GATE EC · EE · IN

Signals

Types, operations, energy, power and averages

This topic sets up the vocabulary for the whole Signals and Systems unit: how signals are classified, the standard building-block signals and the unit impulse, how time shifting, scaling and reversal act on a signal, and how to compute energy, power, average and RMS values. Convolution, Fourier series and transforms, and sampling all lean on these definitions, and the calculations here reward precision: the $1/∣ a ∣$ in $δ (a t)$ , the order of shift and scale in $x (a t + b)$ , the integer-period test for DT sinusoids and merging same-frequency terms before adding powers are where answers go wrong.

18 previous-year questions · 17 solved · 5 practice questions

Classifying signals

A continuous-time (CT) signal $x (t)$ is defined for every real $t$ ; a discrete-time (DT) signal $x [n]$ is defined only at integer $n$ (it is a sequence, with nothing in between the samples). Every classification below has a CT and a DT version, and each one is a test you apply directly to the defining expression.

Class	CT test	DT test	Examples
Periodic	$x (t + T) = x (t)$ for all $t$ , for some $T > 0$ ; the smallest such $T$ is the fundamental period $T_{0}$ (a constant is periodic but has no smallest $T$ )	$x [n + N] = x [n]$ for all $n$ , for some integer $N > 0$ ; the smallest is $N_{0}$ (a constant has $N_{0} = 1$ )	$cos ω_{0} t$ , square wave, $cos (π n /4)$
Aperiodic	no such $T$ exists	no such integer $N$ exists	$rect (t)$ , $e^{- t} u (t)$ , $cos (n)$
Even	$x (- t) = x (t)$	$x [- n] = x [n]$	$cos ω_{0} t$ , $∣ t ∣$ , $rect (t)$ , $δ (t)$
Odd	$x (- t) = - x (t)$ , which forces $x (0) = 0$	$x [- n] = - x [n]$ , which forces $x [0] = 0$	$sin ω_{0} t$ , $t$ , $sgn (t)$
Causal	$x (t) = 0$ for $t < 0$	$x [n] = 0$ for $n < 0$	$u (t)$ , $e^{- 2 t} u (t)$ , $a^{n} u [n]$
Anticausal	$x (t) = 0$ for $t > 0$	$x [n] = 0$ for $n > 0$	$u (- t)$ , $e^{3 t} u (- t)$
Energy signal	$0 < E < \infty$ (then $P = 0$ )	same, with sums	bounded finite-duration pulses, $e^{- a ∣ t ∣}$ with $a > 0$
Power signal	$0 < P < \infty$ (then $E = \infty$ )	same, with sums	sinusoids, square waves, DC, $u (t)$ , $u [n]$
Neither	$E = \infty$ and $P$ is $0$ or $\infty$	same	$t u (t)$ and $e^{t}$ ( $P = \infty$ ); $t^{- 1/2} u (t - 1)$ ( $E = \infty$ , $P = 0$ )

EVEN AND ODD PARTS

x_{e} (t) = \frac{x ( t ) + x ( - t )}{2}, x_{o} (t) = \frac{x ( t ) - x ( - t )}{2}, x (t) = x_{e} (t) + x_{o} (t)

Every signal splits this way, uniquely; the DT form is identical with $n$ in place of $t$ . Example: $u (t) = \frac{1}{2} + \frac{1}{2} sgn (t)$ , so $u_{e} (t) = \frac{1}{2}$ and $u_{o} (t) = \frac{1}{2} sgn (t)$ for $t \neq = 0$ .

Symmetry rules worth using

even × even = even, odd × odd = even, even × odd = odd (same in DT).
$\int_{- a}^{a} x_{o} (t) d t = 0$ and $\int_{- a}^{a} x_{e} (t) d t = 2 \int_{0}^{a} x_{e} (t) d t$ ; in DT, $\sum_{n = - N}^{N} x_{o} [n] = 0$ .
The derivative of an even signal is odd, and the derivative of an odd signal is even.
For any energy signal, real or complex, $E_{x} = E_{x_{e}} + E_{x_{o}}$ : the cross term $2 Re \int x_{e} (t) x_{o}^{*} (t) d t$ is the integral of an odd function, which is zero.
For a causal signal, $x (t) = 2 x_{e} (t) = 2 x_{o} (t)$ for $t > 0$ (because $x (- t) = 0$ there), so either part alone fixes the signal everywhere except at $t = 0$ .
The only signal that is both even and odd is $x = 0$ . The sum of a nonzero even signal and a nonzero odd signal is neither.
For complex signals the analogous split is into a conjugate-symmetric part $\frac{1}{2} [x (t) + x^{*} (- t)]$ , which satisfies $x_{cs} (- t) = x_{cs}^{*} (t)$ , and a conjugate-antisymmetric part $\frac{1}{2} [x (t) - x^{*} (- t)]$ .

