Feedback
Negative feedback
Positive Feedback
Vin = +A If output slightly higher Vo=+A+ε = V-­‐ V+ -­‐V-­‐=-­‐ε è Vo =-­‐gain ε lowers Vo Vo=A-­‐β V+ -­‐V-­‐=+β è Vo =gain β increases Vo Vin = +A If output slightly higher Vo=+A+ε =V+ V+ -­‐V-­‐=ε è Vo =gain ε V+ è large GO TO RAIL + Vo=+A-­‐ε V+=A-­‐ε V+ -­‐V-­‐=-­‐ε V+ è -­‐gε GO TO RAIL -­‐ Tends to draw the two terminals toghether Ball in a valley negative feedback ball on top of a hill positive feed back There are many good resources on line to review the basic ideas of passive circuits and
approaches to understanding. Our goal is to have a basic understanding of the role of
impedance in circuit analysis. There are three categories of analysis that we will
consider:
1. phasors
2. Fourier Transforms
3. LaPlace Transforms
Of course they are all related. For example, my view is that the phasor approach is the technique used to employ the results of Fourier Transform. If you analyze your circuit in terms of frequency response then you can characterize the voltage current relationship between components in terms of vectors in the complex plane. The algebra-­‐trigonometry is then simpler in terms of the phasors. This work well when analyzing certain circuits such as the RLC circuit. One must apply the more complete complex analysis for circuits where impedances appear, for example, in the denominator. For many problems the spectral response of the circuit i.e. the Fourier components are not the optimal approach. The underlying justification of the Fourier approach relies on examining your systems response to an applied frequency that in principle began at -­‐∞ and extend to ∞. This makes this analysis less useful when you have initial conditions. The charging and discharging of a capacitor is an example of such a problem. There are also some functions that grow in time in such a manner that they do not have well behaved Fourier transforms but still can be treated using the LaPlace approach. Both Laplace transforms and Fourier transforms are essential tools to be used in analyzing circuits. Complex numbers and their relationships are necessary to understand both methods with phasors as nice tool to see some of the relationships clearer. Here are a few links that I found useful: http://www.stanford.edu/~boyd/ee102/
General treatment è
http://fourier.eng.hmc.edu/e102/lectures/Laplace_Transform/Laplace_Tr
ansform.html
Derivation of Laplace & Fourier relationship
èhttp://www.cambridge.org/us/features/chau/webnotes/chap2laplace.p
df
-2pager on 2nd order linear DIF. EQ. è
http://www.math.psu.edu/tseng/class/Math251/Notes2nd%20order%20ODE%20pt1.pdf basic spice information for LTspice but perhaps usefule in general
http://ltwiki.org/?title=Main_Page
Block diagrams
Boyd EE102 y=Su systems often denoted by block diagram:
y output of a system S when u is applied as an input PSfrag replacements
u
S
y
• lines with arrows denote signals (not wires)
• boxes denote systems; arrows show inputs & outputs
• special symbols for some systems
To view the problems that we have been dealing with consider the RLC analysis. The input can be taken as the AC supply and the output is the current through or voltage across a component. Systems
2–3
The key in analyzing circuits is to remember that all circuits represent coupled linear diff. eq. obtained by nodal analysis or Kickoff’s Laws. For example some set of components might require two coupled equations that depend upon the voltages generated by R,L,C components in a network where there are power supplies that generate voltages 𝑓! 𝑡 , 𝑓! (𝑡). 𝑑
𝑑
𝛾! 𝐼! 𝑡 + 𝛼! 𝐼! 𝑡 + 𝛿! 𝐼! 𝑡 𝑑𝑡 + 𝛾! 𝐼! 𝑡 + 𝛼! 𝐼! 𝑡 + 𝛿! 𝐼! 𝑡 𝑑𝑡 = 𝑓! (𝑡) 𝑑𝑡
𝑑𝑡
!
!
