Summary
16: Signal Components
• Power absorbed is p(t) = v(t) i(t) for i into + terminal.
• Analysis with KVL, KCL, and device characteristics:
– Sources: v = vS (t) ∀ i; i = iS (t) ∀ v
dv(t)
– Resistor: v = iR, Capacitor i(t) = C dt
v
• For any two nodes of a linear circuit v = vOC − i iOC
Any voltage or current waveform v S (t) can be divided
into a sum of signal components:
SC
– Thévenin and Norton equivalents
• Periodic waveforms repeat every T seconds.
T
T
1
2
– p(t) = T1 0 v(t) i(t) dt, V rms =
T 0 v(t) dt,
T
1
v(t) = T 0 v(t) dt
– Sinusoid s(t) = A cos(ωt + φ) Peak amplitude = A,
Phase = φ, Frequency: ω ; T = 2π/ω ; f = 1/T
– cos(ωt − φ) lags cos ωt by φ rad.
• Complex Exponentials and Impedance
m ejωt
– Complex Exponential s
(t) = A
m
Am with amplitude = |Am| and phase = A
– Impedance Z = R + jX ; R resistance, X reactance
=A
– If v(t) = A cos ωt, then let V
= 1 , Resistor Z
=R
– Capacitor Z
jωC
• Capacitors
2
– Energy w(t) = 12 C (v(t))
– RC circuit charging
or discharging:
vC (t) = As 1 − e−t/RC or vC (t) = As e−t/RC
– RC is the time constant.
ELEC176 - Summary
i
vI (t) = VI + vi (t)
• VI (all upper case) is a constant component
It is the average value vA (t) = VA
This is called the dc component
• va(t) (all lower case) is a purely time-varying component
That is, its average value v a(t) = 0
This is called the signal component
• vA(t) is the composite waveform (mixed-case variables).
+
VS
+
+
vB (t) = VB + vb (t)
vb (t) = Vb cos ωt
ELEC176 - Signal Components
vS (t)
−
C
+
vC (t)
=
vS (t) − vR (t)
=
VS + vs (t) − RC
+
vs(t)
VS
vS (t)
+
−
vC (t)
C
−
vS (t) = VS + vs (t)
−
vC (t) = VS + vs(t) − RC
d vc(t)
dt
Signal component vc (t) = vs(t)−RC
especially when vc (t) changes quickly.
d vc (t)
dt
d vc (t)
dt
is reduced,
The high-frequency signal components in v S (t) are
reduced.
DC component VC = VS across capacitor is same as in v S
The dc component of current is zero, so there is zero dc
component across the resistor.
Rather than transferring the signal from v s (t) to vc (t), it
is bypassed by the low reactance of the capacitor.
VS
vS (t)
VC
t
ELEC176 - Signal Components
+
R
vC (t)
vS (t) = VS + vs (t) is expressed as the sum of a dc
component VS and a signal component vs (t).
KVL
16-1
Bypassing
+
R
−
Example of use of upper-case and lower-case letters:
Sinusoid with dc component
Simple RC circuits affect dc and signal components
differently. This circuit only affects the signal component:
vs(t)
vA (t)
+
VA
Resistor-Capacitor Circuit
+
+
va (t)
16-2
ELEC176 - Signal Components
vC (t)
t
16-3
Response to a Square Wave
+
+
vs(t)
VS
−
If frequency of vs (t) is low — that is
+
R
vS (t)
+
High- and Low-Frequency Square Wave
RC
vs (t)
vC (t)
C
−
vc (t)
Consider vS (t) a square wave:
For most of the time vc (t) ≈ vs (t)
vC (t) = As 1 − e−t/RC
If frequency of vs (t) is high — that is
vS (t)
vc (t)
τ
1
f
1
f
RC
vs (t)
t
to
As
e−(t−to )/RC
there is not enough time for complete charging.
Capacitor charges and discharges with time constant
τ = RC .
Amplitude of vc (t) < amplitude of vs (t).
High-frequency components are reduced in amplitude.
ELEC176 - Signal Components
16-4
ELEC176 - Signal Components
Smoothing Example
16-5
Capacitor-Resistor Circuit
This circuit affects the dc component:
Consider a ‘noisy’ waveform v(t):
+
+
vs(t)
t
VS
vS (t)
+
−
t
t
KVL
so
a sinusoid v(t) at frequency f and a noise waveform v N (t)
vC (t)
=
vR (t)
=
=
An RC filter can be used to reduce the noise if the
1
1
frequencies in the noise are RC
and f RC
:
=
=
+
R
R
vR (t)
−
vS (t) = VS + vs (t) is expressed as the sum of a dc
component VS and a signal component vs (t).
as a sum of
v(t)
+
C
+
C
vC (t)
−
t
vS (t) − vR (t)
d vC (t)
dt
d (vS (t) − vR (t))
RC
dt
d (vs (t) − vR (t))
d VS
+RC
RC
dt
dt
d (vs (t) − vR (t))
0 + RC
dt
RC
Constant VS does not contribute to vR (t).
