Analyzing LTI systems through the frequency response and the transfer function H(z): pole-zero geometry, magnitude and phase, phase delay and group delay, and the classical system classes built from them.
10.1 Frequency Response (FR)
10.1.1 Sinusoids as Eigenfunctions
The impulse response(脈衝響應)$h[n]$ completely describes an LSI (linear shift-invariant) system: the output is always the convolution $y[n] = x[n] \circledast h[n] = \sum_{m} h[m]\,x[n-m]$.
Eigenfunction relation — a complex sinusoid passes through an LSI system unchanged in shape; it is only multiplied by the complex number $H(e^{j\omega_0})$(複數弦波是 LSI 系統的特徵函數).
Feed in a complex sinusoid(複數弦波)$x[n] = e^{j\omega_0 n}$ and watch the convolution factor:
The complex scale factor is the DTFT of the impulse response evaluated at the input frequency. Writing it in polar form $H(e^{j\omega}) = |H(e^{j\omega})|\,e^{j\theta(\omega)}$, the output is
Frequency response
$$y[n] = |H(e^{j\omega_0})|\;e^{j(\omega_0 n + \theta(\omega_0))},
\qquad H(e^{j\omega}) = \mathrm{DTFT}\{h[n]\}$$
Concept — why eigenfunctions matter
A complex sinusoid at $\omega_0$ comes out of any LSI system as the same sinusoid, scaled by $|H(e^{j\omega_0})|$ and phase-shifted by $\arg\{H(e^{j\omega_0})\} = \theta(\omega_0)$. Since any signal can be expressed as a superposition of sinusoids (DTFT), knowing $H(e^{j\omega})$ for all $\omega$ tells us everything about the system:
$|H(e^{j\omega})|$ — the magnitude response(振幅響應) → frequency-dependent gain;
$\arg\{H(e^{j\omega})\}$ — the phase response(相位響應) → frequency-dependent phase shift.
In practice signals are real. Decompose a real cosine into two complex sinusoids and apply the eigenfunction property to each:
Real input decomposition
$$x[n] = A\cos(\omega_0 n + \phi)
= \frac{A}{2}e^{j\phi}e^{j\omega_0 n} + \frac{A}{2}e^{-j\phi}e^{-j\omega_0 n}$$
For a real impulse response $h[n]$ the frequency response is conjugate-symmetric, $H(e^{-j\omega}) = H^{*}(e^{j\omega}) = |H(e^{j\omega})|e^{-j\theta(\omega)}$, so the two terms recombine into a real cosine:
Real sinusoid through a real LSI system
$$A\cos(\omega_0 n + \phi)\;\longrightarrow\;
\boxed{\,y[n] = A\,|H(e^{j\omega_0})|\cos\big(\omega_0 n + \phi + \theta(\omega_0)\big)\,}$$
Key result
A real sinusoid of frequency $\omega_0$ passed through an LSI system with real $h[n]$ keeps its frequency; its gain is modified by $|H(e^{j\omega_0})|$ and its phase is shifted by $\theta(\omega_0)$. This is the single most-used fact for computing sinusoidal steady-state outputs on exams.
10.1.3 Transient and Steady State
Most signals start at a finite time(訊號從有限時間開始), e.g. a gated complex sinusoid $x[n] = e^{j\omega_0 n}\,u[n]$. What does gating do? Split the convolution sum:
The steady-state term is identical to the response to a pure (eternal) sinusoid.
The transient term is the DTFT of the impulse-response tail from time $n$ onwards — a consequence of gating.
For an FIR system of length $N$ the tail is empty for $n \ge N$: the transient is exactly zero after $N$ samples.
For any stable IIR system $h[n] \to 0$, so the transient decays to zero for large $n$.
Fig. 10-1 — Gated sinusoid $x[n]=\cos(0.3n)u[n]$ through the stable one-pole system $y[n]=0.8y[n-1]+0.2x[n]$. The output (orange stems) converges to the steady-state sinusoid $0.599\cos(0.3n-0.787)$ (dashed) after the transient decays like $0.8^n$(暫態以 $0.8^n$ 速度消失,之後只剩穩態弦波).
10.1.4 Example: Moving-Average Filter
Worked Example 10-A — frequency response of the M-point moving average
The MA filter $y[n] = \frac{1}{M}\sum_{\ell=0}^{M-1} x[n-\ell]$ is convolution with $h[n] = \frac{1}{M}$ for $0 \le n \le M-1$ (zero otherwise).
