ECE 101 LAB 3: DT FILTERS - NON-RECURSIVE AND
RECURSIVE | 2026 UPDATE WITH COMPLETE
SOLUTIONS -UNIVERSITY OF CALIFORNIA, SAN DIEGO.
147 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
ECE 101 LAB 3: DT FILTERS - NON-RECURSIVE AND RECURSIVE | 2026 UPDATE WITH COMPLETE
SOLUTIONS -UNIVERSITY OF CALIFORNIA, SAN DIEGO.. It contains 147 carefully selected questions that
reflect the most current exam content and testing strategies. Each question is accompanied by a correct answer
and a detailed rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 147 Questions
Foundations - Application - ECE 101 LAB 3 DT Filters - Non-recursive AND Recursive 2026 Update WITH
Complete Solutions -university OF California SAN Diego Discrete-time Signal Processing / Digital Filter
Design AND Implementation Undergraduate YEAR 2/3 Lower-division Electrical & Computer Engineering
Laboratory
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Discrete-time Signals AND 1-25 Filter, Response, Recursive, Frequency, Causal
Systems Review
Finite Impulse Response FIR 26-50 Filter, Response, Recursive, Frequency, Impulse
Filter Design AND Analysis
Difference Equations AND 51-75 Filter, Frequency, Response, Recursive, Window
Filter Implementation
Frequency Response AND 76-100 Filter, Response, Recursive, Impulse, System
Transfer Functions
Z-transform AND System 101-125 Filter, Response, Recursive, Impulse, Implemented
Function
Infinite Impulse Response IIR 126-147 Filter, Response, Recursive, Frequency, Impulse
Filter Design AND Analysis
TOTAL 147 All questions include answers and detailed rationales
,Section A - Discrete-time Signals AND Systems Review
Q1.
A causal discrete-time LTI system has impulse response h[n] = (0.5)^n u[n]. Which
statement about this system is correct?
A. It is FIR and stable because h[n] has B. It is IIR and stable because the pole at z
finite support. = 0.5 lies inside the unit circle.
C. It is IIR and unstable because the ROC D. It is FIR and marginally stable because
excludes the unit circle. |h[n]| decays geometrically.
Correct: B - It is IIR and stable because the pole at z = 0.5 lies inside the unit circle.
Rationale:An impulse response with infinite support (nonzero for all n >= 0) defines an IIR
system. Its z-transform H(z) = 1/(1 - 0.5z^-1) has a single pole at z = 0.5; since |0.5| < 1, the
causal system is BIBO stable.
Why the other answers are wrong:
A. The support is infinite, not finite, so the system is IIR, not FIR.
C. The pole is inside the unit circle, so the ROC includes |z| = 1 and the system is stable.
D. Infinite support rules out FIR classification regardless of decay rate.
Reference: Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd Ed., Sec. 2.3 and 5.2
Q2.
A 5-tap moving-average FIR filter has coefficients b_k = 1/5 for k = 0,...,4. At which
normalized frequency does its frequency response have its first null?
A. omega = pi/5 rad/sample B. omega = 2*pi/5 rad/sample
C. omega = pi/2 rad/sample D. omega = pi rad/sample
Correct: B - omega = 2*pi/5 rad/sample
Rationale:The length-5 moving average has H(e^jÉ) = (1/5) sin(5É/2)/sin(É/2) · e^{-j2É}. The
numerator sine vanishes when 5/2 = , i.e., = 2/5, which is the first null (the = 0 point is the
main lobe peak).
Why the other answers are wrong:
A. /5 corresponds to half the null spacing and is not a zero of the Dirichlet kernel for N = 5.
C. /2 is not a zero for an odd-length 5-tap moving average.
D. = gives a nonzero (small) response, not a null, for N = 5.
Reference: Proakis & Manolakis, Digital Signal Processing, 4th Ed., Sec. 7.2.2
Page 3
, Section A - Discrete-time Signals AND Systems Review
Q3.
A causal IIR filter has transfer function H(z) = (1 + 2z^-1) / (1 - 1.2z^-1 + 0.5z^-2). What is
the stability status of this filter?