That is, the dc component is blocked.
ELEC176 - Signal Components
16-6
ELEC176 - Signal Components
16-7
DC Blocking
+
vs(t)
VS
+
+
Response to a Square Wave
vS (t)
vs(t)
vR (t)
R
−
+
+
C
VS
−
vS (t) = VS + vs (t)
d vs (t)
d vr (t)
vR (t) = RC
− RC
dt
dt
1
Signal component vr (t) = vs (t) − RC
vR (t)dt is
approximately vs (t), especially when vs (t) changes quickly.
+
+
+
C
vS (t)
−
vR (t)
R
−
Consider vS (t) a square wave:
vr (t) = As e−t/RC
vS (t)
The high-frequency signal components in v S (t) are
replicated across the resistor.
t
The dc component VS is blocked:
VS
vS (t)
−As e−(t−to )/RC
vR (t)
t
Voltage across the resistor is v S (t) − vC (t), which goes
to zero as the capacitor charges or discharges with time
constant τ = RC .
t
ELEC176 - Signal Components
16-8
ELEC176 - Signal Components
High- and Low-Frequency Square Wave
If frequency of vS (t) is low — that is
1
f
16-9
Frequency Selection Example
RC
Consider a waveform v(t):
vs (t)
t
vr (t)
as a sum of
then for most of the time vr (t) ≈ 0
t
If frequency of vS (t) is high — that is
vr (t)
1
f
t
RC
vs (t)
a sinusoid vL(t) at low frequency fl and a sinusoid vH (t)
at high frequency fh .
An RC filter can be used to reduce the low-frequency
1
1
component if fh RC
and fl RC
:
there is not enough time for complete charging.
+
Amplitude of vr (t) ≈ amplitude of vs (t).
v(t)
C
+
R
vR (t)
−
t
High-frequency components are passed.
ELEC176 - Signal Components
16-10
ELEC176 - Signal Components
16-11
Differentiator
17: Sinusoid Excitation
+ vr (t) −
A differentiator can be constructed with a resistor,
capacitor, and a current-controlled voltage source that acts
as a negative resistance.
+
R
+
vc (t)
vs (t)
vI (t)
+
C
iX (t)
R
+
C
−
ic (t)
+
vO (t)
v = −R iX (t)
−
Recall that for
vs (t) = As cos ωt
vI (t)
KVL
vO (t)
KVL
=
vO (t) + vR (t) + vC (t)
=
−R iX (t) + R iX (t) + vC (t)
=
vC (t)
=
−R iX (t)
=
d vC (t)
−R C
dt
vO (t) = −RC
then
vc (t) = √
As
1+(ωRC)2
cos(ωt − tan−1 ωRC)
c (t)
ic (t) = C dvdt
ωC
cos ωt +
π
2
− tan−1 ωRC
ωRC
cos ωt +
π
2
− tan−1 ωRC
= As √
1+(ωRC)2
vr (t) = ic(t)R
d vI (t)
dt
= As √
1+(ωRC)2
Each voltage and current has an amplitude and phase.
ELEC176 - Signal Components
16-12
ELEC176 - Sinusoid Excitation
Networks of Impedances
Solution with Impedances
When vs (t) is a sinusoid with angular frequency ω ,
complex amplitudes and impedances can be used to solve
for amplitudes and angles of voltages and currents.
+
+
r
V
−
r = R
Z
s
V
c = 1
Z
jωC
c
V
Ic
+
−
=
As
Ic
=
s
s
jωC
V
V
s
=
=V
1
R
+
1
+
jωRC
Zr + Zc
jωC
=
s
V
c =
Ic Z
1 + jωRC
=
s jωRC
r = V
Ic Z
1 + jωRC
r
V
ic (t) = |Ic | cos ωt + Ic
c = 1
Z
jωC
=
=
+
R
s
V
o
V
Each voltage and current has an amplitude and phase:
s | cos ωt + V
s
vs (t) = |V
For vs(t) a sinusoid with angular frequency ω and
amplitude Vs, find vo .
