Which linear-phase FIR types can realise each desired response: 22-Elec-B1 Digital Signal Processing, May 2018, Q6

Question 6: Which linear-phase FIR types can realise each desired response (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. A window design $h[n] = h_d[n]w[n]$ on $0 \le n \le M$, four desired responses each carrying the linear-phase factor $e^{-j(M/2)\omega}$, and the four standard linear-phase FIR types.

Find. For each desired response, which of Types I–IV can realise it, with a justification based on the symmetry of $h[n]$ and on the forced zeros of each type.

Approach. Every linear-phase FIR response can be written $H(e^{j\omega}) = A(\omega)e^{-j\omega M/2}$ or $jA(\omega)e^{-j\omega M/2}$ with $A$ real; the presence or absence of the factor $j$ decides symmetric versus antisymmetric $h[n]$, and the parity of $M$ then decides which structural zeros are forced at $\omega = 0$ and $\omega = \pi$. A type is admissible when its symmetry matches and none of its forced zeros falls where the desired response must be non-zero.

Decide symmetric or antisymmetric from the factor in front. Windowing preserves symmetry about $M/2$ provided $w[n]$ is itself symmetric, which every standard window is. So the symmetry of $h[n]$ is inherited from $h_d[n]$: $$H_d = A(\omega)e^{-j\omega M/2} \;\Rightarrow\; h_d[n] = h_d[M-n] \ \text{(symmetric)}, \qquad H_d = jA(\omega)e^{-j\omega M/2} \;\Rightarrow\; h_d[n] = -h_d[M-n].$$ Rows (i) and (ii) carry no $j$ and are therefore symmetric — Type I or II. Rows (iii) and (iv) carry $\pm j$ with an odd sign pattern in $\omega$ and are antisymmetric — Type III or IV.
List the structural zeros of the four types. Writing $A(\omega)$ for the real amplitude function:
• Type I ($M$ even, symmetric): $A$ is a cosine sum, free at both $\omega = 0$ and $\omega = \pi$.
• Type II ($M$ odd, symmetric): the pairs $n$ and $M-n$ have opposite parity, so at $\omega = \pi$ each pair cancels — $A(\pi) = 0$ identically.
• Type III ($M$ even, antisymmetric): $A$ is a sine sum with $A(0) = 0$ and $A(\pi) = 0$ identically.
• Type IV ($M$ odd, antisymmetric): $A(0) = 0$ identically, but $A(\pi)$ is free.
The rule that follows is simple: a type is ruled out only when one of its forced zeros sits where $H_d$ must be non-zero.
Row (i), the ideal lowpass. The response is real-amplitude, so $h[n]$ is symmetric and only Types I and II are candidates. The desired gain is 1 at $\omega = 0$ and 0 at $\omega = \pi$. Type I forces nothing, so it works. Type II forces $A(\pi) = 0$, which the lowpass wants anyway, so it also works — the constraint is free of charge. $$\text{Row (i)} \;\Rightarrow\; \boxed{\text{Type I: YES}, \quad \text{Type II: YES}, \quad \text{Types III, IV: NO}.}$$
Row (ii), the ideal highpass. Again real-amplitude, so Types III and IV are excluded on symmetry alone. The desired gain is now 1 at $\omega = \pi$. Type II forces $A(\pi) = 0$ and therefore cannot pass the very band the filter is meant to pass; only Type I survives. $$\text{Row (ii)} \;\Rightarrow\; \boxed{\text{Type I: YES}, \quad \text{Types II, III, IV: NO}.}$$
Row (iii), the 90-degree-shifted highpass. The $\pm j$ with odd sign in $\omega$ makes $A(\omega)$ an odd function, so $h[n]$ is antisymmetric and only Types III and IV are candidates. The desired amplitude is 0 near $\omega = 0$ but of unit magnitude at $\omega = \pi$. Type III forces $A(\pi) = 0$ and is therefore disqualified; Type IV forces only $A(0) = 0$, which the response wants anyway. $$\text{Row (iii)} \;\Rightarrow\; \boxed{\text{Type IV: YES}, \quad \text{Types I, II, III: NO}.}$$
Row (iv), the 90-degree-shifted bandpass. Antisymmetric again, so Types III and IV are the candidates. This time the desired amplitude is zero both at $\omega = 0$ and at $\omega = \pi$, because $0 \lt \omega_{c1} \lt \omega_{c2} \lt \pi$. Type III's two forced zeros land exactly where they are wanted, and Type IV's single forced zero at $\omega = 0$ likewise; nothing prevents a Type IV design from approximating zero at $\pi$, it simply is not obliged to. $$\text{Row (iv)} \;\Rightarrow\; \boxed{\text{Type III: YES}, \quad \text{Type IV: YES}, \quad \text{Types I, II: NO}.}$$ Type III is the more natural choice of the two, since it enforces the upper-edge null structurally.

