Digital Systems
Basic Circuit Theory Review I
CMPE 650
Properties of Sinusoids
Relationship:
ω = 2πf
f ( t ) = V m cos ( ωt + φ )
f(ωt)
A
ω is the angular frequency in radians/sec
φ is the argument or phase angle (in radians)
Period T = 2π/ω = 1/f
v2 = Vmsin(ωt + φ)
v1 = Vmsin(ωt)
π
2π
3π
φ
-A
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UMBC
+
-
1
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Properties of Sinusoids
Trig identities
sin ( ωt + 90° ) = cos ( ωt )
sin ( ωt – 90° ) = – cos ( ωt )
cos ( ωt + 90° ) = – sin ( ωt )
cos ( ωt – 90° ) = sin ( ωt )
sin ( ωt + 180° ) = – sin ( ωt )
cos ( ωt + 180° ) = – cos ( ωt )
sin ( A + B ) = sin A cos B + cos A sin B
sin ( A – B ) = sin A cos B – cos A sin B
cos ( A + B ) = cos A cos B – sin A sin B
cos ( A – B ) = cos A cos B + sin A sin B
Don’t confuse with complex
plane described below
Adding two sinusoids of same ω
+cos(ωt)
A cos ( ωt ) + B sin ( ωt ) = C cos ( ωt – θ )
C =
2
A +B
2
θ = tan
– 90°
– 1  B
-- A
+sin(ωt)
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UMBC
2
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Complex Numbers
Complex numbers: rectangular form
Im
jy
z = x + jy
j =
–1
1
--- = – j
j
z
2
j = –1
r
θ
3
j = –j
4
x
j = 1
Polar form
z = x + jy = r ∠θ = r cos ( θ ) + jr sin ( θ )
z = z ∠θ
r =
2
2
x +y
–1 y
θ = tan  --
 x
rect-to-polar
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or
UMBC
– 1  y
x = r cos ( θ )
z = x + jy
θ = tan
y = r sin ( θ )
z = -x + jy
θ = 180° – tan
polar-to-rect
(-, -)
3
180 +
(+, -)
- x
– 1  y
- x
360 -
(3/18/08)
Digital Systems
Basic Circuit Theory Review I
CMPE 650
Complex Numbers
Complex conjugate
*
z = x – jy = r ∠– θ = e
– jθ
(Last form is exponential form)
Sum/difference: use rectangular form. Mult/div: use polar form
r1
z1
z 1 z 2 = r 1 r 2 ∠( θ 1 + θ 2 )
----- = ----- ∠( θ 1 – θ 2 )
r2
z2
Exponential form
jθ
(similar to polar form)
z = re
from Euler’s formula
e
e
jθ
– jθ
= cos ( θ ) – j sin ( θ )
Adding:
Subtracting:
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jθ
cos ( θ ) = Re ( e )
= cos ( θ ) + j sin ( θ )
UMBC
jθ
sin ( θ ) = Im ( e )
– jθ
1 jθ
cos ( θ ) = --- ( e + e )
2
1 jθ – jθ
sin ( θ ) = ----- ( e – e )
2j
4
(3/18/08)
Digital Systems
Basic Circuit Theory Review I
CMPE 650
Complex Numbers
Useful identities
Assume z = x + jy = r ∠θ
*
2
2
zz = x + y = r
z =
x + jy =
n
2
re
n
jθ ⁄ 2
r ∠( θ ⁄ 2 )
=
n
n jnθ
n
z = ( x + jy ) = r ∠nθ = r e
jθ
ln ( re ) = ln ( r ) + ln e
e
e
e
jπ
jθ
= ln ( r ) + jθ
= –1
j2π
= e
jπ ⁄ 2
j0
= e
= 1
j90°
= j
(1 + j)
j = ---------------2
– jπ ⁄ 2
j270°
= e
= –j
e
e
jπ ⁄ 4
Re ( e
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= e
j45°
( α + jω )t
Im ( e
M
= r ( cos ( nθ ) + j sin ( nθ ) )
( α + jω )t
UMBC
(IMPORTANT IDENTITIES FOR US)
2 j = (1 + j)
=
αt jwt
1
--- =
j
αt
) = Re ( e e
) = e
) = Im ( e e
) = e
αt jwt
5
αt
(1 – j)
– j = ---------------2
cos ( ωt )
sin ( ωt )
(3/18/08)
Digital Systems
Basic Circuit Theory Review I
CMPE 650
Phasors
Sines and cosines can be expressed in terms of phasors.
A phasor is a complex number that represents the amplitude and phase
of a sinusoid.