Step 1 — DTFT as a geometric sum.
$$H(e^{j\omega}) = \frac{1}{M}\sum_{n=0}^{M-1} e^{-j\omega n}
= \frac{1}{M}\,\frac{1-e^{-j\omega M}}{1-e^{-j\omega}}$$
Step 2 — extract linear phase and a real ratio. Factor out half-angle exponentials:
$$H(e^{j\omega}) = \frac{1}{M}\,e^{-j\omega\frac{M-1}{2}}\,
\frac{\sin(M\omega/2)}{\sin(\omega/2)}$$
Step 3 — magnitude and phase.
$$\big|H(e^{j\omega})\big| = \left|\frac{1}{M}\frac{\sin(M\omega/2)}{\sin(\omega/2)}\right|,
\qquad
\theta(\omega) = -\frac{M-1}{2}\,\omega + \pi r,\quad r = \Big\lfloor \frac{M\omega}{2\pi} \Big\rfloor$$
The $\pi r$ term accounts for the sign changes of the Dirichlet ratio — every time $\sin(M\omega/2)$ changes sign the phase jumps by $\pi$.
Step 4 — response to a two-tone input ($M=5$). For $x[n] = e^{j\omega_0 n} + e^{j\omega_1 n}$ with $\omega_0 = 0.1\pi$, $\omega_1 = 0.5\pi$:
$$H(e^{j0.1\pi}) \approx 0.90\,e^{-j0.2\pi}, \qquad H(e^{j0.5\pi}) = 0.2$$
so $y[n] = H(e^{j\omega_0})e^{j\omega_0 n} + H(e^{j\omega_1})e^{j\omega_1 n}$: the slow component passes nearly unchanged (gain 0.90, delayed 2 samples), the fast component is attenuated to 0.2 — the MA filter is a lowpass smoother.
Fig. 10-2 — $M=5$ moving-average filter: impulse response, magnitude $\frac{1}{5}|\sin(5\omega/2)/\sin(\omega/2)|$ with nulls at $\omega = 2\pi k/5$, and the principal-value phase showing the linear slope $-2\omega$ with $\pi$ jumps at the magnitude nulls(振幅零點處相位跳 $\pi$).
Warning — phase jumps are not discontinuous physics
The $\pi$ jumps in $\theta(\omega)$ come from writing a signed real amplitude $\frac{\sin(M\omega/2)}{\sin(\omega/2)}$ as a positive magnitude times a phase. When the sign flips, the $\pi$ is absorbed into the phase. Plotting tools also wrap phase into $(-\pi,\pi]$ (principal value), causing additional cosmetic $2\pi$ jumps. Always check which kind of jump you are looking at.
10.2 Transfer Function (TF)
10.2.1 From Difference Equation to H(z)
Alternatively, take the z-transform of $y[n] = h[n] \circledast x[n]$: convolution becomes multiplication, $Y(z) = H(z)X(z)$. The transfer function(轉移函數) $H(z)$ can be obtained two equivalent ways:
Transfer function
$$H(z) \;=\;
\begin{cases}
\dfrac{\sum_k p_k z^{-k}}{\sum_k d_k z^{-k}} & \text{if the system has difference-equation form}\\[10px]
\displaystyle\sum_n h[n]\,z^{-n} & \text{directly from the impulse response}
\end{cases}$$
For an FIR filter with $h[n] = \{h_0, h_1, \dots, h_{M-1}\}$ the two views coincide trivially: $p_k = h_k$, $d_0 = 1$, and the difference equation is $y[n] = \sum_{k=0}^{M-1} h_k\,x[n-k]$.
Worked Example 10-B — MA filter as a transfer function
Step 1. $h[n] = \frac{1}{M}$ on $0\le n\le M-1$ gives the geometric sum
$$H(z) = \frac{1}{M}\sum_{\ell=0}^{M-1} z^{-\ell} = \frac{1-z^{-M}}{M(1-z^{-1})}
= \frac{z^{M}-1}{M\,z^{M-1}(z-1)}$$
Step 2 — roots. The numerator $z^M = 1$ has $M$ roots — the $M$-th roots of unity $z = e^{j2\pi r/M}$, $r = 0,\dots,M-1$, equally spaced on the unit circle.
Step 3 — cancellation. The root at $z = 1$ (i.e. $r=0$) cancels the denominator pole at $z=1$, so the filter has $M-1$ zeros on the unit circle and only trivial poles at $z=0$. These $M-1$ zeros are exactly the magnitude nulls of Fig. 10-2.
