continuous time signals basic signals – singularity functions transformations of continuous time...
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Continuous Time Signals
•Basic Signals – Singularity Functions•Transformations of Continuous Time Signals•Signal Characteristics•Common Signals
April 19, 2023 Veton Këpuska 2
Continuous-Time Signals
Assumptions: Functions, x(t), are of the one
independent variable that typically represents time, t.
Time t can assume all real values: -∞ < t < ∞,
Function x(t) is typically a real function.
Singularity Functions
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-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.2
0
0.2
0.4
0.6
0.8
1
u(t)
time [sec]
Unit Sample Signal
Unit Step Function
Unit step function definition:
0,0
0,1
t
ttu
April 19, 2023 Veton Këpuska 4
Unit Step Function Properties Scaling:
Unit step function can be scaled by a real constant K (positive or negative)
Multiplication: Multiplication of any
function, say x(t), by a unit step function u(t) is equivalent to defining the signal x(t) for t≥0.
April 19, 2023 Veton Këpuska 5
tKutf
0, ttxtutx
Unit Ramp Function
Unit Ramp Function is defined as:
April 19, 2023 Veton Këpuska 6
0,0
0,
t
tttr
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.2
0
0.2
0.4
0.6
0.8
1
r(t)
time [sec]
Unit Sample Signal
Unit Ramp Function Properties Scaling:
Unit step function can be scaled by a real constant K (positive or negative)
Integral of the unit step function is equal to the ramp function:
Derivative of the unit ramp function is the unit step function.
April 19, 2023 Veton Këpuska 7
tKrtf
Slope of the straight line
t
dutr
dt
tdrtu
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.2
0
0.2
0.4
0.6
0.8
1
(t)
time [sec]
Unit Sample Signal
Unit Impulse Function
Unit Impulse Function, also know as Dirac delta function, is defined as:
April 19, 2023 Veton Këpuska 8
&01
0,0
0,
d
t
tt
Unit Impulse Function Properties Scaling:
Unit impulse function can be scaled by a real constant K (positive or negative)
Delta function can be approximated by a pulse centered at the origin
April 19, 2023 Veton Këpuska 9
)(lim tdtA
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.2
0
0.2
0.4
0.6
0.8
1
(t)
time [sec]
Unit Sample Signal
A2
1
A2
1
A
Unit Impulse Function Properties Unit impulse function is
related to unit step function:
Conversely:
April 19, 2023 Veton Këpuska 10
dt
tdut
0&
ttdtut
Proof:1. t<0
2. t>0
0 since ,00
ttdtut
0,1 since ,01
dtdtut
Time Transformation of Signals
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Time Reversal:
)()(
&
)()(
,
)()(
00
00
0
txty
txty
tt
txty
1
2
0-1-2
-1
1 2 t
1
2
0-1-2
-1
1 2 t
y(t)=x(-t)
x(t)
April 19, 2023 Veton Këpuska 13
Time Scaling
)()(
&
)()(
,
0
&
)()(
00
000
txa
ty
atxty
a
tt
atxty1
2
0-1-2
-1
1 2 t
1
2
0-1-2
-1
1 2 t
y(t)=x(2t)
x(t)
1
2
0-1-2
-1
1 2 t
y(t)=x(0.5t)
3 4
|a| > 1 – Speed Up
|a| < 1 – Slow Down
April 19, 2023 Veton Këpuska 14
Time Shifting
)0()(
:Note
constant a is ,
)()(
: signal aGiven
0
0
0
xty
t
ttxty
x(t)
1
2
0-1-2
-1
1 2 t
x(t)
1
2
0-1-2
-1
1 2 t
x(t-2)
3 4
1
2
0-1-2
-1
1 2 t
x(t+1)
April 19, 2023 Veton Këpuska 15
Example 1
233cos
23cos
)()(
23cos)(
0
0
0
0
0
ttee
tte
ttxty
tetx
tt
tt
t
April 19, 2023 Veton Këpuska 16
Independent Variable Transformations
a
b
a
tytx
batxtytt
a
b
atbat
babatxty
00
00
0 )(
)()(
, ),()(
April 19, 2023 Veton Këpuska 17
Example 2
2
3
232
)32()(
tt
txty
April 19, 2023 Veton Këpuska 18
Example 3
2
0-1-2
-1
1 2
x()
1
024 -2 t=2-2
2
0-2-4
-1
2 4
y(t)=x(1-t/2)