EXAM TIP

Decide the class before you integrate. A bounded periodic signal that is nonzero over some interval is a power signal: its energy is infinite, so the question wants $P$ . A bounded signal of finite duration is an energy signal: its power is zero. One-sided decaying exponentials such as $e^{- a t} u (t)$ with $a > 0$ are energy signals even though they never reach zero.

Standard signals and the unit impulse

These building blocks write piecewise signals compactly: a pulse of height 2 on $0 < t < 3$ is $2 [u (t) - u (t - 3)] = 2 rect (\frac{t - 1.5}{3})$ , and the unit triangle is $tri (t) = r (t + 1) - 2 r (t) + r (t - 1)$ , with the ramp $r (t)$ defined below.

Convention used throughout these notes: $sinc (x) = \frac{sin ( π x )}{π x}$ , with $sinc (0) = 1$ and zeros at every nonzero integer $x$ . Some texts instead use $Sa (x) = \frac{sin x}{x}$ , so that $sinc (x) = Sa (π x)$ ; check which one a question means before using any energy or transform result. The value of $u (t)$ at $t = 0$ is left undefined (some texts take $\frac{1}{2}$ ). It does not change any ordinary integral, but it does make a product such as $u (t) δ (t)$ ill-defined, because sifting needs the signal to be continuous at the impulse.

UNIT STEP, SIGNUM AND RAMP

u (t) = {1, 0, t > 0 t < 0, sgn (t) = 2 u (t) - 1, r (t) = t u (t) = \int_{- \infty}^{t} u (τ) d τ

$sgn (0) = 0$ , so $sgn (t) = 2 u (t) - 1$ holds for $t \neq = 0$ . Derivative chain: $\frac{d}{d t} r (t) = u (t)$ , $\frac{d}{d t} u (t) = δ (t)$ , $\frac{d}{d t} sgn (t) = 2 δ (t)$ .

RECTANGULAR AND TRIANGULAR PULSES

rect (\frac{t}{τ}) = {1, 0, ∣ t ∣ < τ /2 ∣ t ∣ > τ /2, tri (\frac{t}{τ}) = {1 - ∣ t ∣/ τ, 0, ∣ t ∣ \leq τ ∣ t ∣ > τ

For $τ > 0$ : rect has width $τ$ and area $τ$ ; tri has half-width $τ$ (base $2 τ$ ), peak 1 and area $τ$ . Also $rect (t) * rect (t) = tri (t)$ . Some texts write the triangle with total base $τ$ ; read the definition given.

SINC AND COMPLEX EXPONENTIAL

sinc (t) = \frac{sin ( π t )}{π t}, x (t) = A e^{s t}, s = σ + j ω

$σ < 0$ decays, $σ > 0$ grows, and $σ = 0$ with $ω \neq = 0$ gives the rotating phasor $A e^{j ω t}$ of period $2 π /∣ ω ∣$ . With $A = ∣ A ∣ e^{j θ}$ , the real part of $A e^{s t}$ is $∣ A ∣ e^{σ t} cos (ω t + θ)$ .

SIFTING AND SAMPLING PROPERTIES OF THE IMPULSE

\int_{t_{1}}^{t_{2}} x (t) δ (t - t_{0}) d t = {x (t_{0}), 0, t_{1} < t_{0} < t_{2} t_{0} \in / [t_{1}, t_{2}], x (t) δ (t - t_{0}) = x (t_{0}) δ (t - t_{0})

Requires $x$ continuous at $t_{0}$ . When $t_{0}$ sits exactly on a limit, the value depends on a convention, so that case is left out. The impulse has unit area, $\int_{- \infty}^{\infty} δ (t) d t = 1$ , and $u (t) = \int_{- \infty}^{t} δ (τ) d τ$ .

TIME SCALING OF THE IMPULSE

δ (a t + b) = \frac{1}{∣ a ∣} δ (t + \frac{b}{a}), a \neq = 0

With $a = - 1$ : $δ (- t) = δ (t)$ , so the impulse is even. Hence $\int_{- \infty}^{\infty} x (t) δ (a t + b) d t = \frac{1}{∣ a ∣} x (- \frac{b}{a})$ .

IMPULSE OF A FUNCTION

δ (f (t)) = k \sum \frac{δ ( t - t _{k} )}{∣ f ^{'} ( t _{k} ) ∣}

The sum runs over the roots $t_{k}$ of $f (t) = 0$ , and every root must be simple ( $f^{'} (t_{k}) \neq = 0$ ); at a repeated root $δ (f (t))$ is not defined. Example: $δ (t^{2} - 9) = \frac{1}{6} [δ (t - 3) + δ (t + 3)]$ .

