] x Q[-0.9 = y(11)

0.4 = Q[0.45] = 0.5]- x Q[-0.9 = y(10)

0.5- = Q[-0.45] = 0.5] x Q[-0.9 = y(9)

0.5 = Q[0.54] = 0.6]- x Q[-0.9 = y(8)

0.6- = Q[-0.54] = 0.6] x Q[-0.9 = y(7)

0.6 = Q[0.63] = 0.7]- x Q[-0.9 = y(6)

0.7- = Q[-0.63] = 0.7] x Q[-0.9 = y(5)

0.7 = Q[0.72] = 0.8]- x Q[-0.9 = y(4)

0.8- = Q[-0.72] = 0.8] x Q[-0.9 = y(3)

0.8 = Q[0.81] = 0.9]- x Q[-0.9 = y(2)

0.9- = Q[-0.9] = 1] x Q[-0.9 = y(1)

1=y(0)

27 March 2003 P Baudrenghien AB/RF 15

Bound on Zero Input Limit Cycle

• Let h(n) be the impulse response from the quantization noise e(n) to the filter output f(n), we have:

• This bound is often largely over-estimated.

∑

∑∑

∑

∞

∞- =k

1-b-

∞

∞- =k

b-∞

∞- =kmax max

∞

∞- =k

(Rounding) 2 |h(k)|=

n)(Truncatio 2 |h(k)| = |e| |h(k)| =|f(n)|

k)-e(n h(k) = f(n)

27 March 2003 P Baudrenghien AB/RF 16

Floating Point Format

• IEEE 32 bits standard

• No need for scaling … Dynamic range covers 77 decades!

1.f22 - f0s e7 - e0

8 bits exponent

sign bit

23 bits mantissa

hidden bit

27 March 2003 P Baudrenghien AB/RF 17

Quantization error with float• Quantization error introduced by both additions and

multiplications

• By rounding (or truncating) the mantissa of x(n) to the b+1 bits format, we get Q[x(n)]

where

• The quantization noise is an additive error term

(n)ε x(n).+ x(n)= (n))ε + (1 x(n)= Q[x(n)]-b-b 2 ≤(n)ε<2-

(n) x(n).e(n)

The noise is proportional to the signal (constant SNR) and the overall performance is much better