10.2.2 Poles, Zeros, ROC, and Stability
Worked Example 10-C — third-order recursive system (lecture TF example)
Step 1 — difference equation to H(z).
$$y[n] = x[n-1] - 1.2x[n-2] + x[n-3] + 1.3y[n-1] - 1.04y[n-2] + 0.222y[n-3]$$
Collect $Y$ terms on the left and transform:
$$H(z) = \frac{Y(z)}{X(z)} =
\frac{z^{-1} - 1.2z^{-2} + z^{-3}}{1 - 1.3z^{-1} + 1.04z^{-2} - 0.222z^{-3}}$$
Step 2 — factorize.
$$H(z) = \frac{z^{-1}\,(1-\zeta_0 z^{-1})(1-\zeta_0^{*} z^{-1})}
{(1-\lambda_0 z^{-1})(1-\lambda_1 z^{-1})(1-\lambda_1^{*} z^{-1})},
\qquad
\begin{aligned}
\zeta_0 &= 0.6 + j0.8\\
\lambda_0 &= 0.3\\
\lambda_1 &= 0.5 + j0.7
\end{aligned}$$
Zeros(零點)$\zeta_0, \zeta_0^{*}$ lie on the unit circle ($|\zeta_0| = 1$); poles(極點)at $0.3$ and $0.5 \pm j0.7$ ($|\lambda_1| \approx 0.86$) lie inside it.
Step 3 — ROC and stability. The poles $\lambda_i$ determine the ROC(收斂區域):
causal → ROC is $|z| > \max_i |\lambda_i| = 0.86$;
this ROC includes the unit circle → the system is also stable.
Key — causal and stable
For a causal system the ROC is the exterior $|z| > \max|\lambda_i|$. The system is BIBO-stable iff the ROC contains the unit circle, i.e. iff all poles are strictly inside the unit circle(因果且穩定 ⇔ 所有極點都在單位圓內). Zeros may be anywhere.
10.2.3 From H(z) to Frequency Response
The DTFT is the z-transform evaluated on the unit circle(單位圓):
FR = TF on the unit circle
$$H(e^{j\omega}) = H(z)\Big|_{z = e^{j\omega}}$$
Factor $H(z)$ into first-order terms and evaluate at $z = e^{j\omega}$:
Magnitude and phase from roots
$$\big|H(e^{j\omega})\big| = \Big|\frac{p_0}{d_0}\Big|\,
\frac{\prod_{k=1}^{M}\big|e^{j\omega}-\zeta_k\big|}{\prod_{k=1}^{N}\big|e^{j\omega}-\lambda_k\big|},
\qquad
\theta(\omega) = \arg\Big\{\frac{p_0}{d_0}\Big\} + \omega(N-M)
+ \sum_{k=1}^{M}\arg\big\{e^{j\omega}-\zeta_k\big\}
- \sum_{k=1}^{N}\arg\big\{e^{j\omega}-\lambda_k\big\}$$
10.2.4 Geometric Interpretation of the FR
Each factor $(e^{j\omega} - \nu)$ is a vector drawn from the root $\nu$ (pole or zero) to the point $e^{j\omega}$ moving along the unit circle. The whole frequency response is a product/ratio of these vectors:
Key — geometric rules
Magnitude $|H(e^{j\omega})|$ = (product of lengths of vectors from zeros) ÷ (product of lengths of vectors from poles).
Phase $\theta(\omega)$ = (sum of angles of vectors from zeros) − (sum of angles of vectors from poles), plus the linear term $\omega(N-M)$.
Roots near the unit circle cause rapid changes: a zero near/on the u.c. → magnitude minimum (dips to 0 if on it); a pole near the u.c. → magnitude peak (→ $\infty$ if on it); the relative angle swings quickly → fast phase variation.
A pole and zero near each other cancel when seen from afar; they only affect the response when $e^{j\omega}$ gets close.
直觀解釋(點擊展開)
想像 $e^{j\omega}$ 是一個沿著單位圓行走的觀察點。從每個零點、極點各拉一條橡皮筋到這個觀察點:振幅就是「零點橡皮筋長度的乘積 ÷ 極點橡皮筋長度的乘積」。當觀察點走近一個極點時,分母的某條橡皮筋變得很短,振幅暴增(共振峰);走近零點時分子變短,振幅塌陷(凹谷)。把 $|H(z)|$ 想成 z 平面上的 3D 地形:極點是插向天空的「帳篷柱」,零點是釘住地面的「圖釘」,而頻率響應只是沿單位圓走一圈所看到的地形高度。
Fig. 10-3 — Lecture TF example: zeros $0.6\pm j0.8$ on the unit circle, poles at $0.3$ and $0.5 \pm j0.7$. The on-circle zero (angle $\approx 0.295\pi$) forces $|H| = 0$, while the nearby pole at radius $0.86$ (angle $\approx 0.303\pi$) pushes the magnitude up right next to the null — a sharp notch-and-peak pair, with fast phase variation in the same region(單位圓上的零點造成深谷,旁邊的極點造成尖峰).