DERIVATIVES OF THE IMPULSE

\int_{- \infty}^{\infty} x (t) δ^{(n)} (t - t_{0}) d t = (- 1)^{n} x^{(n)} (t_{0})

Requires $x^{(n)}$ continuous at $t_{0}$ . For $n = 1$ (the doublet): $\int_{- \infty}^{\infty} x (t) δ^{'} (t - t_{0}) d t = - x^{'} (t_{0})$ . The doublet $δ^{'} (t)$ is odd, and $x (t) δ^{'} (t) = x (0) δ^{'} (t) - x^{'} (0) δ (t)$ , not just $x (0) δ^{'} (t)$ .

DISCRETE-TIME IMPULSE AND STEP

δ [n] = u [n] - u [n - 1], u [n] = k = 0 \sum \infty δ [n - k] = m = - \infty \sum n δ [m], n = - \infty \sum \infty x [n] δ [n - n_{0}] = x [n_{0}]

$δ [0] = 1$ is an ordinary finite value, unlike $δ (t)$ . $δ [an] = δ [n]$ for every nonzero integer $a$ : there is no $1/∣ a ∣$ factor in DT.

ANY SIGNAL AS A WEIGHTED SUM OF SHIFTED IMPULSES

x [n] = k = - \infty \sum \infty x [k] δ [n - k], x (t) = \int_{- \infty}^{\infty} x (τ) δ (t - τ) d τ

The DT form holds sample by sample, since only the $k = n$ term survives. The CT form is the sifting property with $δ (t - τ) = δ (τ - t)$ , valid wherever $x$ is continuous. Both are the starting point for convolution.

EXAM TIP

Three impulse slips. (1) Dropping the $1/∣ a ∣$ : $\int_{- \infty}^{\infty} e^{t} δ (2 t - 2) d t = \frac{1}{2} e$ , not $e$ . (2) Sifting an impulse that lies outside the limits: $\int_{0}^{5} cos t δ (t + 1) d t = 0$ . (3) Differentiating a signal with a jump: a jump of height $J$ at $t_{0}$ (value just after minus value just before) puts $J δ (t - t_{0})$ into $d x / d t$ , so a derivative sketch that shows only the slopes is incomplete.

Time shift, reversal and scaling (CT and DT)

For $y (t) = x (t - t_{0})$ the graph moves right (a delay) when $t_{0} > 0$ and left when $t_{0} < 0$ . $y (t) = x (- t)$ reflects it about $t = 0$ . $y (t) = x (a t)$ with $∣ a ∣ > 1$ compresses the signal in time by the factor $∣ a ∣$ (the graph is squeezed towards $t = 0$ ), with $0 < ∣ a ∣ < 1$ stretches it, and $a < 0$ adds a reflection. These operations act on the argument $t$ ; amplitude operations $A x (t) + B$ act on the values and can be done at any stage.

WHERE A POINT OF X LANDS UNDER X(AT+B)

y (t) = x (a t + b), a \neq = 0 : the value x (t_{1}) appears at t = \frac{t _{1} - b}{a}

Map each breakpoint of $x$ this way. A support $[t_{1}, t_{2}]$ becomes $[\frac{t _{1} - b}{a}, \frac{t _{2} - b}{a}]$ , with the ends swapped when $a < 0$ . This check does not depend on any order of operations.

Sketching x(at+b) step by step: two correct orders

Shift, then scale. $v (t) = x (t + b)$ (move left by $b$ if $b > 0$ ); then $w (t) = v (∣ a ∣ t) = x (∣ a ∣ t + b)$ ; then, if $a < 0$ , reflect: $y (t) = w (- t) = x (a t + b)$ .
Scale, then shift. Write $x (a t + b) = x (a (t + b / a))$ . First form $x (a t)$ (scale by $∣ a ∣$ , reflect if $a < 0$ ), then replace $t$ by $t + b / a$ : a shift left by $b / a$ , i.e. right by $- b / a$ .
The usual error is scaling first and then shifting by $b$ instead of $b / a$ . Once the signal has been scaled, the shift is $b / a$ .
Each operation replaces $t$ in the current expression, never in the original $x (t)$ .

