🇮🇳 GATE Biomedical Engineering · flashcards
GATE Biomedical Engineering Signals and Systems Flashcards
51 question-and-answer cards covering Signals and Systems as it is examined in GATE Biomedical Engineering. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Signals and Systems deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
What is the standard form of a second-order LTI system transfer function?
$H(s) = \frac{\omega_n^{2}}{s^{2} + 2\zeta\omega_n s + \omega_n^{2}}$, where $\omega_n$ is the undamped natural frequency and $\zeta$ is the damping ratio.
Classify the response of a second-order system by its damping ratio $\zeta$.
$\zeta=0$: undamped (sustained oscillation); $0<\zeta<1$: underdamped (decaying oscillation); $\zeta=1$: critically damped (fastest non-oscillatory); $\zeta>1$: overdamped (slow, no oscillation).
What is the resonant (peak) frequency of an underdamped second-order system?
$\omega_r = \omega_n\sqrt{1-2\zeta^{2}}$, which exists only for $\zeta < \frac{1}{\sqrt{2}}\approx 0.707$. The damped natural frequency of oscillation is $\omega_d = \omega_n\sqrt{1-\zeta^{2}}$.
Define the continuous-time convolution integral.
$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau)\,d\tau$. It expresses the LTI output as the superposition of shifted, scaled impulse responses.
State the commutative, associative, and distributive properties of convolution.
Commutative: $x*h = h*x$. Associative: $(x*h_1)*h_2 = x*(h_1*h_2)$. Distributive: $x*(h_1+h_2) = x*h_1 + x*h_2$ (parallel systems add).
What is the convolution theorem relating time and frequency domains?
Convolution in time corresponds to multiplication in frequency: $x(t)*h(t) \leftrightarrow X(\omega)H(\omega)$. Conversely, multiplication in time corresponds to convolution in frequency (scaled by $\frac{1}{2\pi}$).
What is the length of the linear convolution of two finite sequences of lengths $N_1$ and $N_2$?
The resulting sequence has length $N_1 + N_2 - 1$.
What is the convolution of any signal with a unit impulse?
It reproduces the signal, possibly shifted: $x(t)*\delta(t) = x(t)$ and $x(t)*\delta(t-t_0) = x(t-t_0)$. The impulse is the identity element of convolution.
Define cross-correlation of two continuous-time signals and state how it differs from convolution.
$R_{xy}(\tau) = \int_{-\infty}^{\infty} x(t) y(t-\tau)\,dt$ (or with conjugate for complex signals). Unlike convolution, correlation does not fold (time-reverse) one signal; it measures similarity versus relative shift.
What is autocorrelation, and what does its value at zero lag represent?
Autocorrelation $R_{xx}(\tau)=\int_{-\infty}^{\infty} x(t)x(t-\tau)\,dt$ measures a signal's similarity with a shifted copy of itself. At zero lag $R_{xx}(0)$ equals the signal's total energy (or power) and is its maximum value.
State the Wiener–Khinchin theorem.
The power spectral density of a signal is the Fourier transform of its autocorrelation function: $S_{xx}(\omega) = \mathcal{F}\{R_{xx}(\tau)\}$.
Define the N-point Discrete Fourier Transform (DFT).
$X[k] = \sum_{n=0}^{N-1} x[n] e^{-j\frac{2\pi}{N}kn}$, for $k=0,1,\dots,N-1$. It samples one period of the DTFT at $N$ equally spaced frequencies.
State the inverse DFT (IDFT) formula.
$x[n] = \frac{1}{N}\sum_{k=0}^{N-1} X[k] e^{j\frac{2\pi}{N}kn}$, for $n=0,1,\dots,N-1$.
How does the DFT differ from the DTFT?
The DTFT $X(e^{j\omega})$ is continuous and periodic in $\omega$, suited to infinite/aperiodic sequences. The DFT is discrete in frequency, computed from a finite $N$-point sequence, and treats the sequence as one period of a periodic signal.
What is the FFT and what computational advantage does it provide?
The Fast Fourier Transform is an efficient algorithm to compute the DFT. It reduces complexity from $O(N^{2})$ direct computation to $O(N\log_2 N)$ by exploiting the symmetry and periodicity of the twiddle factors.
What does circular convolution of two N-point sequences compute via the DFT?
Multiplication of their DFTs corresponds to circular (periodic) convolution: $x_1[n] \circledast x_2[n] \leftrightarrow X_1[k]X_2[k]$. Zero-padding both sequences to length $\geq N_1+N_2-1$ makes circular convolution equal linear convolution.
Define the bilateral Z-transform.
$X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}$, where $z$ is complex. It is the discrete-time counterpart of the Laplace transform.
How do the $s$-plane and $z$-plane relate through sampling?
Via $z = e^{sT}$, where $T$ is the sampling period. The left half of the $s$-plane maps to the interior of the unit circle in the $z$-plane, and the $j\omega$-axis maps to the unit circle $|z|=1$.
State the Z-transform stability condition for a causal discrete-time LTI system.
A causal system is stable iff all poles of $H(z)$ lie strictly inside the unit circle ($|z|<1$), equivalently the ROC includes $|z|=1$.
Give the Z-transforms of $\delta[n]$, $u[n]$, and $a^{n}u[n]$.
$\mathcal{Z}\{\delta[n]\}=1$; $\mathcal{Z}\{u[n]\}=\frac{1}{1-z^{-1}}$, $|z|>1$; $\mathcal{Z}\{a^{n}u[n]\}=\frac{1}{1-az^{-1}}$, $|z|>|a|$.
What is the difference between an FIR and an IIR digital filter?
FIR (Finite Impulse Response) filters have an impulse response of finite duration and use only feedforward terms (all-zero, no feedback). IIR (Infinite Impulse Response) filters have an infinitely long impulse response due to feedback (recursive) terms, with poles and zeros.
Why are FIR filters always stable while IIR filters may be unstable?
FIR filters have no poles other than at the origin (only feedforward terms), so they are inherently stable. IIR filters have feedback-introduced poles that must lie inside the unit circle for stability; poor design can place poles outside it.
What key advantage do FIR filters have regarding phase, and what is the trade-off versus IIR?
FIR filters can achieve exactly linear phase (constant group delay, no phase distortion) when their coefficients are symmetric. The trade-off: FIR filters generally need a much higher order than IIR filters to meet the same magnitude specification, so IIR is more computationally efficient but has nonlinear phase.
What is the general input-output difference equation of an IIR filter, and which terms make it recursive?
$y[n] = \sum_{k=0}^{M} b_k x[n-k] - \sum_{k=1}^{N} a_k y[n-k]$. The feedback terms $a_k y[n-k]$ (past outputs) make it recursive, giving it an infinite impulse response; an FIR filter has all $a_k=0$.
What this deck covers
The Signals and Systems deck follows the GATE Biomedical Engineering Signals and Systems syllabus — 2 chapters and 13 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 25.5 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 184 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Signals and Systems flashcards FAQ
How many Signals and Systems flashcards are in this GATE Biomedical Engineering deck?
51 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these GATE Biomedical Engineering flashcards free?
Yes. The preview here is free to read with no signup, and the full 51-card deck is free inside the Examius app.
What do the Signals and Systems cards cover?
They follow the GATE Biomedical Engineering Signals and Systems syllabus — 2 chapters and 13 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.