1
2
1
2
11)(
)0()212()1()(1
2212
21)(
0
0
0 xxty
yyxtxt
tt
txty
April 19, 2023 Veton Këpuska 19
Independent Variable Transformations
1. Replace t with , on the original plot of the signal.
2. Given the time transformation:
Solve for
3. Draw the transformed t-axis directly below the -axis.
4. Plot y(t) on the t-axis.
bat
a
b
at
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Amplitude Transformations
constants real,
)()(
BA
BtAxty
April 19, 2023 Veton Këpuska 21
Example 4 Consider signal in the
figure. Suppose the signal is applied to an amplifier with the gain of 3 and introduces a bias (a DC value) of -1. That is:
1)(3)( txty
2
0-1-2
-1
1 2 t
x(t)
1
3 +-1
x(t) 3x(t) 3x(t)-1
2
0-1-2
-1
1 2 t
3x(t)-1
1
April 19, 2023 Veton Këpuska 22
Example 5
2
0-1-2
-1
1 2
3x()-1
1
246 0 -2 t
1 t/2 => t=-22
2
-1
3x(1 t/2 )-1
1
26 0 -2 t4
222
1
12
13)(
tt
txty
April 19, 2023 Veton Këpuska 23
Transformations of Signals
Name y(t)
Time reversal x(-t)
Time scaling x(at)
Time shifting x(t-t0)
Amplitude reversal -x(t)
Amplitude scaling Ax(t)
Amplitude shifting x(t)+B
Signal Characteristics
April 19, 2023 Veton Këpuska 25
Even and Odd Signals
Even Functions xe(t)=xe(-t)
Odd Functions xo(t)=-xo(-t)
2A
0-1-2
-A
1 2 t
xe(t)
A
2A
0-1-2
-A
1 2 t
xo(t)
A
April 19, 2023 Veton Këpuska 26
Even and Odd Signals
Any signal can be expressed as the sum of even part and on odd part:
)()(2
1)()()(
2
1)(
)()()(
)()()(
)()()(
txtxtxtxtxtx
txtxtx
txtxtx
txtxtx
oe
oe
oe
oe
April 19, 2023 Veton Këpuska 27
Average Value Average Value of the signal x(t) over a period of time [-T, T] is
defined as:
T
T
eT
x
T
T
T
T
oeT
x
T
T
oeT
x
oe
T
TT
x
dttxT
A
dttxdttxT
A
dttxtxT
A
txtxtx
dttxT
A
)(2
1lim
})()({2
1lim
)]()([2
1lim
)]()([)(
)(2
1lim
The average value of a signal is contained in its even function (why?).
April 19, 2023 Veton Këpuska 28
Properties of even and odd functions
1. The sum of two even functions is even.2. The sum of two odd functions is odd.3. The sum of an even function and an odd
function is neither even nor odd.4. The product of two even functions is even.5. The product of two odd functions is even.6. The product of an even function and an
odd function is odd.
Periodic Signals
April 19, 2023 Veton Këpuska 29
April 19, 2023 Veton Këpuska 30
Periodic Signals
tTTtxtx & 0),()(
Continuous-time signal x(t) is periodic if:
T is period of the signal.
A signal that is not periodic is said to be aperiodic.
April 19, 2023 Veton Këpuska 31
Periodic Signals
If constant T is a period of of a function x(t) than nT is also its period, where T>0 and n is any positive integer.
)()(
)2()(
nTtxtx
TtxTtx
The minimal value of the constant T >0 is a that satisfies the definition x(t)= x(t+ T) is called a fundamental period of a signal and it is denoted by T0.
April 19, 2023 Veton Këpuska 32
Examples of Periodic Signals
0 1 2 3 4 5 6 7 8 9 10
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
y(t)=
cos(
wt)
Time [s]
)cos()( tty
Sinusoidal Signal Properties
ttAtx cos
April 19, 2023 Veton Këpuska 33
• A – Amplitude of the signal
• - is the frequency in rad/sec
• - is phase in radians
-1.5 -1 -0.5 0 0.5 1 1.5
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Periodic Signal
cos(t+ )
Angle [rad]
cos(-t+0.00)
cos(-t+0.52)
Sinusoidal Function Properties
Note:
April 19, 2023 Veton Këpuska 34
sincoscossinsin
sinsincoscoscos
][ ,
rad
sin2sin
cos2cos
][
rad
Periodicity of Sinusoidal Signal
April 19, 2023 Veton Këpuska 35
fT
Ttx
TtA
tA
tA
tAtx
22
cos
2cos
2cos
cos
April 19, 2023 Veton Këpuska 36
Example: Sawtooth Periodic Waveform
-3 -2 -1 0 1 2 3 4
0
0.2
0.4
0.6
0.8
1
y(t)
April 19, 2023 Veton Këpuska 37
Period and Frequency
Fundamental Period T0 – Measured in seconds.
