Speech Technology · Quiz 1 · Jan 2026

**1. Complete Solutions to Offline Paper Section (Questions 1 to 8, 12 Marks):** - **Q1 (1 pt) — Role of Mel-Frequency Cepstral Coefficients (MFCC) in speech processing:** - **Correct choice: (b)** *They represent the short-term power spectrum on a perceptual scale.* MFCCs apply a Mel-spaced filterbank and a discrete cosine transform (DCT) to log filterbank energies to capture the vocal tract spectral envelope on a human-auditory perceptual scale. - **Q2 (1 pt) — The phoneme `/f/` is a:** - **Correct choice: (c) fricative.** `/f/` is a voiceless labiodental fricative produced by forcing air through a narrow constriction between the lower lip and upper teeth. - **Q3 (1 pt) — Short Time Fourier Transform (STFT) is required to analyze speech because speech is:** - **Correct choice: (c) non-stationary (or (a) quasi-stationary over short 10–30 ms windows).** Because the vocal tract shape changes continuously over time, the speech signal is globally non-stationary and only quasi-stationary over short frames, requiring short-time windowed analysis (STFT). - **Q4 (1 pt) — ______ membrane in the ear acts as a frequency analyzer:** - **Answer: Basilar** membrane (located inside the cochlea; its tonotopic organization responds to high frequencies at the base and low frequencies at the apex). - **Q5 (1 pt) — For perfect reconstruction of a bandlimited signal, the sampling frequency must be:** - **Correct choice: (b) greater than or equal to twice the maximum frequency** (Nyquist–Shannon sampling theorem: \( F_s \ge 2 F_{\max} \)). - **Q6 (2 pts) — Minimum sampling frequency of a signal low-pass filtered with a cut-off frequency of \( 10 \text{ kHz} \) to avoid aliasing:** - By the Nyquist theorem, \( F_s \ge 2 f_c = 2 \times 10 \text{ kHz} = 20 \text{ kHz} \). Thus, the minimum sampling frequency is **20 kHz**. - **Q7 (2 pts) — When spectrograms are obtained using longer time-windows, we get ______ frequency-resolution:** - **Correct choice: (a) good** (narrowband spectrogram). By the time-frequency uncertainty principle, a longer window of duration \( T \) has a narrower main-lobe bandwidth \( \Delta f \propto 1/T \) in the frequency domain, yielding fine (good) frequency resolution at the expense of poorer time resolution. - **Q8 (3 pts) — Linearity and time-invariance of the given discrete-time systems:** - **(a) \( y[n] = 2x[n] + 3 \):** **Non-linear** (fails homogeneity/additivity due to the constant offset \( +3 \); zero input gives \( y[n] = 3 \neq 0 \)) and **Time-Invariant** (shifting input \( x[n - n_0] \) produces \( 2x[n - n_0] + 3 = y[n - n_0] \)). - **(b) \( y[n] = x[n]^2 \):** **Non-linear** (scaling input by \( a \) scales output by \( a^2 \neq a \)) and **Time-Invariant** (shifting input by \( n_0 \) shifts output by \( n_0 \)). - **(c) \( y[n] = x[2n] \):** **Linear** (satisfies superposition: \( a x_1[2n] + b x_2[2n] = a y_1[n] + b y_2[n] \)) and **Time-Varying (Not Time-Invariant)** (delaying the output by \( n_0 \) gives \( y[n - n_0] = x[2(n - n_0)] = x[2n - 2n_0] \), whereas delaying the input first gives \( x[2n - n_0] \neq y[n - n_0] \)). **Answer: A — I have answered this question in the offline paper.** A: **Correct:** Confirms completion of Questions 1–8 in the offline answer booklet. B: **Incorrect:** Selecting this indicates the offline questions were skipped. C: **Incorrect:** This question applies to all candidates taking the Speech Technology exam.

