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Signals and Systems Practice Hub

Practice signals and systems with question sets on Fourier transforms, convolution, sampling, and GATE-level problems for ECE students.

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BeginnerEnergy vs powerIntermediateSampling theoremAdvancedGATE aliasing trap

Hub intro

Signals and Systems is a subject that rewards careful attention to definitions and transform properties. This hub builds from basic signal classification and Fourier series at the beginner level through convolution and sampling theorem at the intermediate level to GATE-style problems involving system properties and inverse transforms at the advanced level.

Difficulty levels

BeginnerIntermediateAdvanced

Practice sets

Beginner: Signal Classification and Basics
Beginner
Define an energy signal and a power signal. Give one example of each.
Answer: An energy signal has finite total energy (E < infinity) and zero average power. Example: a rectangular pulse of finite duration. A power signal has finite average power but infinite total energy. Example: a periodic sinusoid.
Total energy E = integral from -inf to +inf of |x(t)|² dt. Average power P = lim(T→inf) (1/2T) × integral from -T to T of |x(t)|² dt. A finite-duration non-zero signal has finite energy and zero power. A periodic signal repeats indefinitely, giving infinite energy but finite average power over one period. A signal cannot be both an energy and a power signal simultaneously.
State the Dirichlet conditions for the existence of the Fourier series of a periodic signal.
Answer: The signal must be absolutely integrable over one period, have a finite number of maxima and minima per period, and have a finite number of discontinuities per period, all of finite magnitude.
These three conditions are sufficient (not necessary) for convergence of the Fourier series. Most practical periodic signals satisfy them. At points of discontinuity, the Fourier series converges to the average of the left and right limits (Gibbs phenomenon causes overshoot near the discontinuity but does not violate convergence in the mean-square sense).
What is the Fourier transform of the unit impulse function delta(t)?
Answer: F{delta(t)} = 1. The impulse has a flat spectrum containing all frequencies with equal amplitude.
By the sifting property, F{delta(t)} = integral from -inf to +inf of delta(t)·e^(-j2πft) dt = e^(0) = 1. The magnitude spectrum is 1 for all frequencies. This means the impulse contains equal energy at all frequencies, which is why it is used to characterise the frequency response of LTI systems: the output spectrum Y(f) = H(f) × X(f) = H(f) × 1 = H(f).
What is meant by a causal system? Give the condition on the impulse response h(t).
Answer: A causal system produces no output before the input is applied. The condition is h(t) = 0 for all t < 0.
Causality means the output at any time t depends only on past and present inputs, not future inputs. In terms of the impulse response, h(t) represents the output when the input is a unit impulse at t=0. If h(t) were nonzero for t<0, the system would respond before the impulse was applied, which is non-causal. All real-time physical systems are causal.
Define linear convolution of two sequences x[n] and h[n]. What is its length if x[n] has length M and h[n] has length N?
Answer: Convolution y[n] = sum over k of x[k]·h[n-k]. The output length is M + N - 1.
Linear convolution combines two finite sequences to produce a longer output. If x[n] spans indices 0 to M-1 (length M) and h[n] spans 0 to N-1 (length N), then y[n] = x[n]*h[n] spans 0 to M+N-2, which has length M+N-1. This is distinct from circular convolution used in the DFT, which wraps around and may alias if the circular length is shorter than M+N-1.
Intermediate: Transform Analysis and System Properties
Intermediate
Find the Fourier transform of x(t) = e^(-at)·u(t) where a > 0 and u(t) is the unit step function.
Answer: X(f) = 1/(a + j2πf), or in terms of angular frequency ω: X(jω) = 1/(a + jω).