First-order building block: a single zero
The factor $H(e^{j\omega}) = 1 - re^{j\theta}e^{-j\omega}$ (one zero at $z = re^{j\theta}$) is the elementary building block — every rational response is a product/ratio of such terms. Its log magnitude, phase and group delay all concentrate their activity around $\omega = \theta$:
Fig. 10-4 — Single zero $1 - re^{j\theta}e^{-j\omega}$ with $r = 0.9$ for $\theta = 0, \pi/2, \pi$ (lecture Examples I & II). The dip in log magnitude, the rapid phase swing, and the negative group-delay dip all sit at $\omega = \theta$; moving $\theta$ just translates the pattern. A pole contributes the reciprocal magnitude and negated phase of the same plots(極點為倒數振幅、相反相位).
10.2.5 Filtering: Shaping the Magnitude Response
Filtering means separating information in frequency with a constructed $H(e^{j\omega})$: given $x[n] = A\cos(\omega_1 n) + B\cos(\omega_2 n)$ where only the $\omega_1$ part is wanted, build $|H(e^{j\omega_1})| \approx 1$ and $|H(e^{j\omega_2})| \approx 0$, so that $y[n] \approx A\cos(\omega_1 n + \theta(\omega_1))$.
Worked Example 10-D — designing a 3-point FIR filter (lecture filtering example)
Step 2 — constraints. Input mixes tones at $\omega_1 = 0.1$ and $\omega_2 = 0.4$ rad/sample; we want to keep the first and remove the second:
$$\beta + 2\alpha\cos(0.1) = 1, \qquad \beta + 2\alpha\cos(0.4) = 0$$
Step 4 — verify. The output is $y[n] \approx \cos(0.1n + \theta(0.1))$ with $\theta(\omega) = -\omega$: the wanted tone survives with a one-sample delay, the unwanted tone is annihilated (Fig. 10-5).
Fig. 10-5 — Designed 3-pt filter $h[n] = \{6.76, -12.46, 6.76\}$: magnitude in dB with unit gain at $\omega_1 = 0.1$ (red line) and a null at $\omega_2 = 0.4$ (blue line); below, the two-tone input (stems) and the output (red), which is the clean $\omega_1$ tone delayed by one sample(輸出只剩想要的低頻弦波,延遲 1 個樣本).
Tip — degrees of freedom
A length-$L$ FIR filter has $L$ free taps, so it can satisfy about $L$ independent magnitude/phase constraints. Here 2 unknowns $(\alpha,\beta)$ met 2 constraints exactly. Real filter design (Chapter on filter design) generalizes this idea to many constraints over whole frequency bands.
10.3 Phase Delay and Group Delay
10.3.1 Phase Delay
For a sinusoidal input $x[n] = \cos(\omega_0 n)$ we found $y[n] = |H(e^{j\omega_0})|\cos(\omega_0 n + \theta(\omega_0))$. The phase shift can be re-read as a time shift by factoring $\omega_0$ out:
The minus sign makes a positive $\tau_p$ mean a delay (causal behavior): a falling phase corresponds to the carrier arriving later.
Worked Example 10-E — phase delay of the 3-pt filter
Step 1. From Example 10-D, $H(e^{j\omega}) = e^{-j\omega}(\beta + 2\alpha\cos\omega)$, so wherever the real factor is positive, $\theta(\omega) = -\omega$.
Step 2.
$$\tau_p(\omega) = -\frac{-\omega}{\omega} = +1 \text{ sample at all frequencies}$$
i.e. every sinusoid is delayed exactly 1 sample — as observed in Fig. 10-5: the filter is linear-phase, the delay equals the index of the symmetry center of $h[n]$.
10.3.2 Group Delay and Modulated Carriers
Phase delay describes a single sinusoid. Real signals carry information in envelopes. Consider a modulated carrier(調變載波):
AM test signal
$$x[n] = A[n]\cos(\omega_c n), \quad A[n] = A\cos(\omega_m n),\ \omega_m \ll \omega_c
\;\;\Longrightarrow\;\;
x[n] = \frac{A}{2}\big[\cos((\omega_c-\omega_m)n) + \cos((\omega_c+\omega_m)n)\big]$$
The signal occupies two close frequencies $\omega_c \pm \omega_m$. Assume $|H(e^{j\omega})| \approx 1$ around $\omega_c$, and call the phases at the two side frequencies $\theta(\omega_c - \omega_m) = \theta_l$ and $\theta(\omega_c + \omega_m) = \theta_u$. The output recombines into
Carrier vs envelope phase
$$y[n] = A\cos\Big(\omega_c n + \tfrac{\theta_u + \theta_l}{2}\Big)\cdot
\cos\Big(\omega_m n + \tfrac{\theta_u - \theta_l}{2}\Big)$$
The average phase shifts the carrier; the difference shifts the envelope. If $\theta(\omega)$ is locally linear around $\omega_c$, i.e. $\theta(\omega_c + \Delta\omega) \approx \theta(\omega_c) + S\,\Delta\omega$ with slope $S = \frac{d\theta}{d\omega}\big|_{\omega_c}$, then:
Carrier phase shift $\frac{\theta_u + \theta_l}{2} = \theta(\omega_c)$ → carrier delayed by $-\theta(\omega_c)/\omega_c = \tau_p(\omega_c)$ — the phase delay.
