A continuous-time LTI system with impulse response h(t) is BIBO stable if and only if:

h(t) tends to zero as t tends to infinity, with no further condition

Which of the following statements about linear time-invariant and causal systems are true?

y(t) = x²(t) is a non-linear system

y[n] = x[−n] is a non-causal system

y(t) = x²(t) is a non-linear system

y[n] = x[−n] is a non-causal system

A periodic square wave of amplitude ±1 has half-wave symmetry, x(t + T/2) = −x(t), and is an odd function. Its Fourier series contains:

only sine terms of odd harmonics

only cosine terms of odd harmonics

sine and cosine terms of all harmonics

Which of the following statements about the Fourier and z transforms are true?

Delaying a signal by t₀ multiplies its Fourier transform by e^{−jωt₀}, leaving the magnitude unchanged

Convolution of two signals in time corresponds to multiplication of their Fourier transforms

A causal signal has a z-transform whose ROC lies outside its outermost pole

Delaying a signal by t₀ multiplies its Fourier transform by e^{−jωt₀}, leaving the magnitude unchanged

Convolution of two signals in time corresponds to multiplication of their Fourier transforms

Compressing a signal in time compresses its spectrum in frequency

A causal signal has a z-transform whose ROC lies outside its outermost pole

A discrete-time system has the transfer function H(z) = z/(z − 0.5) and is causal. Which statement is correct?

It is stable, with the ROC |z| > 0.5 containing the unit circle

It is unstable, because the pole is at z = 0.5

It is stable only if the ROC is |z| < 0.5

Its impulse response is −(0.5)ⁿu[−n − 1]

Signals and Systems: Signals, LTI Systems, Fourier Series, Sampling, and the Fourier, Laplace and z Transforms

Section B1.2 of the Robotics and Automation paper, in Part B1, is one paragraph and a whole subject: the representation of continuous- and discrete-time signals, shifting and scaling, linear time-invariant and causal systems, the Fourier series of periodic signals in both time domains, Shannon’s sampling theorem, the definitions and properties of the Fourier, Laplace and z transforms, and the calculation of R.M.S. and average values. It is the language of the control section that follows and of every sensor signal that is filtered, sampled or averaged. GATE asks it as short calculations: a period, a convolution sample, a Fourier coefficient, a Nyquist rate, an R.M.S. value, a region of convergence.

1. Signals: representation, shifting, scaling and periodicity

The building blocks are few. The unit step u(t) is 0 for t < 0 and 1 after; the unit impulse δ(t) has unit area, ∫δ(t)dt = 1, and the sifting property ∫x(t)δ(t − t₀)dt = x(t₀); δ(at) = δ(t)/|a|. The ramp r(t) = t·u(t) is the integral of the step, and the step the integral of the impulse. In discrete time δ[n] is 1 only at n = 0, u[n] − u[n − 1] = δ[n], and the complex exponential e^jωn replaces e^jωt. Every signal is a sum of these, which is why the response to an impulse or a step describes a whole system.

A signal is even if x(−t) = x(t) and odd if x(−t) = −x(t); every signal is the sum of an even part [x(t) + x(−t)]/2 and an odd part [x(t) − x(−t)]/2. Its energy is E = ∫|x|² dt and its power P = lim (1/2T)∫|x|² dt: an energy signal has finite E and P = 0, a power signal has finite non-zero P and infinite E, and every periodic signal is a power signal. The energy of e^−2tu(t) is ∫₀^∞ e^−4t dt = 1/4; the power of A cos ωt is A²/2.

Shifting, scaling and reversal of x(t)
Operation	Effect on the graph
x(t − t₀)	Delayed by t₀ (moved right for t₀ > 0)
x(t + t₀)	Advanced by t₀ (moved left)
x(−t)	Reversed about t = 0
x(at), \|a\| > 1	Compressed by the factor a
x(at), \|a\| < 1	Stretched by the factor 1/\|a\|

For x(at − b) the safe method is to write it as x(a(t − b/a)): scale by a, then delay by b/a. If x(t) is non-zero for 0 ≤ t ≤ 2, then x(2t − 4) is non-zero where 0 ≤ 2t − 4 ≤ 2, that is 2 ≤ t ≤ 3: the pulse is halved in width and starts at t = 2, not at t = 4.