Fundamental Frequency f0 – Measured in Hz – number of periods (cycles) per second or equivalently in radian frequency rad/s.
s
rad
Tf
HzT
f
000
00
122
][1
April 19, 2023 Veton Këpuska 38
Testing for Periodicity
1.
)()(
2for )sin()sin(
)(
?)(
)sin(
)sin(
)sin(
txeTtx
TtTt
eTtx
etx
t
Tt
t
April 19, 2023 Veton Këpuska 39
Testing for Periodicity
2.
)(
)(
)()(
2for )sin()sin(
)()(
?)(
)sin(
)sin()sin(
)sin(
)sin(
)sin(
tx
Tetx
Tete
eTtTtx
TtTt
eTtTtx
tetx
t
tt
t
Tt
t
April 19, 2023 Veton Këpuska 40
Composite Signals Each signal can be decomposed into a sum of series of
pure periodic signals (Taylor Series Expansion/Fourier Series Expansion)
The sum of continuous-time periodic signals is periodic if and only if the ratios of the periods of the individual signals are ratios of integers.
Composite Signals If a sum of N periodic signals is periodic, the
fundamental period can be found as follows:
1. Convert each period ratio, To1/Toi≤ i ≤ N , to a ratio of integers, where To1 is the period of the first signal considered and Toi is the period of one of the other N-1 signals. If one or more of these ratios is not rational, the sum of signals is not periodic.
2. Eliminate common factors from the numerator and denominator of each ratio of integers.
3. The fundamental period of the sum of signals is To=koTo1 ; kois the least common multiple of the denominators of the individual ratios of integers.
April 19, 2023 Veton Këpuska 41
Composite Signals
If x1(t) is periodic with period T1, and
x2(t) is periodic with period T2, Then
x1(t)+x2(t) is periodic with period equal to the least common multiple (T1, T2) if the ratio of the two periods is a rational number, where k1 and k2 are integers:
April 19, 2023 Veton Këpuska 42
22111
2
2
1 TkTkk
k
T
T
Composite Signals
Let T’= k1T1 = k2T2
y(t) = x1(t)+x2(t)
Then y(t+T’) = x1(t+T’)+x2(t+T’)= x1(t+ k1T1)+x2(t+ k2T2)=
x1(t)+x2(t) = y(t)
April 19, 2023 Veton Këpuska 43
Example 2.7 a)
Assume that v(t) is a sum of periodic signals given below. Determine if the signal is periodic and what its periodicity?
April 19, 2023 Veton Këpuska 44
)6
7cos()(
)2cos()(
)5.3cos()(
)()()()(
3
2
1
321
ttx
ttx
ttx
txtxtxtv
0 10 20 30 40 50 60-4
-2
0
2
4
v(t)
time
0 10 20 30 40 50 60-1
-0.5
0
0.5
1
x1(t)
0 10 20 30 40 50 60-1
-0.5
0
0.5
1
x2(t)
0 10 20 30 40 50 60-1
-0.5
0
0.5
1
x3(t)
Solution
Determine whether v(t) constituent signals have periods with ratios that are integers (rational numbers):
April 19, 2023 Veton Këpuska 45
67
22
2
22
5.3
22
303
202
101
T
T
T
21
7
5.367
67
25.3
2
7
4
5.3
2
22
5.32
03
01
02
01
T
T
T
T
Solution
Ratios of periods are rational numbers thus the composite signal v(t) is periodic.
Elimination of common factors: T01/T02 = 4/7
T01/T03 = 7/21=1/3
Least common multiple of the denominator ratios: n1= 3*7=21
Fundamental period of v(t) is: T0= n1 T01 = 21*2/3.5=12
April 19, 2023 Veton Këpuska 46
Example 2.7 b)
Assume that to v(t) is added a periodic signal x4(t) given below. Determine if the signal is periodic and what its periodicity?