X(jω) = integral from 0 to inf of e^(-at)·e^(-jωt) dt = integral from 0 to inf of e^(-(a+jω)t) dt = [-1/(a+jω)·e^(-(a+jω)t)] from 0 to inf. As t→inf, e^(-(a+jω)t) = e^(-at)·e^(-jωt) → 0 since a>0. At t=0, the term is 1. So X(jω) = 1/(a+jω). Magnitude: 1/√(a²+ω²), phase: -arctan(ω/a). This is a first-order low-pass filter response.
State the Nyquist-Shannon sampling theorem and derive the minimum sampling rate for a bandlimited signal with highest frequency component fm = 4 kHz.
Answer: The sampling theorem states that a bandlimited signal can be perfectly reconstructed if the sampling rate fs ≥ 2·fm. For fm = 4 kHz, fs_min = 8 kHz (8000 samples per second).
When a continuous-time signal x(t) is sampled at rate fs, the spectrum of the sampled signal is periodic with period fs. If fs < 2fm, the spectral replicas overlap, causing aliasing, and the original signal cannot be recovered. At fs = 2fm (Nyquist rate), the replicas just touch without overlapping. For fm=4 kHz: fs_min = 2×4000 = 8000 Hz. In practice, a slightly higher rate is used and an anti-aliasing low-pass filter is applied before sampling.
An LTI system has impulse response h(t) = e^(-2t)·u(t). Determine whether the system is BIBO stable.
Answer: The system is BIBO stable because h(t) is absolutely integrable: integral |h(t)| dt = 1/2 < infinity.
A necessary and sufficient condition for BIBO (Bounded Input, Bounded Output) stability of an LTI system is that the impulse response be absolutely integrable: integral from -inf to +inf of |h(t)| dt < infinity. For h(t) = e^(-2t)·u(t): integral from 0 to inf of e^(-2t) dt = [-1/2·e^(-2t)] from 0 to inf = 0 - (-1/2) = 1/2. Since 1/2 is finite, the system is BIBO stable.
Compute the z-transform of x[n] = a^n·u[n] and state the region of convergence (ROC).
Answer: X(z) = z/(z-a) = 1/(1-az^(-1)), ROC: |z| > |a|.
X(z) = sum from n=0 to inf of a^n·z^(-n) = sum of (az^(-1))^n = 1/(1-az^(-1)) for |az^(-1)| < 1, i.e. |z| > |a|. Rewriting: X(z) = z/(z-a). The ROC is the exterior of a circle of radius |a| in the z-plane. For a causal signal, the ROC always extends outward from the outermost pole. If |a| < 1, the pole is inside the unit circle and the system (if h[n] = a^n·u[n]) is BIBO stable.
A discrete-time signal x[n] = cos(π·n/4) is sampled from a continuous-time sinusoid. What is the fundamental period of x[n]?
Answer: The fundamental period N = 8 samples.
For a discrete-time sinusoid x[n] = cos(ω0·n), the signal is periodic if ω0/(2π) = k/N for integers k and N. Here ω0 = π/4. ω0/2π = (π/4)/(2π) = 1/8. So k/N = 1/8; with k=1 (smallest), N=8. Verify: x[n+8] = cos(π(n+8)/4) = cos(πn/4 + 2π) = cos(πn/4) = x[n]. The fundamental period is 8.
Advanced: GATE-Style Problems
Advanced
The z-transform X(z) = (z+1)/[(z-0.5)(z+0.5)] has ROC |z| > 0.5. Find x[n] using partial fractions.
Answer: x[n] = [2·(0.5)^n - (-0.5)^n]·u[n].
X(z)/z = (z+1)/[z(z-0.5)(z+0.5)]. Partial fractions: X(z)/z = A/z + B/(z-0.5) + C/(z+0.5). Multiply through by z: at z=0, (0+1)/[(0-0.5)(0+0.5)] = 1/(-0.25) = -4 = A. At z=0.5: (0.5+1)/[(0.5)(0.5+0.5)] = 1.5/(0.5) = 3 = B·1/(0.5·1) → B = 1.5/1 = ... careful: B = (z+1)/[z(z+0.5)] at z=0.5 = (1.5)/(0.5×1) = 3. At z=-0.5: C = (-0.5+1)/[(-0.5)(-0.5-0.5)] = 0.5/[(-0.5)(-1)] = 0.5/0.5 = 1. So X(z) = -4·1 + 3·z/(z-0.5) + 1·z/(z+0.5). Hmm, X(z)/z partial fractions: A/z + B/(z-0.5) + C/(z+0.5). Multiply by z(z-0.5)(z+0.5) = (z+1). A(z-0.5)(z+0.5)+Bz(z+0.5)+Cz(z-0.5) = z+1. z=0: A(-0.25) = 1, A=-4. z=0.5: B(0.5)(1) = 1.5, B=3. z=-0.5: C(-0.5)(-1) = 0.5, C=1. X(z) = -4 + 3z/(z-0.5) + z/(z+0.5). Inverse z-transform: -4·delta[n] + 3·(0.5)^n·u[n] + (-0.5)^n·u[n]. Note the delta term and ROC: |z|>0.5 implies causal sequences.
An LTI system has frequency response H(jω) = 1 for |ω| ≤ W and 0 otherwise. This is an ideal low-pass filter. Is it causal? Justify.
Answer: No. The ideal low-pass filter is non-causal because its impulse response h(t) = W·sinc(Wt/π)/π is nonzero for t < 0.