Phasors based on Euler’s identity
e
jθ
= cos ( θ ) + j sin ( θ )
and
e
– jθ
= cos ( θ ) – j sin ( θ )
So cos(θ) and sin(θ) represent the Re and Im parts of ejθ.
We can write a sinusoid
v ( t ) = V m cos ( ωt + φ ) = Re ( V m e
jφ jωt
v ( t ) = Re ( V m e e
)
j ( ωt + φ )
)
Arg includes phase and time part
Defining the phasor as the magnitude and the non-time part of the arg
v ( t ) = Re ( V e
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UMBC
jωt
)
V = V me
6
jφ
= V m ∠φ
(3/18/08)
Digital Systems
Basic Circuit Theory Review I
CMPE 650
Phasors
The plot a rotating phasor: sinor Vejωt
Rotation
at ω
(rad/s)
Vm
Im
Re
φ
Vm
Projection
on real axis
v ( t ) = Re ( V e
jωt
)
t1
t
at t = t1
-Vm
Note value of sinor at t = 0 is phasor V of the sinusoid v(t).
NOTE: the term ejωt is implicitly present whenever a sinusoid is expressed as
a phasor.
This is where the third parameter, ω, is realized -- in the time part.
It is the third constant associated with the sinusoid.
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UMBC
7
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Phasors
To obtain the sinusoid associated with a phasor, multiply phasor V by ejωt
and take the Re part.
v ( t ) = Re ( V e
jωt
) = Re ( V m e
j ( ωt + φ )
) = Re ( V m cos ( ωt + φ ) + jV m sin ( ωt + φ ) )
v ( t ) = V m cos ( ωt + φ )
To obtain the phasor associated with a sinusoid, first write the sinusoid in
cosine form (so it can be written as the real part of a complex number), and
take out the time factor.
v ( t ) = V m cos ( ωt + φ ) = Re ( V m e
j ( ωt + φ )
jφ jωt
) = Re ( V m e e
) = V me
jφ
= V m ∠φ
Phasors are complex, and therefore can be represented in rectangular, polar
or exponential form.
They are vectors (magnitude and direction) and can be plotted as such.
The phasor domain is also known as the frequency domain.
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UMBC
8
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Phasors
Taking the derivative of a sinusoid
dv
------ ( V m cos ( ωt + φ ) ) = – ωV m sin ( ωt + φ ) = ωV m cos ( ωt + φ + 90 )
dt
jωt jφ j90°
jωt
dv
------ ( V m cos ( ωt + φ ) ) = Re ( ωV m e e e
) = Re ( jωV e )
dt
The relationships of derivative and integral
dv
------ ⇔ jωV
dv
V
-----v
d
t
⇔
∫
jω
These relationships are useful in computing the steady-state response (where
we don’t need to know the initial values).
Phasors are also useful for summing sinusoids of the same frequency.
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UMBC
9
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Phasors
Important FACTs to remember:
• v(t) is time dependent, while V is NOT.
• v(t) is ALWAYS real, while V is generally complex.
• Phasor analysis applies only when frequency is constant and allows two or
more sinusoids to be manipulated ONLY if they have the same frequency.
Examples
– sin A = cos ( A + 90° )
v = – 4 sin ( 30t + 50° ) V
v = 4 cos ( 30t + 50° + 90° ) = 4 cos ( 30t + 140° )
V = 4 ∠140° V
I = – 3 + j4 A
I = 5 ∠126.87°
i ( t ) = 5 cos ( ωt + 126.87° ) A
(Phasors can be in any of the three forms)
– j20°
since j = 1 ∠90°
V = j8e
V
V = j8 ∠– 20° = ( 1 ∠90° ) ( 8 ∠– 20° ) = 8 ∠90° – 20° = 8 ∠70°
v ( t ) = 8 cos ( ωt + 70° ) V
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UMBC
10
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Phasors
Used to solve an integrodifferential equation
Solve for i(t)
di
4i + 8 ∫ i dt – 3 ----- = 50 cos ( 2t + 75° )
dt
8I
4I + ------ – 3 jωI = 50 ∠75°
jω
Convert to phasor domain
ω is 2 and- 1/j = -j.
I ( 4 – j4 – j6 ) = 50 ∠75°
50 ∠75°
50 ∠75°
I = ------------------- = --------------------------------------- = 4.642 ∠143.2°
10.77 ∠( – 68.2 )°
4 – j10
2
2
4 + 10 = 10.77
– 1 10
φ = tan  ------ = 360° – 68.2° = 291.8°
 4
m =
φ = 75° – 291.8° = – 216.8 ° + 360 = 143.2°
So far we’ve shown how to represent a voltage and current in phasors.