+
s
V
c
V
17-1
s
V
c
R + RL||Z
RL
−
c
RL||Z
s
V
R+
o
V
RL
1+jωCRL
·
RL
1 + jωCRL
=
s RL
V
R + RL + jωCRL R
=
s RL
V
1
·
R + RL 1 + jωC RLR
R+R
L
c | cos ωt + Vc
vc (t) = |V
ELEC176 - Sinusoid Excitation
17-2
ELEC176 - Sinusoid Excitation
17-3
Frequency Dependence
Amplitudes and Phase
For vs (t) a sinusoid with angular frequency ω :
r
V
+
−
vs (t)
+
As
c =
V
1+jωRC
−
R
+
c = 1
Z
jωC
As
Ic
vc (t)
ic (t)
Amplitude
Phase
s | = As
|V
v
s = 0
|
vc | = v
c = − tan−1 ωRC
As
1+(ωRC)2
lags v
s (t)
ic = π2 − tan−1 ωRC
|ic | = ωCAs
1+(ωRC)2
leads v
c by π/2
|
vc(t)| = |
vs(t)| = As
As
1 + (ωRC)2
v
s(t) = 0
vr (t)
v
c(t) = − tan
−1
v
r = π2 − tan−1 ωRC
|
vr | = ωRCAs
1+(ωRC)2
ωRC
leads v
s and v
c
Phasor Diagram:
• for ω = 0
|
vc(t)| = As and v
c (t) = 0
That is, low frequencies are passed.
• for ω → ∞
c (t) = −π/2
|
vc(t)| → 0 and v
That is, high frequencies are blocked.
1
• for ω = RC
As
|
vc(t)| = √
and v
c (t) = − tan−1 1 = −π/4
2
|vr | = ωRCAs
1+(ωRC)2
tan−1 ωRC
|ic |
|vs | = As
As
In this case, |
vr (t)| = √
.
2
I.e. |
vr (t)| + |
vc (t)| = As.
|vc | = ELEC176 - Sinusoid Excitation
17-4
17-5
Low Frequency
+
vr
1+(ωRC)2
ωRC
1+(ωRC)2
vr (t) = ωRCAs
1+(ωRC)2
cos(ωt − tan−1 ωRC)
cos ωt + π2 − tan−1 ωRC
vr (t)
i(t)
Amplitude
Phase
As
0
As
→ −0
vs (t)
As
1+(ωRC)2
As
C
−
For ω → 0, tan−1 ωRC → 0:
|
vc | = √
vc (t) = vc
ic
|
vs| = As
tan
+
vs
|
vr | = √ ωRCAs
−1
−
R
+
|ic |
As
1+(ωRC)2
ELEC176 - Sinusoid Excitation
Phasors
tan−1 ωRC
tan−1 ωRC
vs (t) = As cos ωt
vc (t)
vc (t)
ic (t)
As vr (t)
As ωRC
1+(ωRC)2
ωC
1+(ωRC)2
1+(ωRC)2
→ π2
leads vc by π/2
→ π2
Amplitude is small
|vr | = As ωRC
1+(ωRC)2
t
vc ≈ vs
tan−1 ωRC
|vs | = As
|vc | = As
1+(ωRC)2
Low-frequency limit, vc (t) = vs(t).
ELEC176 - Sinusoid Excitation
17-6
ELEC176 - Sinusoid Excitation
17-7
18: Inductor
High Frequency
For ω → ∞, tan−1 ωRC → π2 :
Amplitude
Phase
As
0
vs (t)
vc (t)
+ v(t) −
i(t)
As
→ − π2
→0
1+(ωRC)2
ic (t)
As → ARs
1+(ωRC)2
→0
vr (t)
As ωRC
→0
ωC
1+(ωRC)2
L
→ As
v(t) = L
di(t)
dt
or
i(t) =
1
L
t
−∞
v(t)dt
• An inductor induces a potential that opposes change
in current — measured in volts per ampere/second.
(v(t)dt = L di(t), so L = vdt/di [Vs/A])
|vr | = As ωRC
1+(ωRC)2
|vs | = As
• Coefficient L is called inductance.
Unit of inductance:
henry (H)
One henry induces one volt for a rate of change in
current of one ampere per second.
tan−1 ωRC
|vc | = As
1+(ωRC)2
High-frequency limit, vc (t) ≈ 0.
vr (t)
vs (t)
vc (t)
t
ELEC176 - Sinusoid Excitation
17-8
ELEC176 - Inductor
18-1
Energy in an Inductor
Constant Inductance
• Inductance L can vary with time, temperature and
operating conditions.
