GATE EC · EE · IN

Fourier transform, frequency response, group and phase delay

The Fourier transform describes a signal by how much of each frequency it contains, and the frequency response $H (j ω)$ describes how an LTI system scales and phase-shifts each of those frequencies. With a short list of standard pairs and the properties table (shift, scaling, duality, modulation, Parseval) you can find transforms, energies and filter outputs without integrating. Reading the magnitude and phase of $H (j ω)$ then tells you whether a system distorts a signal, and by how much it delays the carrier (phase delay) and the envelope (group delay) of a narrowband signal. The same ideas carry over to sequences through the discrete-time Fourier transform (DTFT).

22 previous-year questions · 22 solved · 6 practice questions

1. Definition, convention and existence

The Fourier transform (FT) resolves a signal $x (t)$ into complex exponentials $e^{j ω t}$ . $X (ω)$ is complex: $∣ X (ω) ∣$ is the magnitude spectrum and $∠ X (ω)$ the phase spectrum. Convention used in these notes: angular frequency $ω$ in rad/s, with the factor $1/2 π$ in the inverse transform. You will also see the same function written $X (j ω)$ ; here $X (ω)$ and $X (j ω)$ mean the same thing. The Hz form, in $f$ , is given alongside every result because questions use both.

FOURIER TRANSFORM PAIR (Ω FORM, USED IN THESE NOTES)

X (ω) = \int_{- \infty}^{\infty} x (t) e^{- j ω t} d t x (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} X (ω) e^{j ω t} d ω

$ω$ in rad/s. The $1/2 π$ sits only in the inverse, so it reappears in duality, the multiplication property, Parseval's theorem and the formula for $x (0)$ .

FOURIER TRANSFORM PAIR (F FORM)

X (f) = \int_{- \infty}^{\infty} x (t) e^{- j 2 π f t} d t x (t) = \int_{- \infty}^{\infty} X (f) e^{j 2 π f t} df

$f$ in Hz; symmetric, with no factor outside the integrals. Convert with $X (f) = X (ω)_{ω = 2 π f}$ , and for impulses use $δ (ω) = δ (f) /2 π$ , so $2 π δ (ω)$ becomes $δ (f)$ .

SINC CONVENTION

sinc (x) = \frac{sin ( π x )}{π x}, sinc (0) = 1

Normalised sinc, with zeros at $x = \pm 1, \pm 2, \dots$ . Some texts use $sin x / x$ under the same name; every sinc result below is also written in explicit $sin$ form, so nothing depends on the convention.

Existence. The Dirichlet conditions are sufficient (not necessary) for $X (ω)$ to exist as an ordinary function:

$x (t)$ is absolutely integrable: $\int_{- \infty}^{\infty} ∣ x (t) ∣ d t < \infty$ .
$x (t)$ has a finite number of maxima and minima in any finite interval.
$x (t)$ has a finite number of finite discontinuities in any finite interval. At a jump, the inverse transform converges to the midpoint of the jump.
A finite-energy signal ( $\int ∣ x ∣^{2} d t < \infty$ ) also has a transform, converging in the mean-square sense. $sin t / t$ is square-integrable but not absolutely integrable, and its transform is a rectangle.
Signals such as $1$ , $u (t)$ , $sgn (t)$ , $cos ω_{0} t$ and every periodic signal have a transform only in the generalised sense, with impulses in $X (ω)$ . Growing signals such as $e^{2 t} u (t)$ have no Fourier transform at all.

LINK WITH THE LAPLACE TRANSFORM

X (ω) = X (s)_{s = j ω} only if the ROC of X (s) contains the j ω axis

$X (s)$ is the bilateral Laplace transform. Example: $e^{- a t} u (t)$ , $a > 0$ , has ROC $Re (s) > - a$ , which contains the axis. If the ROC boundary is the $j ω$ axis (poles on the axis, as for $u (t)$ with ROC $Re (s) > 0$ ), the Fourier transform exists only with impulses and the substitution misses them. If the ROC excludes the axis altogether (as for $e^{2 t} u (t)$ , ROC $Re (s) > 2$ ), there is no Fourier transform.

EXAM TIP

Trap: $L {u (t)} = 1/ s$ , but $F {u (t)} = π δ (ω) + \frac{1}{j ω}$ , not $\frac{1}{j ω}$ . The impulse carries the average value $1/2$ of $u (t)$ . In the same way, $F {cos (ω_{0} t) u (t)} = \frac{π}{2} [δ (ω - ω_{0}) + δ (ω + ω_{0})] + \frac{j ω}{ω _{0}^{2} - ω ^{2}}$ ; putting $s = j ω$ in $s / (s^{2} + ω_{0}^{2})$ gives only the second term.

2. Standard transform pairs

Two pulse shapes recur. $rect (t / T)$ is the unit-height pulse of width $T$ centred at $t = 0$ : it is 1 for $∣ t ∣ < T /2$ and 0 for $∣ t ∣ > T /2$ . $tri (t / T)$ is the unit-height triangle with base from $- T$ to $T$ : it equals $1 - ∣ t ∣/ T$ for $∣ t ∣ \leq T$ and 0 otherwise. $tri (t / T)$ is $\frac{1}{T}$ times $rect (t / T)$ convolved with itself, which is why its transform is a sinc squared.

$x (t)$	$X (ω)$ (rad/s)	$X (f)$ (Hz)	Condition / note
$δ (t)$	$1$	$1$	flat spectrum
$δ (t - t_{0})$	$e^{- j ω t_{0}}$	$e^{- j 2 π f t_{0}}$	unit magnitude, linear phase
$1$	$2 π δ (ω)$	$δ (f)$	dual of $δ (t)$
$e^{j ω_{0} t}$	$2 π δ (ω - ω_{0})$	$δ (f - f_{0})$	$f_{0} = ω_{0} /2 π$
$cos ω_{0} t$	$π [δ (ω - ω_{0}) + δ (ω + ω_{0})]$	$\frac{1}{2} [δ (f - f_{0}) + δ (f + f_{0})]$	real and even
$sin ω_{0} t$	$- j π [δ (ω - ω_{0}) - δ (ω + ω_{0})]$	$\frac{1}{2 j} [δ (f - f_{0}) - δ (f + f_{0})]$	imaginary and odd
$sgn (t)$	$\frac{2}{j ω}$	$\frac{1}{j π f}$	no impulse: $sgn$ has zero average
$u (t)$	$π δ (ω) + \frac{1}{j ω}$	$\frac{1}{2} δ (f) + \frac{1}{j 2 π f}$	$u = \frac{1}{2} + \frac{1}{2} sgn$
$e^{- a t} u (t)$	$\frac{1}{a + j ω}$	$\frac{1}{a + j 2 π f}$	$a > 0$
$t e^{- a t} u (t)$	$\frac{1}{( a + j ω ) ^{2}}$	$\frac{1}{( a + j 2 π f ) ^{2}}$	$a > 0$ ; generally $\frac{t ^{n - 1} e ^{- a t}}{( n - 1 )!} u (t) \leftrightarrow \frac{1}{( a + j ω ) ^{n}}$
$e^{- a t} cos (ω_{0} t) u (t)$	$\frac{a + j ω}{( a + j ω ) ^{2} + ω _{0}^{2}}$	$\frac{a + j 2 π f}{( a + j 2 π f ) ^{2} + ω _{0}^{2}}$	$a > 0$ ; $ω_{0}$ in rad/s
$e^{- a t} sin (ω_{0} t) u (t)$	$\frac{ω _{0}}{( a + j ω ) ^{2} + ω _{0}^{2}}$	$\frac{ω _{0}}{( a + j 2 π f ) ^{2} + ω _{0}^{2}}$	$a > 0$ ; $ω_{0}$ in rad/s
$e^{- a ∣ t ∣}$	$\frac{2 a}{a ^{2} + ω ^{2}}$	$\frac{2 a}{a ^{2} + 4 π ^{2} f ^{2}}$	$a > 0$ ; real, even, positive
$e^{- a t^{2}}$	$π / a e^{- ω^{2} /4 a}$	$π / a e^{- π^{2} f^{2} / a}$	$a > 0$ ; a Gaussian maps to a Gaussian; $e^{- π t^{2}} \leftrightarrow e^{- π f^{2}}$
$rect (t / T)$	$\frac{2 sin ( ω T /2 )}{ω} = T sinc (\frac{ω T}{2 π})$	$T sinc (f T)$	zeros at $ω = 2 π k / T$ ( $f = k / T$ ), $k \neq = 0$
$\frac{sin W t}{π t}$	$1$ for $∣ ω ∣ < W$ , else $0$	$1$ for $∣ f ∣ < B$ , else $0$ ; $B = W /2 π$	ideal LPF impulse response; equals $2 B sinc (2 B t)$
$tri (t / T)$	$T [\frac{sin ( ω T /2 )}{ω T /2}]^{2}$	$T sinc^{2} (f T)$	zeros at $ω = 2 π k / T$ , $k \neq = 0$
$(\frac{sin W t}{π t})^{2}$	$\frac{1}{2 π} (2 W - ∣ ω ∣)$ for $∣ ω ∣ < 2 W$ , else $0$	$2 B - ∣ f ∣$ for $∣ f ∣ < 2 B$ , else $0$ ; $B = W /2 π$	a triangle (dual of the $tri$ pair); total area of $x (t)$ is $X (0) = W / π$
$\sum_{n} δ (t - n T)$	$ω_{s} \sum_{k} δ (ω - k ω_{s})$	$\frac{1}{T} \sum_{k} δ (f - \frac{k}{T})$	$ω_{s} = 2 π / T$ ; an impulse train maps to an impulse train
$\frac{1}{π t}$	$- j sgn (ω)$	$- j sgn (f)$	Hilbert transformer (dual of the $sgn$ pair)

RECTANGULAR PULSE AND SINC

rect (\frac{t}{T}) \leftrightarrow \frac{2 sin ( ω T /2 )}{ω} = T sinc (\frac{ω T}{2 π}) = T sinc (f T)

Pulse of width $T$ and height 1. $X (0) = T$ , the pulse area. First nulls at $ω = \pm 2 π / T$ , i.e. $f = \pm 1/ T$ : a narrower pulse has a wider main lobe.

