GATE GUIDE

GATE ECE Signals and Systems Formula Sheet 2027, with Traps

By MD ANISH AHAMADUpdated 4 Oct 20269 min read
GATE ECE Signals and Systems Formula Sheet 2027, with Traps

Networks, Signals and Systems is the largest technical section of the GATE EC paper, with a mean of 17.0 marks across the five papers counted for the ECE book. Its questions rarely need a new formula. They need the right one, applied under the right convention.

In this guide
  1. Key takeaways
  2. The notation this sheet uses
  3. Phasors, impedance and AC power
  4. Thevenin, Norton and superposition in dc networks
  5. First-order RL and RC transients
  6. Series and parallel resonance
  7. Two-port parameters, reciprocity and symmetry
  8. LTI properties and convolution
  9. Fourier series symmetry rules
  10. Fourier, Laplace and Parseval
  11. Sampling and aliasing
  12. z-transform: ROC, causality and stability
  13. DTFT, DFT and all-pass systems
  14. FIR and IIR filter design
  15. Quick revision

This sheet gives the most-used formulas topic by topic, each with a short worked example and the trap that costs marks. Use it to check recall, not to learn from scratch.

Key takeaways

The notation this sheet uses

Angular frequency ω\omega is in rad/s and frequency ff is in Hz, so ω=2πf\omega = 2\pi f. A voltage or current is rms unless the word peak is written. A phasor is the complex number that carries a sinusoid's amplitude and phase. The passive sign convention holds throughout.

An LTI system is linear and time-invariant, and h(t)h(t) or h[n]h[n] is its impulse response. The ROC is the region of convergence of a Laplace or z-transform. The DFT is unnormalised in the forward direction. For the section's topic ranking, read the GATE ECE important topics guide.

Phasors, impedance and AC power

Turn each element into an impedance and solve as if the circuit were resistive.

Formula Example Watch out for
ZL=jωLZ_L = j\omega L, ZC=1jωC=−jωC\displaystyle Z_C = \frac{1}{j\omega C} = -\frac{j}{\omega C} 10 mH10\text{ mH} at 50 Hz50\text{ Hz}: ω=314\omega = 314, ZL=j3.14 ΩZ_L = j3.14\,\Omega Using ff in place of ω\omega
Vrms=Vm2\displaystyle V_{\text{rms}} = \frac{V_m}{\sqrt{2}} v(t)=1002cos⁡ωtv(t) = 100\sqrt{2}\cos\omega t reads 100 V100\text{ V} on a meter A written source amplitude is peak
P=VrmsIrmscos⁡θP = V_{\text{rms}}I_{\text{rms}}\cos\theta, θ=∠Z\theta = \angle Z 100 V100\text{ V}, 2 A2\text{ A}, θ=60∘\theta = 60^\circ: P=100 WP = 100\text{ W} θ\theta is the angle between VV and II
S=VI∗=P+jQS = VI^{*} = P + jQ V=100∠0∘V = 100\angle 0^\circ, I=2∠−30∘I = 2\angle{-30^\circ}: S=173.2+j100S = 173.2 + j100 Conjugate the current, not the voltage
ZL=Zth∗Z_L = Z_{th}^{*}, Pmax⁡=Vth24Rth\displaystyle P_{\max} = \frac{V_{th}^{2}}{4R_{th}} Vth=10 VV_{th} = 10\text{ V}, Rth=5 ΩR_{th} = 5\,\Omega: Pmax⁡=5 WP_{\max} = 5\text{ W} A purely resistive load needs RL=∣Zth∣R_L = \vert Z_{th}\vert

Trap: Ideal voltmeters and ammeters read rms, while a source written as Vmcos⁡ωtV_m\cos\omega t gives the peak. GATE 2025 Q39 framed a phasor problem through ideal meter readings, so the rms convention was the whole difficulty.

Thevenin, Norton and superposition in dc networks

The Thevenin voltage is the open-circuit voltage at the terminals. The Thevenin resistance is what the terminals see with the independent sources switched off.