The impulse response is the inverse Fourier transform of H(jω): h(t) = (1/2π)·integral from -W to W of e^(jωt) dω = W·sinc(Wt)/π (using the sinc definition sinc(x)=sin(πx)/(πx) for normalized, or sin(x)/x for unnormalized). This sinc function extends from -infinity to +infinity, meaning h(t) ≠ 0 for t < 0. Since the impulse response is nonzero before the impulse is applied, the system is non-causal. All ideal filters (low-pass, high-pass, band-pass with brick-wall characteristics) are non-causal and physically unrealisable.
A signal x(t) = 2·cos(100πt) + 3·sin(200πt) is sampled at fs = 250 Hz. Identify any aliasing and determine the alias frequencies that appear in the sampled signal.
Answer: Component at 50 Hz (100π rad/s): no aliasing, since 50 < fs/2 = 125 Hz. Component at 100 Hz (200π rad/s): no aliasing, since 100 < 125 Hz. No aliasing occurs.
fs/2 = 125 Hz (Nyquist frequency). The two components are at f1 = 50 Hz and f2 = 100 Hz. Both are below 125 Hz, so neither component aliases. Aliasing occurs only for components at f > fs/2. If the sampling rate were reduced to, say, 150 Hz (Nyquist = 75 Hz), then the 100 Hz component would alias to |100 - 150| = 50 Hz, coinciding with the first component. The problem tests whether you correctly compute the Nyquist frequency (half the sampling rate) and compare each spectral component individually.
The output of an LTI system is y[n] = 0.5·y[n-1] + x[n]. Determine the system function H(z), its ROC for a causal system, and whether the system is stable.
Answer: H(z) = 1/(1-0.5z^(-1)) = z/(z-0.5). ROC: |z| > 0.5. Since the pole at z=0.5 is inside the unit circle (|0.5| < 1), the system is stable.
Taking the z-transform: Y(z) = 0.5z^(-1)Y(z) + X(z). Y(z)[1-0.5z^(-1)] = X(z). H(z) = Y(z)/X(z) = 1/(1-0.5z^(-1)) = z/(z-0.5). For a causal system (right-sided impulse response), the ROC is |z| > 0.5 (exterior of the circle passing through the pole). Stability requires the ROC to include the unit circle |z|=1. Since 0.5 < 1, the unit circle is within the ROC, confirming stability. h[n] = (0.5)^n·u[n] decays to 0, consistent with stability.
A continuous-time signal x(t) has a Fourier transform X(f) that is nonzero for |f| ≤ 1000 Hz. It is sampled at 1500 Hz and then passed through an ideal reconstruction filter with cutoff at 750 Hz. Describe the output.
Answer: The output is a distorted version of x(t) with aliasing. The components of x(t) between 750 Hz and 1000 Hz will fold back and appear as aliases between 500 Hz and 750 Hz in the reconstructed signal.
Sampling at 1500 Hz means fs/2 = 750 Hz. Components of x(t) above 750 Hz (up to 1000 Hz) violate the Nyquist criterion. These components alias into the baseband [0, 750 Hz]. Specifically, a component at frequency f aliases to |f - k·fs| for the smallest result in [0, fs/2]. For f = 800 Hz: alias at |800-1500| = 700 Hz. For f = 900 Hz: alias at |900-1500| = 600 Hz. For f = 1000 Hz: alias at |1000-1500| = 500 Hz. These aliases corrupt the reconstructed signal in the range [500, 750] Hz, which overlaps with valid signal content. An anti-aliasing low-pass filter (cutoff at 750 Hz) applied before sampling would have prevented this.

Lab exercises

  • Compute and plot the DTFT of a rectangular window in MATLAB and observe the frequency domain leakage: /labs/dtft-rectangular-window-lab
  • Demonstrate aliasing by sampling a 3 kHz tone at 4 kHz and listening to the alias frequency: /labs/aliasing-audio-demo-lab
  • Verify the convolution theorem by multiplying spectra in the frequency domain and comparing with time-domain convolution: /labs/convolution-theorem-verification-lab

Revision checklist

  • Can you state the difference between an energy signal and a power signal and classify a given signal correctly?
  • Do you know the Fourier transform pairs for at least 6 standard signals (impulse, step, exponential, sinusoid, rectangular pulse, Gaussian)?
  • Can you apply the sampling theorem and predict alias frequencies when fs < 2fm?
  • Can you determine the ROC of a z-transform from the pole locations for both causal and anti-causal sequences?
  • Can you verify BIBO stability of an LTI system from its impulse response or from the pole positions?