6.002
CIRCUITS AND
ELECTRONICS
The Impedance Model
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Review
„
Sinusoidal Steady State (SSS)
Reading 13.1, 13.2
vI = Vi cos ωt +
–
C
+
vO
–
„
Focus on steady state, only care
about vP as vH dies away.
„
Focus on sinusoids.
SSS
„
R
Sinusoidal Steady State (SSS)
Reading 13.1, 13.2
Reading: Section 13.3 from course notes.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
vP
Review
Vi cos ωt
1
usual
circuit
model
sneak
in
Vi e jωt
drive
set
up
DE
complex
algebra
V p cos[ωt + ∠V p ]
nightmare
trig.
Vp
2
The Sneaky Path
3
vH
take 4
real total
part
V p e jω t
Vi
1 + jωRC
Vp contains all the information we need:
Vp
∠V p
Amplitude of output cosine
phase
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Review
vO = V p cos(ωt + ∠V p )
Vp
Vi
Vp
=
1
= H ( jω ) transfer
function
1 + jωRC
remember
demo
1
Vi
1
2
1
1
ωRC
1 + ω 2 R 2C 2
Bode plot
∠
Vp
ω=
Vi
ω
1
ω=
RC
break frequency
0
1
RC
ω
⎛ − ωRC ⎞ π
tan −1 ⎜
⎟ −
⎝ 1 ⎠ 4
−
π
2
The Frequency View
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Is there an even simpler way
to get Vp ?
Vi
Vp =
1 + jωRC
Divide numerator and denominator by jωC.
1
V p = Vi
jω C
1
+R
jω C
Hmmm… looks like a voltage divider
relationship.
ZC
V p = Vi
ZC + R
Let’s explore further…
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Impedance Model
Is there an even simpler way to get Vp ?
Consider:
+
vR
–
+
vC
–
iR
i R = I r e jω t
vR = RiR
R
vR = Vr e jω t
Vr e jω t = RI r e jω t
Resistor
iC
C
iC = I C e
jω t
vC = VC e jω t
Vr = RI r
dvC
iC = C
dt
I C e jω t = CVC jωe jω t
1
IC
j ωC
ZC
di
vL = L L
dt
Capacitor
+
vL
–
iL
i L = I l e jω t
L
vL = Vl e jω t
VC =
Vl e jω t = LI l jωe jω t
Inductor
Vl = jωL I l
ZL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Impedance Model
In other words,
capacitor
Ic
+
Vc
–
ZC
Vc = Z C I c
1
ZC =
j ωC
impedance
inductor
resistor
Il
+
Vl
–
+
Vr
–
ZL
Ir
ZR
Vl = Z l I l
Z l = j ωL
Vr = Z r I r
Zr = R
For a drive of the form Vc e jωt ,
complex amplitude Vc is related to the
complex amplitude Ic algebraically,
by a generalization of Ohm’s Law.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Back to RC example…
R
+
C vC
–
vI +
–
Impedance model:
ZR = R
Ic
+
Vc
–
Vi +
–
1
ZC =
jωC
1
ZC
jωC
Vc =
Vi =
Vi
1
ZC + Z R
+R
jωC
Vc =
1
Vi
1 + jωRC
Done!
All our old friends apply!
KVL, KCL, superposition…
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Another example, recall series RLC:
Remember, we want only the steady-state
response to sinusoid
Ir
L
Vi e jω t
Vi +
–
Vi cos ωt
C
R
Vi Z R
Vr =
Z L + ZC + Z R
+
Vr
–
Vr e jω t
Vr cos(ωt + ∠Vr )
Vi R
Vr =
1
j ωL +
+R
jωC
Vr =
Vi jωCR
− ω 2 LC + 1 + jωCR
We will study this and other functions
in more detail in the next lecture.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Big Picture…
V p cos[ωt + ∠V p ]
Vi cos ωt
usual
circuit
model
set
up
DE
nightmare
trig.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Big Picture…
V p cos[ωt + ∠V p ]
Vi cos ωt
usual
circuit
model
Vi e jωt
drive
set
up
DE
nightmare
trig.
complex
algebra
take
real
part
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Big Picture…
V p cos[ωt + ∠V p ]
Vi cos ωt
usual
circuit
model
Vi e jωt
drive
set
up
DE
nightmare
trig.
complex
algebra
impedance-based
circuit model
take
real
part
complex
algebra
No D.E.s, no trig!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Back to
Ir
Vr
jωRC
=
Vi 1 + jωRC − ω 2 LC
Vi
L
+
–
C
R
+
Vr
–
Let’s study this transfer function
Vr
jωRC
=
Vi 1 + jωRC − ω 2 LC
(
jωRC
1 − ω 2 LC ) − jωRC
=
⋅
2
(1 − ω LC ) + jωRC (1 − ω 2 LC ) − jωRC
Vr
=
Vi
(1 − ω
ωRC
2
LC ) + (ωRC )
2
2
Observe
Low ω : ≈ ωRC
R
High ω : ≈
ωL
ω LC = 1 : ≈ 1
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Graphically
Vr
=
Vi
ωRC
(1 − ω
2
LC ) + (ωRC )
2
2
Low ω : ≈ ωRC
R
High ω : ≈
ωL
ω LC = 1 : ≈ 1
Vr
Vi
“Band Pass”
1
R
ωL
ωRC
1
LC
ω
Remember this trick to sketch the form of
transfer functions quickly.
More next week…
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17