Figure 6.1 — the four linear-phase FIR types. The symmetry of $h[n]$ about $M/2$ (upper row symmetric, lower row antisymmetric) decides which desired responses are reachable at all, and the parity of $M$ decides which amplitude zeros are forced at $\omega = 0$ and $\omega = \pi$.

$H_d(e^{j\omega})$	Type I	Type II	Type III	Type IV
(i) Lowpass, $e^{-j(M/2)\omega}$ for $\|\omega\| \lt \omega_c$	YES	YES	—	—
(ii) Highpass, $e^{-j(M/2)\omega}$ for $\omega_c \lt \|\omega\| \lt \pi$	YES	—	—	—
(iii) 90-degree-shifted highpass, $\pm je^{-j(M/2)\omega}$ for $\omega_c \lt \|\omega\| \lt \pi$	—	—	—	YES
(iv) 90-degree-shifted bandpass, $\pm je^{-j(M/2)\omega}$ for $\omega_{c1} \lt \|\omega\| \lt \omega_{c2}$	—	—	YES	YES

Type	$M$	Symmetry of $h[n]$	Forced amplitude zeros	Natural use
I	even	$h[n] = h[M-n]$	none	any real-amplitude filter: LP, HP, BP, BS
II	odd	$h[n] = h[M-n]$	$A(\pi) = 0$	lowpass and bandpass only
III	even	$h[n] = -h[M-n]$	$A(0) = A(\pi) = 0$	differentiators and bandpass Hilbert transformers
IV	odd	$h[n] = -h[M-n]$	$A(0) = 0$	differentiators and highpass or full-band Hilbert transformers

Topic. The four generalised-linear-phase FIR types and the structural zeros that decide which frequency-selective responses each can realise.

Key relations. Generalised linear phase: $H(e^{j\omega}) = A(\omega)e^{-j\omega M/2}$ (symmetric $h$) or $jA(\omega)e^{-j\omega M/2}$ (antisymmetric $h$), with $A$ real. Structural constraints: Type II has $H(e^{j\pi}) = 0$; Type III has $H(e^{j0}) = H(e^{j\pi}) = 0$; Type IV has $H(e^{j0}) = 0$. Windowing preserves symmetry: if $h_d$ is (anti)symmetric about $M/2$ and $w$ is symmetric on $0 \le n \le M$, then $h = h_d w$ has the same symmetry.

Why this works. The forced zeros are not design failures but algebraic identities. For a symmetric $h$ with $M$ odd, every sample pairs with a partner an odd distance away, so at $\omega = \pi$ the two terms $h[n]e^{-j\pi n}$ and $h[M-n]e^{-j\pi(M-n)}$ have opposite sign and cancel — there is no coefficient choice that avoids it. Equivalently, linear phase forces the zeros of $H(z)$ into reciprocal groups, and a Type II filter is obliged to carry a zero at $z = -1$. Recognising this is what turns a four-by-four table into four short arguments: match the symmetry, then check whether the forced zeros land in a stopband (harmless) or in a passband (fatal).

Common pitfall. Answering row (iv) with Type III alone on the grounds that "the response is zero at both ends". Type IV is not forbidden from approximating a zero at $\omega = \pi$; it is merely not compelled to, and the windowed design will produce a good approximation there because $h_d$ has no energy in that band. The disqualifying test is always the reverse one — a forced zero where the response must be non-zero, which is what kills Type II in row (ii) and Type III in row (iii). A second pitfall is forgetting that an asymmetric window (or a truncation that is not centred on $M/2$) destroys the linear phase entirely, at which point none of the four types applies.

Source. Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd ed., §5.7.3 (generalised linear phase, the four types and their constraints) and §7.5 (FIR design by windowing); Mitra, §7.3.

Question 6 of 6: Which linear-phase FIR types can realise each desired response

Question 6: Which linear-phase FIR types can realise each desired response (12 marks)