EFFECT OF TIME OPERATIONS ON AREA, ENERGY AND POWER

\int_{- \infty}^{\infty} x (a t + b) d t = \frac{1}{∣ a ∣} \int_{- \infty}^{\infty} x (t) d t, E_{x (a t + b)} = \frac{E _{x}}{∣ a ∣}, P_{x (a t + b)} = P_{x} (x periodic), a \neq = 0

Shifts change none of these. Time scaling divides area and energy by $∣ a ∣$ but leaves the power, average and RMS value of a periodic signal unchanged (its period becomes $T_{0} /∣ a ∣$ ). Amplitude scaling by $k$ multiplies $E$ and $P$ by $∣ k ∣^{2}$ .

WORKED EXAMPLE · EXAMPLE 1: SHIFT, SCALE AND REVERSE TOGETHER

Let $x (t) = t$ for $0 \leq t \leq 2$ and $x (t) = 0$ otherwise. For $y (t) = x (- 2 t + 4)$ , find the interval outside which $y (t)$ is zero, the expression for $y (t)$ on that interval, the value $y (1.5)$ , and the energies $E_{x}$ and $E_{y}$ .

Solution

Breakpoint mapping with $a = - 2$ , $b = 4$ : $t_{1} = 0 \to t = \frac{0 - 4}{- 2} = 2$ and $t_{1} = 2 \to t = \frac{2 - 4}{- 2} = 1$ . So $y$ is zero outside $1 \leq t \leq 2$ (ends swapped because $a < 0$ ).
Check by operations: $v (t) = x (t + 4)$ lives on $[- 4, - 2]$ ; $w (t) = v (2 t) = x (2 t + 4)$ lives on $[- 2, - 1]$ ; $y (t) = w (- t) = x (- 2 t + 4)$ lives on $[1, 2]$ .
On that interval $y (t) = - 2 t + 4$ . It equals $2$ at $t = 1$ (the image of $x (2) = 2$ ) and $0$ at $t = 2$ (the image of $x (0) = 0$ ). So $y (1.5) = 4 - 3 = 1$ .
$E_{x} = \int_{0}^{2} t^{2} d t = \frac{8}{3}$ .
$E_{y} = \int_{1}^{2} (4 - 2 t)^{2} d t$ . With $u = 4 - 2 t$ , $d t = - d u /2$ : $E_{y} = \frac{1}{2} \int_{0}^{2} u^{2} d u = \frac{4}{3}$ , which is $E_{x} /∣ a ∣ = \frac{8/3}{2}$ , as the scaling rule says.

Answer: $y (t) = 4 - 2 t$ for $1 \leq t \leq 2$ and zero outside that interval; $y (1.5) = 1$ ; $E_{x} = 8/3$ and $E_{y} = 4/3$ .

DT operation	Definition	What it does	Reversible?
Shift	$y [n] = x [n - n_{0}]$ , integer $n_{0}$	delays by $n_{0}$ samples if $n_{0} > 0$	yes
Reversal	$y [n] = x [- n]$	reflects about $n = 0$	yes
Downsampling (decimation) by $M$	$y [n] = x [M n]$	keeps every $M$ -th sample and discards the rest; $y [1] = x [M]$	no, in general
Upsampling (zero insertion) by $L$	$y [n] = x [n / L]$ if $L$ divides $n$ , else $0$	inserts $L - 1$ zeros between consecutive samples	yes: downsample by $L$

EXAM TIP

In DT the order matters and samples can be lost. Upsample by 2 and then downsample by 2 and you get $x [n]$ back; downsample first and the odd-indexed samples are gone for good. For $y [n] = x [2 n - 3]$ , shift first and then decimate: $y [0] = x [- 3]$ , $y [1] = x [- 1]$ , $y [2] = x [1]$ , so only odd-indexed samples of $x$ survive. Zero insertion keeps the energy $\sum ∣ x [n] ∣^{2}$ unchanged, while decimation can only keep or reduce it; the CT rule $E /∣ a ∣$ has no DT counterpart.

Periodicity of sums of sinusoids (CT and DT)

CT SINUSOID AND SUMS OF CT PERIODIC SIGNALS

T = \frac{2 π}{ω _{0}} (ω_{0} > 0); \frac{T _{1}}{T _{2}} = \frac{p}{q} rational \Rightarrow x_{1} + x_{2} periodic, with period LCM (T_{1}, T_{2}) = q T_{1} = p T_{2}

Take $p / q$ in lowest terms. For non-constant $x_{1}$ , $x_{2}$ with fundamental periods $T_{1}$ , $T_{2}$ the converse also holds: if $T_{1} / T_{2}$ is irrational, $x_{1} + x_{2}$ is not periodic. The LCM is always a period of the sum, and it is the fundamental period unless components cancel: $x_{1} = cos t + cos 2 t$ and $x_{2} = - cos t$ both have period $2 π$ , but their sum $cos 2 t$ has period $π$ . For a sum of sinusoids with nonzero amplitudes at distinct positive frequencies, the fundamental angular frequency is the GCD of the component frequencies, $ω_{0} = g cd (ω_{1}, ω_{2}, \dots)$ , and $T_{0} = 2 π / ω_{0}$ ; a DC term does not change this.

