Filter structures

One filter can be wired up in several ways. Below, a biquad’s block diagram rearranges itself: its two halves swap places, and then their delays merge. Watch the delay count fall while the multipliers stay at five.

Swap the halves, share the delays

A filter’s equation says what to compute. It does not say how to wire it up, and there are several ways. They all give the same output, but they keep different numbers inside. In this lesson I build one filter five ways and look inside each.

The filter is the biquad of Audio equalisers and biquads (20.5):

I write $B(z)$ for its top and $A(z)$ for its bottom, so $H(z)=B(z)/A(z)$. By Transfer functions, poles & zeros (16.3), the roots of $B$ are the zeros and the roots of $A$ are the poles. Read backwards, the same page turns $H(z)$ into its difference equation:

In Difference equations (6.1), every term became a wire into an adder, with a delay box for each step back and a gain triangle for its number. Do that here. The three $x$ terms hang off a column of two delay boxes on the input, and the two $y$ terms off a second column on the output.

That wiring is called direct form I. Its first half computes $B(z)$ on the input and its past: I call it the zeros’ half. Its second half divides by $A(z)$, using past outputs: the poles’ half. It needs four delays and five multipliers.

Now the trick. The two halves are two LTI systems in a row, and by Properties of LTI systems (5.4) their order does not matter. So swap them: poles first, zeros second. In the z-domain it is one line:

After the swap, both delay columns are fed by the same signal, so they hold the same numbers. Think of two people each keeping a copy of the same notebook: one copy, shared, is enough.

The picture at the top of the page makes the swap, and then shares the delays. Notice what changed and what did not. The multipliers stay five; only the delays halve. The output depends on two steps of the past, so two delays is the fewest possible. A structure with as many delays as the filter’s order is called canonical, and this one is direct form II.

What direct form II stores

Name the signal on the shared column $w[n]$. It is the input passed through the poles alone, so $W(z)=X(z)/A(z)$. Then the zeros’ half builds the output from it:

This $w[n]$ is not a window, even though windows also use the letter; here it is a signal inside a filter. The two delays hold ${w[n-1]}$ and ${w[n-2]}$, and nothing else.

There is a third way to wire a biquad with two delays, the transposed form. I will show where it comes from after the next instrument. Its delays hold two running sums, $s_1[n]$ and $s_2[n]$:

Each output is $b_0x[n]$ plus the sum saved one step ago. Then both sums are brought up to date for the next step. SciPy’s lfilter works this way.

Same output, different insides

Do the three structures really agree, and what do they store? Let’s feed one sine to each.

The filter is a piece of the elliptic filter of Choosing FIR or IIR (20.6). Its order is 5, so it has five poles: one on the real axis and two mirror pairs. The pair nearest the unit circle, at radius 0.9527 and angles ±45.64°, makes the sharpest peak. With its two zeros, it is one biquad:

Its zeros sit on the unit circle at ±57.44°, which is 1276 Hz at $f_s=8$ kHz. The input is a 1 kHz sine of amplitude 1, $x[n]=\sin(2\pi\cdot1000n/8000)=\sin(\pi n/4)$: eight samples a cycle. It starts from rest, so every stored value starts at 0.

Three kitchens can make the same soup, one of them in a pot that nearly overflows. Watch for the pot. Many chips store numbers in fixed point: a fixed number of bits and a fixed range, as in Quantization & noise (11.1). A stored value that outgrows the range overflows.

After the clip, choose a structure with its button to compare their insides.

The output panel never changes. In floating point the three outputs differ by less than $10^{-14}$, rounding noise in the last digits. What differs is the inside: direct form I and the transposed form keep numbers no bigger than the output, while direct form II’s $w[n]$ is about three times bigger.

Why? By Properties of LTI systems (5.4), once the start-up has died away, a sine that goes in comes out as a sine at the same frequency, scaled by the gain there. At 1 kHz the biquad’s gain is $\lvert H\rvert=5.010$, but the poles alone give $\lvert 1/A\rvert=14.83$. And $w[n]$ is the input through the poles alone.

The zeros pull the output back down. They sit on the circle at 1276 Hz, not far from 1 kHz, so $\lvert B\rvert$ is only 0.3378 there. So $w$ is bigger than $y$ by $1/\lvert B\rvert=2.960$, and the peaks on screen have the same ratio. They are a little lower than 5.010 and 14.83 because here the eight samples of a cycle never land exactly on a crest.