Inductors are devices with nearly constant inductance.
p(t) = i(t) v(t) = i(t) L di(t)
dt
• Power
• Energy
• For an inductor with constant L, v(t) = L di(t)
dt .
If v(t) = A cos ωt,
A
A
then i(t) = ωL
sin ωt = ωL
cos(ωt − π/2)
Thus i(t) lags v(t) by π/2.
w(t)
=
jωL
i(t)
−∞
i(t) L
i(t) di(t) = L
i(−∞)
x2
L
2
t
i(t)
i(−∞)
di(t)
dt
dt
i(t)
x dx
i(−∞)
i(−∞))2
i(t)2
−
=L
2
2
A jωt
jωL e
• Assuming i(−∞) = 0 gives
ωL
I = −π/2, so I lags A by π/2.
ELEC176 - Inductor
p(t) dt =
L
=
l = jωL
• The impedance of the inductor is Z
because
if v
(t) = Aejωt then i(t) = L1 Aejωt dt =
jωt and A = Z
l I,
Thus if i(t) = Ie
A
then I =
= − jA .
−∞
=
t
w(t) =
18-2
ELEC176 - Inductor
1
2
L i(t)
2
18-3
LR Impedance Network
LR Circuit
+
+
vL
L
+
iL
vS
=
=
=
diL
dt
Time constant =
r /R, and the inductor V
l = jωL I
I = V
so
I
=
r
V
=
cf:
vS
R
L
+
C
+
vC
−
iC
18-4
s
V
+
=
=
=
=
s
V
π
4
I
ELEC176 - Inductor
+
r
V
18-5
vs
R
L
C
−
+
vo (t)
−
To analyse this circuit, let the complex amplitude of the
s , with an angular
voltage source vs be denoted by V
frequency ω , so that the impedances of the resistor,
r = R, Z
c = 1/jωC , and
capacitor and inductor are Z
l = jωL.
Z
s
V
L
1 + jω R
t
Z
s
V
1+j
s (1 − j)
V
2
s
s
V
V
−j
2
2
t
Z
=
r + Z
l ||Z
c = Z
r + Zl Zc
Z
c
Zl + Z
=
R+
jωL/jωC
1
jωL + jωC
=
R+
jωL
1 − ω 2LC
=
Rt + jXt
vs (t)
vr (t)
t
r
V
ELEC176 - Inductor
L
1 + jω R
LRC Example
R
L
r
V
s
V
−
R
r
V
For ω =
L
R + jωL
r → 0, i.e. high frequencies are blocked.
If ω → ∞ then V
Mid Frequency
Vl
R
s , i.e. low frequencies are passed.
r → V
If ω → 0 then V
ELEC176 - Inductor
+
r + Z
l
Z
sR
V
t
R
vS
r + Z
l
Z
s
V
=
vR
vL
+ RCvC = RCvS in:
s
V
=
+ vR −
dvC
dt
−
In terms of impedances at frequency ω :
vS − vL
diL
vS − L
dt
L dvR
vS −
R dt
Vs
+
r
V
R
Vs t ≥ 0
If vS =
0 t < 0
vR = Vs 1 − e−tL/R
I
L
R
dvR
R
+ vR = vS
dt
L
L
Thus
−
−
For the resistor iL = vR /R, and the inductor vL = L
vR
+
s
V
vR
R
KVL:
l
V
+
−
18-6
Total
Resistor
Capacitor
Inductor
ELEC176 - Inductor
Impedance
t
Z
R
1
jωC
jωL
Resistance
t)
(Z
R
0
0
Reactance
t)
(Z
0
1
Xc = ωC
Xl = ωL
18-7
LRC Example
R
vs
L
Tuned Circuit
+
vo (t)
−
C
R
vs
L
C
+
vo (t)
−
Complex amplitude of the output v o (t) is then:
Vo
=
=
=
s
V
Z
Z
l c
+Z
c
Z
l
r
+Z
×
c
l Z
Z
l + Z
c
Z
=
1 + R jωC +
1
jωL
For ω 18-8
√1 ,
LC
+
v1
−
=
v2
=
s ,
V
Vo ≈
−j
ωRC
s ,
V
ELEC176 - Inductor
18-9
Ideal Transformer
i2
L1
L2
+
v2
−
M
v1
jωL
R
1
Circuit passes sinusoids at a frequency of ω o = √LC
,
and reduces the amplitudes of those at other frequencies.
Transformer
i1
Vo ≈
o = V
s , that is: vo (t) = vs .
V
that is: |vo (t)| |vs | and vo lags by π/2.
s , that is: vo (t) = vs .
