Which statements about FIR and IIR digital filters are correct?

an FIR filter is always BIBO stable

an FIR filter with symmetric coefficients has linear phase

an IIR filter usually meets a given magnitude specification with fewer coefficients

an FIR filter is always BIBO stable

an FIR filter with symmetric coefficients has linear phase

an IIR filter usually meets a given magnitude specification with fewer coefficients

an IIR filter can never be unstable

Which statements about the autocorrelation R_xx(τ) of a real signal are correct?

R_xx(τ) is an even function of τ

its Fourier transform is the power spectral density

R_xx(τ) is an even function of τ

its Fourier transform is the power spectral density

the autocorrelation of a periodic signal is aperiodic

Signals and Systems: LTI Analysis, Laplace and Fourier Transforms, Sampling, DFT, z-Transform and Digital Filters

Section 3 of the GATE Biomedical Engineering (BM) paper, Signals and Systems, in one chapter. The syllabus splits it into a continuous-and-discrete half and a discrete-time half, and the chapter follows that split. First the signals themselves — periodic, aperiodic and impulse — and the linear time-invariant (LTI) system that is characterised completely by its impulse response. Then convolution and correlation, which are the same integral asking different questions. Then the Laplace and Fourier transforms, the transfer function, and the frequency response of first- and second-order LTI systems. Then the sampling theorem, which decides whether a digitised ECG or EEG still contains the signal. It closes with discrete-time systems: impulse and frequency response, the DFT, the z-transform, and the basics of FIR and IIR filters, including the 50 Hz notch every biopotential recorder needs.

1. Periodic, aperiodic and impulse signals; LTI systems

A signal is periodic if x(t) = x(t + T) for some smallest T > 0, the fundamental period; a sum of periodic signals is periodic only if the ratio of their periods is rational, and then the fundamental period is the least common multiple. Anything else is aperiodic. The unit impulse δ(t) is zero except at t = 0 and has unit area; its defining sifting property is ∫x(t)δ(t − t₀)dt = x(t₀), and the unit step u(t) is its running integral. A signal with finite total energy is an energy signal (a single ECG beat); one with finite average power but infinite energy is a power signal (a sinusoid).

System properties and their tests
Property	Test	Counter-example
Linear	a x₁ + b x₂ → a y₁ + b y₂ (additivity and homogeneity)	y = x + 1 (fails at zero input)
Time-invariant	x(t − t₀) → y(t − t₀)	y(t) = t·x(t)
Causal	output depends on present and past input only; h(t) = 0 for t < 0	y(t) = x(t + 1)
BIBO stable	∫\|h(t)\|dt < ∞; all poles in the left half s-plane	an ideal integrator, h(t) = u(t)

For an LTI system the impulse response h(t) is everything: the output to any input is the convolution y = x * h, and the transfer function is the transform of h. A cascade of LTI systems has impulse response h₁ * h₂ and a parallel connection h₁ + h₂.

2. Convolution and correlation

Convolution y(t) = ∫x(τ)h(t − τ)dτ flips h, slides it across x and integrates the overlap; in discrete time y[n] = Σ x[k]h[n − k]. It is commutative, associative and distributive; convolving with δ(t − t₀) simply delays by t₀; and the output’s duration is the sum of the input durations (for sequences of lengths N and M, N + M − 1 samples). Two rectangular pulses of equal width convolve into a triangle.

Cross-correlation R_xy(τ) = ∫x(t)y(t + τ)dt measures how alike two signals are as one is shifted — the same integral as convolution without the flip. The autocorrelation R_xx(τ) is even, is largest at τ = 0 where it equals the signal’s energy (or power), and keeps the period of a periodic signal. By the Wiener–Khinchin theorem the Fourier transform of the autocorrelation is the power (or energy) spectral density. Correlation is how a template QRS complex is found in a noisy ECG and how a periodic evoked response is pulled out of background EEG.

⚠️ Convolution flips; correlation does not

For a symmetric kernel the two coincide, which is why the difference is easy to forget. For an asymmetric one — a decaying exponential, say — convolution and correlation give mirror-image results, and only convolution gives a system’s output.