𝛾! !" 𝐼! 𝑡 + 𝛼! 𝐼! 𝑡 + 𝛿! 𝐼! 𝑡 𝑑𝑡 + 𝛾! !! 𝐼! 𝑡 + 𝛼! 𝐼! 𝑡 + 𝛿! 𝐼! 𝑡 𝑑𝑡 = 𝑓! (𝑡) In considering this problem as a Fourier or spectral analysis problem my approach is to remember that in projectile motion problems we decouple the horizontal and vertical parts of the projectiles trajectory. We simply find the horizontal motion by propagating the projectile horizontally using the horizontal velocity. We treat the vertical motion by considering the gravitational acceleration and the initial vertical velocity. The simplicity and correctness of this approach relies on the fact that the motion is described in terms of orthogonal directions. The horizontal motion is completely independent of the vertical. This very same feature is present in the spectral analysis. The amount that a circuit oscillates at any frequency ω1 is completely independent of its oscillation at frequency ω2. You are then free to find the way the circuit responds to each frequency separately. Given some arbitrary driving or input 𝑓! 𝑡 on finds the part of this signal that is oscillating at ω1 ,solve the problem with this much simpler function then proceed to find the response to all the frequencies present in 𝑓! 𝑡 and add them together. This topic was discussed briefly when reviewing test1 and some of these comments are
available on the web site
So far we have examined the response of some circuit networks to a fixed AC signal and
recognized that networks and/or devices could be characterized by frequency response.
The next step in our formal approach to networks will be using the LaPlace transform to
solve the diff. eq. that govern the network behavior.
Consider the derivative and integral as expressed using the Fourier and Laplace
transforms
Fourier
!
=
𝜈 𝑖𝑠 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑖𝑛 𝐻𝑧
𝐹 𝑠 = ℒ 𝑓 𝑡 (𝑠)
!
𝑓 𝑡 𝑒 !!!!"# 𝑑𝑡
=
𝑓 𝑡 𝑒 !!" 𝑑𝑡
!
!!
𝑠 = 𝜎 + 𝑗𝜔
𝐻 𝜔 = ℱ ℎ 𝑡 (𝜔)
!
1
=
ℎ 𝑡 𝑒 !!"# 𝑑𝑡
2𝜋 !!
𝑓 𝑡 = ℱ !! 𝐹 𝜈 (𝑡)
notation: Let f, F be functions when using ν Let h, H “””” ω h(t) = f(t) 𝑑!
𝑑𝑡 !
Laplace
𝐹 𝜈 = ℱ 𝑓 𝑡 (𝜈)
𝜔 = 2𝜋𝜈 !
=
𝑓 𝑡 = ℒ !! 𝑓 𝑠 (𝑡)
1 !!!!
=
𝑓 𝑠 𝑒 !" 𝑑𝑠
2𝜋𝑗 !!!!
𝐹 𝜈 𝑒 !!!"# 𝑑𝜈
!!
ℎ 𝑡 = ℱ !! 𝐻 𝜔 (𝑡)
!
1
=
𝐻 𝜔 𝑒 !"# 𝑑𝜔
2𝜋 !!
ℱ ℎ!! 𝑡 𝜔
= 𝑗𝜔 ! ℱ ℎ 𝑡
= 𝑗𝜔 ! 𝐻(𝜔)
𝑑
→ 𝑠 𝐹 𝑠 − 𝑓 0
𝑑𝑡
𝜔
require lim!→±! ℎ 𝑥 = 0
No initial conditions since the
integration starts at -∞
Note the Fourier function form
orthogonal basis
𝑒
!!
!" !
→ 𝑠 ! 𝐹 𝑠 − 𝑠𝑓 0 - 𝑓′ 0
𝑠 ! 𝐹 𝑠 − 𝑠 !!! 𝑓 0
𝑑𝑓 𝑑𝑓 !!!
− 𝑠 !!! 0 … − 𝑠 ! !!!
0
𝑑𝑡
𝑑𝑡
Note that the LaPlace is similar but
requires a knowledge of the driving
function and its derivatives at t=0
LaPlace
𝑒 !!" = 𝑒 !(!!!")!
!!!!"#
!
𝑑𝑡
𝑔 𝑡 =
𝐺 𝜔 =
1
𝐹(𝜔)
𝑗𝜔
𝑓 𝜏 𝑑𝜏
!