**1. Complete Solutions to Offline Paper Section (Questions 9 to 11, 13 Marks):** - **Q9 (3 pts) — Weather HMM Viterbi Decoding for \( O = (\text{Walk}, \text{Shop}, \text{Clean}) \):** - Initial probabilities: \( \pi_S = 0.6,\ \pi_R = 0.4 \). - Transitions: \( P(S\to S)=0.8, P(S\to R)=0.2, P(R\to S)=0.4, P(R\to R)=0.6 \). - Emissions: \( B_S = [0.6, 0.3, 0.1] \), \( B_R = [0.1, 0.4, 0.5] \). - **Step \( t = 1 \) (`Walk`):** - \( v_1(S) = \pi_S B_S(\text{Walk}) = 0.6 \times 0.6 = 0.36 \) - \( v_1(R) = \pi_R B_R(\text{Walk}) = 0.4 \times 0.1 = 0.04 \) - **Step \( t = 2 \) (`Shop`):** - \( v_2(S) = \max(0.36 \times 0.8,\ 0.04 \times 0.4) \times 0.3 = \max(0.288,\ 0.016) \times 0.3 = 0.0864 \) (from \( S \)) - \( v_2(R) = \max(0.36 \times 0.2,\ 0.04 \times 0.6) \times 0.4 = \max(0.072,\ 0.024) \times 0.4 = 0.0288 \) (from \( S \)) - **Step \( t = 3 \) (`Clean`):** - \( v_3(S) = \max(0.0864 \times 0.8,\ 0.0288 \times 0.4) \times 0.1 = \max(0.06912,\ 0.01152) \times 0.1 = 0.006912 \) (from \( S \)) - \( v_3(R) = \max(0.0864 \times 0.2,\ 0.0288 \times 0.6) \times 0.5 = \max(0.01728,\ 0.01728) \times 0.5 = 0.00864 \) (tie from \( S \) or \( R \)) - **Backtracking:** Since \( v_3(R) = 0.00864 > v_3(S) = 0.006912 \), the final state is \( R \). Tracing back gives **\( (S, S, R) \)** or **\( (S, R, R) \)** (both achieve the maximum joint probability \( 0.00864 \)). - **Q10 (5 pts) — 2-Component 1D GMM at \( x = 2.0 \):** - Component 1: \( \mu_1 = 1.0, \sigma_1^2 = 1.0, w_1 = 0.6 \); Component 2: \( \mu_2 = 3.0, \sigma_2^2 = 1.0, w_2 = 0.4 \). - Both component densities at \( x = 2.0 \) have \( |x - \mu_k| = 1.0 \) and \( \sigma_k = 1.0 \): \[ \mathcal{N}(2.0; 1.0, 1.0) = \mathcal{N}(2.0; 3.0, 1.0) = \frac{1}{\sqrt{2\pi}} e^{-1/2} \approx 0.24197 \] - **(a) Mixture density \( p(x) \):** \[ p(2.0) = 0.6 \mathcal{N}(2; 1, 1) + 0.4 \mathcal{N}(2; 3, 1) = (0.6 + 0.4) \frac{e^{-0.5}}{\sqrt{2\pi}} = \frac{e^{-0.5}}{\sqrt{2\pi}} \approx 0.242 \] - **(b) Posterior probability of Component 1:** \[ P(C_1 \mid x = 2.0) = \frac{w_1 \mathcal{N}(2; 1, 1)}{p(2.0)} = \frac{0.6 \times 0.24197}{1.0 \times 0.24197} = 0.6 \] - **Q11 (5 pts) — POS Tagging HMM for "Time flies fast" (with "Time" = Noun):** - **(a) Greedy approach:** - For `flies` (given previous tag `Noun`): compare \( P(T \mid \text{Noun}) P(\text{flies} \mid T) \): - `Noun`: \( 0.3 \times 0.1 = 0.03 \) - `Verb`: \( 0.5 \times 0.8 = 0.40 \) (max \( \implies \) pick **Verb**) - `Adjective`: \( 0.2 \times 0.1 = 0.02 \) - For `fast` (given previous tag `Verb`): compare \( P(T \mid \text{Verb}) P(\text{fast} \mid T) \): - `Noun`: \( 0.4 \times 0.2 = 0.08 \) - `Verb`: \( 0.4 \times 0.3 = 0.12 \) (max \( \implies \) pick **Verb**) - `Adjective`: \( 0.2 \times 0.5 = 0.10 \) - Note: if greedy chooses transitions first or joint step-by-step, step-by-step greedy gives **(Verb, Verb)** with path score \( 0.40 \times 0.12 = 0.048 \) (or if transition-only greedy: Verb then Noun/Verb). - **(b) Viterbi algorithm:** - At `flies`: \( V(\text{Noun}) = 0.03 \), \( V(\text{Verb}) = 0.40 \), \( V(\text{Adj}) = 0.02 \). - At `fast`: - \( V(\text{Noun}) = \max(0.03\times 0.3,\ 0.40\times 0.4,\ 0.02\times 0.2) \times 0.2 = 0.16 \times 0.2 = 0.032 \) (from `Verb`) - \( V(\text{Verb}) = \max(0.03\times 0.5,\ 0.40\times 0.4,\ 0.02\times 0.3) \times 0.3 = 0.16 \times 0.3 = 0.048 \) (from `Verb`) - \( V(\text{Adj}) = \max(0.03\times 0.2,\ 0.40\times 0.2,\ 0.02\times 0.5) \times 0.5 = 0.08 \times 0.5 = 0.040 \) (from `Verb`) - Optimal Viterbi tag sequence for (`flies`, `fast`): **(Verb, Verb)** with probability \( 0.048 \). **Answer: A — I have answered this question in the offline paper.** A: **Correct:** Confirms completion of Questions 9–11 in the offline answer booklet. B: **Incorrect:** Selecting this indicates the offline questions were skipped. C: **Incorrect:** This question applies to all candidates taking the Speech Technology exam.