FOUNDATION · CHAPTER 6 OF 8

From series to transform: one family, five names

This chapter shows that the Fourier series, Fourier transform, Laplace, z, DTFT and DFT are one idea with different freedoms. You need it to choose the right transform and to read a region of convergence correctly.

Updated 12 Oct 2026

What you will be able to do

You will see why the Fourier transform is the Fourier series of a signal whose period grows without limit. You will add a decay factor to reach the Laplace transform and read its region of convergence. You will sample to reach the z-transform and the DTFT. You will place five transforms in one table and use duality to get one pair from another.

Before you start

Chapter 2, The integrals, sums and limits that keep coming back
Chapter 5, Why sinusoids: eigenfunctions and the frequency idea (its Fourier series coefficients $a_{k}$ )
The unit step $u (t)$ , the unit impulse $δ (t)$ and the sifting property $\int δ (t - t_{0}) f (t) d t = f (t_{0})$
Riemann sums: a sum of $f (x) Δ x$ becomes an integral as $Δ x \to 0$
The modulus and angle of a complex number, and $e^{j θ} = cos θ + j sin θ$

From the series to the transform, then to Laplace

Chapter 5 wrote a periodic signal as lines at the harmonics $k ω_{0}$ , with $ω_{0} = 2 π / T_{0}$ . Repeat one pulse $x (t)$ , zero outside $∣ t ∣ < T_{1}$ , every $T_{0} > 2 T_{1}$ seconds to get $x_{p} (t)$ . As $T_{0}$ grows, the lines crowd together and shrink, because $a_{k}$ averages over a longer period. In the limit they merge into a curve and the sum becomes an integral. Round brackets mean continuous time, $x (t)$ , square brackets discrete time, $x [n]$ ; $j$ is the imaginary unit.

THE SCALED COEFFICIENTS SAMPLE THE TRANSFORM

X (ω) = \int_{- \infty}^{\infty} x (t) e^{- j ω t} d t, T_{0} a_{k} = \int_{- T_{0} /2}^{T_{0} /2} x_{p} (t) e^{- j k ω_{0} t} d t = X (k ω_{0})

The middle term is chapter 5's $a_{k} = \frac{1}{T _{0}} \int_{- T_{0} /2}^{T_{0} /2} x_{p} (t) e^{- j k ω_{0} t} d t$ times $T_{0}$ . Since $T_{0} > 2 T_{1}$ , $x_{p} = x$ on that interval and $x = 0$ outside it, so the integral is $X (k ω_{0})$ . $ω$ is in rad/s; $X (ω)$ is also written $X (j ω)$ and certainly exists when $\int ∣ x (t) ∣ d t$ is finite.

THE SERIES BECOMES THE INVERSE TRANSFORM

x_{p} (t) = k \sum a_{k} e^{j k ω_{0} t} = \frac{1}{2 π} k \sum X (k ω_{0}) e^{j k ω_{0} t} ω_{0} T_{0} \to \infty x (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} X (ω) e^{j ω t} d ω

Use $a_{k} = X (k ω_{0}) / T_{0} = X (k ω_{0}) ω_{0} /2 π$ . With $ω_{k} = k ω_{0}$ and $Δ ω = ω_{0}$ the sum is $\sum_{k} X (ω_{k}) e^{j ω_{k} t} Δ ω$ , a Riemann sum; $Δ ω \to 0$ makes it the integral, and $x_{p} (t) \to x (t)$ as the other copies move away. So $1/2 π$ sits only in the inverse. The hertz form $X (f) = \int x (t) e^{- j 2 π f t} d t$ , with no factor either way, is also used.

A signal such as $e^{2 t} u (t)$ grows, so $\int ∣ x (t) ∣ d t$ diverges and it has no Fourier transform. Multiply it by $e^{- σ t}$ , with $σ$ real: for $σ > 2$ the product $e^{(2 - σ) t} u (t)$ decays and its Fourier transform exists. Pair $σ$ with $ω$ as one complex variable, $s = σ + j ω$ . The set of $s$ for which this works is the region of convergence (ROC).

LAPLACE TRANSFORM AS A DAMPED FOURIER TRANSFORM

X (s) = \int_{- \infty}^{\infty} x (t) e^{- s t} d t = \int_{- \infty}^{\infty} [x (t) e^{- σ t}] e^{- j ω t} d t, s = σ + j ω

Bilateral form; $σ$ in $s^{- 1}$ , $ω$ in rad/s. Since $∣ e^{- s t} ∣ = e^{- σ t}$ , the ROC, the set of $s$ with $\int ∣ x (t) ∣ e^{- σ t} d t$ finite, depends on $σ$ alone. So it is a vertical strip in the $s$ plane, which may be a half-plane, the whole plane or empty.