3. Laplace and Fourier transforms; first- and second-order frequency response

The Laplace transform X(s) = ∫x(t)e^−stdt turns convolution into multiplication and derivatives into factors of s, so the transfer function H(s) = Y(s)/X(s) with zero initial conditions is the Laplace transform of h(t). Its poles fix the natural modes: a causal system is stable when every pole has a negative real part. The Fourier transform X(jω) = ∫x(t)e^−jωtdt is the Laplace transform on the imaginary axis (when that axis lies in the region of convergence), and H(jω) is the frequency response — the gain and phase applied to a sinusoid of frequency ω.

Pairs and properties used most
Time domain	Laplace	Fourier
δ(t)	1	1
u(t)	1/s	πδ(ω) + 1/(jω)
e^−atu(t), a > 0	1/(s + a)	1/(a + jω)
rectangular pulse of width T	(1 − e^−sT)/s (starting at 0)	T sinc — zeros at multiples of 1/T Hz
x(t − t₀)	e^−st₀X(s)	e^−jωt₀X(jω) — pure phase
x * h	X(s)H(s)	X(jω)H(jω)

A first-order system H(s) = K/(1 + sτ) has a step response K(1 − e^−t/τ) — 63.2% at t = τ, a 10–90% rise time of 2.2τ — and a −3 dB bandwidth of ω = 1/τ (f = 1/(2πτ)). A second-order system H(s) = ω_n²/(s² + 2ζω_n s + ω_n²) is underdamped for 0 < ζ < 1, with damped frequency ω_d = ω_n√(1 − ζ²), peak time t_p = π/ω_d, percentage overshoot 100·exp(−ζπ/√(1 − ζ²)) and 2% settling time about 4/(ζω_n). Its frequency response has a resonant peak only when ζ < 0.707, at ω_r = ω_n√(1 − 2ζ²) with height M_r = 1/(2ζ√(1 − ζ²)).

🎯 Why an overdamped blood-pressure catheter still distorts

A fluid-filled catheter and its transducer behave as a second-order system. With too little damping the resonant peak amplifies the harmonics of the pressure wave and overstates systolic pressure; with too much, the bandwidth shrinks and the waveform is smoothed. A damping ratio around 0.6–0.7 keeps the magnitude flattest over the band that matters.

4. The sampling theorem and aliasing

Sampling x(t) every T_s seconds multiplies it by an impulse train, which in frequency replicates the spectrum at every multiple of f_s = 1/T_s. If x is band-limited to f_max, the copies do not overlap provided f_s > 2f_max; 2f_max is the Nyquist rate, and x is then recovered exactly by an ideal low-pass filter of cut-off f_s/2 (sinc interpolation). If f_s is too low the copies overlap and a component at f appears at the alias |f − k f_s|, folded into the band 0 to f_s/2 — and nothing after the sampler can undo it.

An anti-aliasing analog low-pass filter before the ADC removes everything above f_s/2 that the signal does not need, including muscle noise and radio pickup.
Real filters roll off gradually, so practice samples at several times the Nyquist rate — a diagnostic ECG with content to about 150 Hz is typically sampled at 500 Hz or more.
For a sum of sinusoids the Nyquist rate is twice the highest frequency present, not twice the lowest or the sum of them.

🧠 Find an alias by folding

Subtract the nearest multiple of f_s and take the magnitude: 70 Hz sampled at 100 Hz appears at |70 − 100| = 30 Hz; 130 Hz sampled at 100 Hz appears at 30 Hz too. The alias always lands between 0 and f_s/2.

5. Discrete-time systems: DFT, z-transform, FIR and IIR filters

A discrete-time LTI system is described by its impulse response h[n] or a difference equation, and its frequency response H(e^jω) is periodic in ω with period 2π, the digital frequency ω = 2πf/f_s. The N-point DFT X[k] = Σ_n=0^N−1 x[n]e^−j2πkn/N samples that spectrum at N equally spaced points: bin k corresponds to f = k·f_s/N, so the frequency resolution is f_s/N — set by the record length N/f_s, not by the sampling rate alone. Multiplying DFTs gives circular, not linear, convolution unless the sequences are zero-padded to at least N + M − 1. The FFT computes the same DFT in the order of N log₂N operations instead of N².

The z-transform X(z) = Σx[n]z⁻ⁿ does for sequences what the Laplace transform does for signals: a delay by one sample is a factor z⁻¹, and a^n u[n] ↔ 1/(1 − az⁻¹) with region of convergence |z| > |a|. A causal system is stable exactly when all its poles lie inside the unit circle, and the frequency response is H(z) on |z| = 1.