DT SINUSOID PERIODICITY

cos (Ω_{0} n + θ) periodic ⟺ \frac{Ω _{0}}{2 π} = \frac{m}{N} rational; N_{0} = N with \frac{m}{N} in lowest terms, N > 0

$Ω_{0}$ is in rad/sample. Equivalently, $N_{0}$ is the smallest positive integer $N$ for which $Ω_{0} N$ is a multiple of $2 π$ ; for $Ω_{0} \neq = 0$ that is $N_{0} = 2 π m / Ω_{0}$ with the smallest positive integer $m$ that makes it an integer. The same test applies to $e^{j Ω_{0} n}$ . The one exception is a sequence that is identically zero, such as $sin (π n)$ , which trivially has period 1. A sum of periodic sequences is always periodic: $LCM (N_{1}, N_{2})$ is a period, and it is the fundamental period unless components cancel.

DT FREQUENCY IS DEFINED ONLY MODULO 2Π

e^{j (Ω_{0} + 2 π k) n} = e^{j Ω_{0} n}, cos ((2 π - Ω_{0}) n) = cos (Ω_{0} n)

For integers $k$ and $n$ . Distinct DT sinusoids need only $Ω \in (- π, π]$ (or $[0, π]$ for real cosines). The fastest oscillation is at $Ω = π$ : $cos (π n) = (- 1)^{n}$ .

Property	CT sinusoid $cos (ω_{0} t)$	DT sinusoid $cos (Ω_{0} n)$
Always periodic?	yes (for $ω_{0} > 0$ ), $T = 2 π / ω_{0}$	only if $Ω_{0} /2 π$ is rational
Fundamental period	$2 π / ω_{0}$	the smallest integer $N > 0$ with $Ω_{0} N$ a multiple of $2 π$ ; equal to $2 π / Ω_{0}$ only when that is an integer
Distinct frequencies	every $ω_{0} \geq 0$ gives a different signal	$Ω_{0}$ and $Ω_{0} + 2 π k$ give the same signal
Oscillation rate	grows without limit as $ω_{0}$ grows	grows up to $Ω_{0} = π$ , then falls back to zero at $Ω_{0} = 2 π$
Sum of two periodic signals	periodic if $T_{1} / T_{2}$ is rational (and, for non-constant signals, only then)	always periodic; $LCM (N_{1}, N_{2})$ is a period
Average power, amplitude $A$	$A^{2} /2$ for $ω_{0} \neq = 0$	$A^{2} /2$ if $Ω_{0}$ is not a multiple of $π$ ; $A^{2}$ if it is (every sample is then $\pm A$ )

WORKED EXAMPLE · EXAMPLE 2: FUNDAMENTAL PERIOD AND POWER

(a) Find the fundamental period and the average power of $x (t) = 3 + 4 cos (6 π t) cos (4 π t)$ . (b) Find the fundamental period and the average power of $x [n] = cos (\frac{3 π n}{10}) + 2 sin (\frac{π n}{4})$ . (c) Is $g [n] = cos (n /4)$ periodic?

Solution

(a) Product to sum: $4 cos (6 π t) cos (4 π t) = 2 cos (10 π t) + 2 cos (2 π t)$ , so $x (t) = 3 + 2 cos (2 π t) + 2 cos (10 π t)$ , with components at $1$ Hz and $5$ Hz.
Their periods are $1$ s and $0.2$ s, ratio $5$ (rational), so $T_{0} = LCM (1, 0.2) = 1$ s. Equivalently $g cd (2 π, 10 π) = 2 π$ rad/s.
The DC term and the two distinct-frequency cosines are orthogonal, so their powers add: $P = 3^{2} + \frac{2 ^{2}}{2} + \frac{2 ^{2}}{2} = 9 + 2 + 2 = 13$ (RMS $= 13 \approx 3.61$ ).
(b) $\frac{Ω _{1}}{2 π} = \frac{3}{20}$ in lowest terms, so $N_{1} = 20$ ; $\frac{Ω _{2}}{2 π} = \frac{1}{8}$ , so $N_{2} = 8$ . Hence $N_{0} = LCM (20, 8) = 40$ .
The frequencies $3 π /10$ and $π /4$ are distinct points of $[0, π]$ and neither is $0$ or $π$ , so $P = \frac{1 ^{2}}{2} + \frac{2 ^{2}}{2} = 2.5$ .
(c) $\frac{Ω _{0}}{2 π} = \frac{1/4}{2 π} = \frac{1}{8 π}$ is irrational, so no integer $N$ works: $g [n]$ is aperiodic, even though the CT signal $cos (t /4)$ has period $8 π$ .