Three more ways to build it

The same biquad can be built in three more ways, and the figure draws each with its numbers. The transposed form is the one from the equations above. The parallel form splits $H(z)$ into a sum of simpler filters that all see the same input.

The lattice builds the poles’ half, $1/A(z)$, from two stages. One wire runs forwards from $x[n]$ to $y[n]$, and one runs back through the delays. Each stage has one number, its reflection coefficient: $k_2$ for the stage the input enters, $k_1$ for the next. In stage 1 two gains, $k_1$ and $-k_1$, cross between the wires; stage 2 needs only $-k_2$.

This biquad has $b_0=b_2=1$. As in 6.1, a gain of 1 is not drawn, so two wires of (a) have no triangle.

Transposing. A branch point is a dot where one wire splits into several. To transpose a block diagram, reverse every arrow, turn every adder into a branch point and every branch point into an adder, and swap input and output. The result has the same $H(z)$.

Transposing direct form II gives (a). In direct form II each tap of the delay column is a branch point, feeding a $b$ gain and an $a$ gain. In (a) those taps have become adders, which is why its delays hold sums. The Linear Algebra note below gives a one-line reason the rule works.

Parallel. In Properties and the inverse z-transform (16.2), residuez split a ratio of polynomials into a direct term plus one fraction per pole. Do that here, then add the two fractions of the mirror pair back together, so that the coefficients are real:

That is (b): a constant beside a second-order branch, both fed $x[n]$, their outputs added. A longer filter gets one real branch per pair of poles, two delays each.

Cascade. Or keep the biquads one after another, the output of one feeding the next. SciPy’s sosfilt runs 20.6’s whole filter this way, and it is the subject of Second-order sections (21.2).

Lattice. Follow the wires of (c) back from $y[n]$, stage by stage, and the input comes out as

That is the equation of $1/A(z)$ when $a_1=k_1(1+k_2)$ and $a_2=k_2$. Turned round, $k_2=a_2$ and $k_1=a_1/(1+a_2)$.

Now the reason to use it. Stability and causality (16.4) said both poles are inside the circle when $\lvert a_2\rvert < 1$, $1+a_1+a_2 > 0$ and $1-a_1+a_2 > 0$. In terms of the $k$‘s, the first is $\lvert k_2\rvert < 1$, which makes $1+k_2$ positive. The last two are $(1+k_2)(1+k_1) > 0$ and $(1+k_2)(1-k_1) > 0$.

So the stability triangle is just $\lvert k_1\rvert < 1$ and $\lvert k_2\rvert < 1$. You can read stability straight off the lattice’s gains. Here $k_2=0.9076$ and $k_1=-0.6984$, so the biquad is stable.

Worked example

1. Counting. For a biquad, direct form I has 4 delays, direct form II has 2 and the transposed form has 2. All three have 5 multipliers, or 4 when $b_0=1$. This lesson’s biquad also has $b_2=1$, so it needs only 3.

2. Three samples by hand. The input starts $x[0]=0$, $x[1]=0.7071$, $x[2]=1$. Direct form II gives $w[0]=0$ and $w[1]=0.7071$, then $w[2]=1+1.3322\cdot0.7071=1.9420$. The output is $y[1]=w[1]=0.7071$ and $y[2]=1.9420-1.0764\cdot0.7071=1.1809$. Already $w$ is bigger than $y$.

3. The parallel form. residuez(b, a) gives the residues $-0.05091\mp0.23762j$ at the poles $0.66612\pm0.68108j$, and the direct term 1.10182. Combined, the branch numerator is $-0.10182+0.39150z^{-1}$. Check at $z=1$. There the branch’s top is $-0.10182+0.39150=0.28968$ and its bottom is $A(1)=0.57535$, so $H(1)=1.10182+0.28968/0.57535=1.6053$, the gain at DC.

4. The lattice. $k_2=a_2=0.90759$ and $k_1=a_1/(1+a_2)=-1.33224/1.90759=-0.69839$. Both are between −1 and 1, so the biquad is stable.

Where you’ll meet this

Audio code often runs the transposed form in floating point: it has the fewest stored values and good numerics. Fixed-point DSP chips often use direct form I. Its five products add up in one wide running total, the accumulator, and it stores only inputs and outputs, so nothing it stores outgrows them.

Speech coders use lattices, because their reflection coefficients show stability at a glance (Parametric models and linear prediction, 25.3). Next come cascades of biquads in Second-order sections (21.2), and overflow and word length in Finite word-length effects (21.3). Polyphase structures for changing the sample rate are in Polyphase structures (22.4).

Reference card