FIR and IIR filters
Feature	FIR (finite impulse response)	IIR (infinite impulse response)
Structure	non-recursive: y[n] = Σb_k x[n − k]	recursive: past outputs fed back
Poles	only at z = 0 — always stable	anywhere — stable only if inside the unit circle
Phase	exactly linear if the coefficients are symmetric	non-linear in general
Order for a given sharpness	high	low — far fewer coefficients
Typical design	window method, frequency sampling	bilinear transform or impulse invariance of an analog prototype

ℹ️ The 50 Hz notch

Mains interference is removed by placing a pair of zeros on the unit circle at z = e^±jω₀, with ω₀ = 2π × 50/f_s, and a pair of poles just inside them at the same angle so that the notch stays narrow. At f_s = 500 Hz the zeros sit at ±36° (0.2π rad). An FIR moving average of length N places zeros at multiples of f_s/N instead, and doubles as a crude low-pass.

Key takeaways

An LTI system is its impulse response: y = x * h, and H(s) is the Laplace transform of h.
Autocorrelation is even, peaks at zero lag, and its Fourier transform is the power spectral density.
First order: 63.2% at τ, bandwidth 1/τ; second order: overshoot exp(−ζπ/√(1 − ζ²)), resonant peak only for ζ < 0.707.
f_s > 2f_max; a tone at f aliases to |f − kf_s|; filter before sampling, never after.
DFT resolution is f_s/N; causal stability means poles inside |z| = 1; FIR is always stable and can be exactly linear-phase.

Practice questions (14)

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.