IDEAL LOWPASS SPECTRUM AND ITS TIME FUNCTION

\frac{sin W t}{π t} = \frac{W}{π} sinc (\frac{W t}{π}) \leftrightarrow {1, 0, ∣ ω ∣ < W ∣ ω ∣ > W

The dual of the rect–sinc pair. Peak value $W / π$ at $t = 0$ , zeros at $t = k π / W$ ( $k \neq = 0$ ). This is $h (t)$ of the ideal lowpass filter with cutoff $W$ rad/s.

ONE-SIDED AND TWO-SIDED EXPONENTIALS

e^{- a t} u (t) \leftrightarrow \frac{1}{a + j ω}, e^{- a ∣ t ∣} \leftrightarrow \frac{2 a}{a ^{2} + ω ^{2}} (a > 0)

$e^{- a ∣ t ∣} = e^{- a t} u (t) + e^{a t} u (- t)$ , so its transform is $\frac{1}{a + j ω} + \frac{1}{a - j ω}$ . For $a < 0$ both signals grow and have no Fourier transform; $a = 0$ gives $u (t)$ or $1$ , which need impulses.

UNIT STEP AND SIGNUM

u (t) \leftrightarrow π δ (ω) + \frac{1}{j ω}, sgn (t) \leftrightarrow \frac{2}{j ω}

The impulse in $F {u (t)}$ is the transform of its average value: $\frac{1}{2} \cdot 2 π δ (ω) = π δ (ω)$ . $sgn (t)$ has zero average, so it has no impulse.

GAUSSIAN

e^{- a t^{2}} \leftrightarrow \frac{π}{a} e^{- ω^{2} / (4 a)} (a > 0)

A narrow Gaussian in time (large $a$ ) gives a wide Gaussian in frequency. In the $f$ form, $e^{- π t^{2}} \leftrightarrow e^{- π f^{2}}$ exactly.

PERIODIC IMPULSE TRAIN

n = - \infty \sum \infty δ (t - n T) \leftrightarrow ω_{s} k = - \infty \sum \infty δ (ω - k ω_{s}), ω_{s} = \frac{2 π}{T}

Every Fourier-series coefficient of the impulse train equals $1/ T$ , which gives impulses of area $2 π / T$ . This pair is the basis of sampling theory.

EXAM TIP

Width and bandwidth move in opposite directions. Halving the pulse width from $T$ to $T /2$ moves the first null from $2 π / T$ to $4 π / T$ rad/s and halves $X (0)$ . This is the scaling property $x (a t) \leftrightarrow \frac{1}{∣ a ∣} X (ω / a)$ at work.

3. Properties of the Fourier transform

Take $x (t) \leftrightarrow X (ω)$ and $y (t) \leftrightarrow Y (ω)$ . The table gives each property in both conventions. The $2 π$ factors are where the two forms differ, and they are where most factor slips happen.

Property	Time domain	$ω$ form	$f$ form
Linearity	$a x (t) + b y (t)$	$a X (ω) + bY (ω)$	$a X (f) + bY (f)$
Time shift	$x (t - t_{0})$	$e^{- j ω t_{0}} X (ω)$	$e^{- j 2 π f t_{0}} X (f)$
Frequency shift	$e^{j ω_{0} t} x (t)$	$X (ω - ω_{0})$	$X (f - f_{0})$
Modulation	$x (t) cos ω_{0} t$	$\frac{1}{2} [X (ω - ω_{0}) + X (ω + ω_{0})]$	$\frac{1}{2} [X (f - f_{0}) + X (f + f_{0})]$
Time scaling	$x (a t)$ , real $a \neq = 0$	$\frac{1}{∣ a ∣} X (\frac{ω}{a})$	$\frac{1}{∣ a ∣} X (\frac{f}{a})$
Time reversal	$x (- t)$	$X (- ω)$	$X (- f)$
Duality	$X (t)$	$2 π x (- ω)$	$x (- f)$
Conjugation	$x^{*} (t)$	$X^{*} (- ω)$	$X^{*} (- f)$
Differentiation in time	$\frac{d ^{n} x}{d t ^{n}}$	$(j ω)^{n} X (ω)$	$(j 2 π f)^{n} X (f)$
Differentiation in frequency	$t x (t)$	$j \frac{d X ( ω )}{d ω}$	$\frac{j}{2 π} \frac{d X ( f )}{df}$
Integration	$\int_{- \infty}^{t} x (τ) d τ$	$\frac{X ( ω )}{j ω} + π X (0) δ (ω)$	$\frac{X ( f )}{j 2 π f} + \frac{1}{2} X (0) δ (f)$
Convolution	$x (t) * y (t)$	$X (ω) Y (ω)$	$X (f) Y (f)$
Multiplication	$x (t) y (t)$	$\frac{1}{2 π} X (ω) * Y (ω)$	$X (f) * Y (f)$

TIME SCALING (WITH SHIFT)

x (a t) \leftrightarrow \frac{1}{∣ a ∣} X (\frac{ω}{a}), x (a t - b) \leftrightarrow \frac{1}{∣ a ∣} e^{- j ω b / a} X (\frac{ω}{a})

Real $a \neq = 0$ . Compressing in time ( $∣ a ∣ > 1$ ) stretches the spectrum and lowers it. Since $x (a t - b) = x (a (t - b / a))$ , the delay is $b / a$ , not $b$ .

DUALITY

if x (t) \leftrightarrow X (ω) then X (t) \leftrightarrow 2 π x (- ω)

In the $f$ form there is no $2 π$ : $X (t) \leftrightarrow x (- f)$ . For an even $x$ , $x (- ω) = x (ω)$ .

MODULATION

x (t) cos ω_{0} t \leftrightarrow \frac{1}{2} [X (ω - ω_{0}) + X (ω + ω_{0})]

Each shifted copy has half the height. With $sin ω_{0} t$ instead: $\frac{1}{2 j} [X (ω - ω_{0}) - X (ω + ω_{0})]$ . If the copies do not overlap, the energy is halved.

DIFFERENTIATION IN TIME AND IN FREQUENCY

\frac{d x}{d t} \leftrightarrow j ω X (ω), t x (t) \leftrightarrow j \frac{d X ( ω )}{d ω}

Differentiation removes any DC impulse: $\frac{d}{d t} u (t) = δ (t)$ , and $j ω [π δ (ω) + \frac{1}{j ω}] = 1$ because $ω δ (ω) = 0$ .

INTEGRATION

\int_{- \infty}^{t} x (τ) d τ \leftrightarrow \frac{X ( ω )}{j ω} + π X (0) δ (ω)

The impulse term vanishes only when $X (0) = \int x d t = 0$ . Dividing by $j ω$ alone is the classic error.

CONVOLUTION AND MULTIPLICATION

x (t) * y (t) \leftrightarrow X (ω) Y (ω), x (t) y (t) \leftrightarrow \frac{1}{2 π} X (ω) * Y (ω)

The $1/2 π$ belongs to the multiplication property in the $ω$ form only. The convolution property is why an LTI output is $Y (ω) = H (ω) X (ω)$ .

AREA AND VALUE THEOREMS

X (0) = \int_{- \infty}^{\infty} x (t) d t, x (0) = \frac{1}{2 π} \int_{- \infty}^{\infty} X (ω) d ω

Put $ω = 0$ in the forward integral, or $t = 0$ in the inverse. In the $f$ form, $x (0) = \int X (f) df$ . If $x$ jumps at $t = 0$ , the second integral gives the midpoint of the jump.

Symmetry rules for a real signal $x (t)$ :

$X (- ω) = X^{*} (ω)$ (conjugate symmetry), so the negative-frequency half of the spectrum carries no new information.
$∣ X (ω) ∣$ and $Re X (ω)$ are even; $∠ X (ω)$ and $Im X (ω)$ are odd.
Real and even $x (t)$ gives a real and even $X (ω)$ . Real and odd $x (t)$ gives a purely imaginary and odd $X (ω)$ .
Even part $x_{e} (t) \leftrightarrow Re X (ω)$ ; odd part $x_{o} (t) \leftrightarrow j Im X (ω)$ .

EXAM TIP

Use duality instead of integrating. From $e^{- ∣ t ∣} \leftrightarrow \frac{2}{1 + ω ^{2}}$ , duality gives $\frac{2}{1 + t ^{2}} \leftrightarrow 2 π e^{- ∣ ω ∣}$ , so $\frac{1}{1 + t ^{2}} \leftrightarrow π e^{- ∣ ω ∣}$ . Then $\int_{- \infty}^{\infty} \frac{d t}{1 + t ^{2}} = X (0) = π$ with no calculus.