Formula Example Watch out for
Vth=VocV_{th} = V_{oc}, Rth=VocIsc\displaystyle R_{th} = \frac{V_{oc}}{I_{sc}}, IN=VthRth\displaystyle I_N = \frac{V_{th}}{R_{th}} Voc=6 VV_{oc} = 6\text{ V}, Isc=2 AI_{sc} = 2\text{ A}: Rth=3 ΩR_{th} = 3\,\Omega Using Voc/IscV_{oc}/I_{sc} with a source still shorted
Deactivate: voltage source short, current source open Two 6 Ω6\,\Omega resistors in parallel: Rth=3 ΩR_{th} = 3\,\Omega Dependent sources are never deactivated
Test source: Rth=VtestItest\displaystyle R_{th} = \frac{V_{\text{test}}}{I_{\text{test}}} Apply 1 A1\text{ A}, measure 4 V4\text{ V}: Rth=4 ΩR_{th} = 4\,\Omega Needed whenever a dependent source is present
Superposition: v=∑kvkv = \sum_k v_k Contributions 2 V2\text{ V} and 3 V3\text{ V} give 5 V5\text{ V} Power does not add: 25≠4+925 \ne 4 + 9

DC network analysis appeared in four of the five counted papers (2010 Q34, 2020 Q19, 2025 Q19, 2026 Q30). The 2019 paper had none.

First-order RL and RC transients

Every first-order response moves from its initial value to its final value along one exponential:

x(t)=x(∞)+[x(0+)−x(∞)]e−t/τ,τ=RC   or   LRx(t) = x(\infty) + \big[x(0^{+}) - x(\infty)\big]e^{-t/\tau}, \qquad \tau = RC \;\text{ or }\; \frac{L}{R}

Here xx is any voltage or current, and RR is the Thevenin resistance seen by the capacitor or inductor. Take R=1 kΩR = 1\text{ k}\Omega and C=1 μFC = 1\,\mu\text{F} charging from 00 to 10 V10\text{ V}. Then τ=1 ms\tau = 1\text{ ms} and vC(1 ms)=10(1−e−1)=6.32 Vv_C(1\text{ ms}) = 10(1 - e^{-1}) = 6.32\text{ V}.

The trap is continuity: vCv_C and iLi_L do not jump, unless switching forms a capacitor loop or an inductor cut-set. Then conserve charge or flux.

Series and parallel resonance

Formula Example Watch out for
ω0=1LC\displaystyle \omega_0 = \frac{1}{\sqrt{LC}} L=1 mHL = 1\text{ mH}, C=1 μFC = 1\,\mu\text{F}: ω0=31623 rad/s\omega_0 = 31623\text{ rad/s} Converting to Hz needs 2π2\pi
Series: Q=ω0LR\displaystyle Q = \frac{\omega_0 L}{R} R=10 ΩR = 10\,\Omega: Q=3.16Q = 3.16 Impedance is minimum, equal to RR
Parallel: Q=Rω0L=RCL\displaystyle Q = \frac{R}{\omega_0 L} = R\sqrt{\frac{C}{L}} R=1 kΩR = 1\text{ k}\Omega: Q=31.6Q = 31.6 It is the reciprocal of the series form
BW=ω0Q\displaystyle \text{BW} = \frac{\omega_0}{Q} Series case: BW=104 rad/s\text{BW} = 10^4\text{ rad/s} Bandwidth in rad/s unless converted

Two-port parameters, reciprocity and symmetry

Each parameter set opens or shorts one port. For example, z11=V1I1∣I2=0\displaystyle z_{11} = \frac{V_1}{I_1}\Big\vert_{I_2 = 0} and y11=I1V1∣V2=0\displaystyle y_{11} = \frac{I_1}{V_1}\Big\vert_{V_2 = 0}.

Condition z y h ABCD Watch out for
Reciprocal z12=z21z_{12} = z_{21} y12=y21y_{12} = y_{21} h12=−h21h_{12} = -h_{21} AD−BC=1AD - BC = 1 Only h carries the minus sign
Symmetric z11=z22z_{11} = z_{22} y11=y22y_{11} = y_{22} h11h22−h12h21=1h_{11}h_{22} - h_{12}h_{21} = 1 A=DA = D Symmetry does not imply reciprocity
Interconnection Series: zz adds Parallel: yy adds Series-parallel: hh adds Cascade: ABCD multiplies Keep the matrix order in a cascade

A single series impedance ZZ has ABCD matrix [1Z01]\begin{bmatrix} 1 & Z \\ 0 & 1 \end{bmatrix}. Then AD−BC=1AD - BC = 1 and A=DA = D, so it is reciprocal and symmetric.

Remember: A two-port question appeared exactly once in each of the five counted papers, but the set kept changing. 2019 Q14 named no matrix and stated reciprocity as excitation and response interchanged.

LTI properties and convolution

Test linearity with two checks, additivity and scaling. Test time invariance by delaying the input and comparing.