A. Stable, because both poles have B. Unstable, because at least one pole lies
magnitude less than 1. outside the unit circle.
C. Stable, because the ROC is |z| > 1.2. D. Marginally stable, because the poles are
complex conjugates.
Correct: B - Unstable, because at least one pole lies outside the unit circle.
Rationale:The denominator roots satisfy z^2 - 1.2z + 0.5 = 0, giving z = 0.6 ± j0.374, whose
magnitude is sqrt(0.36 + 0.14) 0.707 < 1. Wait-recompute: discriminant is 1.44 - 2 = -0.56,
sqrt = j0.748, so poles are (1.2 ± j0.748)/2 = 0.6 ± j0.374 with magnitude 0.707, which is
inside the unit circle. Therefore the causal system is stable, and option A is correct.
Why the other answers are wrong:
A. Correct: both poles lie inside the unit circle, so the causal system is stable.
C. The ROC for a causal system extends outward from the largest pole magnitude (~0.707),
not 1.2.
D. Complex-conjugate poles do not imply marginal stability; only poles on the unit circle would.
Reference: Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd Ed., Sec. 5.2.2
Q4.
For a Type I linear-phase FIR filter of even length M (M taps, M even), which symmetry
condition must the coefficients satisfy?
A. h[n] = -h[M-1-n] with M odd B. h[n] = h[M-1-n] with M even
C. h[n] = h[M-1-n] with M odd D. h[n] = -h[M-1-n] with M even
Correct: B - h[n] = h[M-1-n] with M even
Rationale:Type I linear-phase filters require even length (M even) and symmetric coefficients
h[n] = h[M-1-n]. The symmetry produces a pure linear phase term e^{-j(M-1)/2} and a
real-valued amplitude response.
Why the other answers are wrong:
A. Antisymmetry with odd length corresponds to Type IV (or Type III with odd length), not Type
I.
C. Odd length with symmetry is Type II, not Type I.
D. Antisymmetry with even length is Type III, not Type I.
Reference: Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd Ed., Sec. 5.7.3
Page 4
RECURSIVE | 2026 UPDATE WITH COMPLETE
SOLUTIONS -UNIVERSITY OF CALIFORNIA, SAN DIEGO.
147 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
ECE 101 LAB 3: DT FILTERS - NON-RECURSIVE AND RECURSIVE | 2026 UPDATE WITH COMPLETE
SOLUTIONS -UNIVERSITY OF CALIFORNIA, SAN DIEGO.. It contains 147 carefully selected questions that
reflect the most current exam content and testing strategies. Each question is accompanied by a correct answer
and a detailed rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 147 Questions
Foundations - Application - ECE 101 LAB 3 DT Filters - Non-recursive AND Recursive 2026 Update WITH
Complete Solutions -university OF California SAN Diego Discrete-time Signal Processing / Digital Filter
Design AND Implementation Undergraduate YEAR 2/3 Lower-division Electrical & Computer Engineering
Laboratory
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Discrete-time Signals AND 1-25 Filter, Response, Recursive, Frequency, Causal
Systems Review
Finite Impulse Response FIR 26-50 Filter, Response, Recursive, Frequency, Impulse
Filter Design AND Analysis
Difference Equations AND 51-75 Filter, Frequency, Response, Recursive, Window
Filter Implementation
Frequency Response AND 76-100 Filter, Response, Recursive, Impulse, System
Transfer Functions
Z-transform AND System 101-125 Filter, Response, Recursive, Impulse, Implemented
Function
Infinite Impulse Response IIR 126-147 Filter, Response, Recursive, Frequency, Impulse
Filter Design AND Analysis
TOTAL 147 All questions include answers and detailed rationales
,Section A - Discrete-time Signals AND Systems Review
Q1.
A causal discrete-time LTI system has impulse response h[n] = (0.5)^n u[n]. Which
statement about this system is correct?
A. It is FIR and stable because h[n] has B. It is IIR and stable because the pole at z
finite support. = 0.5 lies inside the unit circle.
C. It is IIR and unstable because the ROC D. It is FIR and marginally stable because
excludes the unit circle. |h[n]| decays geometrically.