Answer: (a) $T_{0} = 1$ s, $P = 13$ . (b) $N_{0} = 40$ , $P = 2.5$ . (c) No, $g [n]$ is aperiodic.

EXAM TIP

Traps. $cos t + cos (π t)$ is not periodic (the period ratio $π$ is irrational), though it is still a power signal with $P = 1$ . More generally, a sum of sinusoids with nonzero amplitudes at distinct nonzero frequencies is periodic exactly when every ratio of two of its frequencies is rational. $cos^{2} (ω_{0} t)$ and $∣ sin (ω_{0} t) ∣$ have period $π / ω_{0}$ , not $2 π / ω_{0}$ . Sampling a periodic CT signal need not give a periodic sequence: $cos (ω_{0} n T_{s})$ is periodic only if $ω_{0} T_{s} /2 π$ is rational. In DT, $A cos (π n + θ) = A cos θ (- 1)^{n}$ has power $A^{2} cos^{2} θ$ , not $A^{2} /2$ ; in particular $sin (π n) = 0$ for every integer $n$ .

Energy and power

ENERGY

E = \int_{- \infty}^{\infty} ∣ x (t) ∣^{2} d t, E = n = - \infty \sum \infty ∣ x [n] ∣^{2}

Normalised energy, as if $x$ were a voltage across $1 Ω$ . Use $∣ x ∣^{2}$ , not $x^{2}$ , for complex signals.

AVERAGE POWER

P = T \to \infty lim \frac{1}{2 T} \int_{- T}^{T} ∣ x (t) ∣^{2} d t, P = N \to \infty lim \frac{1}{2 N + 1} n = - N \sum N ∣ x [n] ∣^{2}

The window is symmetric about zero, which is why $A u (t)$ has power $A^{2} /2$ . If $E < \infty$ then $P = 0$ ; if $0 < P < \infty$ then $E = \infty$ .

POWER OF A PERIODIC SIGNAL

P = \frac{1}{T _{0}} \int_{t_{0}}^{t_{0} + T_{0}} ∣ x (t) ∣^{2} d t, P = \frac{1}{N _{0}} n = n_{0} \sum n_{0} + N_{0} - 1 ∣ x [n] ∣^{2}

Any one full period, from any starting point. For a periodic signal, $P = X_{rms}^{2}$ .

TIME AVERAGE (DC VALUE)

⟨ x ⟩ = T \to \infty lim \frac{1}{2 T} \int_{- T}^{T} x (t) d t, ⟨ x ⟩ = N \to \infty lim \frac{1}{2 N + 1} n = - N \sum N x [n]

For a periodic signal this reduces to the average over one period, $\frac{1}{T _{0}} \int_{t_{0}}^{t_{0} + T_{0}} x (t) d t$ or $\frac{1}{N _{0}} \sum_{n = n_{0}}^{n_{0} + N_{0} - 1} x [n]$ . With the symmetric window, $⟨ u (t)⟩ = \frac{1}{2}$ and $⟨ sgn (t)⟩ = 0$ .

POWER OF DC PLUS SINUSOIDS

x (t) = A_{0} + k \sum A_{k} cos (ω_{k} t + θ_{k}) (ω_{k} distinct and positive) \Rightarrow P = A_{0}^{2} + k \sum \frac{A _{k}^{2}}{2}

For a finite sum with real amplitudes, whether or not $x$ is periodic. Terms at the same frequency must first be merged into one sinusoid; a negative frequency is the same frequency, since $cos (- ω t + θ) = cos (ω t - θ)$ . For $\sum_{k} C_{k} e^{j ω_{k} t}$ with distinct $ω_{k}$ (here $ω$ and $- ω$ do count as different), $P = \sum_{k} ∣ C_{k} ∣^{2}$ .

MERGING TWO SINUSOIDS OF THE SAME FREQUENCY

A_{1} cos (ω t + θ_{1}) + A_{2} cos (ω t + θ_{2}) = A cos (ω t + ϕ), A^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos (θ_{1} - θ_{2})

This is phasor addition, $A ∠ ϕ = A_{1} ∠ θ_{1} + A_{2} ∠ θ_{2}$ . The powers add directly only when $cos (θ_{1} - θ_{2}) = 0$ ; for example $a cos ω t + b sin ω t$ has amplitude $a^{2} + b^{2}$ .