𝐺 𝑠 =
1
𝐹(𝑠)
𝑠
f is assumed to be a differentiable function, and its derivative is assumed to be of exponential type. This can then be obtained by integration by parts
evaluate at the point just before 0. Differentiation
derivative formula for discontinuous functions
disc. at t=0 if signal f is discontinuous at t = 0, then
Second Differentiation
L(f !) = sF (s) − f (0−)
f is assumed twice differentiable and the second derivative to be of exponential type. Follows by applying the Differentiation property to f′(t).
f is assumed to be n-­‐times differentiable, with nth derivative of exponential type. Follow by mathematical induction example: f is unit step, so f !(t) = δ(t)
L(f !) = s
General Differentiation ! "
1
−0=1
s
Linearity
convolutions f(t) and g(t) are extended by zero for t <
0 in the definition of the convolution.
theIntegration
Laplace transform is linear : if f and g are any signals, and a is
any
the
Laplace
u(t) is the Heaviside step function. Note (u scalar,
we
have
∗ f)(t) is the convolution of u(t) and f(t).
scalar, we have
linear The Laplace transform
L(af ) = aF,
L(f + g) = F + G
Time scaling
i.e., homogeneity
homogeneity & superposition
i.e.,
hold
3–22
transform of a function multiplied by a constant is the transform mult. by same constant transform of the sum of two functions is the sum of the transforms if you change scales the transform changes as shown time scaling define signal g by g(t) = f (at), where a > 0; then
example:
example:
Time
delay
Exponential
!
G(s)
="(1/a)F
(s/a) scaling t
)
L 3δ(t) − 2et = 3L(δ(t)) − 2L(e
exponential at
f bescaled
a signalbyand
a scalar, and define
g(t) = e f (t); then makes sense: timesletare
a,afrequencies
by21/a
scaling = 3−
s−1
let’s
check:
G(s)
= F (s − a)
let
f be
a signal and T > 0; define the signal
g as
3s
−
5
! ∞=
translate !
in t∞
ime g is same function but T later !
−st
−
0
< T1 dτ = (1/a)F (s/a)
f (at)e
dt = (1/a)
f0(τ≤)est−(s/a)τ
G(s) =
let’s check:
0
g(t) =
f (t − T0 ) t ≥ T
∞ e−(s−a)tf (t) dt = F (s − a)
where
τ=
athave G(s) = e
then
we
F
(s)
(g is f , delayed by T seconds &0 ‘zero-padded’ up 0to T )
The Laplace
Laplace transform
transform t
The
L(e ) = 1/(s − 1) so
example:
derivation:
!
∞
G(s) = −sT e−steatf (t) dt =
f (t)
PSfrag replacements
example: L(cos t) = s/(s2 + 1), and hence
g(t)
!
!
∞
G(s) =
−t
t
The Laplace transform
(s + 1)T + 1
t=
=T
!
= e
The Laplace transform
3–10
3–10
1
∞1
=
s+1
s
+
1
e L(eg(t)
=1 s2 −ea−st
(s/a)
= f2 (t − T ) dt
cosdt
t) =−
L(eat) = −st
(1/a)
0
The Laplace transform
!
∞
0
−sT
s + 2s + 2
t
e−s(τ +T )f (τ ) dτ
F (s)
3–16
3–14
3–15
SUMMARY OF LAPLACE TRANSFORMS
3
where (of course) g(t) does not depend of f (t).
Short Table of Laplace Transforms
f (t)
1
s
1
s2
n!
1
t
tn
eat
tn eat
cos(!t)
sin(!t)
eat cos(!t)
eat sin(!t)
t cos(!t)
t sin(!t)
sin(!t)
F (s)
!t cos(!t)
sn+1
1
s a
n!
(s a)n+1
s
2
s + !2
!
2
s + !2
s a
(s a)2 + ! 2
!
(s a)2 + ! 2
s2 ! 2
(s2 + ! 2 )2
2!s
2
(s + ! 2 )2
2! 3
(s2 + ! 2 )2
Department of Mathematics and Statistics, Texas Tech University, Lubbock, TX
79409-1042
E-mail address: lance.drager@ttu.edu
Example
Do some transforms-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐ let’s find Laplace transform of f (t) = et:
! ∞
! ∞
t −st
e e
dt =
e(1−s)t dt =
F (s) =
0
0
"∞
1 (1−s)t ""
1
e
=
"
1−s
s−1
0
provided we can say e(1−s)t → 0 as t → ∞, which is true for $s > 1:
"
" "
" "
"
" (1−s)t" " −j(#s)t" " (1−$s)t"
e
=
e
e
" "
" "
" = e(1−$s)t
"
# $% &
=1
• the integral defining F makes sense for all s ∈ C with $s > 1 (the
‘region of convergence’ of F )
• but the resulting formula for F makes sense for all s ∈ C except s = 1
More examples
constant: (or unit step) f (t) = 1 (for t ≥ 0)
F (s) =
!