Periodicity. A sum of two periodic continuous-time signals is periodic if the ratio of their periods is rational, and the fundamental period is the least common multiple: cos 2πt has T₁ = 1, sin 3πt has T₂ = 2/3, the ratio is 3/2, and the sum repeats every 2. A discrete-time sinusoid cos(ωn) is periodic only if ω/2π is rational, equal to m/N in lowest terms, and then its period is N: for cos(3πn/7), ω/2π = 3/14, so the period is 14, not 7/3. Unlike the continuous case, discrete frequencies ω and ω + 2π are the same signal.

⚠️ Shift first or scale first?

Applying the delay to t and then the scaling to the result gives a different signal from the reverse order. x(2t − 4) is a delay of 2 after compression by 2 — not a delay of 4. Substitute the end points of the pulse to check: they must map back to the original support.

2. Linear time-invariant and causal systems, and convolution

A system is linear if the response to ax₁ + bx₂ is ay₁ + by₂ (superposition, which includes zero output for zero input); time-invariant if delaying the input by t₀ only delays the output by t₀; causal if the output at time t depends on the input at t and earlier only; memoryless if it depends on the present input only; and BIBO stable if every bounded input gives a bounded output. Tested on examples: y = t·x(t) is linear but time-varying (the gain t moves with the clock); y = x² is non-linear; y(t) = x(t + 1) looks ahead and is non-causal; y[n] = x[−n] is linear but time-varying and non-causal; y[n] = x[n] − x[n − 1] is linear, time-invariant and causal.

An LTI system is completely described by its impulse response h, and the output is the convolution y = x ∗ h: y(t) = ∫x(τ)h(t − τ)dτ, and in discrete time y[n] = Σ x[k]h[n − k]. Convolution is commutative, associative and distributive, so cascaded systems convolve their impulse responses (multiply their transfer functions) and parallel systems add; δ ∗ x = x, and convolving with δ(t − t₀) delays by t₀. The step response is the integral of h, and h is the derivative of the step response.

What the impulse response says
Property	Condition on h
Causal	h(t) = 0 for t < 0; h[n] = 0 for n < 0
BIBO stable	∫\|h(t)\| dt < ∞; Σ\|h[n]\| < ∞
Memoryless	h(t) = Kδ(t)
Example: e^−tu(t)	causal and stable
Example: u(t), the integrator	causal but not stable

Worked convolutions. Discrete: {1, 2, 3} (starting at n = 0) with {1, 1} gives y[0] = 1, y[1] = 1 + 2 = 3, y[2] = 2 + 3 = 5, y[3] = 3: {1, 3, 5, 3}, of length 3 + 2 − 1 = 4. Continuous: e^−tu(t) ∗ e^−2tu(t) = e^−2t∫₀ᵗ e^τdτ = e^−t − e^−2t for t ≥ 0, which at t = ln 2 is 0.5 − 0.25 = 0.25; and u(t) ∗ u(t) = t·u(t), the ramp.

3. Fourier series of periodic signals, and R.M.S. and average values

A periodic continuous-time signal of period T and ω₀ = 2π/T is x(t) = Σ c_k e^jkω₀t, with c_k = (1/T)∫_T x(t) e^−jkω₀t dt. The trigonometric form x = a₀ + Σ(a_k cos kω₀t + b_k sin kω₀t) has c₀ = a₀ and c_k = (a_k − jb_k)/2. Symmetry saves work: an even x has only cosine terms, an odd x only sine terms, and a half-wave symmetric x, x(t + T/2) = −x(t), only odd harmonics. The square wave of amplitude ±1 is odd with half-wave symmetry, so x = (4/π)(sin ω₀t + sin 3ω₀t/3 + sin 5ω₀t/5 + …).

Parseval’s relation gives the average power from the coefficients: P = (1/T)∫|x|² dt = Σ|c_k|² = a₀² + Σ(a_k² + b_k²)/2. For x = 3 + 4 cos ω₀t + 2 sin 2ω₀t, P = 9 + 16/2 + 4/2 = 19, so the R.M.S. value is √19 ≈ 4.36. The DC term is the average value and each sinusoid contributes half its squared amplitude to the mean square; the check on the ±1 square wave is Σ over odd k of ½(4/kπ)² = (8/π²)(π²/8) = 1, the mean square of a wave that is always ±1.