WORKED EXAMPLE

Let $x (t) = \frac{sin 2 t}{π t} cos 10 t$ . (a) Find $X (ω)$ . (b) Find the energy of $x (t)$ . (c) Find $\int_{- \infty}^{\infty} x (t) d t$ , and check $x (0)$ from the spectrum. (d) $x (t)$ is applied to an ideal highpass filter of unit gain with cutoff 10 rad/s. Find the output energy.

Solution

From the pair $\frac{sin W t}{π t} \leftrightarrow 1$ for $∣ ω ∣ < W$ (0 otherwise) with $W = 2$ , the factor $\frac{sin 2 t}{π t}$ has a flat spectrum of height 1 on $∣ ω ∣ < 2$ .
Modulation by $cos 10 t$ gives two copies of half height centred at $\pm 10$ rad/s: $X (ω) = \frac{1}{2}$ for $8 < ∣ ω ∣ < 12$ and $0$ elsewhere. It is real and even, as it must be for a real, even $x (t)$ .
Energy by Parseval: $E = \frac{1}{2 π} \int ∣ X (ω) ∣^{2} d ω = \frac{1}{2 π} (2 bands \times 4 rad/s \times \frac{1}{4}) = \frac{1}{2 π} \times 2 = \frac{1}{π} \approx 0.318$ .
Area: $\int x d t = X (0) = 0$ , since $ω = 0$ lies outside both bands.
Check: $x (0) = \frac{1}{2 π} \int X d ω = \frac{1}{2 π} (2 \times 4 \times \frac{1}{2}) = \frac{2}{π}$ . Directly, $\frac{sin 2 t}{π t} \to \frac{2}{π}$ as $t \to 0$ and $cos 0 = 1$ . The two agree.
The highpass filter keeps $10 < ∣ ω ∣ < 12$ , the upper half of each band. Output energy $= \frac{1}{2 π} (2 \times 2 \times \frac{1}{4}) = \frac{1}{2 π} \approx 0.159$ , half of the input energy.

Answer: $X (ω) = \frac{1}{2}$ for $8 < ∣ ω ∣ < 12$ rad/s, else 0; $E = 1/ π \approx 0.318$ ; $\int x d t = 0$ and $x (0) = 2/ π$ ; output energy $1/ (2 π) \approx 0.159$ .

4. Parseval's theorem, energy spectral density and periodic signals

Energy can be computed in either domain. For the Fourier transform this is Parseval's theorem, also called Rayleigh's energy theorem. It turns awkward time integrals such as $\int (sin t / t)^{2} d t$ into areas of rectangles in frequency, and it gives the energy passed by a filter directly from $∣ H ∣^{2} ∣ X ∣^{2}$ .

PARSEVAL / RAYLEIGH ENERGY THEOREM

E = \int_{- \infty}^{\infty} ∣ x (t) ∣^{2} d t = \frac{1}{2 π} \int_{- \infty}^{\infty} ∣ X (ω) ∣^{2} d ω = \int_{- \infty}^{\infty} ∣ X (f) ∣^{2} df

For energy signals. Generalised form: $\int x (t) y^{*} (t) d t = \frac{1}{2 π} \int X (ω) Y^{*} (ω) d ω$ .

ENERGY SPECTRAL DENSITY (ESD) AND ENERGY IN A BAND

Ψ_{x} (ω) = ∣ X (ω) ∣^{2}, E_{ω_{1} \leq ∣ ω ∣ \leq ω_{2}} = \frac{1}{2 π} \cdot 2 \int_{ω_{1}}^{ω_{2}} ∣ X (ω) ∣^{2} d ω

The band formula is for real $x (t)$ . The factor 2 counts the matching negative-frequency band, which carries the same energy because $∣ X ∣$ is even.

ESD THROUGH AN LTI SYSTEM

Ψ_{y} (ω) = ∣ H (ω) ∣^{2} Ψ_{x} (ω), E_{y} = \frac{1}{2 π} \int_{- \infty}^{\infty} ∣ H (ω) ∣^{2} ∣ X (ω) ∣^{2} d ω

The phase of $H$ has no effect on output energy.

ENERGY AUTOCORRELATION AND ESD

R_{x} (τ) = \int_{- \infty}^{\infty} x (t) x^{*} (t - τ) d t \leftrightarrow ∣ X (ω) ∣^{2}, R_{x} (0) = E

$R_{x} (τ) = x (τ) * x^{*} (- τ)$ . The ESD is the transform of the autocorrelation and carries no phase information, so different signals can share one ESD.

EXAM TIP

Two integrals follow in one line each from the pair $\frac{sin t}{π t} \leftrightarrow 1$ for $∣ ω ∣ < 1$ . Area theorem: $\int_{- \infty}^{\infty} \frac{sin t}{t} d t = π X (0) = π$ . Parseval: $\int_{- \infty}^{\infty} \frac{sin ^{2} t}{t ^{2}} d t = π^{2} \cdot \frac{1}{2 π} \int_{- 1}^{1} d ω = π$ .

A periodic signal has infinite energy, so its ordinary Fourier integral diverges. In the generalised sense its transform is a train of impulses at the harmonics $k ω_{0}$ , with areas $2 π$ times the Fourier-series coefficients $c_{k}$ .

FOURIER TRANSFORM OF A PERIODIC SIGNAL

x (t) = k = - \infty \sum \infty c_{k} e^{j k ω_{0} t} \leftrightarrow X (ω) = 2 π k = - \infty \sum \infty c_{k} δ (ω - k ω_{0})

$ω_{0} = 2 π / T_{0}$ . In the $f$ form, $X (f) = \sum_{k} c_{k} δ (f - k f_{0})$ with $f_{0} = 1/ T_{0}$ .

FOURIER-SERIES COEFFICIENTS FROM THE TRANSFORM OF ONE PERIOD

c_{k} = \frac{1}{T _{0}} X_{1} (ω)_{ω = k ω_{0}}

$X_{1} (ω)$ is the transform of one period of $x (t)$ (taken as zero outside that period). The coefficients are samples of $X_{1} (ω) / T_{0}$ .

AVERAGE POWER OF A PERIODIC SIGNAL (PARSEVAL FOR FOURIER SERIES)

P = \frac{1}{T _{0}} \int_{T_{0}} ∣ x (t) ∣^{2} d t = k = - \infty \sum \infty ∣ c_{k} ∣^{2}

Periodic signals are power signals: infinite energy, finite power. A periodic input to a stable LTI system gives $y (t) = \sum_{k} c_{k} H (k ω_{0}) e^{j k ω_{0} t}$ .

EXAM TIP

A rectangular pulse train of height 1, pulse width $τ$ and period $T_{0}$ has $X_{1} (ω) = \frac{2 sin ( ω τ /2 )}{ω}$ , so $c_{k} = \frac{τ}{T _{0}} sinc (\frac{k τ}{T _{0}})$ . With duty cycle $τ / T_{0} = 1/4$ , the harmonics $k = \pm 4, \pm 8, \dots$ are absent because they fall on the sinc nulls.

5. Frequency response, distortionless transmission, phase and group delay

For a stable LTI system with impulse response $h (t)$ , the input $e^{j ω t}$ produces the output $H (ω) e^{j ω t}$ : complex exponentials are eigenfunctions, and $H (ω)$ is the eigenvalue. $H (ω) = F {h (t)}$ is the frequency response, usually written $H (j ω)$ because it equals the transfer function $H (s)$ at $s = j ω$ . Write $H (ω) = ∣ H (ω) ∣ e^{j θ (ω)}$ : $∣ H ∣$ is the gain and $θ (ω) = ∠ H (ω)$ the phase shift at each frequency. For a real $h (t)$ , $∣ H ∣$ is even and $θ$ is odd.

FREQUENCY RESPONSE

H (j ω) = \int_{- \infty}^{\infty} h (t) e^{- j ω t} d t = H (s)_{s = j ω}, Y (ω) = H (ω) X (ω)

Putting $s = j ω$ is valid when the system is BIBO stable ( $\int ∣ h ∣ d t < \infty$ ), so the ROC contains the $j ω$ axis. For a stable system described by $\sum a_{k} y^{(k)} = \sum b_{k} x^{(k)}$ , $H (j ω) = \frac{\sum b _{k} ( j ω ) ^{k}}{\sum a _{k} ( j ω ) ^{k}}$ .

SINUSOIDAL STEADY STATE

A cos (ω_{0} t + ϕ) ⟶ A ∣ H (j ω_{0}) ∣ cos (ω_{0} t + ϕ + θ (ω_{0}))

Real $h (t)$ , stable system, input applied for all time. The frequency is unchanged; only amplitude and phase change. Superpose for several tones.

Distortionless transmission. The output is distortionless if it is a scaled, delayed copy of the input, $y (t) = K x (t - t_{d})$ . Taking transforms, $Y (ω) = K e^{- j ω t_{d}} X (ω)$ . So over the band the signal occupies, the gain must be constant and the phase must be a straight line through the origin.

