5 1 Sinusoidal signals
.
>
-
AC is mostly used (i e. power system , telecoms)
.
Sinusoidal AC Voltage (UCt)) :
MATHEMATICALLY :
V (H) =
Um COS (wt + 0
> relative phase shift langle
↓ ↓ angular frequency (O)
Amplitude/peak Crad(s)
Value (V)
GRAPHICALLY
* for uniformity for sinusoidal waveforms
Sin
cut + of90 % = (OS ( Wh + O)
* O can be determined by finding the peak value
>
-
T: period of sinusoidal waveform : WT I =
>
-
f : Frequency of sinusoidal waveform
: amount of cycles per second
inverse of period f
=
:
>
-
w : angular frequency =
2 = f
F
Root-Mean-Square (RMS) values / effective value
>
-
When U(t) is placed across a resistance (R) ,
Power (P(H) is :
D(t) = 12t
-
R
>
-
P(t) is also periodic and sinusoidal
(Pavy)
Ms Im
>
-
The average power =
At = =
,-Roman-squareof voltage: urm the
>
-
RMS value is an . This means Ac voltages current will
equivalent DC value of the AC Waveform
deliver the same power AC Voltage current would deliver if the the DC value was equal to RMS value
of AC Voltage / current
Power (AC Voltage /Current) Power (ocvoltage
= /Current) C Voltages Current = RMS of Ac Voltage
current
=m Irms-m
>
-
Urms and
Amplitude mplitude
,Exercise 51 .
Importanavatioens
Given UCH 100 COSCO) V applied over a 500 resistor.
=
Sketch uct) and Calculate
C and Paug t
Paug Vrms = = Irms R
R
finding PCH) > P(t) = V(t)2
R
WT 2
UCH
=
(i) P(t) UCH2= =
Um COS (W ++ O)
R
NB
=
(100 OS (100))2
50 (852X
1 + 1 COSC(X)
=
a
=
1002 (OS (100)
50
=
200 ( + Y2 (OS (2001)
=
100 + 100COS (200)
BE
finding Pava IN RAD
Ch 100COS(IOUE) Curms
fo(00<LOOCOSCIO)At
= =
W =
100
Urms = 70 /V.
CiCUT It =
Urms"
T 0 825
=
.
(ICH =
= 1 = 10 o
draw UCH
V(t) (U)
A
I
V (H) =
Um Cos (cut + O) T
Willo
>
-
Um = 100 100
<
>
-
W = 100
>
T 0 82S
Pens
-
=
.
>
-
0 =
00 tmax O
, draw PC
P(t) (W)
P(+) =
100 + 100 COS(200) A
T
.i
<
200
upward by 100W ↑
Pm = 100 , PMT 100 + 100 200
= =
W = 200
T= 0 81s
Fim)
.
max
.
>
-
AC is mostly used (i e. power system , telecoms)
.
Sinusoidal AC Voltage (UCt)) :
MATHEMATICALLY :
V (H) =
Um COS (wt + 0
> relative phase shift langle
↓ ↓ angular frequency (O)
Amplitude/peak Crad(s)
Value (V)
GRAPHICALLY
* for uniformity for sinusoidal waveforms
Sin
cut + of90 % = (OS ( Wh + O)
* O can be determined by finding the peak value
>
-
T: period of sinusoidal waveform : WT I =
>
-
f : Frequency of sinusoidal waveform
: amount of cycles per second
inverse of period f
=
:
>
-
w : angular frequency =
2 = f
F
Root-Mean-Square (RMS) values / effective value
>
-
When U(t) is placed across a resistance (R) ,
Power (P(H) is :
D(t) = 12t
-
R
>
-
P(t) is also periodic and sinusoidal
(Pavy)
Ms Im
>
-
The average power =
At = =
,-Roman-squareof voltage: urm the
>
-
RMS value is an . This means Ac voltages current will
equivalent DC value of the AC Waveform
deliver the same power AC Voltage current would deliver if the the DC value was equal to RMS value
of AC Voltage / current
Power (AC Voltage /Current) Power (ocvoltage
= /Current) C Voltages Current = RMS of Ac Voltage
current
=m Irms-m
>
-
Urms and
Amplitude mplitude
,Exercise 51 .
Importanavatioens
Given UCH 100 COSCO) V applied over a 500 resistor.
=
Sketch uct) and Calculate
C and Paug t
Paug Vrms = = Irms R
R
finding PCH) > P(t) = V(t)2
R
WT 2
UCH
=
(i) P(t) UCH2= =
Um COS (W ++ O)
R
NB
=
(100 OS (100))2
50 (852X
1 + 1 COSC(X)
=
a
=
1002 (OS (100)
50
=
200 ( + Y2 (OS (2001)
=
100 + 100COS (200)
BE
finding Pava IN RAD
Ch 100COS(IOUE) Curms
fo(00<LOOCOSCIO)At
= =
W =
100
Urms = 70 /V.
CiCUT It =
Urms"
T 0 825
=
.
(ICH =
= 1 = 10 o
draw UCH
V(t) (U)
A
I
V (H) =
Um Cos (cut + O) T
Willo
>
-
Um = 100 100
<
>
-
W = 100
>
T 0 82S
Pens
-
=
.
>
-
0 =
00 tmax O
, draw PC
P(t) (W)
P(+) =
100 + 100 COS(200) A
T
.i
<
200
upward by 100W ↑
Pm = 100 , PMT 100 + 100 200
= =
W = 200
T= 0 81s
Fim)
.
max