For any signal x(t) continuous at t₀, ∫_−∞^∞ x(t)δ(t − t₀) dt equals
1. x(t₀)
2. x(0)
3. 1
4. 0
Show answer
Answer: A — x(t₀)
This is the sifting property: the impulse picks out the value of x at the instant where it sits, t₀. x(0) would be correct only for an impulse at the origin, and 1 is the area of the impulse itself, not of its product with x.
The fundamental period of x(t) = cos(2πt/3) + sin(2πt/5) is
1. 15 s
2. 8 s
3. 1.875 s
4. the signal is not periodic
Show answer
Answer: A — 15 s
The periods are 3 s and 5 s; their ratio 3/5 is rational, so the sum is periodic with the least common multiple, 15 s. Adding the periods gives 8 s, and the product over the sum gives 1.875 s — neither repeats both terms.
A 70 Hz sinusoid is sampled at 100 samples per second. The frequency (in Hz) at which it appears in the sampled signal, between 0 and 50 Hz, is ____.
Numerical answer — type the value.
Show answer
Answer: 30
70 Hz lies above f_s/2 = 50 Hz, so it folds to |70 − 100| = 30 Hz. Answering 70 ignores aliasing; answering 170 adds f_s instead of folding into the base band.
The Nyquist rate (in Hz) for x(t) = cos(200πt) + cos(600πt) is ____.
Numerical answer — type the value.
Show answer
Answer: 600
The components are at 100 Hz and 300 Hz, so f_max = 300 Hz and the Nyquist rate is 2 × 300 = 600 Hz. Reading 600π as 600 Hz gives 1200 Hz, and doubling the lower component gives 200 Hz.
An ECG sampled at 500 Hz is analysed with a 1000-point DFT. The bin index k (counting from k = 0) at which the 50 Hz mains interference appears is ____.
Numerical answer — type the value.
Show answer
Answer: 100
Bin spacing is f_s/N = 500/1000 = 0.5 Hz, so 50 Hz falls at k = 50/0.5 = 100. Taking the spacing as 1 Hz gives 50; forgetting that the bins run in steps of f_s/N rather than 1/N gives nonsense such as 0.1.
A second-order system has damping ratio ζ = 0.5. The percentage overshoot of its unit-step response, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 16.3
M_p = exp(−ζπ/√(1 − ζ²)) = exp(−0.5π/0.866) = exp(−1.814) = 0.163, i.e. 16.3%. Leaving out the square root gives exp(−π/2) = 20.8%, and using the resonant-peak formula instead of the overshoot formula gives 1.155.
A first-order system has transfer function 1/(1 + 2s). Its −3 dB bandwidth is
1. 0.5 rad/s
2. 2 rad/s
3. 0.5 Hz
4. 3.14 rad/s
Show answer
Answer: A — 0.5 rad/s
With τ = 2 s the corner is at ω = 1/τ = 0.5 rad/s (about 0.08 Hz). 2 rad/s inverts the relation, and 0.5 Hz puts the rad/s value in the wrong unit.
Which statements about FIR and IIR digital filters are correct?
1. an FIR filter is always BIBO stable
2. an FIR filter with symmetric coefficients has linear phase
3. an IIR filter usually meets a given magnitude specification with fewer coefficients
4. an IIR filter can never be unstable
Show answer
Answer: A — an FIR filter is always BIBO stable; B — an FIR filter with symmetric coefficients has linear phase; C — an IIR filter usually meets a given magnitude specification with fewer coefficients
An FIR filter has all its poles at the origin, so it cannot be unstable, and symmetric taps give exactly linear phase. IIR feedback buys sharpness with few coefficients, but its poles can move outside the unit circle — through design or through coefficient quantisation — and then it is unstable.
Which causal system is unstable?
1. H(z) = 1/(1 − 1.2z⁻¹)
2. H(z) = 1/(1 − 0.5z⁻¹)
3. H(z) = 1 + z⁻¹ + z⁻²
4. H(z) = 1/(1 + 0.9z⁻¹)
Show answer
Answer: A — H(z) = 1/(1 − 1.2z^{−1})
A causal system is stable only if every pole lies inside the unit circle. The pole at z = 1.2 lies outside, so h[n] = 1.2^n u[n] grows without bound. Poles at 0.5 and −0.9 are inside, and the FIR case has poles only at the origin.
x[n] = {1, 2, 3} for n = 0, 1, 2 is convolved with h[n] = {1, 1} for n = 0, 1. The output sample y[2] is ____.
Numerical answer — type the value.
Show answer
Answer: 5
y[n] = x[n] + x[n − 1], giving y = {1, 3, 5, 3}; so y[2] = x[2] + x[1] = 3 + 2 = 5. Multiplying samples point by point instead of convolving gives 3, and summing all of x gives 6.
Which statements about the autocorrelation R_xx(τ) of a real signal are correct?
1. R_xx(τ) is an even function of τ
2. |R_xx(τ)| ≤ R_xx(0)
3. its Fourier transform is the power spectral density
4. the autocorrelation of a periodic signal is aperiodic
Show answer
Answer: A — R_xx(τ) is an even function of τ; B — |R_xx(τ)| ≤ R_xx(0); C — its Fourier transform is the power spectral density
Shifting x against itself by τ or −τ gives the same overlap, so R_xx is even; the overlap is greatest with no shift, where it equals the power; and the Wiener–Khinchin theorem makes its transform the PSD. A periodic signal matches itself again every period, so its autocorrelation is periodic with the same period.
A digital notch filter must reject 50 Hz mains in an ECG sampled at 500 Hz. The angle (in degrees) of its zeros on the unit circle, measured from the positive real axis, is ± ____.
Numerical answer — type the value.
Show answer
Answer: 36
ω₀ = 2π f/f_s = 2π × 50/500 = 0.2π rad, which is 0.2 × 180° = 36°. Using ω₀ = 2πf = 314 rad forgets the division by f_s; 18° comes from taking π instead of 2π.
The inverse Laplace transform of 1/[s(s + 2)] is
1. ½(1 − e^−2t)u(t)
2. (1 − e^−2t)u(t)
3. ½e^−2tu(t)
4. (1 + e^−2t)u(t)/2
Show answer
Answer: A — ½(1 − e^{−2t})u(t)
Partial fractions: 1/[s(s + 2)] = (1/2)/s − (1/2)/(s + 2), so the inverse is ½(1 − e^−2t)u(t). The final value theorem checks it: lim s·F(s) = ½. Dropping the factor ½ gives a final value of 1, which the transform does not have.
A second-order system has ζ = 0.4. The peak of its normalised frequency-response magnitude, M_r, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.36
M_r = 1/(2ζ√(1 − ζ²)) = 1/(0.8 × √0.84) = 1/(0.8 × 0.9165) = 1.36, occurring at ω_r = ω_n√(1 − 2ζ²) = 0.825ω_n. Taking 1/(2ζ) = 1.25 is the value at ω = ω_n, not the peak, and the overshoot 25.4% belongs to the step response.