April 19, 2023 Veton Këpuska 47
)5cos(3)(
)6
7cos()(
)2cos()(
)5.3cos()(
)()()()()(
4
3
2
1
4321
ttx
ttx
ttx
ttx
txtxtxtxtv
0 10 20 30 40 50 60-6
-4
-2
0
2
4
6
v(t)
time
Solution
Since ratio of the x1(t) and x4(t) periods is not a rational number the v(t) is not periodic.
April 19, 2023 Veton Këpuska 48
5
22
404 T
7
10
5.3
5
52
5.32
04
01 T
T
Homework #1:
April 19, 2023 Veton Këpuska 49
1. For x(t)=Acos(t+) find What are its maximum and minimum values?
What are corresponding times when they occur?
What is the value of the function when it crosses vertical y- axis (ordinate) and horizontal x-axis (abscissa)?
At what time instances the function becomes zero?
Indicate all the above point values in a plot.
Homework #1 Use the following MATLAB
script to test your calculations and plot the function:
function pfunc(A, f, th1, th2)%% Periodic Sine Function% A - gain (1)% f - frequency (1)% th1 - phase of the first signal (0)% th2 - phase of the second signal
(pi/6)% w = 2.*pi.*f; % radial frequencyfs = 0.0001*f;mint = -pi*f/2;maxt = pi*f/2;miny = -1.2*A;maxy = 1.2*A;
t = mint:fs:maxt; % time axisy = A*cos(w*t+th1); plot(t, y, 'b', 'LineWidth',2);
title('Periodic Signal'); grid on; hold; axis([mint maxt miny
maxy]);y = A*cos(w*t+th2);plot(t, y, 'r', 'LineWidth',2);ylabel('cos(\omegat+\theta)');
xlabel('Angle x\pi [rad]'); grid on; hold; axis([mint maxt miny
maxy]);x=-0.8; text(x,A*cos(w*x+th1),sprintf('%s+
%3.2f)','\leftarrow cos(-\pit',th1),...
'HorizontalAlignment','left',... 'BackgroundColor','b');x=-0.6; text(x,A*cos(w*x+th2),sprintf('%s+
%3.2f)','\leftarrow cos(-\pit',th2),...
'HorizontalAlignment','left',... 'BackgroundColor','r');
April 19, 2023 Veton Këpuska 50
April 19, 2023 Veton Këpuska 51
Homework #1
Problems 2.1, 2.2, 2.9, 2.10, 2.13, 2.14, 2.20.
April 19, 2023 Veton Këpuska 52
Example
2
0-1-2
-1
1 2 t
x(t)
1
2
0-1-2
-1
1 2 t
x(-t)
1
2
0-1-2
-1
1 2 t
xe(t)
1
2
0-1-2 1 2 t
xo(t)
1
-1
Consider the signal to the left and its time reversed version. The signal is decomposed into its even and odd functions:
)()(2
1)()()(
2
1)( txtxtxtxtxtx oe
Common Signals in Engineering
April 19, 2023 Veton Këpuska 53
Common Signals
Continuous-time physical systems are typically modeled with ordinary linear differential equations with constant coefficients.
April 19, 2023 Veton Këpuska 54
0,)0()(
:Solution
constant),()(
textx
ataxdt
tdx
at
Exponential Signals
Useful Complex Exponential Relations
April 19, 2023 Veton Këpuska 55
partImaginary
part Real
constantscomplex becan &,)(
ja
aCCetx at
Euler’s Formula
April 19, 2023 Veton Këpuska 56
j
eeee
je
je
jjjj
j
j
2sin&
2cos
sinsin& coscos
sincos
sincos
Example of Exponential Functions
1. C and a real, x(t)=Ceat a= Increasing Exponential:
Chemical Reactions, Uninhibited growth of bacteria, human population?
a= Decaying Exponential: Radioactive decay, response of an RC circuit, damped mechanical
system. a= Constant (DC) signal.
April 19, 2023 Veton Këpuska 57
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
0
0.5
1
1.5
2
2.5
Aet
Ae+/
-t
Time [sec]
Aet, >0Ae
t, <0
Aet, =0
Time Constant of the Exponential Function
The constant parameter t is called the time constant in of the exponential function presented below.