Signal (real $A$ unless stated)	Energy $E$	Power $P$
$A rect (t / τ)$	$A^{2} τ$	$0$
$A tri (t / τ)$ , half-width $τ$	$\frac{2}{3} A^{2} τ$	$0$
$A e^{- a t} u (t)$ , $a > 0$	$\frac{A ^{2}}{2 a}$	$0$
$A e^{- a ∣ t ∣}$ , $a > 0$	$\frac{A ^{2}}{a}$	$0$
$A sinc (B t)$ , $B > 0$	$\frac{A ^{2}}{B}$	$0$
$A \frac{s i n ( W t )}{π t}$ , $W > 0$	$\frac{A ^{2} W}{π}$	$0$
$A$ (DC)	$\infty$	$A^{2}$
$A u (t)$	$\infty$	$\frac{A ^{2}}{2}$
$A cos (ω_{0} t + θ)$ , $ω_{0} \neq = 0$	$\infty$	$\frac{A ^{2}}{2}$
$A e^{j (ω_{0} t + θ)}$ , $A$ may be complex	$\infty$	$∣ A ∣^{2}$
Square wave between $+ A$ and $- A$ , any duty cycle	$\infty$	$A^{2}$
$a^{n} u [n]$ , $∣ a ∣ < 1$	$\frac{1}{1 - ∣ a ∣ ^{2}}$	$0$
$u [n]$	$\infty$	$\frac{1}{2}$
$t u (t)$	$\infty$	$\infty$ (neither)

EXAM TIP

Two checks worth doing. Scaling: $x (a t)$ has energy $E_{x} /∣ a ∣$ , so compressing a pulse lowers its energy, and $x (t - t_{0})$ has the same energy as $x (t)$ . Same frequency: in $v (t) = 2 cos (ω t) + 2 cos (ω t + 12 0^{\circ})$ the powers do not add ( $2 + 2 = 4$ is wrong); the merged amplitude is $4 + 4 + 8 cos 12 0^{\circ} = 2$ , so $P = 2$ .

Average and RMS values of periodic waveforms

AVERAGE (DC) VALUE AND RMS VALUE

V_{avg} = \frac{1}{T} \int_{t_{0}}^{t_{0} + T} v (t) d t, V_{rms} = \frac{1}{T} \int_{t_{0}}^{t_{0} + T} v^{2} (t) d t

Over any one full period of a real waveform $v (t)$ . $V_{rms}^{2}$ is the average power in $1 Ω$ : the RMS value is the DC level that would dissipate the same power in a resistor.

DC AND AC PARTS

V_{rms}^{2} = V_{avg}^{2} + V_{ac,rms}^{2}

The AC part $v (t) - V_{avg}$ has zero mean, so it is orthogonal to the DC part. For $v = V_{0} + \sum_{k} V_{k} cos (ω_{k} t + θ_{k})$ with distinct positive $ω_{k}$ : $V_{rms} = V_{0}^{2} + \sum_{k} V_{k}^{2} /2$ . For a waveform with nonzero mean, the ratio $V_{ac,rms} / V_{avg} = FF^{2} - 1$ (FF as defined below, with the full-period average) is the ripple factor used for rectifier outputs.

STRAIGHT-LINE SEGMENT SHORTCUT

segment from a to b : mean = \frac{a + b}{2}, mean square = \frac{a ^{2} + ab + b ^{2}}{3}

Valid for any straight-line piece, whatever its slope or duration. Weight each segment's mean (or mean square) by its fraction of the period and add. A flat piece at level $c$ is the case $a = b = c$ .

FORM FACTOR AND CREST (PEAK) FACTOR

FF = \frac{V _{rms}}{V _{avg}}, CF = \frac{V _{peak}}{V _{rms}}

For a waveform with nonzero mean, $V_{avg}$ is the full-period average. For a zero-mean AC wave that average is zero, so $V_{avg}$ in FF means the average of $∣ v (t) ∣$ over a period, which for a sine, square or symmetric triangle wave equals the average over a positive half-cycle. Sine: FF $= \frac{π}{2 2} \approx 1.11$ , CF $= 2 \approx 1.414$ . Square: FF $=$ CF $= 1$ . Triangle: FF $= \frac{2}{3} \approx 1.155$ , CF $= 3 \approx 1.732$ . Half-wave rectified sine: FF $= \frac{π}{2} \approx 1.571$ , CF $= 2$ . Full-wave rectified sine: the same as the sine, FF $\approx 1.11$ , CF $= 2$ .