Envelope phase shift $\frac{\theta_u - \theta_l}{2} = \omega_m S$ → envelope delayed by $-S$ — the group delay.
Group delay(群延遲)
$$\boxed{\;\tau_g(\omega_c) = -\frac{d\theta(\omega)}{d\omega}\bigg|_{\omega=\omega_c}\;}$$
Fig. 10-6 — AM signal $\cos(0.02\pi n)\cos(0.4\pi n)$ through an all-pass filter with poles at $0.9e^{\pm j0.4\pi}$. The sample-by-sample carrier shifts only by $\tau_p \approx 2.5$ samples, but the information-bearing envelope (red dashed) arrives $\tau_g \approx 19$ samples late — the group delay evaluated at $\omega_c$ (bottom panel)(載波延遲由 $\tau_p$ 決定,包絡延遲由 $\tau_g$ 決定).
Phase delay $\tau_p(\omega)$
Group delay $\tau_g(\omega)$
Definition
$-\theta(\omega)/\omega$
$-d\theta(\omega)/d\omega$
Geometric meaning
(negative) slope of the chord from origin to $(\omega, \theta(\omega))$
(negative) local tangent slope of $\theta(\omega)$
What it delays
the carrier / a single sinusoid(載波)
the envelope / information(包絡、訊息)
Equal when
$\theta(\omega) = -\tau\omega$ exactly linear through the origin → $\tau_p = \tau_g = \tau$
10.3.3 Phase Distortion
If $\theta(\omega)$ is not linear across the band occupied by $A[n]$, different envelope frequencies suffer different delays and the envelope is smeared — phase distortion(相位失真). The lecture closes with a striking example: a signal made of pulse bursts at different carrier frequencies passes through a system whose magnitude is nearly flat but whose group delay differs strongly between the two carrier bands — at the output the bursts arrive at different times and overlap differently, even though "the magnitude response barely touched them".
Warning — flat magnitude does not mean distortion-free
A system can have $|H(e^{j\omega})| \equiv 1$ (all-pass, next section) and still mangle a waveform badly through its nonlinear phase. Checking only the magnitude response is a classic exam (and engineering) trap. Distortionless transmission requires both constant magnitude and linear phase over the signal band.
(Supplementary to the slides.) Three classical families of systems are defined directly by their pole-zero geometry, and together they organize everything in 10.2–10.3: all-pass systems carry only phase, minimum-phase systems carry the least possible phase for a given magnitude, and linear-phase FIR systems carry a pure delay phase.
10.4.1 All-Pass Systems
An all-pass system(全通系統) has $|H_{ap}(e^{j\omega})| = $ constant for all $\omega$. The construction: pair every pole $a$ with a zero at its conjugate reciprocal $1/a^{*}$:
All-pass building block
$$H_{ap}(z) = \frac{z^{-1} - a^{*}}{1 - a z^{-1}}
\qquad\Longrightarrow\qquad
\big|H_{ap}(e^{j\omega})\big| = 1 \ \ \forall \omega$$
On the unit circle the vector to the zero at $1/a^{*}$ is always proportional in length to the vector to the pole at $a$ — the ratio of lengths is constant, so the magnitude never moves. Only the phase (and hence the group delay) varies.
Fig. 10-7 — First-order all-pass with pole $a = 0.8$ and zero $1/a = 1.25$: the magnitude is exactly 1 everywhere, while the group delay is positive and peaks near the pole angle ($\omega = 0$ here). For a stable causal all-pass, $\tau_g(\omega) > 0$ always(穩定因果全通系統的群延遲恆為正).
Key facts about all-pass systems
Poles inside the circle, zeros at conjugate-reciprocal positions outside.
$\tau_g(\omega) > 0$ for every $\omega$ (stable causal case) — an all-pass can only add delay.
The continuous phase $\arg H_{ap}(e^{j\omega})$ is monotonically decreasing.
Used to equalize phase without touching magnitude, and as the "phase carrier" in the decomposition of 10.4.2.
10.4.2 Minimum-Phase and Inverse Systems
A causal stable system is minimum-phase(最小相位) if all its zeros (as well as poles) are inside the unit circle. Then the inverse system $1/H(z)$ — whose poles are the zeros of $H(z)$ — is also causal and stable: minimum-phase systems are exactly the ones that can be undone by a realizable filter (e.g. channel equalization).