Discrete time. A periodic x[n] of period N is x[n] = Σ c_k e^jk(2π/N)n over any N consecutive k, with c_k = (1/N)Σ_n=0^N−1 x[n] e^{−jk(2π/N)n}. Unlike the continuous case the series is finite and the coefficients themselves are periodic, c_k+N = c_k, so there are exactly N distinct ones. For x[n] = {1, 0, 1, 0} repeating, N = 4: c₀ = 1/2, c₁ = ¼(1 + e^−jπ) = 0, c₂ = ¼(1 + e^−j2π) = 1/2, c₃ = 0. Parseval is (1/N)Σ|x[n]|² = Σ|c_k|²: both sides equal 1/2.

Average and R.M.S. values of common waveforms, peak V_p
Waveform	Average	R.M.S.
Sine (over a full cycle)	0	V_p/√2 ≈ 0.707 V_p
Full-wave rectified sine	2V_p/π ≈ 0.637 V_p	V_p/√2
Half-wave rectified sine	V_p/π ≈ 0.318 V_p	V_p/2
Square wave, ±V_p	0	V_p
Triangular wave, ±V_p	0	V_p/√3

Both are integrals over one period: average = (1/T)∫x dt and R.M.S. = √[(1/T)∫x² dt]. A half-wave rectified sine of 10 V peak has an average of 10/π = 3.18 V and an R.M.S. value of 5 V: the negative half-cycle contributes nothing to either, so the mean square is half that of the full sine, V_p²/4.

4. Sampling and Shannon’s theorem

Ideal sampling multiplies x(t) by an impulse train of period T_s, and in the frequency domain replicates the spectrum about every multiple of the sampling frequency f_s = 1/T_s, scaled by f_s. Shannon’s sampling theorem: a signal band-limited to a highest frequency f_m is completely determined by its samples, and recoverable, if it is sampled at f_s > 2f_m; 2f_m is the Nyquist rate. Recovery is by an ideal low-pass filter with cut-off between f_m and f_s − f_m, which in the time domain is sinc interpolation.

Below the Nyquist rate the replicas overlap and a component at f aliases to |f − kf_s| for the k that brings it into the band ±f_s/2: a 7 kHz tone sampled at 10 kHz is indistinguishable from 3 kHz. The cure is an anti-aliasing low-pass filter before the sampler. In discrete time the sampled frequency is ω = 2πf/f_s radians per sample, and the unambiguous range is |ω| ≤ π.

Bandwidth after combining band-limited signals (bandwidths B₁ and B₂)
Operation	Bandwidth of the result
x₁(t) + x₂(t)	max(B₁, B₂)
x₁(t) · x₂(t)	B₁ + B₂
x₁(t) ∗ x₂(t)	min(B₁, B₂)
x(at)	a · B
x²(t)	2B

Worked: x(t) = cos(2π·1000t) cos(2π·2000t) = ½[cos(2π·3000t) + cos(2π·1000t)] has a highest frequency of 3000 Hz, so its Nyquist rate is 6000 Hz; the bandwidth rule for a product gives the same, 1000 + 2000. Sampling exactly at 2f_m is a borderline: a sine at f_m sampled at its zero crossings would be lost, which is why the theorem says strictly greater.

5. The Fourier, Laplace and z transforms: definitions and properties

The Fourier transform X(jω) = ∫x(t)e^−jωtdt exists for absolutely integrable signals and takes a non-repeating signal to its spectrum. The pairs to know: δ(t) ↔ 1; e^−atu(t) ↔ 1/(a + jω); e^−a|t| ↔ 2a/(a² + ω²); a rectangular pulse of width 2T ↔ 2 sin ωT/ω; 1 ↔ 2πδ(ω); cos ω₀t ↔ π[δ(ω − ω₀) + δ(ω + ω₀)]. The properties do most of the work: linearity; time shift x(t − t₀) ↔ e^−jωt₀X (magnitude unchanged, phase −ωt₀); frequency shift e^jω₀tx ↔ X(ω − ω₀); scaling x(at) ↔ X(ω/a)/|a|, so narrow in time is wide in frequency; differentiation dx/dt ↔ jωX; convolution in time is multiplication in frequency, and the reverse (with 1/2π); and Parseval, E = (1/2π)∫|X|²dω.