∞
−st
e
0
"∞
1 −st ""
1
dt = − e
" =s
s
0
t→
provided
wefirst
can express
say e−stf →
sinusoid:
(t) 0=ascos
ωt ∞,
as which is true for %s > 0 since
" −st" "" −j(#s)t"" "" −($s)t""
−($s)t
"e "f=
"e + (1/2)e
" = e−jωt
" jωt
(t)"e= (1/2)e
# $% &
=1
can find
F as F makes sense for all s with %s > 0
•now
the we
integral
defining
! ∞
• but the resulting formula
for
" F makes sense for all s# except s = 0
e−st (1/2)ejωt + (1/2)e−jωt dt
F (s) =
0
= (1/2)
!
∞
Impulses
at t = 0
(−s+jω)t
dt + (1/2)
e
0
!
∞
e(−s−jω)t dt
0
if f contains impulses at t = 0 we choose to include them in the integral
1
1
The Laplace transform
3–6
defining F :
+! (1/2)
= (1/2)
∞
s − jω
s + jω
f (t)e−st dt
s F (s) =
= 2
0−
s + ω2
(you can also choose to not include them, but this changes some formulas
(validseefor&"s
> 0; final formula OK for s #= ±jω)
we’ll
use)
example: impulse function, f = δ
! ∞
"
The Laplace transform
δ(t)e−st dt = e−st"t=0 = 1
F (s) =
3–7
0−
derivation of derivative formula: start from the defining integral
Derivative transform
similarly for f = δ (k) we have
F (s) =
!
∞
0−
δ (k)(t)e−st
!
∞
"
"
d −st"" f " (t)ek −st
dt
G(s)
k=
dt = (−1) k e "
= s e−st"t=0 = sk
dt 0 t=0
k
integration by parts yields
The Laplace transform
G(s) =
=
3–9
"∞
e−stf (t)" −
0
!
∞
f (t)(−se−st) dt
0
lim f (t)e−st − f (0) + sF (s)
t→∞
for "s large enough the limit is zero, and we recover the formula
G(s) = sF (s) − f (0)
Example: RC circuit
1Ω
PSfrag replacements
y
1F
u
𝑄
−𝑢
𝑡
+
𝐼𝑅
+
= 0 apacitor is uncharged at t = 0, i.e., y(0)
𝐶 =0
(t) is a unit step
m last lecture,
!"
!
𝑢 𝑡 = 𝑅 !" + !
𝑢 𝑡 = 𝑉! [𝑠𝑡𝑒𝑝 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛]
!"
(t)! [𝑠𝑡𝑒𝑝]
+ y(t)==𝑅𝐶
u(t) + Q
y !𝐶𝑉
!"
Laplace transform, term by term:
𝑈 𝑠 = 𝐶𝑉! 1
𝑠
sY𝑑𝑄
(s)→+𝑠𝑌
Y (s)
𝑠 −=𝑦 1/s
0 = 𝑠𝑌 𝑠
𝑑𝑡
𝑄 → 𝑌(𝑠)
ng y(0) = 0 and U (s) = 1/s)
place transform
!
𝐶𝑉! ! = 𝑅𝐶𝑠𝑌 𝑠 + Y(s)= Y(s)(1+RCs)
𝑌 𝑠 =
𝐶𝑉! 𝑅𝐶
3–23
𝐶𝑉! 1
= 𝐶𝑉! 𝑅𝐶
; 𝐴 = 𝑅𝐶𝑠
𝑠(1 + 𝑅𝐶𝑠)
𝐴(1 + 𝐴)
!
!
− !!! = 𝐶𝑉! 𝑅𝐶
!
𝑌 𝑠 =
!!! !"
!"#
−
!!!
!!! !"
!!!"#
= 𝐶𝑉!
RULE af (t) è aF(s)
𝐶𝑉! 1 − ℒ !!
!
− !(!!!) = 𝐶𝑉! 𝑅𝐶
!(!!!)
!"
(!!!"#)
;
!
!
−
!
!(!!!)
!"
(!!!"#)
!"
(!!!"#)
=
!