Waveform (peak $V_{m}$ )	Average	RMS
Sinusoid $V_{m} sin ω t$	$0$ over a full cycle; $\frac{2 V _{m}}{π}$ over a half-cycle	$\frac{V _{m}}{2}$
Half-wave rectified sine	$\frac{V _{m}}{π}$	$\frac{V _{m}}{2}$
Full-wave rectified sine	$\frac{2 V _{m}}{π}$	$\frac{V _{m}}{2}$
Square wave $\pm V_{m}$ , 50% duty	$0$	$V_{m}$
Pulse train between $0$ and $V_{m}$ , duty cycle $D$	$D V_{m}$	$V_{m} D$
Symmetric triangle, $- V_{m}$ to $V_{m}$	$0$	$\frac{V _{m}}{3}$
Sawtooth or triangle, $0$ to $V_{m}$ (any rise/fall split)	$\frac{V _{m}}{2}$	$\frac{V _{m}}{3}$
Sawtooth, $- V_{m}$ to $V_{m}$	$0$	$\frac{V _{m}}{3}$
Two-level wave: $V_{1}$ for $D T$ , $V_{2}$ for $(1 - D) T$	$D V_{1} + (1 - D) V_{2}$	$D V_{1}^{2} + (1 - D) V_{2}^{2}$
$V_{0} + \sum_{k} V_{k} cos (ω_{k} t + θ_{k})$ , distinct $ω_{k} > 0$	$V_{0}$	$V_{0}^{2} + \sum_{k} \frac{V _{k}^{2}}{2}$

WORKED EXAMPLE · EXAMPLE 3: AVERAGE, RMS AND FORM FACTOR OF A PIECEWISE WAVEFORM

A periodic voltage has period $T = 4$ ms. In each period it rises linearly from $0$ to $6$ V during $0 < t < 2$ ms, stays at $6$ V for $2 < t < 3$ ms, and is $0$ V for $3 < t < 4$ ms. Find its average value, RMS value, form factor, crest factor, and the RMS value of its AC part.

Solution

Average: weight each segment's mean by its duration. $V_{avg} = \frac{1}{4} [\frac{0 + 6}{2} (2) + 6 (1) + 0 (1)] = \frac{6 + 6}{4} = 3$ V.
Mean square: the ramp from $0$ to $6$ has mean square $\frac{0 + 0 + 36}{3} = 12$ , and the flat part has $36$ . So $\overline{v^{2}} = \frac{1}{4} [12 (2) + 36 (1) + 0 (1)] = \frac{60}{4} = 15 V^{2}$ .
$V_{rms} = 15 \approx 3.873$ V.
Form factor $= \frac{3.873}{3} \approx 1.291$ ; crest factor $= \frac{6}{3.873} \approx 1.549$ .
AC part: $V_{ac,rms} = V_{rms}^{2} - V_{avg}^{2} = 15 - 9 = 6 \approx 2.449$ V.

Answer: $V_{avg} = 3$ V, $V_{rms} = 15 \approx 3.87$ V, form factor $\approx 1.29$ , crest factor $\approx 1.55$ , AC RMS $= 6 \approx 2.45$ V.

EXAM TIP

RMS values do not add. The RMS of a sum is $\sum_{k} V_{k, rms}^{2}$ when the components are mutually orthogonal, for example DC plus sinusoids of distinct frequencies. A half-wave rectified sine keeps half of the full-wave's $\overline{v^{2}}$ , so its RMS is $V_{m} /2$ , not $V_{m} / (22)$ ; its average is half the full-wave's, $V_{m} / π$ . Shifting or time-scaling a periodic waveform, or changing the rise/fall split of a triangle, leaves both its average and its RMS unchanged.

PREVIOUS-YEAR QUESTIONS

18 questions from the EC, EE and IN papers

Try each question first, then open the answer. One shows the official answer only. We placed each question under a topic ourselves; the exam paper does not label topics.

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GATE 2026 EC Q.43

GATE 2026 · ECQ.43 · 2 marks · MCQ

Consider a real signal $x (t)$ , $- \infty < t < \infty$ , such that

$x (t) = 0$ for $t < 0$ , $x (t) = 2$ for $0 \leq t < 1$ and $x (t) = 0$ for $t \geq 1$ .

Let $E [x (t)] = \int_{- \infty}^{\infty} [x (t)]^{2} d t$ .

Which of the following options correctly represents the ratio,

$E [x (t)] / E [3 x (- 3 t + 5)]$ ?

(A)
$3$
(B)
$1$
(C)
$1/3$
(D)
$1/9$

Answer and solution

GATE 2026 EC Q.43: answer and solution

See the question in the paper

Signals

Classifying signals

Standard signals and the unit impulse

Time shift, reversal and scaling (CT and DT)

Periodicity of sums of sinusoids (CT and DT)

Energy and power

Average and RMS values of periodic waveforms

18 questions from the EC, EE and IN papers

5 questions to test yourself