DISTORTIONLESS SYSTEM

H (j ω) = K e^{- j ω t_{d}} : ∣ H (j ω) ∣ = K, θ (ω) = - ω t_{d}

Here $K > 0$ . The conditions are needed only across the band the signal occupies. Adding a multiple of $2 π$ to the phase changes nothing, and a phase of $- ω t_{d} \pm π$ gives $y (t) = - K x (t - t_{d})$ , an inverted but undistorted copy. Any other constant phase offset changes the waveform (for a narrowband bandpass signal it shifts only the carrier phase; see below).

Failing either condition distorts the waveform:

Amplitude (magnitude) distortion: $∣ H ∣$ is not constant over the band, so components are scaled unequally. An RC lowpass filter rounds the edges of a pulse whose spectrum extends well past $1/ R C$ rad/s.
Phase (delay) distortion: $θ (ω)$ is not of the form $- ω t_{d}$ (plus a multiple of $π$ ), so components are delayed by different amounts. With a flat gain $K$ , $∣ Y ∣ = K ∣ X ∣$ , so the energy changes only by the factor $K^{2}$ , yet the waveform shape changes.
Both are linear distortions: no new frequencies appear. A nonlinear system (such as $y = x^{2}$ ) creates new frequencies, which no $H (ω)$ can describe.

PHASE DELAY AND GROUP DELAY

τ_{p} (ω) = - \frac{θ ( ω )}{ω}, τ_{g} (ω) = - \frac{d θ ( ω )}{d ω}

$θ$ in radians and $ω$ in rad/s give seconds. Use the unwrapped phase for $τ_{p}$ . For $θ = - ω t_{d}$ , $τ_{p} = τ_{g} = t_{d}$ at every frequency; conversely, $τ_{p} = τ_{g}$ at every frequency only when $θ = - ω t_{d}$ for some constant $t_{d}$ . In terms of $f$ : $τ_{p} = - \frac{θ}{2 π f}$ and $τ_{g} = - \frac{1}{2 π} \frac{d θ}{df}$ .

What each delay means. Pass a narrowband signal $x (t) = m (t) cos ω_{c} t$ through a system with real $h (t)$ whose gain is nearly constant over $ω_{c} \pm W$ , where the envelope $m (t)$ occupies $∣ ω ∣ < W$ and $W ≪ ω_{c}$ . Expand the phase to first order about the carrier: $θ (ω) \approx θ (ω_{c}) + (ω - ω_{c}) θ^{'} (ω_{c})$ for $ω > 0$ . Then:

NARROWBAND OUTPUT: ENVELOPE AND CARRIER DELAYS

y (t) \approx ∣ H (j ω_{c}) ∣ m (t - τ_{g}) cos (ω_{c} (t - τ_{p})), τ_{g} = τ_{g} (ω_{c}), τ_{p} = τ_{p} (ω_{c})

The envelope (the information) is delayed by the group delay, also called envelope delay. The carrier is delayed by the phase delay. If $τ_{g}$ varies across the band, the envelope is dispersed.

$H (j ω)$	$θ (ω)$ for $ω > 0$	$τ_{p} (ω)$	$τ_{g} (ω)$
$e^{- j ω t_{d}}$ (pure delay)	$- ω t_{d}$	$t_{d}$	$t_{d}$
linear phase with intercept, $∣ H ∣ = 1$	$- ω t_{d} - ϕ_{0}$	$t_{d} + ϕ_{0} / ω$	$t_{d}$
$\frac{1}{1 + j ω τ}$ (RC lowpass, $τ = R C$ )	$- tan^{- 1} ω τ$	$\frac{tan ^{- 1} ω τ}{ω}$	$\frac{τ}{1 + ω ^{2} τ ^{2}}$
$\frac{j ω τ}{1 + j ω τ}$ (RC highpass)	$\frac{π}{2} - tan^{- 1} ω τ$	$\frac{tan ^{- 1} ω τ - π /2}{ω}$ (negative)	$\frac{τ}{1 + ω ^{2} τ ^{2}}$
$\frac{a - j ω}{a + j ω}$ (first-order all-pass, $a > 0$ )	$- 2 tan^{- 1} (ω / a)$	$\frac{2 tan ^{- 1} ( ω / a )}{ω}$	$\frac{2 a}{a ^{2} + ω ^{2}}$
cascade $H_{1} H_{2}$	$θ_{1} + θ_{2}$	$τ_{p 1} + τ_{p 2}$	$τ_{g 1} + τ_{g 2}$

EXAM TIP

Group delays add in a cascade, so a rational $H (s)$ with real poles and zeros can be read factor by factor (each $a > 0$ ): a pole at $s = - a$ adds $\frac{a}{a ^{2} + ω ^{2}}$ ; a zero at $s = - a$ subtracts $\frac{a}{a ^{2} + ω ^{2}}$ ; a zero at $s = + a$ adds $\frac{a}{a ^{2} + ω ^{2}}$ ; a delay $e^{- s T}$ adds $T$ ; a constant gain adds nothing. Example: $\frac{1 - s}{1 + s} e^{- 2 s}$ has $τ_{g} = \frac{2}{1 + ω ^{2}} + 2$ , which is 4 s at $ω = 0$ .

WORKED EXAMPLE

The input $x (t) = cos 9 t + cos 11 t$ is applied to the all-pass system $H (j ω) = \frac{10 - j ω}{10 + j ω}$ . Find the exact output, write it as envelope times carrier, and compare the envelope and carrier delays with $τ_{g}$ and $τ_{p}$ at $ω_{c} = 10$ rad/s.

Solution

$∣ H (j ω) ∣ = 1$ at every frequency and $θ (ω) = - 2 tan^{- 1} (ω /10)$ . So $θ (9) = - 2 tan^{- 1} 0.9 = - 1.4656$ rad and $θ (11) = - 2 tan^{- 1} 1.1 = - 1.6660$ rad.
Each tone keeps its amplitude and is phase-shifted: $y (t) = cos (9 t - 1.4656) + cos (11 t - 1.6660)$ .
With $cos A + cos B = 2 cos \frac{A + B}{2} cos \frac{A - B}{2}$ : $y (t) = 2 cos (t - 0.1002) cos (10 t - 1.5658)$ .
The input is $x (t) = 2 cos t cos 10 t$ . So the envelope $cos t$ is delayed by $0.1002/1 = 0.1002$ s and the carrier $cos 10 t$ by $1.5658/10 = 0.1566$ s.
At $ω_{c} = 10$ : $τ_{g} = \frac{2 a}{a ^{2} + ω ^{2}} = \frac{20}{200} = 0.1$ s and $τ_{p} = \frac{2 tan ^{- 1} 1}{10} = \frac{π}{20} \approx 0.1571$ s. These match the exact envelope and carrier delays to within the first-order approximation.
The gain is flat, so there is no amplitude distortion. But $τ_{p} \neq = τ_{g}$ because the phase is not linear, so a wideband signal through this system would suffer phase distortion.

Answer: $y (t) = 2 cos (t - 0.1002) cos (10 t - 1.5658)$ . Envelope delay $0.1002$ s, close to $τ_{g} (10) = 0.1$ s; carrier delay $0.1566$ s, close to $τ_{p} (10) = π /20 \approx 0.157$ s.

Discrete time. The same ideas hold for a stable DT LTI system with frequency response $H (e^{j Ω}) = \sum_{n} h [n] e^{- j Ω n}$ (the DTFT of $h [n]$ ; see section 7), where $Ω$ is the normalised frequency in rad/sample and $H$ is periodic in $Ω$ with period $2 π$ . Distortionless now means $H (e^{j Ω}) = K e^{- j Ω n_{0}}$ with integer $n_{0}$ , which gives $y [n] = K x [n - n_{0}]$ . Phase delay $- θ (Ω) /Ω$ and group delay $- d θ / d Ω$ come out in samples.

DISCRETE-TIME GROUP DELAY; LINEAR-PHASE FIR

τ_{g} (Ω) = - \frac{d θ ( Ω )}{d Ω} samples; h [n] = h [N - 1 - n] \Rightarrow τ_{g} = \frac{N - 1}{2}

A length- $N$ FIR filter with symmetric coefficients has $H (e^{j Ω}) = e^{- j Ω (N - 1) /2} A (Ω)$ with $A$ real, so its phase is linear (apart from jumps of $π$ where $A$ changes sign). Antisymmetric coefficients, $h [n] = - h [N - 1 - n]$ , give $H (e^{j Ω}) = j e^{- j Ω (N - 1) /2} A (Ω)$ : the same group delay $(N - 1) /2$ with a constant $π /2$ phase offset. Example: $h [n] = {1, 2, 1}$ for $n = 0, 1, 2$ gives $e^{- j Ω} (2 + 2 cos Ω)$ , a delay of 1 sample.

EXAM TIP

Unwrap the phase before computing $τ_{p}$ . For $H (j ω) = e^{- j 5 ω}$ at $ω = 1$ rad/s, a calculator returns the principal angle $- 5 + 2 π = 1.283$ rad, which would give a negative phase delay; the true phase is $- 5$ rad and $τ_{p} = 5$ s. Group delay is unaffected by $2 π$ jumps because it uses the slope.

EXAM TIP

If the phase is given in degrees, convert to radians before dividing by $ω$ or differentiating. A phase of $- 9 0^{\circ}$ at $ω = 100$ rad/s is a phase delay of $(π /2) /100 \approx 15.7$ ms, not $0.9$ s.