Property Example Watch out for
Linear: additive and homogeneous y=x+2y = x + 2 fails: zero input gives 22 y(t)=x(t)cos⁡ω0ty(t) = x(t)\cos\omega_0 t is linear but time-varying
Time-invariant y(t)=x(2t)y(t) = x(2t) is time-varying Time scaling is linear and still time-varying
Causal: h(t)=0h(t) = 0 for t<0t < 0 h[n]=u[n+1]h[n] = u[n + 1] is not causal Check the sample at n=−1n = -1
BIBO stable: ∫∣h(t)∣ dt<∞\int \vert h(t)\vert\,dt < \infty, ∑∣h[n]∣<∞\sum \vert h[n]\vert < \infty h[n]=0.5nu[n]h[n] = 0.5^{n}u[n] sums to 22 u[n]u[n] alone is not stable
Cascade h1∗h2h_1 * h_2; parallel h1+h2h_1 + h_2 Lengths 44 and 33 convolve to length 66 Length is N+M−1N + M - 1

An interconnection is LTI only if every block in it is LTI.

Fourier series symmetry rules

Symmetry Condition What vanishes Watch out for
Even x(t)=x(−t)x(t) = x(-t) Sine terms, bn=0b_n = 0 Kills terms, not harmonics
Odd x(t)=−x(−t)x(t) = -x(-t) a0a_0 and cosine terms The dc term goes too
Half-wave x(t)=−x(t±T/2)x(t) = -x(t \pm T/2) All even harmonics Kills harmonics, not terms
Time scaling x(at)x(at) Nothing; coefficients unchanged Fundamental frequency multiplies by aa

A zero-mean square wave centred on t=0t = 0 is even and half-wave symmetric, so it holds only cosine terms at odd harmonics.

This sheet keeps to the most-used rows. The GATE ECE 2027 book has a last-minute revision sheet covering all nine sections, ranks 78 recurring concepts, and includes 925 questions with worked solutions and 10 full mock tests.

Fourier, Laplace and Parseval

Formula Example Watch out for
y(t)=A∣H(jω0)∣cos⁡(ω0t+ϕ+∠H(jω0))y(t) = A\vert H(j\omega_0)\vert\cos\big(\omega_0 t + \phi + \angle H(j\omega_0)\big) H(s)=1s+1\displaystyle H(s) = \frac{1}{s + 1}, input cos⁡t\cos t: y=0.707cos⁡(t−45∘)y = 0.707\cos(t - 45^\circ) Add the phase of HH
f(∞)=lim⁡s→0sF(s)f(\infty) = \lim_{s \to 0} sF(s) F(s)=5s(s+2)\displaystyle F(s) = \frac{5}{s(s + 2)}: f(∞)=2.5f(\infty) = 2.5 Poles of sF(s)sF(s) must be in the open left half-plane
E=∫∣x(t)∣2dt=12π∫∣X(jω)∣2dω\displaystyle E = \int \vert x(t)\vert^{2}dt = \frac{1}{2\pi}\int \vert X(j\omega)\vert^{2}d\omega x=e−tu(t)x = e^{-t}u(t): E=0.5E = 0.5 The 12π\displaystyle \frac{1}{2\pi} goes with ω\omega, not ff
Energy of x(at)x(at) is E∣a∣\displaystyle \frac{E}{\vert a\vert} x(2t)x(2t) has half the energy Shifting does not change energy

Apply the final value theorem to F(s)=1s2+1\displaystyle F(s) = \frac{1}{s^{2} + 1} and you get 00, yet f(t)=sin⁡tf(t) = \sin t never settles.

Sampling and aliasing

Sample faster than twice the highest frequency: fs>2fmax⁡f_s > 2f_{\max}. The Nyquist rate is 2fmax⁡2f_{\max}; the Nyquist frequency is fs2\displaystyle \frac{f_s}{2}.

