Fundamentals of Signals

Valentina Hubeika, Jan Cernocky

DCGM FIT BUT Brno, {ihubeika|cernocky}@fit.vutbr.cz

• clasification of signals

• transformation of time axis

• energy and power of a signal

• periodic signals

• harmonic signals

• unit step and impulse signal

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Signals

• arbitrary physical values.

• one or more independent axis (usually time), one dependent variable.

• Example: acoustic preasure generated by humans, gray scales of BW video frames,

ratio of CZK/EURo.

2

0 5 10 15 2031.6

32

32.4

prac. dnyC

ZK

/EU

R −

lede

n 20

02

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Mathematic view of signals ( 1 dependent variable)

According to character of T , we divide signals:

• Continuous time signals: t ∈ ℜ, s(t).

• Discrete time signals: n ∈ Z, s[n].

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Deterministic and non-deterministic signals

Deterministic signals can be described by equation. Example 1: continuous-time square

impuls:

x(t) =

2 pro − 2 ≤ t ≤ 2

0 jinde

Example 2: discrete unit impulse:

δ[n] =

1 pro n = 0

0 jinde

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Non-deterministic signals cannot be described by equation.

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Transformation of independent variable – modification of time axis

Examples for continuous time signals:

Turn of time axis: s(−t).

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Delay of signals: s(t − τ) for positive τ .

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Turn of time axis with shifting

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Examples for discrete time signals – discrete unit step:

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. . . Try to check it!11

Modification of time scale

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Energy and power

p(t) = u(t)i(t) = u2(t)/R = i2(t)R.

current power:

p(t) = |s(t)|2

For discrete signals:

p[n] = |s[n]|2

We are interested in interval [t1, t2]:

∫ t2

t1

p(t)dt =

∫ t2

t1

|s(t)|2dt1

t2 − t1

∫ t2

t1

p(t)dt =1

t2 − t1

∫ t2

t1

|s(t)|2dt

similarly for discrete signals:

n2∑

n1

p[n] =

n2∑

n1

|s[n]|2 1

n2 − n1

n2∑

n1

p[n] =1

n2 − n1

n2∑

n1

|s[n]|2

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Absolute energy in interval−∞ to ∞:

E∞ = limT→∞

∫ T

−T

|s(t)|2dt =

∫ +∞

−∞

|s(t)|2dt E∞ = limN→∞

∞∑

−∞

|s[n]|2 =

N∑

−N

|s[n]|2

Signals with finite energy and with infinite energy.

Overall mean power:

P∞ = limT→∞

1

2T

∫ T

−T

|s(t)|2dt P∞ = limN→∞

1

2N + 1

N∑

−N

|s[n]|2

Pokud je P∞ nenulovy, je E∞ = ∞.

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Non-periodic and periodic signals

For periodic signals, we can find such T or N :

s(t + T ) = s(t) continuous time (1)

s[n + N ] = s(n) discrete time, (2)

For example:

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For discrete signals:

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Harmonic signals

s(t) = C1 cos(ω1t + φ1) (3)

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• C1 is– amplitude.

• ω1 – frequency [rad/s]. ω1 = 2πf1. Fundamental period T1 = 1

f1

= 2πω1

.

• φ1 is phase [rad]. For t = 0 s(0) = C1 cos φ1.

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Power of periodic signals

For an arbitrary signal the average power is :

P∞ = limT→∞

1

2T

∫ T

−T

|s(t)|2dt

For a periodic signal it is enough to integrate over one period:

Ps =1

T1

∫ T1/2

−T1/2

|s(t)|2dt =1

T1

∫ T1

0

|s(t)|2dt = . . .

Root mean quare (quadratic mean) is the value of a constant signal with the average

power Ps:

Cef =√

Ps.

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For harmonic signals: s(t) = C1 cos(ω1t).

if we use: cos2 α = 1+cos 2α2

:

Ps =1

T1

∫ T1

0

[C1 cos(ω1t)]2dt =

1

T1

∫ T1

0

C12 1

2(1 + cos 2ω1t)dt

Ps =1

T1

∫ T1

0

C12 1

2dt =

1

T1

C12

2T1 =

C12

2

Cef =√

Ps =C1√

2,

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Harmonic signals for discrete time

s[n] = C1 cos(ω1n + φ1) (4)

• C1 – amplitude.

• ω1 – angular frequency. ω1 is measured only [rad] (time n is dimensionless).

• φ1 phase [rad]. For n = 0 je s[0] = C1 cos φ1.

Example: s[n] = 5 cos(2πn/12), ω1 = π/6.

−20 −10 0 10 20−5

0

5

n

s[n]

The basic period N1 cannot be calculated as for continuous time signals N1 = 2π/ω1 as

we need N1 result into an integer number.

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Example 1: s[n] = 5 cos(2πn/12), ω1 = π/6. Find the basic period N1.π6N1 = k2π, solution: k = 1, N1 = 12 (see picture).

For continuous time signal: T1 = 2ππ/6

= 12. Same result.

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Example 2: s[n] = cos(8πn/31), ω1 = 8π/31.8π31

N1 = k2π, k = 4

31N1, 31k = 4N1. The solution is : k = 4, N1 = 31.

−20 −10 0 10 20−1

−0.5

0

0.5

1

n

s[n]

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Example 3: s[n] = cos(n/6), ω1 = 1/6.1

6N1 = k2π, N1 = k12π. No solution,non-periodic signals

−20 −10 0 10 20−1

−0.5

0

0.5

1

n

s[n]

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Some interesting signals – discrete time

Unit step signal and impuls:

σ[n] =

1 for n ≥ 0

0 elsewhereδ[n] =

1 for n = 0

0 elsewhere

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We get the impulse signal by differentiating the unit step signal:

δ[n] = σ[n] − σ[n − 1]

Shifted unit impuls δ[n − k]:

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Some interesting signals – continuous time

Unit step: σ(t) =

1 for t ≥ 0

0 elsewhere

(Dirac impuls): We derivate the unit step function with respect to time:

δ(t) =dσ(t)

dt.

We cannot derivate at t = 0 (non-continuity). Thus we use a trick:

δ∆(t) =dσ∆(t)

dt.

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Now, the slope of δ(t) between 0 and ∆ is 1/∆ (thus δ has derivation). Note the area of

the rectangle defined by δ∆(t) is equal to 1. In limit, ∆ → 0

δ(t) = lim∆→0

δ∆(t).

The area is still 1∫ +∞

−∞

δ(t)dt = 1.

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