EXAM TIP

Phase delay can be negative, as for the RC highpass (a phase lead), while the group delay is positive. A negative $τ_{p}$ describes only the steady-state phase of a sinusoid; it does not mean the output appears before the input.

6. Ideal filters, bandwidth and the Hilbert transform

An ideal filter has unit gain in its passband, zero gain in its stopband, and (with delay $t_{d}$ ) linear phase $- ω t_{d}$ in the passband. The impulse responses below are for $t_{d} = 0$ ; a delay replaces $t$ by $t - t_{d}$ . For the band filters, $B = ω_{2} - ω_{1}$ and $ω_{0} = (ω_{1} + ω_{2}) /2$ .

Filter	$H (j ω)$	$h (t)$
Lowpass, cutoff $ω_{c}$	1 for $∣ ω ∣ < ω_{c}$ , else 0	$\frac{sin ω _{c} t}{π t}$
Highpass, cutoff $ω_{c}$	0 for $∣ ω ∣ < ω_{c}$ , else 1	$δ (t) - \frac{sin ω _{c} t}{π t}$
Bandpass, $ω_{1} < ∣ ω ∣ < ω_{2}$	1 in the band, else 0	$\frac{sin ω _{2} t - sin ω _{1} t}{π t} = \frac{2 sin ( B t /2 )}{π t} cos ω_{0} t$
Bandstop, $ω_{1} < ∣ ω ∣ < ω_{2}$	0 in the band, else 1	$δ (t) - \frac{sin ω _{2} t - sin ω _{1} t}{π t}$

Why ideal filters cannot be built:

Non-causal: $h (t) = \frac{sin ω _{c} t}{π t}$ is nonzero for $t < 0$ . A delay $t_{d}$ shifts it right, but the sinc tails extend to $- \infty$ , so $h (t - t_{d}) \neq = 0$ for some $t < 0$ whatever finite $t_{d}$ is chosen. The same holds for the HPF, BPF and BSF.
Paley–Wiener criterion: a square-integrable gain $∣ H (j ω) ∣$ can belong to a causal filter only if $\int_{- \infty}^{\infty} \frac{ln ∣ H ( j ω ) ∣}{1 + ω ^{2}} d ω < \infty$ . A gain that is exactly zero over a band makes $ln ∣ H ∣ = - \infty$ across that band, so the integral diverges. A causal filter can have zero gain at isolated frequencies, but not over any band, however narrow.
Not BIBO stable: $\int ∣ sin ω_{c} t / (π t) ∣ d t$ diverges, although $h (t)$ has finite energy.
Practical filters approximate the ideal with a finite transition band and a phase that is only approximately linear in the passband.

EXAM TIP

The step response of the ideal LPF (zero delay) is $\frac{1}{2} + \frac{1}{π} Si (ω_{c} t)$ , where $Si (x) = \int_{0}^{x} \frac{s i n u}{u} d u$ . It overshoots to about $1.09$ (about 9%) whatever the cutoff; raising $ω_{c}$ only makes the ringing faster. This is the Gibbs phenomenon.

Bandwidth has several definitions, and a question should say which it means. For a real signal the spectrum is symmetric about $ω = 0$ , so bandwidth is normally quoted from the positive-frequency half: a lowpass spectrum occupying $∣ ω ∣ < W$ has bandwidth $W$ . The null-to-null bandwidth of a lowpass pulse is quoted both as the full width of the main lobe and as the one-sided width to the first null, so check which is meant. A nonzero signal cannot be both time-limited and band-limited: a pulse of finite duration has infinite absolute bandwidth, and a strictly band-limited signal lasts for all time.

Definition	Meaning	Example
Absolute	range outside which $X$ or $H$ is exactly zero	$\frac{sin W t}{π t}$ has absolute bandwidth $W$ rad/s; a time-limited pulse has infinite absolute bandwidth
3-dB (half-power)	range over which $∣ H ∣ \geq ∣ H ∣_{m a x} / 2$ , i.e. $∣ H ∣^{2} \geq \frac{1}{2} ∣ H ∣_{m a x}^{2}$	RC lowpass: $ω_{3 dB} = 1/ R C$ rad/s, $f_{3 dB} = 1/ (2 π R C)$ Hz; a bandpass response: $ω_{2} - ω_{1}$ between its two half-power points
Null-to-null (main lobe)	width between the first zeros on either side of the spectral peak	$rect (t / T)$ : full main lobe $4 π / T$ rad/s ( $2/ T$ Hz); one-sided, to the first null, $2 π / T$ rad/s ( $1/ T$ Hz)
Energy (essential)	band holding a stated fraction of the energy	$e^{- a t} u (t)$ : the fraction in $∣ ω ∣ < W$ is $\frac{2}{π} tan^{- 1} (W / a)$

3-DB BANDWIDTH OF A FIRST-ORDER LOWPASS

H (j ω) = \frac{1}{1 + j ω R C} : ∣ H ∣ = \frac{1}{2} at ω_{3 dB} = \frac{1}{R C} rad/s (f_{3 dB} = \frac{1}{2 π R C} Hz)

At the 3-dB point the phase is $- 4 5^{\circ}$ . The 10–90% step rise time is $t_{r} = R C ln 9 \approx 2.2 R C$ , so $t_{r} f_{3 dB} \approx 0.35$ .

The Hilbert transformer is the LTI system that shifts every positive-frequency component by $- 9 0^{\circ}$ and every negative-frequency component by $+ 9 0^{\circ}$ , with unit gain. It is used to build single-sideband signals and analytic signals.

HILBERT TRANSFORM

\overset{x}{^} (t) = x (t) * \frac{1}{π t}, H (j ω) = - j sgn (ω) = {- j, + j, ω > 0 ω < 0

Gain 1 at every $ω \neq = 0$ , phase $- 9 0^{\circ}$ for $ω > 0$ . Its $h (t) = 1/ (π t)$ is non-causal, so the ideal Hilbert transformer is also unrealisable.

Hilbert transform facts:

$cos ω_{0} t \to sin ω_{0} t$ and $sin ω_{0} t \to - cos ω_{0} t$ (for $ω_{0} > 0$ ). A constant maps to 0.
Applying the transform twice gives $- x (t)$ for a signal with no DC component, because $(- j sgn ω)^{2} = - 1$ for $ω \neq = 0$ .
$x$ and $\overset{x}{^}$ have the same magnitude spectrum (apart from $ω = 0$ ) and the same energy, and for a real finite-energy $x$ they are orthogonal: $\int x \overset{x}{^} d t = 0$ .
The analytic signal $x (t) + j \overset{x}{^} (t)$ has spectrum $2 X (ω)$ for $ω > 0$ and $0$ for $ω < 0$ .
If $m (t)$ is lowpass with bandwidth below $ω_{c}$ , the Hilbert transform of $m (t) cos ω_{c} t$ is $m (t) sin ω_{c} t$ .

EXAM TIP

Do not confuse the Hilbert transformer with a differentiator. $\frac{d}{d t}$ has $H = j ω$ : phase $+ 9 0^{\circ}$ for $ω > 0$ and gain $∣ ω ∣$ . The Hilbert transformer has phase $- 9 0^{\circ}$ for $ω > 0$ and gain 1.

7. The discrete-time Fourier transform (DTFT)

Convention. For a sequence $x [n]$ the DTFT is written $X (e^{j Ω})$ , with $Ω$ the normalised frequency in rad/sample. It is periodic in $Ω$ with period $2 π$ , so every integral over frequency runs over one period, usually $- π$ to $π$ . $Ω = 0$ (and every multiple of $2 π$ ) is the lowest frequency and $Ω = \pm π$ the highest: $(- 1)^{n} = e^{j π n}$ is the fastest-varying sequence. If $x [n] = x_{c} (n T_{s})$ comes from sampling, $Ω = ω T_{s}$ . The DTFT exists as an ordinary function when $\sum ∣ x [n] ∣ < \infty$ , in the mean-square sense when $\sum ∣ x [n] ∣^{2} < \infty$ , and only with impulses for sequences such as $1$ , $u [n]$ and $cos Ω_{0} n$ .

DTFT PAIR

X (e^{j Ω}) = n = - \infty \sum \infty x [n] e^{- j Ω n} x [n] = \frac{1}{2 π} \int_{- π}^{π} X (e^{j Ω}) e^{j Ω n} d Ω

Any interval of length $2 π$ may replace $- π$ to $π$ . The DTFT is the $z$ -transform on the unit circle, $X (e^{j Ω}) = X (z)_{z = e^{j Ω}}$ , valid when the ROC of $X (z)$ contains $∣ z ∣ = 1$ .