Reflecting a zero $\zeta$ to $1/\zeta^{*}$ multiplies the response by an all-pass factor, which leaves $|H(e^{j\omega})|$ unchanged. Hence many systems share one magnitude, and:
Every rational $H(z)$ is a minimum-phase system (carrying all of the magnitude) cascaded with an all-pass (carrying the excess phase). Among all systems with the same $|H(e^{j\omega})|$, the minimum-phase one has:
the smallest (continuous) phase lag and the smallest group delay at every $\omega$;
the most front-loaded impulse response: partial energy $\sum_{n=0}^{m}|h[n]|^2$ is maximal for every $m$ (minimum energy delay).
Fig. 10-8 — Minimum-phase FIR $H_{min}$ (zeros $0.8e^{\pm j0.6\pi}$) vs the maximum-phase mirror $H_{max}$ (zeros reflected to $1.25e^{\pm j0.6\pi}$): identical magnitude (left, curves overlap), but $H_{max}$ accumulates more phase lag and a larger group delay everywhere(相同振幅、不同相位/群延遲——最小相位版本延遲最小).
10.4.3 Generalized Linear-Phase FIR Systems
A system has (generalized) linear phase(線性相位) if $H(e^{j\omega}) = A(\omega)\,e^{-j(\alpha\omega - \beta)}$ with $A(\omega)$ real. Then $\tau_g(\omega) = \alpha$ is constant — no phase distortion, every envelope is delayed by the same $\alpha$ samples. For causal FIR filters of length $M$ this happens exactly when the impulse response is symmetric or antisymmetric:
The symmetry constrains the zero locations: $z^{-(M-1)}H(1/z) = \pm H(z)$, so if $\zeta$ is a zero, then $1/\zeta$, $\zeta^{*}$, $1/\zeta^{*}$ are all zeros. The zero constellation therefore consists of:
reciprocal-conjugate quadruples off the circle and off the real axis ($\zeta, \zeta^{*}, 1/\zeta, 1/\zeta^{*}$);
conjugate pairs on the unit circle (self-reciprocal);
reciprocal pairs on the real axis ($r, 1/r$);
possible single zeros at $z = \pm 1$ (their own reciprocal and conjugate).
Fig. 10-9 — A symmetric length-8 FIR ($h[n] = h[7-n]$, left) and its zero constellation (right): a quadruple $0.7e^{\pm j\pi/2}$, $1.43e^{\pm j\pi/2}$, a conjugate pair on the unit circle at $e^{\pm j0.8\pi}$, and a zero at $z = -1$(線性相位 ⇒ 零點成「倒數共軛」對稱組).
Warning — linear phase vs minimum phase vs causality
A linear-phase FIR cannot be minimum-phase (its zeros come in $\zeta, 1/\zeta$ pairs, so some lie on/outside the circle, except in degenerate cases) — the price of zero phase distortion is the fixed delay $(M-1)/2$. More generally, causality forces a nonzero phase response: a causal system cannot have $\theta(\omega) \equiv 0$ unless $h[n] = c\,\delta[n]$. Zero-phase filtering is only possible offline (non-causal processing, e.g. MATLAB filtfilt).
$e^{j\omega_0 n}$ is an eigenfunction: output $= H(e^{j\omega_0})\,e^{j\omega_0 n}$. Real sinusoid → gain $|H|$, phase shift $\theta(\omega_0)$. Gated input = steady state − transient (DTFT of the $h[n]$ tail); transient vanishes after $N$ samples for FIR, decays for stable IIR.
Transfer function
$Y(z) = H(z)X(z)$; $H(z)$ from the difference equation ($\sum p_k z^{-k} / \sum d_k z^{-k}$) or from $h[n]$. Causal → ROC outside the largest pole; stable ⇔ poles inside the unit circle. FR $=$ TF on the unit circle.
Pole-zero geometry
$|H|$ = product of zero-vector lengths ÷ pole-vector lengths; $\theta$ = zero angles − pole angles $+\ \omega(N-M)$. Zeros near the u.c. → dips; poles near the u.c. → peaks; close pole-zero pairs cancel from afar.
Phase delay vs group delay
$\tau_p = -\theta(\omega)/\omega$ delays the carrier; $\tau_g = -d\theta/d\omega$ delays the envelope (information). Nonlinear $\theta(\omega)$ over the signal band → phase distortion even with flat magnitude.
System classes
All-pass: $|H| \equiv 1$, pole $a$ ↔ zero $1/a^{*}$, $\tau_g > 0$. Minimum-phase: all roots inside, invertible by a causal stable filter, least delay for its magnitude. Any $H = H_{min}H_{ap}$.