To relate to the time constant the following is necessary:
0,
t
at CeCetx
CbC
abaty
Ce
CCxt
eC
Cedt
d
dt
tdx
t
t
tt
;
0,0
0,
0
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
0
0.5
1
1.5
2
2.5
Aet
Ae
t
Time [sec]
0.368
Example of Exponential Functions
2. C complex, a imaginary, x(t)=Ceat a=j C=Aej – A and are real:
For C – real (=0)
2. x(t) is periodic:
3. Why x(t) is periodic?
April 19, 2023 Veton Këpuska 59
tjAtAAeeAeCetx tjtjjat sincos
tjAtAAeCetx tjat sincos
period theis TTtxtx
Periodicity of Complex Exponential
April 19, 2023 Veton Këpuska 60
txtxeTtx
jje
jTjT
Tff
T
TjTe
txeeCeCeTtx
Cetx
Tj
Tj
Tj
TjTjtjTtj
tj
~
1012sin2cos
2sin
2cossincos
2 2&
1
sincos
~
Example of Complex Exponentials
3. C complex, a complex, x(t)=Ceat C=Aej A and are real; a=j are also real.
April 19, 2023 Veton Këpuska 61
ImRe
Factor Damping
sincos
sincos
tAejtAe
tjtAe
eAeeAeAe
eAeCetx
tt
t
tjtjtjtjtjt
tjjat
Time-Shifted Signals
Time-Shifted Impulse Function Shift (t) by t0:
“Sifting” Property of the Impulse Function:
April 19, 2023 Veton Këpuska 62
0
00 0
,1
tt
tttt
00 tfdtf
Examples of Impulse Functions
April 19, 2023 Veton Këpuska 63
Properties of the Unit Impulse Function
April 19, 2023 Veton Këpuska 64
tt
dta
tt
adttat
t
ttdtttu
ttudt
dtt
tttftttftttf
tttftfdttttf
tttftfdttttf
t
.7
1.6
0,0
,1.5
.4
at continuous .3
at continuous ,.2
at continuous ,.1
-
0
-
0
-
000
00
0000
0
-
00
0
-
00
Shifted Unit Step Function
April 19, 2023 Veton Këpuska 65
-10 -8 -6 -4 -2 0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
u(t) & u(t-td)
u(t)
& u
(t-t
d)
Time [sec]
u(t) u(t-td)
-10 -8 -6 -4 -2 0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
u(t) & u(t-td)
u(t)
& u
(t-t
d)
Time [sec]
u(t)u(t-td)
4& tutu
5& tutu
0
00 0
,1
tt
ttttu
Continuous & Piece-wise Continuous Functions
-10 -8 -6 -4 -2 0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
p(t)
p (t)
Time [sec]
p(t)
-/2 +/2
Rectangular Function
April 19, 2023 Veton Këpuska 66
elsewhere
ttptp0
22,1,
Rectangular Function
April 19, 2023 Veton Këpuska 67
-10 -8 -6 -4 -2 0 2 4 6 8 10
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
p(t)=u(t+td)-u(t-td)
p (t)
, u(t)
& u
(t-t
d)
Time [sec]
u(t)
u(t-td)
p(t)
-/2 +/2
22,
tututptp
Triangular Pulse Function
April 19, 2023 Veton Këpuska 68
201
2
02
,12
tt
tt
tp tr
-10 -8 -6 -4 -2 0 2 4 6 8 10-0.2
0
0.2
0.4
0.6
0.8
1
ptr(t)
p tr(t)
Time [sec]
ptr(t), t<0 p
tr(t), t>0
ptr(t)
-/2 +/2
Straight Line Equation
00 xxmyy
[x0,y0] - a point on the line
[x1,y1] - a point on the line
[x2,y2] - a point on the line
m = (y2-y1)/(x2-x1) - slope
April 19, 2023 Veton Këpuska 69
Composite Signal from Straight Lines
April 19, 2023 Veton Këpuska 70
3222111100
322222
211100
211111
10
10000
,
,
,
,
tttxttuttmtxttuttmtxtx
tttttuttmtx
tttxttuttmtxtx
tttttuttmtx
tttxtx
ttttuttmtx
Example of Composite Signal
April 19, 2023 Veton Këpuska 71
25113 tututttututf
-10 -8 -6 -4 -2 0 2 4 6 8 10
-5
-4
-3
-2
-1
0
1
2
3
4
5
f(t)=
3u(t)
+tu(
t)-(t-
1)u(
t-1)-5
u(t-2
)
Time [sec]
u(t)
tu(t)
-(t-1)u(t-1)
-5u(t-2)
f(t)=3u(t)+tu(t)-(t-1)u(t-1)-5u(t-2)