$x [n]$	$X (e^{j Ω})$	Condition / note
$δ [n]$	$1$	flat spectrum
$δ [n - n_{0}]$	$e^{- j Ω n_{0}}$	integer $n_{0}$
$a^{n} u [n]$	$\frac{1}{1 - a e ^{- j Ω}}$	$∣ a ∣ < 1$
$(n + 1) a^{n} u [n]$	$\frac{1}{( 1 - a e ^{- j Ω} ) ^{2}}$	$∣ a ∣ < 1$
$a^{∣ n ∣}$	$\frac{1 - a ^{2}}{1 - 2 a cos Ω + a ^{2}}$	$∣ a ∣ < 1$ ; real and even
$1$ for $0 \leq n \leq N - 1$ , else $0$	$e^{- j Ω (N - 1) /2} \frac{sin ( N Ω/2 )}{sin ( Ω/2 )}$	value $N$ at $Ω = 0$ ; zeros at $Ω = 2 π k / N$ , $k$ not a multiple of $N$
$1$ for $∣ n ∣ \leq N_{1}$ , else $0$	$\frac{sin ( ( 2 N _{1} + 1 ) Ω/2 )}{sin ( Ω/2 )}$	real and even; value $2 N_{1} + 1$ at $Ω = 0$
$\frac{sin W n}{π n}$ (value $W / π$ at $n = 0$ )	$1$ for $∣Ω∣ < W$ , $0$ for $W < ∣Ω∣ \leq π$ , repeated every $2 π$	$0 < W < π$ ; ideal DT lowpass filter
$1$	$2 π \sum_{k} δ (Ω - 2 π k)$	impulses at every multiple of $2 π$
$e^{j Ω_{0} n}$	$2 π \sum_{k} δ (Ω - Ω_{0} - 2 π k)$
$cos Ω_{0} n$	$π \sum_{k} [δ (Ω - Ω_{0} - 2 π k) + δ (Ω + Ω_{0} - 2 π k)]$
$u [n]$	$\frac{1}{1 - e ^{- j Ω}} + π \sum_{k} δ (Ω - 2 π k)$

Property	$x [n]$	$X (e^{j Ω})$
Time shift	$x [n - n_{0}]$ , integer $n_{0}$	$e^{- j Ω n_{0}} X (e^{j Ω})$
Frequency shift	$e^{j Ω_{0} n} x [n]$	$X (e^{j (Ω - Ω_{0})})$
Multiplication by $(- 1)^{n}$	$(- 1)^{n} x [n]$	$X (e^{j (Ω - π)})$ : turns a lowpass response into a highpass one
Time reversal	$x [- n]$	$X (e^{- j Ω})$
Conjugation	$x^{*} [n]$	$X^{*} (e^{- j Ω})$
Time expansion by integer $L$	$x [n / L]$ when $n$ is a multiple of $L$ , else $0$	$X (e^{j L Ω})$
Differentiation in frequency	$n x [n]$	$j \frac{d X ( e ^{j Ω} )}{d Ω}$
First difference	$x [n] - x [n - 1]$	$(1 - e^{- j Ω}) X (e^{j Ω})$
Accumulation	$\sum_{k = - \infty}^{n} x [k]$	$\frac{X ( e ^{j Ω} )}{1 - e ^{- j Ω}} + π X (e^{j 0}) \sum_{k} δ (Ω - 2 π k)$
Convolution	$x [n] * y [n]$	$X (e^{j Ω}) Y (e^{j Ω})$
Multiplication	$x [n] y [n]$	$\frac{1}{2 π} \int_{- π}^{π} X (e^{j λ}) Y (e^{j (Ω - λ)}) d λ$ (periodic convolution)

DTFT VALUE THEOREMS

X (e^{j 0}) = n \sum x [n], X (e^{j π}) = n \sum (- 1)^{n} x [n], x [0] = \frac{1}{2 π} \int_{- π}^{π} X (e^{j Ω}) d Ω

Put $Ω = 0$ or $Ω = π$ in the forward sum, or $n = 0$ in the inverse integral. More generally $\frac{1}{2 π} \int_{- π}^{π} X (e^{j Ω}) e^{j Ω k} d Ω = x [k]$ .

PARSEVAL'S THEOREM FOR THE DTFT

n = - \infty \sum \infty ∣ x [n] ∣^{2} = \frac{1}{2 π} \int_{- π}^{π} ∣ X (e^{j Ω}) ∣^{2} d Ω

The frequency integral spans one period, $2 π$ wide, never $- \infty$ to $\infty$ .

DISCRETE-TIME SINUSOIDAL STEADY STATE

A cos (Ω_{0} n + ϕ) ⟶ A ∣ H (e^{j Ω_{0}}) ∣ cos (Ω_{0} n + ϕ + θ (Ω_{0}))

Real $h [n]$ , stable system ( $\sum ∣ h [n] ∣ < \infty$ ), input applied for all $n$ . $Ω_{0}$ and $Ω_{0} + 2 π$ give the same sequence and the same response.

WORKED EXAMPLE

A causal LTI system has $h [n] = (0.5)^{n} u [n]$ . (a) Find $H (e^{j Ω})$ and the gain at $Ω = 0$ and at $Ω = π$ . (b) Find the output for $x [n] = cos (π n /2)$ . (c) Evaluate $\frac{1}{2 π} \int_{- π}^{π} ∣ H (e^{j Ω}) ∣^{2} d Ω$ .

Solution

From $a^{n} u [n] \leftrightarrow \frac{1}{1 - a e ^{- j Ω}}$ with $a = 0.5$ : $H (e^{j Ω}) = \frac{1}{1 - 0.5 e ^{- j Ω}}$ .
At $Ω = 0$ : $H = \frac{1}{1 - 0.5} = 2$ . At $Ω = π$ : $e^{- j π} = - 1$ , so $H = \frac{1}{1.5} = \frac{2}{3}$ . The gain falls from 2 to $2/3$ across the band: a lowpass response.
At $Ω_{0} = π /2$ : $e^{- j π /2} = - j$ , so $H = \frac{1}{1 + 0.5 j}$ . Then $∣ H ∣ = \frac{1}{1.25} = \frac{2}{5} \approx 0.894$ and $θ = - tan^{- 1} 0.5 \approx - 0.464$ rad.
So $y [n] = 0.894 cos (π n /2 - 0.464)$ .
By Parseval, $\frac{1}{2 π} \int_{- π}^{π} ∣ H ∣^{2} d Ω = \sum_{n = 0}^{\infty} (0.25)^{n} = \frac{1}{1 - 0.25} = \frac{4}{3} \approx 1.333$ .

Answer: $H (e^{j Ω}) = 1/ (1 - 0.5 e^{- j Ω})$ , with gain 2 at $Ω = 0$ and $2/3$ at $Ω = π$ ; $y [n] = \frac{2}{5} cos (π n /2 - tan^{- 1} 0.5) \approx 0.894 cos (π n /2 - 0.464)$ ; the integral is $4/3 \approx 1.333$ .

PREVIOUS-YEAR QUESTIONS

22 questions from the EC, EE and IN papers

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

See these papers in full: GATE ECE 2026 · GATE ECE 2025 · GATE ECE 2024 · GATE ECE 2023 · GATE ECE 2022 · GATE ECE 2020 · GATE ECE 2018 · GATE EE 2025 · GATE EE 2024 · GATE EE 2023 · GATE EE 2022 · GATE EE 2018 · GATE IN 2025 · GATE IN 2023 · GATE IN 2022 · GATE IN 2020 · GATE IN 2018

GATE 2026 EC Q.46

GATE 2026 · ECQ.46 · 2 marks · MCQ

Consider a real baseband signal $x (t) = e^{- 2 t}$ , for $t$ (in seconds) $\geq 0$ .

If 99% of energy of $x (t)$ lies within $B$ Hz, then which of the following options is TRUE for the value of $B$ ?

(A)
$B > 1 kHz$
(B)
$63/ π Hz < B < 64/ π Hz$
(C)
$126/ π Hz < B < 128/ π Hz$
(D)
$B < 1 Hz$

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GATE 2026 EC Q.46: answer and solution

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GATE 2025 EC Q.18

GATE 2025 · ECQ.18 · 1 mark · MCQ

Consider a continuous-time, real-valued signal $f (t)$ whose Fourier transform $F (ω) = \int_{- \infty}^{\infty} f (t) exp (- j ω t) d t$ exists.

Which one of the following statements is always TRUE?

(A)
$∣ F (ω) ∣ \leq \int_{- \infty}^{\infty} ∣ f (t) ∣ d t$
(B)
$∣ F (ω) ∣ > \int_{- \infty}^{\infty} ∣ f (t) ∣ d t$
(C)
$∣ F (ω) ∣ \leq \int_{- \infty}^{\infty} f (t) d t$
(D)
$∣ F (ω) ∣ \geq \int_{- \infty}^{\infty} f (t) d t$

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GATE 2025 EC Q.18: answer and solution

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GATE 2025 EE Q.61

GATE 2025 · EEQ.61 · 2 marks · NAT

An ideal low pass filter has frequency response given by
$H (j ω) = {1, 0, ∣ ω ∣ \leq 200 π otherwise$
Let $h (t)$ be its time domain representation. Then $h (0) =$ ______ (round off to the nearest integer).

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GATE 2025 IN Q.48

GATE 2025 · INQ.48 · 2 marks · MCQ

Let $f (t)$ and $g (t)$ represent continuous-time real-valued signals. If $h (t)$ denotes cross-correlation between $f (t)$ and $g (- t)$ , its continuous-time Fourier transform $H (j ω)$ equals

Note: $F (j ω)$ and $G (j ω)$ denote the continuous-time Fourier transforms of $f (t)$ and $g (t)$ , respectively.