Linear phase
FIR with $h[n] = \pm h[M-1-n]$ → $\theta$ linear, $\tau_g = (M-1)/2$ constant, zero constellation in reciprocal-conjugate quadruples / unit-circle pairs / $z=\pm1$. No phase distortion, fixed bulk delay.
Exercises
Exercise 1 — Steady-State Response to a Real Sinusoid
Problem: A 3-point moving-average filter $h[n] = \{\tfrac13, \tfrac13, \tfrac13\}$ (for $n = 0,1,2$) is driven by $x[n] = \cos\!\big(\tfrac{\pi}{4}n + \tfrac{\pi}{6}\big)$. Find the steady-state output $y[n]$ exactly, and interpret the phase shift as a time delay.
Click to reveal solution
Frequency response (Example 10-A with $M = 3$, or directly):
The phase shift $-\pi/4$ at frequency $\pi/4$ is exactly a one-sample delay ($\tau_p = \frac{\pi/4}{\pi/4} = 1$), consistent with linear phase $\theta(\omega) = -\omega$ and the symmetry center of $h[n]$ at $n = 1$.
Exercise 2 — Moving-Average Frequency Response and Nulls
Problem: (a) Derive the closed-form frequency response of the $M$-point moving average $y[n] = \frac1M\sum_{\ell=0}^{M-1}x[n-\ell]$. (b) For $M = 4$, list all frequencies in $(0, \pi]$ at which the response is exactly zero, and explain them with the zero locations of $H(z)$. (c) Which input sinusoids are completely blocked?
In the z-plane, $H(z) = \frac{z^4 - 1}{4z^3(z-1)}$: the 4th roots of unity $\{1, j, -1, -j\}$ minus the cancelled root at $z=1$ leave zeros at $z = j, -1, -j$, i.e. on the unit circle at angles $\pm\pi/2$ and $\pi$ — exactly the magnitude nulls (the vector from an on-circle zero to $e^{j\omega}$ has length zero there).
(c) Any sinusoid $A\cos(\omega n + \phi)$ with $\omega = \pi/2$ or $\omega = \pi$ produces zero steady-state output — e.g. $\{\dots, 1, 0, -1, 0, \dots\}$ and $\{\dots, 1, -1, 1, -1, \dots\}$: every 4 consecutive samples average to 0.
Exercise 3 — Transfer Function, Poles, ROC, and Stability
Problem: A causal system satisfies
$$y[n] = x[n] - 0.5x[n-1] + 1.1y[n-1] - 0.3y[n-2].$$
(a) Find $H(z)$ and factor it. (b) Locate poles and zeros. (c) Give the ROC and decide stability. (d) Could a different (non-causal) system with the same $H(z)$ expression be stable?
The denominator factors via $\lambda^2 - 1.1\lambda + 0.3 = (\lambda - 0.5)(\lambda - 0.6)$.
(b) Zero at $z = 0.5$, poles at $z = 0.5$ and $z = 0.6$. The pole at $0.5$ is cancelled by the zero (a pole-zero cancellation — "near poles and zeros cancel", here exactly). Effective system: single pole at $z = 0.6$, $h[n] = (0.6)^n u[n]$.
(c) Causal → ROC $|z| > 0.6$. This includes the unit circle → stable.
(d) The other ROC choice $|z| < 0.6$ (left-sided $h[n] = -(0.6)^n u[-n-1]$) does not include the unit circle → unstable. So for this $H(z)$ only the causal version is stable. (In general a stable non-causal choice exists only when some pole lies outside the unit circle and the annular ROC between pole radii still contains $|z| = 1$.)
Exercise 4 — Geometric Interpretation of a Resonator
Problem: Consider the two-pole resonator
$$H(z) = \frac{1}{(1 - 0.9e^{j\pi/3}z^{-1})(1 - 0.9e^{-j\pi/3}z^{-1})}.$$
Using only pole-vector geometry (no full evaluation): (a) near which frequency does $|H(e^{j\omega})|$ peak, and why? (b) Estimate $|H|$ at $\omega = \pi/3$. (c) What happens to the peak as the pole radius $r \to 1$? (d) Sketch the expected phase behavior near the peak.
Click to reveal solution
(a) $|H(e^{j\omega})| = \dfrac{1}{|e^{j\omega} - 0.9e^{j\pi/3}|\;|e^{j\omega} - 0.9e^{-j\pi/3}|}$. The first pole-vector length is minimized when $e^{j\omega}$ passes closest to the pole, i.e. $\omega \approx \pi/3$ — the magnitude peaks there (the true maximum is a hair below $\pi/3$ because the conjugate pole's vector also shrinks slightly as $\omega$ decreases, but for $r = 0.9$ the shift is tiny).