What about the passive elements, R, L and C
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UMBC
11
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Phasors Applied to R, L and C circuits
Need to transform the current/voltage relationships for R, L and C into the
frequency domain.
For R
i = I m cos ( ωt + φ )
v = iR = RI m cos ( ωt + φ )
V = RI m ∠φ
I = I m ∠φ
V = RI
Assume current through a resistance R
And voltage across it
The phasor for this voltage
Phasor for the current
Current and voltage are in phase
Im
V
I
Re
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UMBC
12
(3/18/08)
Digital Systems
Basic Circuit Theory Review I
CMPE 650
Phasors Applied to R, L and C circuits
For L
i = I m cos ( ωt + φ )
di
v = L ----- = – ωLI m sin ( ωt + φ )
dt
di
v = L ----- = ωLI m cos ( ωt + φ + 90° )
dt
V = ωLI m e
j ( φ + 90° )
Assume current through inductor L
Voltage across it
– sin ( A ) = cos ( A + 90° )
jφ j90°
= ωLI m e e
V = jωLI
Into a phasor
e
j90°
= j
Im
Voltage leads the current
by 90 degrees for inductor
V
I
Re
The j term translates into a phase shift in the time domain.
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UMBC
13
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Impedance
For C
I
V = ----------jωC
dv
i = C -----dt
or
I = jωCV
Here, current leads voltage by 90 degrees
The first form relates V and I similar to resistance.
These can be written in terms of their phasor voltage to phasor current
V
---- = R
I
V
---- = jωL
I
V
1
---- = ----------jωC
I
V
Z = ---I
or
V
–j
---- = -------I
ωC
Impedance Z is the ratio of
phasor voltage to current
in Ωs
NOTE: the impedance Z is NOT a phasor -- it does not correspond to a sinusoidally varying quanity.
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UMBC
14
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Impedance
Example
v = 12 cos ( 60t + 45° )
L = 0.1 H
V = jωLI
Find steady-state current
with w = 60 rad/s and V = 12 ∠45°
12 ∠45°
12 ∠45°
V
I = ---------- = --------------------- = ------------------- = 2 ∠– 45 ° A
j60 ( 0.1 )
6 ∠90°
jωL
So we see the magnitude of the impedance is 6 Ωs and the inductor causes a
phase shift in the current (it lags by 90 degrees).
Z = R + jX
Z = R – jX
R = Re ( Z )
X = Im ( Z )
resistance
reactance
Reactance can be negative (capacitive, current lags) or positive (inductive).
Other forms
Z = R + jX = Z ∠θ
Z =
2
R +X
2
R = Z cos ( θ )
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UMBC
15
θ = tan
–1  X
--- R
X = Z sin ( θ )
(3/18/08)
Digital Systems
Basic Circuit Theory Review I
CMPE 650
Admittance
Admittance Y is the reciprocal of impedance, measured in siemens (S)
1
I
Y = --- = ---Z
V
Y = G + jB
G = Re(Y) is called conductance (mohms)
B = Im(Y) is called susceptance
1
G + jB = ---------------R + jX
1 R – jX
R – jX
G + jB = ---------------- ---------------- = -------------------2
2
R + jX R – jX
R +X
R
G = -------------------2
2
R +X
Relating impedance with
admittance
–X
B = -------------------2
2
R +X
So, G does NOT equal 1/R as was true for resistive circuits.
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1966
UMBC
16
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Digital Systems
Basic Circuit Theory Review I
CMPE 650
Examples
5Ω
V s = 10 ( ∠0° )V
v s = 10 cos ( 4t ) +
-
0.1F
The impedance is
1
1
Z = R + ----------- = 5 + ------------------ = 5 – j2.5 Ω
j4 ( 0.1 )
jωC
And current
Vs
10 ( 5 + j2.5 )
10 ∠0°
I = ------ = ------------------- = ------------------------------ = 1.6 + j0.8 = 1.789 ∠26.57°
2
2
5 – j2.5
Z
5 + 2.5
OR
Vs
10 ∠0°
I = ------ = ----------------------------- = 1.789 ∠26.57°
5.6 ∠– 26.57
Z
Voltage across capacitor
I
1.789 ∠26.57°
1.789 ∠26.57°
V = IZ C = ----------- = ---------------------------------- = ---------------------------------- = 4.47 ∠– 63.43°
jωC
j4 ( 0.1 )
0.4 ∠90°
Time domain
i ( t ) = 1.789 cos ( 4t + 26.57° )
v ( t ) = 4.47 cos ( 4t – 63.43° )
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UMBC
17
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