(A)
$F (j ω) G (j ω)$
(B)
$F (- j ω) G (j ω)$
(C)
$F (j ω) G (- j ω)$
(D)
$- F (j ω) G (- j ω)$

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GATE 2024 EC Q.44

GATE 2024 · ECQ.44 · 2 marks · MCQ

Consider two continuous time signals $x (t)$ and $y (t)$ as shown below

[Figure: $x (t)$ and $y (t)$ ]

If $X (f)$ denotes the Fourier transform of $x (t)$ , then the Fourier transform of $y (t)$ is ______.

GATE 2024 EC question 44: Two sketches side by side. — Figure from the GATE 2024 EC paper. It is part of the question.

(A)
$- 4 X (4 f) e^{- j π f}$
(B)
$- 4 X (4 f) e^{- j 4 π f}$
(C)
$- \frac{1}{4} X (f /4) e^{- j π f}$
(D)
$- \frac{1}{4} X (f /4) e^{- j 4 π f}$

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GATE 2024 EC Q.44: answer and solution

GATE 2024 EE Q.33

GATE 2024 · EEQ.33 · 1 mark · NAT

Let $X (ω)$ be the Fourier transform of the signal

x (t) = e^{- t^{4}} cos t, - \infty < t < \infty.

The value of the derivative of $X (ω)$ at $ω = 0$ is ________ (rounded off to 1 decimal place).

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GATE 2023 EC Q.27

GATE 2023 · ECQ.27 · 1 mark · MCQ

Let $m (t)$ be a strictly band-limited signal with bandwidth $B$ and energy $E$ . Assuming $ω_{0} = 10 B$ , the energy in the signal $m (t) cos ω_{0} t$ is

(A)
$\frac{E}{4}$
(B)
$\frac{E}{2}$
(C)
$E$
(D)
$2 E$

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GATE 2023 EC Q.27: answer and solution

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GATE 2023 EC Q.28

GATE 2023 · ECQ.28 · 1 mark · MCQ

The Fourier transform $X (ω)$ of $x (t) = e^{- t^{2}}$ is

Note: $\int_{- \infty}^{\infty} e^{- y^{2}} d y = π$

(A)
$π e^{\frac{ω ^{2}}{2}}$
(B)
$\frac{e ^{- \frac{ω ^{2}}{4}}}{2 π}$
(C)
$π e^{- \frac{ω ^{2}}{4}}$
(D)
$π e^{- \frac{ω ^{2}}{2}}$

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GATE 2023 EC Q.28: answer and solution

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GATE 2023 EC Q.29

GATE 2023 · ECQ.29 · 1 mark · MCQ

In the table shown below, match the signal type with its spectral characteristics.

| Signal type | Spectral characteristics |
|---|---|
| (i) Continuous, aperiodic | (a) Continuous, aperiodic |
| (ii) Continuous, periodic | (b) Continuous, periodic |
| (iii) Discrete, aperiodic | (c) Discrete, aperiodic |
| (iv) Discrete, periodic | (d) Discrete, periodic |

(A)
$(i) \to (a), (ii) \to (b), (iii) \to (c), (i v) \to (d)$
(B)
$(i) \to (a), (ii) \to (c), (iii) \to (b), (i v) \to (d)$
(C)
$(i) \to (d), (ii) \to (b), (iii) \to (c), (i v) \to (a)$
(D)
$(i) \to (a), (ii) \to (c), (iii) \to (d), (i v) \to (b)$

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GATE 2023 EC Q.29: answer and solution

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GATE 2023 EC Q.49

GATE 2023 · ECQ.49 · 2 marks · MCQ

Let $x (t) = 10 cos (10.5 W t)$ be passed through an LTI system having impulse response $h (t) = π (\frac{sin W t}{π t})^{2} cos 10 W t$ . The output of the system is

(A)
$(\frac{15 W}{4}) cos (10.5 W t)$
(B)
$(\frac{15 W}{2}) cos (10.5 W t)$
(C)
$(\frac{15 W}{8}) cos (10.5 W t)$
(D)
$(15 W) cos (10.5 W t)$

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GATE 2023 EE Q.18

GATE 2023 · EEQ.18 · 1 mark · MCQ

The Fourier transform $X (ω)$ of the signal $x (t)$ is given by

X (ω) = 1, for ∣ ω ∣ < W_{0} = 0, for ∣ ω ∣ > W_{0}

Which one of the following statements is true?

(A)
$x (t)$ tends to be an impulse as $W_{0} \to \infty$ .
(B)
$x (0)$ decreases as $W_{0}$ increases.
(C)
At $t = \frac{π}{2 W _{0}}$ , $x (t) = - \frac{1}{π}$
(D)
At $t = \frac{π}{2 W _{0}}$ , $x (t) = \frac{1}{π}$

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GATE 2023 EE Q.18: answer and solution

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GATE 2023 IN Q.44

GATE 2023 · INQ.44 · 2 marks · MSQ

A continuous real-valued signal $x (t)$ has finite positive energy and $x (t) = 0, \forall t < 0$ . From the list given below, select ALL the signals whose continuous-time Fourier transform is purely imaginary.

(A)
$x (t) + x (- t)$
(B)
$x (t) - x (- t)$
(C)
$j (x (t) + x (- t))$
(D)
$j (x (t) - x (- t))$

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GATE 2022 EC Q.15

GATE 2022 · ECQ.15 · 1 mark · MCQ

The Fourier transform $X (j ω)$ of the signal
$x (t) = \frac{t}{( 1 + t ^{2} ) ^{2}}$
is ________.

(A)
$\frac{π}{2 j} ω e^{- ∣ ω ∣}$
(B)
$\frac{π}{2} ω e^{- ∣ ω ∣}$
(C)
$\frac{π}{2 j} e^{- ∣ ω ∣}$
(D)
$\frac{π}{2} e^{- ∣ ω ∣}$

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GATE 2022 EC Q.15: answer and solution

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GATE 2022 EC Q.23

GATE 2022 · ECQ.23 · 1 mark · MCQ

The frequency response $H (f)$ of a linear time-invariant system has magnitude as shown in the figure.

Statement I: The system is necessarily a pure delay system for inputs which are bandlimited to $- α \leq f \leq α$ .

Statement II: For any wide-sense stationary input process with power spectral density $S_{X} (f)$ , the output power spectral density $S_{Y} (f)$ obeys $S_{Y} (f) = S_{X} (f)$ for $- α \leq f \leq α$ .

Which one of the following combinations is true?

GATE 2022 EC question 23: Plot of |H(f)| (vertical axis, labelled |H(f)|) against f (horizontal axis, labelled f (Hz)). — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
Statement I is correct, Statement II is correct
(B)
Statement I is correct, Statement II is incorrect
(C)
Statement I is incorrect, Statement II is correct
(D)
Statement I is incorrect, Statement II is incorrect

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GATE 2022 EC Q.23: answer and solution

GATE 2022 EE Q.47

GATE 2022 · EEQ.47 · 2 marks · MCQ

Let an input $x (t) = 2 sin (10 π t) + 5 cos (15 π t) + 7 sin (42 π t) + 4 cos (45 π t)$ is passed through an LTI system having an impulse response,

h (t) = 2 (\frac{sin ( 10 π t )}{π t}) cos (40 π t) .

The output of the system is

(A)
$2 sin (10 π t) + 5 cos (15 π t)$
(B)
$5 cos (15 π t) + 7 sin (42 π t)$
(C)
$7 sin (42 π t) + 4 cos (45 π t)$
(D)
$2 sin (10 π t) + 4 cos (45 π t)$

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GATE 2022 IN Q.18

GATE 2022 · INQ.18 · 1 mark · MCQ

A signal $x (t)$ is band-limited between 100 Hz and 200 Hz. A signal $y (t)$ is related to $x (t)$ as follows:

y (t) = x (2 t - 5)

The statement that is always true is ________

(A)
$y (t)$ is band-limited between 50 Hz and 100 Hz
(B)
$y (t)$ is band-limited between 100 Hz and 200 Hz
(C)
$y (t)$ is band-limited between 200 Hz and 400 Hz
(D)
$y (t)$ is not band-limited

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GATE 2020 EC Q.52

GATE 2020 · ECQ.52 · 2 marks · NAT

$X (ω)$ is the Fourier transform of $x (t)$ shown below. The value of $\int_{- \infty}^{\infty} ∣ X (ω) ∣^{2} d ω$ (rounded off to two decimal places) is ______.

GATE 2020 EC question 52: Plot of x(t) against t (t-axis marked -3, -2, -1, 0, 1, 2, 3, 4; — Figure from the GATE 2020 EC paper. It is part of the question.

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GATE 2020 EC Q.52: answer and solution

GATE 2020 IN Q.5

GATE 2020 · INQ.5 · 1 mark · MCQ

A differentiator has a transfer function whose

(A)
phase increases linearly with frequency
(B)
magnitude remains constant
(C)
magnitude increases linearly with frequency
(D)
magnitude decreases linearly with frequency

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GATE 2020 IN Q.5: answer and solution

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GATE 2020 IN Q.19

GATE 2020 · INQ.19 · 1 mark · NAT

Consider the signal $x (t) = e^{- ∣ t ∣}$ . Let $X (j ω) = \int_{- \infty}^{\infty} x (t) e^{- j ω t} d t$ be the Fourier transform of $x (t)$ . The value of $X (j 0)$ is ____

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GATE 2020 IN Q.19: answer and solution

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GATE 2018 EC Q.39

GATE 2018 · ECQ.39 · 2 marks · NAT

The input $4 sinc (2 t)$ is fed to a Hilbert transformer to obtain $y (t)$ , as shown in the figure below:

[Figure]

Here $sinc (x) = \frac{sin ( π x )}{π x}$ . The value (accurate to two decimal places) of $\int_{- \infty}^{\infty} ∣ y (t) ∣^{2} d t$ is ______.