(
!
!!)
!"
= !
(!!
!
)
!"
𝑅𝑈𝐿𝐸 scaling by a generates 1/a in the transform or
change the form of the fraction as above and read off
the transform
𝐶𝑉! 1 −
!
𝑅𝐶 ! !
𝑒 !" = 𝑄! 1 − 𝑒 !!"
𝑅𝐶
Examples
simple
discharge RC circuit:
PSfrag replacements
R
circuit equation: RCv
v
C
solution: v(t) = v(0)e
𝑄
𝐼𝑅 + = 0 𝐶
population
dynamics:
𝑑𝑄
→ 𝑠𝑌 𝑠 − 𝑦 0
𝑑𝑡
𝑄
𝑌(𝑠) is population
• →y(t)
of some bacteria at time t
0•=growth
𝑅𝐶 𝑠𝑌 𝑠 −
𝑦 0decay
+ 𝑌 𝑠 if negative)
(or
0 = 𝑅𝐶 𝑌 𝑠 𝑠 + 1/𝑅𝐶 − 𝑅𝐶𝑦(0)
d is death
𝑦(0) rate
𝑌 𝑠 =
𝑠 + 1/𝑅𝐶
(b−d)t
• y(t) = y(0)e
𝐼 0 𝑒
!
rate is y ! = by − dy
(grows if b > d; decays if b < d
!
!"
Natural response of first and second order systems
Second
order systems
a(s2Y (s) − sy(0) − y !(0)) + b(sY (s) − y(0)) + cY (s) = 0
!
"#
$
L(y !! )
!
solve for Y (just!!algebra!)!
ay + by + cy = 0
"#
L(y ! )
$
αs + β
(0)χ+ by(0)
asy(0)
+ ay of
Roots
solution of ay !! + byY! (s)
+ cy
= 2
= = 0 is2
as + bs + −1)
c
as + bs + c
a > 0 (otherwise
multiply
equation
(two) roots
of characteristic
polynomial by
χ are
!√
"
!
(0)
+
by(0)
where α = ay(0) and β = ay−b
2
± b αs
− 4ac+ β
=
by Laplace transform:y(t) =λ1,2
L−1
2a 2
as
+ bs
c 4ac)
Natural response
of first and distinct
second order systems
Real
roots
(b2+>
!
as roots dependent he ccY
onstants −denominator y (0)) + hb(sY
(s)
− y(0)on ) t+
(s) = 0
a(s Y (s) − sy(0)
cases:
!
"#
$
!
"#
$
2 three
+• roots
bs
c realisand
called
characteristic
polynomial of the
χ(s) = as L(y
!! ) +
2
> 4ac
are
distinct:
roots χ(s) Real = a(s
− λ1b)(s
−L(y
λ2!)) (λ1, λ2 real)
2
i.e., we have as ! + bs + c = a(s − λ1)(s − λ2)
4–7
2
λ1 =
−b +
√
b2 − 4ac
,
2a
λ2 =
−b −
system
√
b2 − 4ac
2a
form of y = L (Y ) depends on roots of characteristic polynomial χ
mY page
(just4-6,
algebra!)
2
• roots are real and equal: b = 4ac
αs + β
coefficients of numerator
αs
+
β
come
Y (s) = λ1 = λ2 = −b/(2a)from initial conditions
a(s
− λ1)(s −αs
λ2+
) β
by(0)
asy(0) + ay !(0) +
= 2
Y (s) =
2
as
+
bs
+
c
as
+
bs
+
c
ere α, β depend on initial conditions
−1
Natural response of first and second order systems
press
Y asand β =
= ay(0)
4–9
write this as a product of {Partial fractions} 𝑥
𝑦
!
+
(0)
by(0)
ay
(𝑠 −
𝜆1)+ (𝑠
− 𝜆2)
r2
r1
+
Y (s) =
Find the transforms s −from λ1 the table. s − λ2
Partial Fractions ere r1 and r2 are found
from
R(x) = ƒ(x)/g(x) se of first and second order systems
ich yields
4–7
r1 + r2 = α/a,
−λ
2 r1 − λ1 r2 = β/a
λ α+β
,
r1 = √
2
b − 4ac
al response of first and second order systems
1
al response of first and second order systems
−λ2α − β
r2 = √
b2 − 4ac
4–
4–1