1 Signals

These notes were written by Danylo Malyuta for the Robotics, Systems and Control Master atETH Zurich. The aim is to review the basic concepts necessary to understand System Identificationconcepts more easily.

1.1 Fourier transform

1.1.1 Continuous Fourier Transform

The forward continuous time Fourier transform is defined as:

X() =

x(t)ejtdt (1)

The inverse continuous time Fourier transform is define as:

x(t) =

X()ejtd (2)

Where we have = 2pif is the angular frequency while f is the frequency [Hz].

1.1.2 Continuous Time Fourier series (CTFS)

The continuous time Fourier series attempts to reconstruct any periodic signal by a sum ofsines and cosines (i.e. sum of sinusoids) of frequencies f = f0, 2f0, 3f0, . . . ,f0 where f0 = 1/Tis the fundamental frequency and T is the original signals period. The formula is:

x(t) = Re

[a02

+

n=1

{Cne

jnt}]

a0 =2

T

T0

f(t)dt

Cn =2

T

T0

f(t)ejntdt

(3)

Then an amplitude plot may be obtained by plotting |Cn| for n = 1, 2, . . . , and a phase plotmay be obtained by plotting (Cn) = arctan

(Im[Cn]Re[Cn]

)for n = 1, 2, . . . ,.

1.1.3 Discrete Time Fourier series (DTFS)

The DTFS does the job of the CTFS, but in discrete time. Suppose we are given a discretesignal x[n] for n = 0, . . . , N 1 - in other words this is nothing more than an N -row table ofpoints. A very important side-note: the discrete signal x[n] must be periodic and the N pointsyou give must be those of a single period!!!

By direct analogy with (3) (i.e. replace integrals with sums) we obtain the DTFS:Xk =

1

N

N1n=0

{x[n]ej

2piN nk

}x[n] =

N1k=0

{Xke

j 2piN nk} (4)

Then an amplitude plot may be obtained by plotting 2 |Xk| for n = 0, 1, . . . , N1 and a phaseplot may be obtained by plotting (Xk) = arctan

(Im[Xk]Re[Xk]

)for n = 0, 1, . . . , N 1.

As far as I can tell, its the amplitude plot that is most useful as the distinct peaks show directlythe frequencies present in the signal - despite even severe amounts of noise!

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1.2 Laplace transform

The Laplace transform may be derived directly following from the Fourier transform (1). Thephysical world is governed by differential equations - these have as their solution exponentials andsinusoids. Thats just how the world works. We can add therefore an exponential term into theFourier transform, giving us:

X(, ) =

(x(t)et

)ejtdt

=

x(t)e(+j)tdt

=

x(t)estdt

(5)

Where we defined:

s = + j (6)

The last integral in (5) is the Laplace transform:

X(s) =

x(t)estdt (7)

Here is the very simple - and easy to remember - relationship between the Laplace and theFourier transforms, which also helps understand a lot about what is happening in the s-plane thatwe shall talk about next:

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Figure 1: Taken from The Scientist and Engineers Guide to Digital Signal Processing, freelyavailable online.

1.2.1 Properties of the s-plane

The Laplace transform is very easy to understand: in the s-plane, the imaginary axis (where = 0) is simply the Fourier transform (1) while the right-half plane corresponds to multiplyingthe function x(t) by a decreasing exponential (because there we have > 0 and recall that wemultiply by et) while the left-half plane corresponds to multiplying by increasing exponentials!

A pole in the s-plane corresponds to a condition on {, } which makes the integral (7) barely-

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infinite while a zero corresponds to a conditions on {, } which makes the integral (7) evaluateto zero. Here is a good example of this idea in action:

Figure 2: Taken from The Scientist and Engineers Guide to Digital Signal Processing, freelyavailable online.

Finally lets show some properties about a second-order system that can be read off the s planejust by taking a quick look:

4

j

0

0

1 2

= 0

Figure 3: Properties we can read directly off the s-plane.

We have for a typical mass-spring-damper oscillator:

mx+ bx+ cx = F (8)

0 =

k

md = 0

1 2 =

0(9)

Referring to Figure 4 we have:

= arccos() (10)

Therefore lines of constant are lines of constant damping ratio . These lines radiate fromthe origin of the s-plane. This is how the poles of a second order system travel in the s-plane:

j

= 0 no damping

= 1

> 1

0

Lines of constant dampingratio

Figure 4: Motion of the two poles of a second-order system in the s-plane.

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Finally also for second order systems we have:

The more we in the direction of , the faster the signal attenuation. The more we in the direction of +, the slower the signal attenuation (and its a signal

amplification if we go into the right hald plane).

The more we go away from the horizontal axis, the faster the oscillation.

1.3 The z transform

The z transform comes into play for discrete systems - therefore, today more than ever, it is ofutmost importance. Fortunately, its very easy to understand. Its formulation follows by directanalogy from the continuous-time Laplace transform:

X(, ) =

x(t)etejtdt X(, ) =

n=

{x[n]enejn

}(11)

Let us define z as:

z := eej := rej (12)

Therefore, the z transform of a discrete signal x[n] is defined as:

X(z) =

n=

{x[n]zn

}(13)

1.3.1 z transform properties

Let us summarize side-by-side the definition of s and z:{s = + j

z = rej(14)

From these definitions, we see that s is defined in a rectangular coordinate system while z isdefined in a polar coordinate system! Therefore, lines in the s domain appear as circles in the zdomain. This leads to a very important property about stability in continuous vs. discrete time.

Recall that in continuous time, stable systems have their poles strictly to the left of the imag-inary axis (the = 0 axis) in the s plane. For = 0, z = ej which is the unit circle in the zplane. Correspondingly, we get circles smaller than the unit circle in the z plane for the left halfof the s plane and we get circles larger than the unit circle in the z plane for the right half of theplane. Identifying this with the stability property in the s plane, we conclude that stable discretesystems have their poles strictly inside the unit circle if in the z plane! Here is an illustration:

6

< 0 = 0 > 0

Marginally stable

Stable

Unstable

s-plane z-planej

r = 1

Figure 5: Stability in the s vs. the z planes.

Another important point can be uncovered: because, in the z plane, the natural frequency represents the phase (in polar coordinates), it follows that by definitions pi < < pi. Let usrelate the discrete samples n to a moment in time nTs where Ts is the sampling time intervalin seconds. We may pass this extra Ts into our definition of z to give:

z = eTsejTs (15)

For the frequency response (discussed later) we pass a pure, unattenuated sinusoid, therefore = 0 and we get:

z = ejTs (16)

Here, Ts represents the phase and we have concluded that pi < Ts < pi, therefore since := 2pif we have after simplification:

fs2< f