GATE 2018 EC question 39: A single block labelled 'Hilbert Transform'. — Figure from the GATE 2018 EC paper. It is part of the question.

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GATE 2018 EC Q.39: answer and solution

GATE 2018 EE Q.40

GATE 2018 · EEQ.40 · 2 marks · NAT

The Fourier transform of a continuous-time signal $x (t)$ is given by $X (ω) = \frac{1}{( 10 + j ω ) ^{2}}, - \infty < ω < \infty,$ where $j = - 1$ and $ω$ denotes frequency. Then the value of $∣ ln x (t) ∣$ at $t = 1$ is ________ (up to 1 decimal place). ( $ln$ denotes the logarithm to base $e$ )

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GATE 2018 IN Q.33

GATE 2018 · INQ.33 · 2 marks · MCQ

The Fourier transform of a signal $x (t)$ , denoted by $X (j ω)$ , is shown in the figure.

Let $y (t) = x (t) + e^{j t} x (t)$ . The value of Fourier transform of $y (t)$ evaluated at the angular frequency $ω = 0.5$ rad/s is

GATE 2018 IN question 33: Plot of X(j ω) (vertical axis) against ω in rad/s (horizontal axis). — Figure from the GATE 2018 IN paper. It is part of the question.

(A)
0.5
(B)
1
(C)
1.5
(D)
2.5

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GATE 2018 IN Q.33: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 11 mark · MCQ

A continuous-time signal $x (t)$ has Fourier transform $X (ω) = \int_{- \infty}^{\infty} x (t) e^{- j ω t} d t$ . The Fourier transform of $y (t) = x (2 t - 3)$ is

(A)
$\frac{1}{2} e^{- j 3 ω /2} X (ω /2)$
(B)
$\frac{1}{2} e^{- j 3 ω} X (ω /2)$
(C)
$2 e^{- j 3 ω /2} X (2 ω)$
(D)
$\frac{1}{2} e^{- j 6 ω} X (ω /2)$

Answer and solution

Answer: (A)

Write $y (t) = x (2 (t - 1.5))$ : scale first, then delay by $1.5$ s.
$x (2 t) \leftrightarrow \frac{1}{2} X (ω /2)$ , and a delay of $1.5$ s multiplies by $e^{- j 1.5 ω}$ .
Check by the other order: $x (t - 3) \leftrightarrow e^{- j 3 ω} X (ω)$ ; replacing $t$ by $2 t$ scales every $ω$ to $ω /2$ and divides by 2, giving $\frac{1}{2} e^{- j 3 ω /2} X (ω /2)$ . Option B forgets that the shift is scaled too.

PRACTICE 22 marks · NAT

The signal $x (t) = e^{- t} u (t)$ is applied to an ideal lowpass filter with $H (j ω) = 1$ for $∣ ω ∣ < 3$ rad/s and $H (j ω) = 0$ otherwise. The energy of the output signal, $\int_{- \infty}^{\infty} ∣ y (t) ∣^{2} d t$ , is ______ (rounded off to three decimal places).

Answer and solution

Answer: $0.333$ (accepted range $\pm 0.002$ )

$X (ω) = \frac{1}{1 + j ω}$ , so $∣ X (ω) ∣^{2} = \frac{1}{1 + ω ^{2}}$ .
By Parseval, $E_{y} = \frac{1}{2 π} \int_{- 3}^{3} \frac{d ω}{1 + ω ^{2}} = \frac{1}{2 π} \cdot 2 tan^{- 1} 3 = \frac{1}{2 π} \cdot \frac{2 π}{3} = \frac{1}{3}$ .
Check: the input energy is $\int_{0}^{\infty} e^{- 2 t} d t = \frac{1}{2}$ and the fraction passed is $\frac{2}{π} tan^{- 1} 3 = \frac{2}{3}$ , so $E_{y} = \frac{1}{3} \approx 0.333$ .

PRACTICE 31 mark · MCQ

The signal $x (t) = 3 cos 2 t + sin 5 t$ is passed through an ideal Hilbert transformer with frequency response $H (j ω) = - j sgn (ω)$ . The output is

(A)
$3 sin 2 t - cos 5 t$
(B)
$- 3 sin 2 t + cos 5 t$
(C)
$3 sin 2 t + cos 5 t$
(D)
$- 3 sin 2 t - cos 5 t$

Answer and solution

Answer: (A)

The transformer has unit gain and shifts every positive-frequency component by $- 9 0^{\circ}$ , so a real sinusoid $cos (ω_{0} t + ϕ)$ with $ω_{0} > 0$ becomes $cos (ω_{0} t + ϕ - 9 0^{\circ}) = sin (ω_{0} t + ϕ)$ .
$3 cos 2 t \to 3 sin 2 t$ . Also $sin 5 t = cos (5 t - 9 0^{\circ}) \to cos (5 t - 18 0^{\circ}) = - cos 5 t$ .
The output is $3 sin 2 t - cos 5 t$ .

PRACTICE 41 mark · NAT

A real signal $x (t)$ has Fourier transform $X (ω) = \int_{- \infty}^{\infty} x (t) e^{- j ω t} d t$ given by $X (ω) = 1 - \frac{∣ ω ∣}{4}$ for $∣ ω ∣ \leq 4$ rad/s and $X (ω) = 0$ otherwise. The value of $x (0)$ is ______ (rounded off to three decimal places).

Answer and solution

Answer: $0.637$ (accepted range $\pm 0.002$ )

$x (0) = \frac{1}{2 π} \int_{- \infty}^{\infty} X (ω) d ω$ : the area under the spectrum divided by $2 π$ .
The spectrum is a triangle of base 8 rad/s and height 1, with area $\frac{1}{2} \times 8 \times 1 = 4$ .
$x (0) = \frac{4}{2 π} = \frac{2}{π} \approx 0.637$ .

PRACTICE 52 marks · MSQ

A sequence has $x [n] = 1, 2, 3, 2, 1$ for $n = 0, 1, 2, 3, 4$ respectively, and $x [n] = 0$ for every other $n$ . Its DTFT is $X (e^{j Ω}) = \sum_{n} x [n] e^{- j Ω n}$ . Which of the following statements is/are correct?

(A)
$X (e^{j 0}) = 9$
(B)
$X (e^{j π}) = 1$
(C)
$\frac{1}{2 π} \int_{- π}^{π} ∣ X (e^{j Ω}) ∣^{2} d Ω = 19$
(D)
$\frac{1}{2 π} \int_{- π}^{π} X (e^{j Ω}) d Ω = 3$

Answer and solution

Answer: (A), (B), (C)

$X (e^{j 0}) = \sum x [n] = 1 + 2 + 3 + 2 + 1 = 9$ . A is correct.
$X (e^{j π}) = \sum (- 1)^{n} x [n] = 1 - 2 + 3 - 2 + 1 = 1$ . B is correct.
By Parseval the integral equals $\sum ∣ x [n] ∣^{2} = 1 + 4 + 9 + 4 + 1 = 19$ . C is correct.
$\frac{1}{2 π} \int_{- π}^{π} X (e^{j Ω}) d Ω = x [0] = 1$ , not 3. The value 3 is $x [2]$ , which needs the factor $e^{j 2Ω}$ inside the integral. D is wrong.

PRACTICE 62 marks · MSQ

A stable LTI system has transfer function $H (s) = \frac{4 - s}{4 + s}$ . Phase delay is $τ_{p} = - θ (ω) / ω$ and group delay is $τ_{g} = - d θ / d ω$ , where $θ (ω) = ∠ H (j ω)$ is the unwrapped phase. Which of the following statements is/are correct?

(A)
$∣ H (j ω) ∣ = 1$ for every $ω$ .
(B)
The group delay at $ω = 0$ is $0.5$ s.
(C)
The phase delay equals the group delay at $ω = 4$ rad/s.
(D)
The phase delay at $ω = 4$ rad/s is $π /8$ s.

Answer and solution

Answer: (A), (B), (D)

$∣4 - j ω ∣ = ∣4 + j ω ∣$ , so the gain is 1 at every frequency (an all-pass). A is correct.
$θ (ω) = - 2 tan^{- 1} (ω /4)$ , so $τ_{g} = \frac{2 \cdot 4}{16 + ω ^{2}} = \frac{8}{16 + ω ^{2}}$ . At $ω = 0$ , $τ_{g} = 0.5$ s. B is correct.
At $ω = 4$ : $θ = - 2 \cdot \frac{π}{4} = - \frac{π}{2}$ , so $τ_{p} = \frac{π /2}{4} = \frac{π}{8} \approx 0.393$ s. D is correct.
At $ω = 4$ , $τ_{g} = 8/32 = 0.25$ s, which is not $0.393$ s, so C is wrong. The phase is not linear, so the two delays differ.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

frequency response, group delay, phase delay … Fourier transform

GATE EE

Applications of Fourier Transform for continuous and discrete time signals

GATE IN

Fourier … transforms … frequency response

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