Question 6 of 6: IIR Design Choices and Linear-Phase FIR Types

Question 6: IIR Design Choices and Linear-Phase FIR Types (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Twelve prototype poles from a squared-magnitude Butterworth design, an impulse-invariance design that has returned a non-integer order, and the four linear-phase FIR types.

Find. The pole subset that defines $H(s)$ and the mapping to $H(z)$; the next design step after a fractional order and which band to meet exactly; and the impossible type/response combinations with their justification.

Approach. Recognise the twelve poles as the $2N$ roots of the Butterworth squared magnitude and split them by half-plane; then handle the fractional order by rounding up and re-solving one of the two band equations; finally test each FIR type against its forced zeros at $z=\pm1$.

1. Part (a)(i): identify the order and split the poles. The Butterworth squared magnitude $|H(j\Omega)|^{2}=H(s)H(-s)|_{s=j\Omega}$ has $2N$ poles equally spaced on a circle of radius $\Omega_c$. Twelve poles therefore mean $N=6$ and $\Omega_c=0.71$. The set is symmetric about both axes, and only the six poles with negative real part — those at angles $7\pi/12$ through $17\pi/12$, i.e. $$\boxed{p_4,\ p_5,\ p_6,\ p_7,\ p_8,\ p_9}$$ belong to $H(s)$.
2. Justify the choice. Assigning the left-half-plane poles makes $H(s)$ simultaneously causal and stable, since a causal analogue system is stable only when every pole has $\mathrm{Re}\{s\} \lt 0$. The discarded six are the mirror images $-p$ that belong to $H(-s)$, so no information is lost: the product of the two sets reproduces the given squared magnitude exactly. The chosen six also occur in conjugate pairs ($p_4$ with $p_9$, $p_5$ with $p_8$, $p_6$ with $p_7$), which guarantees that $H(s)$ has real coefficients, and none of them lies on the $j\Omega$ axis, as required for a genuine Butterworth response.
3. Part (a)(ii): map to the z-plane. Substitute the bilinear transformation $$s=\frac{2}{T_d}\,\frac{1-z^{-1}}{1+z^{-1}}\qquad\Longrightarrow\qquad H(z)=H(s)\Big|_{s=\frac{2}{T_d}\frac{1-z^{-1}}{1+z^{-1}}} .$$ This maps the entire $j\Omega$ axis onto the unit circle once, so it cannot alias; the price is the non-linear frequency warping $\Omega=\frac{2}{T_d}\tan(\omega/2)$. Consequently every critical digital frequency must be pre-warped to its analogue counterpart before the prototype is designed, and the left-half plane maps into the interior of the unit circle, so the stability established in part (i) is carried over automatically.
4. Part (b)(i): round the order up. A filter order must be an integer, and the two design equations were solved simultaneously to give $N=5.305$. Rounding up to $$\boxed{N=6}$$ is the next step: with the extra pole the filter exceeds both specifications, whereas rounding down to 5 would fail at least one of them. Having fixed $N=6$, the pair of equations is over-determined, so $\Omega_c$ must be re-computed from one of them — the printed value 0.815 belongs to the fractional-order solution and is no longer exact.
5. Part (b)(ii): meet the pass band exactly. Re-solving the pass-band equation gives a smaller cutoff than re-solving the stop-band equation (raising $N$ lowers $\Omega_c$ obtained from the pass band and raises the one obtained from the stop band), so any cutoff between the two satisfies both bands. Choosing the pass-band-exact value therefore places the response as deep into the stop band as the specifications allow: $$\boxed{\text{meet the PASS-band corner exactly; the stop band is then over-satisfied}}$$ The reason this matters is specific to impulse invariance: the transformation $h[n]=T_d\,h_a(nT_d)$ aliases the analogue tail, and the aliased energy adds to the response where the response is smallest, i.e. in the stop band. Reserve is therefore needed in the stop band, not in the pass band; meeting the stop-band corner exactly would leave no margin at all for the aliasing error. (The desired tolerances themselves are not printed on the paper, so this argument is stated structurally rather than numerically.)
6. Part (c): list the forced zeros of each FIR type. Linear phase requires $h[n]=\pm h[M-n]$, and the sign together with the parity of $M$ forces zeros of the amplitude response at $\omega=0$ (i.e. $z=+1$) and/or $\omega=\pi$ (i.e. $z=-1$):

   Forced zeros of the four linear-phase types

   Type	Length / symmetry	Forced zero at $z=+1$ ($\omega=0$)	Forced zero at $z=-1$ ($\omega=\pi$)
   I	$M$ even, symmetric	no	no
   II	$M$ odd, symmetric	no	yes
   III	$M$ even, antisymmetric	yes	yes
   IV	$M$ odd, antisymmetric	yes	no

7. Cross each forced zero against what the response needs. A low-pass and a band-stop response must be non-zero at $\omega=0$; a high-pass and a band-stop must be non-zero at $\omega=\pi$; a band-pass needs neither. A type is impossible exactly when one of its forced zeros lands where the response must not vanish:

   X marks the combinations that cannot be realised

   FIR filter	Lowpass	Highpass	Bandpass	Bandstop
   Type I
   Type II		X		X
   Type III	X	X		X
   Type IV	X			X

   In words: Type I has no forced zero and realises all four; Type II vanishes at $\pi$ so it cannot be high-pass or band-stop; Type III vanishes at both $0$ and $\pi$ so only band-pass survives; Type IV vanishes at $0$ so it cannot be low-pass or band-stop. Note that a forced zero sitting inside a stop band is harmless — it is only fatal when it falls in a band that must pass.