The Laplace transform X(s) = ∫x(t)e^−stdt (the one-sided form starts at 0⁻) generalises the Fourier transform to complex s = σ + jω, and it exists only in a region of convergence (ROC). Pairs: δ ↔ 1; u ↔ 1/s; t·u ↔ 1/s²; e^−atu ↔ 1/(s + a); cos ω₀t·u ↔ s/(s² + ω₀²); e^−atcos ω₀t·u ↔ (s + a)/[(s + a)² + ω₀²]. Properties: dx/dt ↔ sX − x(0⁻); ∫x ↔ X/s; delay by T multiplies by e^−sT; multiplying by e^−at shifts s to s + a; initial value x(0⁺) = lim_s→∞ sX(s); final value x(∞) = lim_s→0 sX(s), valid only if every pole of sX(s) lies in the left half-plane. For X(s) = 5(s + 2)/[s(s + 5)] the final value is 5 × 2/5 = 2.

The z-transform X(z) = Σ x[n]z⁻ⁿ is the discrete-time counterpart. Pairs: δ[n] ↔ 1; u[n] ↔ z/(z − 1), |z| > 1; aⁿu[n] ↔ z/(z − a), |z| > |a|; n aⁿ u[n] ↔ az/(z − a)²; delay by k multiplies by z^−k; convolution becomes multiplication; final value x[∞] = lim_z→1(z − 1)X(z) when it exists. The link to the Laplace variable is z = e^sT_s, so the left half of the s-plane maps inside the unit circle. Stability: a Laplace ROC that includes the jω axis, or a z-ROC that includes the unit circle. A causal signal has an ROC outside its outermost pole; a causal, stable system therefore has every pole in the left half-plane, or inside |z| = 1.

⚠️ The expression is not the signal

z/(z − a) is the transform of aⁿu[n] for |z| > |a| and of −aⁿu[−n − 1] for |z| < |a|: the same algebra, two different signals, told apart only by the region of convergence. The same holds for 1/(s + a): e^−atu(t) for Re s > −a, and −e^−atu(−t) for Re s < −a. Never invert without the ROC.

Key takeaways

Shift and scale in order: x(at − b) = x(a(t − b/a)); a DT sinusoid is periodic only if ω/2π is rational, with period N from m/N.
An LTI system is its impulse response: y = x ∗ h; causal iff h = 0 for t < 0, BIBO stable iff h is absolutely integrable.
Parseval: P = Σ|c_k|² = a₀² + Σ(a_k² + b_k²)/2; DC term is the average, √P the R.M.S. value; a DT series has N distinct coefficients.
Shannon: f_s > 2f_m; a product of signals adds bandwidths, x² doubles them; a tone above f_s/2 aliases to |f − kf_s|.
Convolution in time is multiplication in the transform domain; FVT needs poles of sX(s) in the left half-plane; a transform is not a signal without its ROC.