Signal	$X (s)$	ROC
$e^{- a t} u (t)$	$\frac{1}{s + a}$	$Re s > - a$
$- e^{- a t} u (- t)$	$\frac{1}{s + a}$	$Re s < - a$

Both rows follow from chapter 2's $\int e^{- (s + a) t} d t$ with $a$ real: over $t > 0$ it needs $σ > - a$ , over $t < 0$ it needs $σ < - a$ , and there the signal's minus sign restores $1/ (s + a)$ .

WATCH OUT FOR

One expression, two signals: $1/ (s + a)$ alone names no signal. Test: read the ROC. For a rational $X (s)$ , a right-sided signal (zero before some instant) has its ROC to the right of its rightmost pole (a root of the denominator).

WORKED EXAMPLE

Find the Laplace transform and ROC of $x (t) = e^{- 2 t} u (t) + e^{t} u (- t)$ . Does its Fourier transform exist?

Solution

Transform each part separately: the integral of a sum is the sum of the integrals.
Right part: $\int_{0}^{\infty} e^{- (s + 2) t} d t = \frac{1}{s + 2}$ for $σ > - 2$ , the table's first row with $a = 2$ .
Left part: $\int_{- \infty}^{0} e^{(1 - s) t} d t = [\frac{e ^{(1 - s) t}}{1 - s}]_{- \infty}^{0} = \frac{1}{1 - s} - 0$ . The lower limit gives 0 only if $e^{(1 - σ) t} \to 0$ as $t \to - \infty$ , that is for $σ < 1$ .
The sum exists only where both parts exist, so the ROC is the overlap, $- 2 < σ < 1$ .
Add the parts: $\frac{1}{s + 2} + \frac{1}{1 - s} = \frac{( 1 - s ) + ( s + 2 )}{( s + 2 ) ( 1 - s )} = \frac{3}{( s + 2 ) ( 1 - s )}$ .
The strip contains $σ = 0$ , so the $j ω$ axis is inside the ROC and $s = j ω$ gives the Fourier transform: $X (ω) = \frac{3}{( 2 + j ω ) ( 1 - j ω )}$ .

Answer: $X (s) = \frac{3}{( s + 2 ) ( 1 - s )}$ with ROC $- 2 < Re s < 1$ . The Fourier transform exists.

Sampling: from Laplace to z and the DTFT

Sample $x (t)$ every $T_{s}$ seconds: $f_{s} = 1/ T_{s}$ in Hz, $ω_{s} = 2 π / T_{s}$ in rad/s, $x [n] = x (n T_{s})$ . Model the samples as impulses of area $x [n]$ at $t = n T_{s}$ . By sifting, an impulse at $t_{0}$ has Laplace transform $e^{- s t_{0}}$ , so sample $n$ contributes $x [n] e^{- s n T_{s}} = x [n] (e^{s T_{s}})^{- n}$ . So $s$ enters only through $e^{s T_{s}}$ . Name it $z$ ; $T_{s}$ is also written $T$ .

THE Z-TRANSFORM IS THE LAPLACE TRANSFORM OF THE SAMPLES

X_{s} (s) = n = - \infty \sum \infty x [n] e^{- s n T_{s}} = n \sum x [n] z^{- n} = X (z), z = e^{s T_{s}}

$X_{s} (s)$ is the Laplace transform of $x_{s} (t) = \sum_{n} x [n] δ (t - n T_{s})$ . The middle sum is the same sum with $e^{s T_{s}}$ renamed $z$ ; it defines $X (z)$ , the bilateral z-transform. Its ROC is a ring centred on the origin; for a causal sequence, $∣ z ∣ > r$ .

WHERE S LANDS IN THE Z PLANE

∣ z ∣ = e^{σ T_{s}}, ∠ z = ω T_{s} = Ω

From $z = e^{σ T_{s}} e^{j ω T_{s}}$ , angle modulo $2 π$ . $Ω = ω T_{s} = 2 π f / f_{s}$ is in rad/sample, with no seconds in it, unlike $ω$ in rad/s. Raising $ω$ by $ω_{s}$ leaves $z$ unchanged, so $Ω$ has period $2 π$ .

