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BASIC CONCEPTS
•Signal Waveforms
Continuous/Discontinuous
SINUSOIDS
x(t )  X M sin  t
Adimensional plot
As function of time
X M  amplitude or maximum value
  angular frequency (rads/sec)
 t  argument (radians)
T
2

" leads by  "
 Period  x (t )  x (t  T ), t
1 

 frequency in Hertz (cycle/sec)
T 2
  2 f
f 
" lags by  "
BASIC TRIGONOMETRY
ESSENTIAL IDENTITIES
sin(   )  sin  cos   cos sin 
cos(   )  cos cos   sin  sin 
sin(  )   sin 
cos( )  cos
SOME DERIVED IDENTITIES
sin(   )  sin  cos   cos sin 
cos(   )  cos cos   sin  sin 
1
1
sin  cos   sin(   )  sin(   )
2
2
1
1
cos cos   cos(   )  cos(   )
2
2
APPLICATIO NS

cos t  sin( t  )
2

sin  t  cos( t  )
2
cos t   cos( t   )
sin  t   sin( t   )
RADIANS AND DEGREES
2 radians  360 degrees
180
 (rads) 
 (degrees)

ACCEPTED EE CONVENTION

sin( t  )  sin( t  90)
2
Derivative and Integral
a  A sin( t )
da

 A cos(t )  A sin( t  )
dt
2
A
A

 A sin( t )dt   cos(t )  k   sin( t  )  k


2
Example
cos( t  45)
cos( t  45  360)
 cos( t  45)
cos( t  45  180)
cos( t )
Leads by 45 degrees
Lags by 315
Leads by 225 or lags by 135
Example
v1 (t )  12 sin(1000t  60), v2 (t )  6 cos(1000t  30)
FIND FREQUENCY AND PHASE ANGLE BETWEEN VOLTAGES
Frequency in radians per second is the factor of the time variable
f ( Hz) 

 159.2 Hz
2
  1000 sec1
To find phase angle we must express both sinusoids using the same
trigonometric function; either sine or cosine with positive amplitude
take care of minus sign with cos( )   cos(  180)
 6 cos(1000t  30)  6 cos(1000t  30  180)
Change sine into cosine with cos( )  sin(  90)
6 cos(1000t  210)  6 sin(1000t  210  90)
We like to have the phase shifts less than 180 in absolute value
6 sin(1000t  300)  6 sin(1000t  60)
v1 (t )  12 sin(1000t  60) (1000t  60)  (1000t  60)  120
v2 (t )  6 sin(1000t  60)
(1000t  60)  (1000t  60)  120
v1 leads v2 by 120
v2 lags v1 by 120
Unit step
For positive time functions
f ( t )  f ( t ) u( t )
Using square pulses to approximate
an arbitrary function
Pulse width = T, Pulse height = 1
Using the unit step to build a pulse function
THE IMPULSE FUNCTION
(Good model for impact, lightning, and
other well known phenomena)
Approximations
to the impulse
Height is proportional
to area
Representation of the impulse
Periodic Functions
To
2
vA
for 0  t 
v -A
To
 t  To
2
3To
for To  t 
2
vA
for
etc...
Square Waveforms
Triangle Wavefrom
Sawtooth Waveform
EFFECTIVE OR RMS VALUES
Instantaneous power
i (t )
p( t )  i 2 ( t ) R
R
If the current is sinusoidal the average
power is known to be
1 2
Pav  I M
R
1 2
2
2
 I eff
 IM
2
The effective value is the equivalent DC
value that supplies the same average power
If current is periodic with period T
1
Pav 
T
 1 t0 T 2


 p(t )dt  R T  i (t )dt 
t0
 t0

t 0 T
2
Pdc  RI dc
2
I eff

1
T
t 0 T
2
i
 (t )dt
t0
I eff 
1
T
t 0 T
i
the effective value is
X
X eff  M
2
1
For sinusoidal case Pav  VM I M cos( v   i )
2
I eff : Pav  Pdc
If current is DC (i (t )  I dc ) then
For a sinusoidal signal
x (t )  X M cos( t   )
Pav  Veff Ieff cos( v   i )
effective  rms (root mean square)
2
(t )dt
t0
Definition is valid for ANY periodic
signal with period T
EXAMPLE
Compute the rms value of the voltage waveform
T 3
X rms
One period
v
1
2
3
(t )dt   (4t ) dt   (4(t  2))2 dt
0
2
0
2
1
3
16 3  32
v
(
t
)
dt

2


 3 t   3
0
0
2
Vrms 
t 0 T
x
2
(t )dt
t0
The two integrals have the
same value
 4t 0  t  1

v (t )  
0 1 t  2
  4( t  2) 2  t  3

T
1

T
1 32
  1.89(V )
3 3
EXAMPLE
Compute the rms value of the voltage waveform and
use it todetermine the average power supplied to the resistor
T  4( s)
i (t )
R  2
R
i (t )  16; 0  t  4
2
X rms
I rms  4( A)
2
Pav  RI rms
 32(W )
1

T
t 0 T
x
t0
2
(t )dt