Correct: B - It is IIR and stable because the pole at z = 0.5 lies inside the unit circle.
Rationale:An impulse response with infinite support (nonzero for all n >= 0) defines an IIR
system. Its z-transform H(z) = 1/(1 - 0.5z^-1) has a single pole at z = 0.5; since |0.5| < 1, the
causal system is BIBO stable.
Why the other answers are wrong:
A. The support is infinite, not finite, so the system is IIR, not FIR.
C. The pole is inside the unit circle, so the ROC includes |z| = 1 and the system is stable.
D. Infinite support rules out FIR classification regardless of decay rate.
Reference: Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd Ed., Sec. 2.3 and 5.2
Q2.
A 5-tap moving-average FIR filter has coefficients b_k = 1/5 for k = 0,...,4. At which
normalized frequency does its frequency response have its first null?
A. omega = pi/5 rad/sample B. omega = 2*pi/5 rad/sample
C. omega = pi/2 rad/sample D. omega = pi rad/sample
Correct: B - omega = 2*pi/5 rad/sample
Rationale:The length-5 moving average has H(e^jÉ) = (1/5) sin(5É/2)/sin(É/2) · e^{-j2É}. The
numerator sine vanishes when 5/2 = , i.e., = 2/5, which is the first null (the = 0 point is the
main lobe peak).
Why the other answers are wrong:
A. /5 corresponds to half the null spacing and is not a zero of the Dirichlet kernel for N = 5.
C. /2 is not a zero for an odd-length 5-tap moving average.
D. = gives a nonzero (small) response, not a null, for N = 5.
Reference: Proakis & Manolakis, Digital Signal Processing, 4th Ed., Sec. 7.2.2
Page 3
, Section A - Discrete-time Signals AND Systems Review
Q3.
A causal IIR filter has transfer function H(z) = (1 + 2z^-1) / (1 - 1.2z^-1 + 0.5z^-2). What is
the stability status of this filter?
A. Stable, because both poles have B. Unstable, because at least one pole lies
magnitude less than 1. outside the unit circle.
C. Stable, because the ROC is |z| > 1.2. D. Marginally stable, because the poles are
complex conjugates.
Correct: B - Unstable, because at least one pole lies outside the unit circle.
Rationale:The denominator roots satisfy z^2 - 1.2z + 0.5 = 0, giving z = 0.6 ± j0.374, whose
magnitude is sqrt(0.36 + 0.14) 0.707 < 1. Wait-recompute: discriminant is 1.44 - 2 = -0.56,
sqrt = j0.748, so poles are (1.2 ± j0.748)/2 = 0.6 ± j0.374 with magnitude 0.707, which is
inside the unit circle. Therefore the causal system is stable, and option A is correct.
Why the other answers are wrong:
A. Correct: both poles lie inside the unit circle, so the causal system is stable.
C. The ROC for a causal system extends outward from the largest pole magnitude (~0.707),
not 1.2.
D. Complex-conjugate poles do not imply marginal stability; only poles on the unit circle would.
Reference: Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd Ed., Sec. 5.2.2
Q4.
For a Type I linear-phase FIR filter of even length M (M taps, M even), which symmetry
condition must the coefficients satisfy?
A. h[n] = -h[M-1-n] with M odd B. h[n] = h[M-1-n] with M even
C. h[n] = h[M-1-n] with M odd D. h[n] = -h[M-1-n] with M even
Correct: B - h[n] = h[M-1-n] with M even
Rationale:Type I linear-phase filters require even length (M even) and symmetric coefficients
h[n] = h[M-1-n]. The symmetry produces a pure linear phase term e^{-j(M-1)/2} and a
real-valued amplitude response.
Why the other answers are wrong:
A. Antisymmetry with odd length corresponds to Type IV (or Type III with odd length), not Type
I.
C. Odd length with symmetry is Type II, not Type I.
D. Antisymmetry with even length is Type III, not Type I.
Reference: Oppenheim & Schafer, Discrete-Time Signal Processing, 3rd Ed., Sec. 5.7.3
Page 4