So $σ < 0$ gives $∣ z ∣ < 1$ , $σ > 0$ gives $∣ z ∣ > 1$ , and the line $σ = 0$ becomes the unit circle $∣ z ∣ = 1$ . Putting $z = e^{j Ω}$ in $X (z)$ gives the DTFT, $X (e^{j Ω}) = \sum_{n} x [n] e^{- j Ω n}$ , just as $s = j ω$ gave the Fourier transform. In DTFT work $Ω$ is often written as lower-case $ω$ ; here $ω$ stays in rad/s.

WATCH OUT FOR

$s = j ω$ or $z = e^{j Ω}$ gives the Fourier transform or DTFT only if the ROC contains the $j ω$ axis or unit circle. Test: for $u (t)$ (ROC $σ > 0$ ) and $u [n]$ (ROC $∣ z ∣ > 1$ ) it does not, so $1/ (j ω)$ is not the transform of $u (t)$ .

WORKED EXAMPLE

A signal is sampled with $T_{s} = 0.1$ s, and $z = e^{s T_{s}}$ . Where does $s = - 5 + j 10 π s^{- 1}$ land in the z plane, and which other $s$ land at the same point?

Solution

The modulus depends only on $σ$ : $∣ z ∣ = e^{σ T_{s}} = e^{- 5 \times 0.1} = e^{- 0.5} \approx 0.607$ .
The angle depends only on $ω$ : $ω T_{s} = 10 π rad/s \times 0.1 s = π$ rad.
So $z \approx 0.607 e^{j π} = - 0.607$ : on the negative real axis, inside the unit circle.
$ω_{s} = 2 π /0.1 = 20 π$ rad/s. Adding $ω_{s}$ to $ω$ adds $2 π$ to the angle and leaves $z$ unchanged, while $∣ z ∣$ fixes $σ = - 5$ .
Hence exactly the points $s = - 5 + j (10 π + 20 π k)$ , $k$ an integer, give this $z$ .

Answer: $z \approx - 0.607$ . Exactly the points $s = - 5 + j (10 π + 20 π k) s^{- 1}$ , $k$ an integer, map to it.

One family, and duality as a picture

The CTFT, Laplace, DTFT and z-transform differ in two choices. Is time continuous or discrete? Is $σ$ allowed to be non-zero? The DFT adds a third: for a length- $N$ sequence (zero outside $0 \leq n \leq N - 1$ ) it keeps $N$ samples of the DTFT, so $N$ numbers in time give $N$ numbers in frequency.

Name	Variable	What it is for	Link to the others
Fourier (CTFT)	$ω$ , real, rad/s	Spectrum of a continuous-time signal	Laplace on $s = j ω$
Laplace	$s = σ + j ω$ , complex	Continuous systems, stability, differential equations, always with the ROC	Fourier of $x (t) e^{- σ t}$
DTFT	$Ω$ , real, rad/sample, period $2 π$	Spectrum of a sequence	z on $z = e^{j Ω}$ , with $Ω = ω T_{s}$
z	$z = e^{s T_{s}}$ , complex	Discrete systems, difference equations, always with the ROC	Laplace of the sampled signal
DFT	$k = 0, \dots, N - 1$ , integer	Computing a spectrum from $N$ numbers	DTFT at $Ω = 2 π k / N$

THE DFT IS THE DTFT SAMPLED

X [k] = n = 0 \sum N - 1 x [n] W_{N}^{k n} = X (e^{j Ω})_{Ω = 2 π k / N}, W_{N} = e^{- j 2 π / N}

For a length- $N$ sequence the DTFT sum has only these $N$ terms, and $W_{N}^{k n} = e^{- j (2 π k / N) n}$ . $k = 0, \dots, N - 1$ is an integer; the samples are $2 π / N$ rad/sample apart. Only the inverse DFT carries a factor, $1/ N$ .

The CTFT and its inverse differ only in the sign of the exponent and the factor $1/2 π$ . So a pair read backwards, with time and frequency swapped, is another pair. This is duality.

DUALITY

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

Line 1: put $- t$ for $t$ in the inverse transform and multiply by $2 π$ . Line 2: rename the integration variable $ω$ as $t$ (any name will do) and the free variable $t$ as $ω$ . The right side is now the transform of the time function $X (t)$ . Both integrals must converge.

RECTANGLE AND SINC

\int_{- T /2}^{T /2} e^{- j ω t} d t = \frac{e ^{j ω T /2} - e ^{- j ω T /2}}{j ω} = \frac{2 sin ( ω T /2 )}{ω} = T sinc (\frac{ω T}{2 π})

For a height-1 rectangle on $∣ t ∣ < T /2$ , zero outside, $T$ in seconds, $ω \neq = 0$ ; at $ω = 0$ the value is $T$ . $sinc (x) = sin (π x) / (π x)$ , with $sinc (0) = 1$ . Peak $T$ at $ω = 0$ ; first zeros at $ω = \pm 2 π / T$ .