Practice questions (19)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A discrete-time signal x[n] = cos(ωn) is periodic if and only if:
1. ω/2π is a rational number
2. ω is a rational number
3. ω is an integer
4. ω < π
Show answer
Answer: A — ω/2π is a rational number
Periodicity needs an integer N with ωN a multiple of 2π, so ω/2π = m/N is rational. ω = 1 rad/sample is rational but cos n never repeats, because 1/2π is irrational.
The fundamental period of the discrete-time signal x[n] = cos(3πn/7) is ____ samples.
Numerical answer — type the value.
Show answer
Answer: 14
ω/2π = (3π/7)/(2π) = 3/14, in lowest terms, so N = 14: 3πN/7 = 6π is the first multiple of 2π that N makes an integer. 7 gives 3π, an odd multiple of π, and the cosine changes sign.
The fundamental period of x(t) = cos(2πt) + sin(3πt), with t in seconds, is ____ s.
Numerical answer — type the value.
Show answer
Answer: 2
T₁ = 2π/2π = 1 s and T₂ = 2π/3π = 2/3 s. The ratio T₁/T₂ = 3/2 is rational, and the least common multiple of 1 and 2/3 is 2 s: 2 = 2 × T₁ = 3 × T₂. Taking the larger period, 1, is wrong because sin(3πt) does not repeat every second.
The energy of the signal x(t) = e^−2tu(t) is ____.
Numerical answer — type the value.
Show answer
Answer: 0.25
E = ∫₀^∞ e^−4t dt = 1/4 = 0.25. The signal decays, so its energy is finite and its average power is zero: it is an energy signal. Integrating e^−2t instead of its square gives 0.5.
A signal x(t) is non-zero only for 0 ≤ t ≤ 2. The signal y(t) = x(2t − 4) is non-zero only for:
1. 2 ≤ t ≤ 3
2. 4 ≤ t ≤ 6
3. 0 ≤ t ≤ 1
4. 2 ≤ t ≤ 4
Show answer
Answer: A — 2 ≤ t ≤ 3
Solve 0 ≤ 2t − 4 ≤ 2: 4 ≤ 2t ≤ 6, so 2 ≤ t ≤ 3. The pulse is compressed to half its width and delayed to start at 2. 4 ≤ t ≤ 6 is what a pure delay of 4, x(t − 4), would give — the compression by 2 is forgotten.
The system y(t) = t · x(t) is:
1. linear and time-varying
2. linear and time-invariant
3. non-linear and time-invariant
4. non-linear and time-varying
Show answer
Answer: A — linear and time-varying
Superposition holds because t is only a multiplier of the input. Delaying the input by t₀ gives t·x(t − t₀), whereas delaying the output gives (t − t₀)·x(t − t₀), so the system is time-varying.
The discrete-time sequences x[n] = {1, 2, 3} for n = 0, 1, 2 and h[n] = {1, 1} for n = 0, 1 are convolved. The value of y[2] is ____.
Numerical answer — type the value.
Show answer
Answer: 5
y[2] = Σ x[k]h[2 − k] = x[1]h[1] + x[2]h[0] = 2·1 + 3·1 = 5; x[0]h[2] is zero because h[2] = 0. The full output is {1, 3, 5, 3}.
An LTI system has impulse response h(t) = e^−2tu(t) and is driven by x(t) = e^−tu(t). The output y(t) at t = ln 2 is ____.
Numerical answer — type the value.
Show answer
Answer: 0.25
y(t) = ∫₀ᵗ e^−τe^{−2(t − τ)}dτ = e^−2t(eᵗ − 1) = e^−t − e^−2t. At t = ln 2, e^−t = 1/2 and e^−2t = 1/4, so y = 0.25. Evaluating x(t) or h(t) alone at ln 2 (0.5 or 0.25) confuses a sample with the convolution.
A continuous-time LTI system with impulse response h(t) is BIBO stable if and only if:
1. ∫|h(t)| dt is finite
2. h(t) = 0 for t < 0
3. h(t) tends to zero as t tends to infinity, with no further condition
4. ∫h(t) dt = 0
Show answer
Answer: A — ∫|h(t)| dt is finite
Absolute integrability of h is both necessary and sufficient. h(t) = 0 for t < 0 is the condition for causality, and a decaying h such as 1/t for t ≥ 1 tends to zero yet has an infinite integral, so decay alone is not enough.
Which of the following statements about linear time-invariant and causal systems are true?
1. y(t) = x²(t) is a non-linear system
2. y[n] = x[n] − x[n − 1] is linear, time-invariant and causal
3. y(t) = x(t + 1) is causal
4. y[n] = x[−n] is a non-causal system
Show answer
Answer: A — y(t) = x²(t) is a non-linear system; B — y[n] = x[n] − x[n − 1] is linear, time-invariant and causal; D — y[n] = x[−n] is a non-causal system
(A) Doubling x quadruples y, so homogeneity fails. (B) A first difference has h[n] = δ[n] − δ[n − 1], which is zero for n < 0. (C) False: it needs the input one second ahead of the output. (D) At n = −1, y[−1] = x[1], a future input, and the same holds for every n < 0.
A periodic signal is x(t) = 3 + 4 cos(ω₀t) + 2 sin(2ω₀t). Its average power is ____ W across 1 Ω.
Numerical answer — type the value.
Show answer
Answer: 19