(b) At $\omega = \pi/3$:
distance to the near pole: $|e^{j\pi/3} - 0.9e^{j\pi/3}| = 1 - 0.9 = 0.1$;
distance to the conjugate pole: $|e^{j\pi/3} - 0.9e^{-j\pi/3}|$. With $e^{j\pi/3} = 0.5 + j0.866$, $0.9e^{-j\pi/3} = 0.45 - j0.779$: difference $= 0.05 + j1.645$, length $\approx 1.646$.
(c) As $r \to 1$ the near-pole distance $\to 0$, so the peak height $\to \infty$ (pole on the unit circle = oscillator) and the peak bandwidth shrinks $\propto (1-r)$.
(d) Phase = $-$(sum of pole-vector angles)$+2\omega$. As $e^{j\omega}$ sweeps past the near pole, that pole's vector angle rotates rapidly by nearly $\pi$, so $\theta(\omega)$ drops steeply through the resonance — and correspondingly $\tau_g = -d\theta/d\omega$ has a tall positive peak at $\omega \approx \pi/3$.
Exercise 5 — Phase Delay vs Group Delay
Problem: For the symmetric 3-pt filter $h[n] = \{\alpha, \beta, \alpha\}$ with $H(e^{j\omega}) = e^{-j\omega}(\beta + 2\alpha\cos\omega)$:
(a) Show that wherever $\beta + 2\alpha\cos\omega > 0$, $\tau_p(\omega) = \tau_g(\omega) = 1$ sample, and explain why this filter causes no phase distortion.
(b) What happens to $\theta(\omega)$, $\tau_p$, and $\tau_g$ at a frequency $\omega_z$ where $\beta + 2\alpha\cos\omega$ changes sign?
(c) An AM signal $A[n]\cos(\omega_c n)$ with narrowband $A[n]$ passes through a system with $\theta(\omega_c) = -1.2$ rad, $\frac{d\theta}{d\omega}|_{\omega_c} = -8$, $\omega_c = 0.3$. By how many samples are the carrier and the envelope delayed?
Click to reveal solution
(a) Where the real factor is positive it contributes no phase, so $\theta(\omega) = -\omega$:
Both delays are constant and equal: every sinusoid and every envelope is delayed exactly 1 sample, so the output waveform shape is preserved (only amplitude-shaped by $|\beta + 2\alpha\cos\omega|$). Constant $\tau_g$ ⇔ linear phase ⇔ no phase distortion.
(b) At a sign change the phase jumps by $\pi$ (the magnitude $|\beta + 2\alpha\cos\omega|$ has a corner touching 0). $\tau_p$ becomes $\frac{\omega + \pi \cdot(\text{accumulated jumps})}{\omega}$-like — ill-defined/discontinuous at $\omega_z$ — and $\tau_g$, the derivative of a step, has an impulse there: group delay is simply not meaningful at frequencies where the response is zero (nothing gets through to be delayed).
(c) Carrier delay $\tau_p(\omega_c) = -\theta(\omega_c)/\omega_c = 1.2/0.3 = 4$ samples. Envelope delay $\tau_g(\omega_c) = -d\theta/d\omega = 8$ samples. The carrier slips 4 samples while the information envelope arrives 8 samples late.
Exercise 6 — MATLAB: freqz, grpdelay, zplane on the Lecture TF Example
Problem: For the lecture system
$$H(z) = \frac{z^{-1} - 1.2z^{-2} + z^{-3}}{1 - 1.3z^{-1} + 1.04z^{-2} - 0.222z^{-3}},$$
write MATLAB code to (a) plot the log-magnitude, principal-value phase, and group delay over $0 \le \omega \le \pi$, and (b) draw the pole-zero plot. (c) From the plots, explain the deep magnitude notch and the adjacent peak near $\omega \approx 0.3\pi$, and the sign of the group delay around the notch.
Regenerated result (numpy/scipy equivalent of freqz/grpdelay):
Fig. 10-E6 — Log magnitude, principal phase, and group delay of the lecture transfer function.
(c) The zeros $0.6 \pm j0.8$ sit exactly on the unit circle at angle $0.295\pi$ → $|H| = 0$ there: $20\log_{10}|H| \to -\infty$, the deep notch. The complex poles $0.5 \pm j0.7$ sit at radius $0.86$ at almost the same angle ($0.303\pi$) → a short pole vector boosts the magnitude immediately around the notch, producing the adjacent peak. Around the on-circle zero the phase jumps by $\pi$ (sign change of the zero factor), so the group delay shows a large negative spike right at the notch — group delay is not meaningful at a frequency where nothing is transmitted; just outside the notch the pole dominates and $\tau_g$ is large and positive.