A tone at ff appears at ∣f−kfs∣\vert f - kf_s\vert for the integer kk that brings it below fs2\displaystyle \frac{f_s}{2}. So 7 kHz7\text{ kHz} sampled at 10 kHz10\text{ kHz} appears at 3 kHz3\text{ kHz}. Squaring a signal doubles its bandwidth: a 4 kHz4\text{ kHz} signal squared needs a Nyquist rate of 16 kHz16\text{ kHz}.

z-transform: ROC, causality and stability

Rule Example Watch out for
Right-sided: ROC ∣z∣>∣p∣max⁡\vert z\vert > \vert p\vert_{\max} anu[n]↔11−az−1\displaystyle a^{n}u[n] \leftrightarrow \frac{1}{1 - az^{-1}}, ∣z∣>∣a∣\vert z\vert > \vert a\vert The ROC never contains a pole
Left-sided: ROC ∣z∣<∣p∣min⁡\vert z\vert < \vert p\vert_{\min} −anu[−n−1]-a^{n}u[-n-1] has the same X(z)X(z), ∣z∣<∣a∣\vert z\vert < \vert a\vert Same formula, different sequence
Two-sided: an annulus Poles 0.50.5 and 22: 0.5<∣z∣<20.5 < \vert z\vert < 2 is stable, non-causal Causal and stable cannot both hold here
Stable iff ROC contains ∣z∣=1\vert z\vert = 1 ∣z∣>2\vert z\vert > 2 is causal but unstable Pole-zero cancellation stabilises nothing
Finite length: whole plane except possibly z=0z = 0 or ∞\infty δ[n−1]\delta[n - 1] excludes z=0z = 0 Check both ends

In one line: One H(z)H(z) can describe several sequences; the ROC picks the sequence, and the unit circle inside it means stability.

DTFT, DFT and all-pass systems

The DTFT is X(ejω)=∑nx[n]e−jωnX(e^{j\omega}) = \sum_n x[n]e^{-j\omega n}, periodic in ω\omega with period 2π2\pi. It exists when ∑∣x[n]∣<∞\sum \vert x[n]\vert < \infty, that is, when the ROC contains the unit circle. 2nu[n]2^{n}u[n] has a z-transform but no DTFT.

The NN-point DFT is:

X[k]=∑n=0N−1x[n] e−j2πkn/N,∑n=0N−1∣x[n]∣2=1N∑k=0N−1∣X[k]∣2X[k] = \sum_{n=0}^{N-1} x[n]\,e^{-j2\pi kn/N}, \qquad \sum_{n=0}^{N-1}\vert x[n]\vert^{2} = \frac{1}{N}\sum_{k=0}^{N-1}\vert X[k]\vert^{2}

For x=[1,1,1,1]x = [1, 1, 1, 1], X=[4,0,0,0]X = [4, 0, 0, 0], and both energy sums give 44. A cosine cos⁡(2πk0n/N)\cos(2\pi k_0 n/N) fills only bins k0k_0 and N−k0N - k_0. That holds only for a whole number of periods; otherwise leakage spreads into every bin.

An all-pass system has ∣H∣=1\vert H\vert = 1 at every frequency. In continuous time, H(s)=s−as+a\displaystyle H(s) = \frac{s - a}{s + a} mirrors its zero and pole about the jωj\omega axis. In discrete time, with real aa, H(z)=z−1−a1−az−1\displaystyle H(z) = \frac{z^{-1} - a}{1 - az^{-1}} puts the zero at 1a\displaystyle \frac{1}{a}. Its phase is not linear, so an all-pass system still distorts. Group delay is τg=−dθdω\displaystyle \tau_g = -\frac{d\theta}{d\omega}.

FIR and IIR filter design

The 2027 syllabus names FIR and IIR filter design for the first time. GATE 2019 Q39 and Q54, set by IIT Madras, were both filter-design questions, 4 of that paper's 14 Networks marks. They asked design by zero placement and by least-squares approximation. One paper is a precedent, not a rate; the 2027 syllabus changes guide covers the full comparison.

Idea Formula Example Watch out for
FIR form H(z)=∑n=0Mh[n]z−nH(z) = \sum_{n=0}^{M} h[n]z^{-n} Length M+1M + 1, order MM Length is order plus one
Null at ω0\omega_0, real taps H(z)=1−2cos⁡ω0 z−1+z−2H(z) = 1 - 2\cos\omega_0\,z^{-1} + z^{-2} ω0=π3\displaystyle \omega_0 = \frac{\pi}{3}: h=[1,−1,1]h = [1, -1, 1] The zero must sit on the unit circle, with its conjugate
Least-squares truncation error energy=∑droppedh2[n]\text{error energy} = \sum_{\text{dropped}} h^{2}[n] Drop taps 0.250.25 and 0.1250.125: error 0.0780.078 Keep the retained taps unchanged
Linear-phase FIR h[n]=±h[M−n]h[n] = \pm h[M - n] M=4M = 4: delay of 22 samples Group delay is M2\displaystyle \frac{M}{2}
Bilinear transformation s=2T 1−z−11+z−1\displaystyle s = \frac{2}{T}\,\frac{1 - z^{-1}}{1 + z^{-1}}, Ω=2Ttan⁡ω2\displaystyle \Omega = \frac{2}{T}\tan\frac{\omega}{2} ω=π2\displaystyle \omega = \frac{\pi}{2}, T=1T = 1: Ω=2\Omega = 2 Pre-warp the edges; no aliasing
Impulse invariance h[n]=T ha(nT)h[n] = T\,h_a(nT), pole p→epTp \to e^{pT} p=−1p = -1, T=0.1T = 0.1: z=0.905z = 0.905 Aliases the analogue response