Now read it backwards. Duality with $W = T /2$ gives $2 sin (W t) / t ⟷ 2 π$ for $∣ ω ∣ < W$ and $0$ for $∣ ω ∣ > W$ . Divide by $2 π$ : $sin (W t) / (π t) = (W / π) sinc (W t / π)$ has transform $1$ for $∣ ω ∣ < W$ and $0$ for $∣ ω ∣ > W$ . That integral converges, though not absolutely, by chapter 2's sinc integral. Narrow in one domain means wide in the other: a wider rectangle pulls the sinc's first zeros in to $\pm 2 π / T$ .

IN ONE LINE

Discrete in one domain goes with periodic in the other: samples in time give a DTFT of period $2 π$ , a periodic signal gives lines at $k ω_{0}$ , and the DFT, discrete in both, has $X [k + N] = X [k]$ .

Check yourself

CHECK YOURSELF 1

A pulse $x (t) = 1$ for $∣ t ∣ < 1$ s, zero elsewhere, has $X (ω) = 2 sin ω / ω$ with $ω$ in rad/s. It is repeated every $T_{0} = 4$ s. Find the Fourier series coefficient $a_{1}$ to three decimal places.

Answer and solution

Answer: $0.318$ (accepted range $\pm 0.002$ )

$T_{0} = 4$ s $> 2 T_{1} = 2$ s, so $T_{0} a_{k} = X (k ω_{0})$ holds.
$ω_{0} = 2 π /4 = π /2$ rad/s, and $X (π /2) = 2 sin (π /2) / (π /2) = 4/ π$ .
$a_{1} = X (ω_{0}) / T_{0} = (4/ π) /4 = 1/ π \approx 0.318$ .

CHECK YOURSELF 2

The signal is $x (t) = e^{- 3 t} u (t)$ . Which pair gives its bilateral Laplace transform and ROC?

(A)
$\frac{1}{s + 3}$ , $Re s > - 3$
(B)
$\frac{1}{s + 3}$ , $Re s < - 3$
(C)
$\frac{1}{s - 3}$ , $Re s > 3$
(D)
$\frac{1}{s - 3}$ , $Re s < 3$

Answer and solution

Answer: (A)

$\int_{0}^{\infty} e^{- 3 t} e^{- s t} d t = \int_{0}^{\infty} e^{- (s + 3) t} d t$ .
This converges when $σ + 3 > 0$ , that is $Re s > - 3$ , and equals $1/ (s + 3)$ .
The signal is right-sided, so the ROC lies to the right of the pole at $- 3$ . Answer A.

CHECK YOURSELF 3One or more options are correct

Which statements are correct?

(A)
$X (s)$ at $s = j ω$ is the Fourier transform when the ROC contains the $j ω$ axis.
(B)
With $z = e^{s T_{s}}$ , the $j ω$ axis maps onto the unit circle.
(C)
With $z = e^{s T_{s}}$ , the left half of the $s$ plane maps to $∣ z ∣ > 1$ .
(D)
Two different signals can have the same $X (s)$ with different ROCs.

Answer and solution

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

A is true: $s = j ω$ means $σ = 0$ , which must lie in the ROC.
B is true: $σ = 0$ gives $∣ z ∣ = e^{0} = 1$ .
C is false: $σ < 0$ gives $∣ z ∣ = e^{σ T_{s}} < 1$ .
D is true: with $a$ real, $e^{- a t} u (t)$ and $- e^{- a t} u (- t)$ both give $1/ (s + a)$ with different ROCs.

CHECK YOURSELF 4

The sampling period is $T_{s} = 0.5$ ms and $z = e^{s T_{s}}$ . For $s = - 2000 + j 3000 s^{- 1}$ , find $∣ z ∣$ to three decimal places.

Answer and solution

Answer: $0.368$ (accepted range $\pm 0.002$ )

Convert: $T_{s} = 0.5 ms = 5 \times 1 0^{- 4}$ s.
$∣ z ∣ = e^{σ T_{s}}$ , so only $σ = - 2000 s^{- 1}$ matters.
$σ T_{s} = - 2000 \times 5 \times 1 0^{- 4} = - 1$ , so $∣ z ∣ = e^{- 1} \approx 0.368$ .

From series to transform: one family, five names

What you will be able to do

Before you start

From the series to the transform, then to Laplace

Sampling: from Laplace to z and the DTFT

One family, and duality as a picture

Check yourself

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