Vo = V
ELEC176 - Inductor
√1 ,
LC
√1 ,
LC
that is: |vo (t)| |vs | and vo leads by π/2.
R
1 + j ωL
(ω 2 LC − 1)
√1 ,
LC
− 1)
s
s
−j V
jωLV
=
R
2
jωL − ω LRC + R
ωRC − ωL
−j
For ω s
V
Note that, for ω =
1+
Vs
R
j ωL (ω 2 LC
Note that, for ω =
l
s
V
=
=
l Z
c
Z
r Z
l + Z
c
Zl Zc + Zr Z
s
V
r 1 + 1
1+Z
Z
Z
s
V
c
=
Vo
di1
di2
L1
+ M
dt
dt
di2
di1
M
+ L2
dt
dt
di1
dt
=
v2
=
i1
=
M di2
v1
+
L1 dt
L1
M
M 2 di2
+
v1 + L2 −
L1
L1
dt
M
v1 dt
−
i2 +
L1
L1
−
In an ideal transformer, the coupling between
the
√
inductors is at a maximum,
which
gives
M
=
L
L
,
and
1
2
L
Where L1 and L2 are the inductances of the two
transformer windings, and M is the Mutual Inductance.
2
the factor LM =
L1 = N is called the turns ratio of the
1
transformer — it is the ratio of the number of turns in the two
windings of the transformer.
The inductances L1 and L2 are proportional to the
square of the number of turns in each winding. The mutual
inductance M is increased when the windings share more
of the same magnetic field.
In this ideal case L2 − M
L1 = 0. Also, in an ideal
transformer,
the
inductance
of
each winding is large, so
v1 dt
≈
0
,
and
therefore:
L
2
1
v2 = N v1 and i1 = −N i2
ELEC176 - Inductor
18-10
ELEC176 - Inductor
18-11
Example
Summary
Consider the following circuit where the turns ratio of the
transformer is 20:1, that is N = 0.05.
• Power absorbed is p(t) = v(t) i(t) for i into + terminal.
• Analysis with KVL, KCL, and device characteristics: v =
di
vS ∀ i, i = iS ∀ v , v = iR, i = C dv
dt , v = L dt
RS
i1
i2
1:N
+
240 Vrms
+
R L v2
−
v1
−
• Thévenin and Norton equivalents for v = vOC − i
Applying KVL to the left-hand side v 1 = 240 − RS × i1
For the transformer v2 = N v1 and i1 = −N i2
RS
400 i2
Vrms.
N 2 RS
i2
+
RL
RL v2 = 12
RS
RL + 400
−
12 Vrms
• Periodic waveforms repeat every T seconds.
T
T
1
2
p(t) = T1 0 v(t) i(t) dt, V rms =
T 0 v(t) dt,
T
v(t) = T1 0 v(t) dt
• Sinusoid s(t) = A cos(ωt + φ) Peak amplitude = A,
Phase = φ, Frequency: ω ; T = 2π/ω ; f = 1/T
cos(ωt − φ) lags cos ωt by φ rad.
Thus v2 = 240 N − N RS i1 = 240N + N 2 RS i2 .
That is: v2 = 12 +
vOC
iSC
m ejωt
• Complex Exponential s
(t) = A
m
Am with amplitude = |Am| and phase = A
Phasor Diagram shows complex magnitudes.
= R + jX ; R resistance, X reactance
• Impedance Z
= 1 , Inductor Z
= jωL,
– Capacitor Z
jωC
Resistor Z = R
=A
for complex
– If v(t) = A cos(ωt + φ), then let V
analysis.
ELEC176 - Inductor
18-12
• Capacitors and Inductors
di
– Inductance unit: henry [H], v = L dt
+ v
−
i
2
L
2
– Energy w(t) = 12 C (v(t)) , w(t) = 12 L (i(t))
– RC or RL circuit
charging or
discharging:
vC (t) = As 1 − e−t/RC or vC (t) = As e−t/RC
vR (t) = As 1 − e−tL/R or vR (t) = As e−tL/R
– RC or R/L is the time constant.
• Voltage or current waveform can be divided into dc and
signal components: vI (t) = VI + vi (t)
• Simple RC, RL or RLC circuits affect dc and signal
components differently. Signal bypassing, dc blocking,
and frequency selection are possible.
• Transformer
i1
+
v1
−
i2
L1
L2
+
v2
−
M
In an ideal transformer: v2 = N v1 and i1 = −N i2
ELEC176 - Inductor
18-14
ELEC176 - Inductor
18-13