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CIRCUITS and
SYSTEMS – part I
Prof. dr hab. Stanisław Osowski
Electrical Engineering (B.Sc.)
Projekt współfinansowany przez Unię Europejską w ramach Europejskiego Funduszu Społecznego. Publikacja dystrybuowana jest bezpłatnie
Lecture 2
Analysis of circuits in steady
state at sinusoidal excitation
Sinusoidal signal
u(t )  Um sin( t  )
u(t)
Um

- instantaneous value of signal
- maximum value (magnitude) of signal
- initial phase (phase corresponding to t=0)
 t+
f=1/T
T
- phase angle at time t
- frequency in Hz
- period of sinusoidal signal

- angular frequency measured in radians per second
RMS value of signal
1
F
T
to T

f 2 (t )dt
t0
For sinusoidal signal
• voltage
u(t )  U m sin( t   )

U
Um
2
• current
i (t )  I m sin(t  )

I
Im
2
Steady state of the circuit
Steady state of the circuit is the state in which the character
of the circuit response is the same as the excitation. It means
that at sinusidal excitation the response is also sinusidal of the
same frequency.
For the need of steady state analysis we introduce the so
symbolic method of complex numbers. This method converts
all differential and integral equations into algebraic equations
of complex character.
Symbolic method for RLC circuit
The RLC circuit under analysis
The circuit equation in time domain
1
di
U m sin( t  )  Ri   idt  L
C
dt
General solution of circuit
The general solution of the circuit in time domain is composed of two
components: x(t)=xs(t)+xt(t)
•Steady state component – part xs(t) of general solution for which
the signal has the same character as excitation (at sinusidal excitation
the response is also sinusidal of the same frequency). This state is
theoretically achieved after intinite time (in practice this time is
finite).
•Transient component - part xt(t) of general solution for which the
signal may take different form from excitation (for example at DC
excitation it may be sinusoidal or exponential). The general solution
is just the sum of these two parts
x(t)=xs(t)+xt(t)
Solution in steady state
Symbolic represenation of voltage excitation
u(t )  U m sin( t  )

U (t )  U m e j e jt
Symbolic represenation of current response
i(t )  I m sin( t  )

I (t )  I me j e jt
Symbolic equation of circuit
dI (t ) 1
U (t )  RI (t )  L
  I (t )dt
dt
C
Solution in steady state (cont.)
After performing the appropriate manipulations we get
U m j
I
I
1 I m j i
e  R m e j i  jL m e j i 
e
jC 2
2
2
2
The complex RMS notations of current and voltage
U m j
U
e ,
2
I m j i
I
e
2
The complex RMS equation of the circuit
U  RI  jLI 
1
I
jC
Complex represenation of the RLC
elements
Resistor
U R  RI
 ZR  R
Inductor
U L  jLI  ZL  jL
Capacitor
1
1
1
UC 
I  ZC 
j
jC
jC
C
Complex impedances
• Reactance of inductor
X L  L
• Reactance of capacitor
1
XC 
C
• Impedance of inductor
Z L  jX L
• Impedance of capacitor
Z C   jX C
Final solution of RLC circuit
• Complex algebraic equation of RLC circuit
U  RI  Z L I  ZC I  ZI
• Complex current
I
U e j
U
U


e
2
2
Z R  j L  1 /(C 
R  (L  1 /(C ))
• Magnitude RMS value of current
I 
U
Z

U
R 2  (L  1 /(C )) 2
• Phase of current
 i    arctg
L  1 /(C )
R
L 1 /(C ) 

j  arctg

R


Kirchhoff’s laws for complex
representation
• KCL
I
k
0
k
• KVL
U
k
0
k
• Ohm’s complex law
U  ZI  I  YU
Y=1/Z - complex admittance
Symbolic method - summary
• Conversion: time-complex representation of sources
U m j u
u (t )  U m sin( t  u ) 
e
2
I m j i
i (t )  I m sin( t  i ) 
e
2
• Complex represenattion of RLC elements
• Kirchhoff’s laws for complex values
• Solution of complex equations -> complex currents & voltages.
Example
Determine the currents in in steady state of the circuit at the following
values of parameters: R=10Ω, C=0,0001F, L=5mH,
i(t)=7.07sin(1000t) A.
Circuit structure
Solution
Complex symbolic values of parameters:
ω = 1000
I = 5ej0 = 5
ZL = jωL = j5
ZC = -j/(ωC) = -j10
Admittance and impedance of the circuit
1
1
1
1 10 j 45o
Y 

 0,1  j 0,1  Z  
e
R Z L ZC
Y
2
Solution (cont.)
Voltage and currents
50 j 45o
e
U  ZI 
2
5 j 45o
U
e
IR  
R
2
10  j 45o
U
e

IL 
ZL
2
5 j135o
U
e

IC 
ZC
2
Solution (cont.)
Time representation of the signals
u (t )  50 sin( 1000t  45 o )
i R (t )  5 sin( 1000t  45 o )
i L (t )  10 sin( 1000t  45 o )
iC (t )  5 sin( 1000t  135 o )
Phasor diagram for resistor
Equation
U R  RI R  RI R e
j 0o
Phasor diagram for inductor
Equation
U L  jLI L  LI L e
j 90o
Phasor diagram for capacitor
Equation
1
1
 j 90o
UC   j
IC 
IC e
C
C
Phasor diagram for RLC circuit
• The construction starts from the farest branch from the source. For
series connected elements of this branch start from current; for
parallel connected elements start from voltage. Next we draw
alternatingly the currents and voltages for the succeeding branches,
approaching in this way the source.
• The relation of the input voltage towards the input current
determines the reactive character of the circuit.
– If the input voltage leads its current the character is inductive.
– If (opposite) the input voltage lags its current the character of the circuit is
capacitive.
– When the voltage is in phase with current – the circuit is of resistive character.
Example
Draw the phasor diagram for the circuit
RLC circuit structure
Construction of phasor diagram