An FIR filter has poles only at the origin, so it is always stable and can have exact linear phase. An IIR filter needs a lower order, but a causal, stable one cannot have exactly linear phase. The last two rows have no counted instance; learn them from the syllabus wording.

Trap: To reject a frequency, a filter needs a zero on the unit circle at that frequency, never a pole. With real taps, nulling one frequency costs a conjugate pair of zeros and three taps.

On exam day the marking scheme, which is common to every GATE paper, puts negative marks only on MCQs.

More formula sheets: all of GATE ECE · Analog Circuits · Communications · Electromagnetics

Quick revision

  1. ω=2πf\omega = 2\pi f, meters read rms, and Vrms=Vm2\displaystyle V_{\text{rms}} = \frac{V_m}{\sqrt{2}}.
  2. Never deactivate a dependent source; use a test source, and never superpose power.
  3. Series Q=ω0LR\displaystyle Q = \frac{\omega_0 L}{R}; parallel Q=Rω0L\displaystyle Q = \frac{R}{\omega_0 L}; BW=ω0Q\displaystyle \text{BW} = \frac{\omega_0}{Q}.
  4. Reciprocity: z12=z21z_{12} = z_{21}, y12=y21y_{12} = y_{21}, AD−BC=1AD - BC = 1, and h12=−h21h_{12} = -h_{21}.
  5. Even kills sine terms, odd kills cosines and dc, half-wave kills even harmonics.
  6. Final value theorem only with every pole of sF(s)sF(s) in the open left half-plane.
  7. Stable iff the ROC contains ∣z∣=1\vert z\vert = 1; causal and stable iff every pole is inside it.
  8. To null ω0\omega_0 with real taps, place zeros at e±jω0e^{\pm j\omega_0} on the unit circle.

Frequently asked questions

Which formulas matter most for GATE ECE Networks, Signals and Systems?

Start with the families present in all five counted papers, 2010, 2019, 2020, 2025 and 2026. Those are phasor analysis with impedance and rms values, the z-transform with its region of convergence, and two-port parameters with the reciprocity and symmetry conditions. LTI properties, Fourier series symmetry and dc network theorems follow, each present in four of the five papers.

Is FIR and IIR filter design in the GATE ECE 2027 syllabus?

Yes. The phrase is newly named in the 2027 syllabus, but the idea is not new. GATE 2019 Q39 and Q54, set by IIT Madras, were both filter-design questions worth 2 marks each. They used zero placement and least-squares approximation. Windows, the bilinear transformation and impulse invariance have no counted instance, so prepare them from the syllabus wording alone.

When can I use the final value theorem in GATE?

Only when every pole of sF(s)sF(s) lies strictly in the left half of the s-plane. Then f(∞)=lim⁡s→0sF(s)f(\infty) = \lim_{s \to 0} sF(s). If F(s)F(s) has poles on the imaginary axis other than a single pole at the origin, or any pole in the right half-plane, the limit still returns a tidy number. That number is wrong, because f(t)f(t) oscillates or grows.

How do I read causality and stability from a z-transform ROC?

A right-sided sequence has its ROC outside the outermost pole, and a causal system is right-sided. A system is stable when its ROC contains the unit circle. Combining the two, a causal rational system is stable only when every pole lies strictly inside the unit circle. The same H(z)H(z) with a different ROC is a different sequence, so always ask which ROC the question gives.

How many marks does Networks, Signals and Systems carry in GATE ECE?

Counted question by question across the 2010, 2019, 2020, 2025 and 2026 EC papers, the section averaged 17.0 marks, the highest of any technical section. It is not fixed. It fell to 13 marks in 2025 and reached 21 in 2010. Plan for the middle of that range and check the current paper pattern on the official GATE 2027 site.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.

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