P = a₀² + Σ(amplitude²)/2 = 9 + 16/2 + 4/2 = 9 + 8 + 2 = 19. Adding the squared amplitudes 9 + 16 + 4 = 29 forgets the ½ that a sinusoid’s mean square carries; the R.M.S. value is √19 ≈ 4.36.
A periodic square wave of amplitude ±1 has half-wave symmetry, x(t + T/2) = −x(t), and is an odd function. Its Fourier series contains:
1. only sine terms of odd harmonics
2. only cosine terms of odd harmonics
3. sine and cosine terms of all harmonics
4. only a DC term
Show answer
Answer: A — only sine terms of odd harmonics
Oddness removes the DC term and every cosine, and half-wave symmetry removes the even harmonics, leaving b_k = 4/(kπ) for odd k. A square wave that sits between 0 and 1 would gain a DC term of 1/2 but keep only odd harmonics of the alternating part.
A half-wave rectified sine wave has a peak of 10 V. Its R.M.S. value is ____ V.
Numerical answer — type the value.
Show answer
Answer: 5
Mean square = (1/2π)∫₀^π 100 sin²θ dθ = (100/2π)(π/2) = 25, so R.M.S. = 5 V. The average is 10/π ≈ 3.18 V, and 10/√2 ≈ 7.07 V is the value for the full, unrectified sine.
The signal x(t) = cos(2π·1000t) cos(2π·2000t) is to be sampled without aliasing. Its Nyquist rate is ____ Hz.
Numerical answer — type the value.
Show answer
Answer: 6000
The product equals ½[cos(2π·3000t) + cos(2π·1000t)], whose highest frequency is 3000 Hz, so the Nyquist rate is 2 × 3000 = 6000 Hz. Doubling the larger factor’s frequency (4000 Hz) ignores the sum frequency that the product creates.
A 7 kHz sinusoid is sampled at 10 kHz and reconstructed by an ideal low-pass filter with a cut-off of 5 kHz. The frequency of the reconstructed sinusoid is ____ kHz.
Numerical answer — type the value.
Show answer
Answer: 3
The 7 kHz component is above f_s/2 = 5 kHz and folds to |7 − 10| = 3 kHz, which the filter passes; the original 7 kHz is removed. The reconstruction is a clean 3 kHz tone, indistinguishable from a genuine 3 kHz input.
The Laplace transform of e^−2t cos(3t) u(t) is:
1. (s + 2)/[(s + 2)² + 9]
2. 3/[(s + 2)² + 9]
3. s/(s² + 9)
4. (s − 2)/[(s − 2)² + 9]
Show answer
Answer: A — (s + 2)/[(s + 2)² + 9]
Multiplying by e^−2t shifts s to s + 2 in the transform of cos 3t, s/(s² + 9), giving (s + 2)/[(s + 2)² + 9]. The numerator 3 belongs to the sine, and s − 2 is the shift for a growing exponential e^+2t.
A stable signal has the Laplace transform X(s) = 5(s + 2)/[s(s + 5)]. Its final value x(∞) is ____.
Numerical answer — type the value.
Show answer
Answer: 2
sX(s) = 5(s + 2)/(s + 5) has its only pole at s = −5, in the left half-plane, so the theorem applies: x(∞) = lim_s→0 5(s + 2)/(s + 5) = 5 × 2/5 = 2. The partial fractions 2/s + 3/(s + 5) confirm it, since the second term decays.
Which of the following statements about the Fourier and z transforms are true?
1. Delaying a signal by t₀ multiplies its Fourier transform by e^−jωt₀, leaving the magnitude unchanged
2. Convolution of two signals in time corresponds to multiplication of their Fourier transforms
3. Compressing a signal in time compresses its spectrum in frequency
4. A causal signal has a z-transform whose ROC lies outside its outermost pole
Show answer
Answer: A — Delaying a signal by t₀ multiplies its Fourier transform by e^{−jωt₀}, leaving the magnitude unchanged; B — Convolution of two signals in time corresponds to multiplication of their Fourier transforms; D — A causal signal has a z-transform whose ROC lies outside its outermost pole
(A) The time-shift property. (B) The convolution property. (C) False: x(at) ↔ X(ω/a)/|a|, so compressing in time (a > 1) spreads the spectrum. (D) A right-sided sequence converges for large |z|, outside every pole.
A discrete-time system has the transfer function H(z) = z/(z − 0.5) and is causal. Which statement is correct?
1. It is stable, with the ROC |z| > 0.5 containing the unit circle
2. It is unstable, because the pole is at z = 0.5
3. It is stable only if the ROC is |z| < 0.5
4. Its impulse response is −(0.5)ⁿu[−n − 1]
Show answer
Answer: A — It is stable, with the ROC |z| > 0.5 containing the unit circle
A causal system has its ROC outside the outermost pole, |z| > 0.5, and the unit circle |z| = 1 lies inside that region, so it is stable; h[n] = (0.5)ⁿu[n], which is absolutely summable. The left-sided sequence in the last option is what |z| < 0.5 would give.