Applications of the Laplace Transform - Vis Centercheung/courses/ee422g/lecture6.pdf · EE 422G...

EE 422G Notes: Chapter 6 Instructor: Cheung

Page 6-1

Applications of the Laplace Transform Application in Circuit Analysis 1. Review of Resistive Network

1) Elements

2) Superposition

1

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Page 6-2

3) KVL and KCL – Select a node for ground. Watch out for signs!

4) Equivalent Circuits

Open Circuit Voltages OCV V= =

Rs = Equivalent Resistance

Short Circuit Currents SCI I= =

Thevenin Equivalent Circuit

Norton Equivalent Circuit

Rs = Same as before



Page 6-3

5) Nodal Analysis and Mesh Analysis

Mesh analysis (use KVL)

++=−+−+=

224111

213212111 )()(

SS

S

VIRIRV

IIRIIRIRV

Solve for I1 and I2. 2. Characteristics of Dynamic Networks 1) Inductor

2) Capacitor

∫ ∞−=

=

t

CC

CC

diC

tv

tvdt

dCti

ττ )(1

)(or

)()(

∫ ∞−=

=

t

LL

LL

dvL

ti

tidt

dLtv

ττ )(1

)(or

)()(

(Use KCL)



Page 6-4

3) Operation Amplifier

A general op-amp model is described above. In practice, the input resistance, Rin, is very large (> 1012 Ω) and the gain, A, is very large (>105). Thus, we will use the ideal model in the analysis:

1. Input current Ii(t) = 0 (due to the large input impedence) 2. Input voltage difference vi(t) = 0 and output voltage vo(t) is dictated by

external circuit (due to the large gain)

Based on the ideal op-amp model,

v2(t) = v1(t) (1) Also, as the op-amp does not have any input current, applying KCL at the inverting port, we have

v2(t)/Ra= (vo(t)-v2(t))/Rb

vo(t)/v2(t) = 1+Rb/Ra

Plug in (1), we have vo(t)/v1(t) = 1+Rb/Ra

This circuit is called Non-Inverting Amplifier.

-

+

≡

+

- +

-

vo(t)

vi(t)

+

-

vi(t)

-

+ - Avi(t)

Rin

I i(t)

Inverting input

Non-Inverting input

- +

+ - v1(t)

Ra Rb

vo(t) io(t)

v2(t) +

-

+

-

vo(t)

Example:



Page 6-5

4) Mutual Inductor – used in transformer. Two separate circuits with coupling currents.

To link the two circuits together, introduce a combined current term (i1+i2):

dt

diMLii

dt

dM

dt

diM

dt

diL

dt

diM

dt

diMtv

iidt

dM

dt

diML

dt

diM

dt

diM

dt

diM

dt

diLtv

2221

222

212

211

1

211111

)()(

)(

)()(

)(

−++=

−++=

++−=

++−=

Equivalent circuit:

Make sure both i1 and i2 point either away or toward the polarity marks to make the mutual inductance M positive.



-6

Example : Apply mesh analysis to the following circuit

Using Laplace Transform

)()(1

)()(

)()0()(1

))0()((

)()()()(

sRIsICs

sILs

sRIs

v

s

sI

CissIL

sVsVsVsV

C

RCL

++=

+

++−=

++=

−

Define ‘Generalized Resistors’ (Impedances)

)()()()( sZsIsVLssZ LLL =⇒=

)()()(1

)( sZsIsVCs

sZ CCC =⇒=

Both capacitor and inductor behave exactly like a resistor!

RsZsZ

sVsI

sRIsIsZsIsZsV

CL

s

CLs

++=⇒

++=⇒

)()(

)()(

)()()()()()(

Everything we know about resistive network can be applied to dynamic network in Laplace domain:

analysismesh and analysis Nodal

circuit Equivalent

KCL and KVL

ionsuperposit

Law Ohms dGeneralize



Page 6-7

3. Laplace transform models of circuit elements. What if the initial conditions are not zero? 1) Capacitor

Alternatively, you can also represent it as an impedance and a parallel current source (Norton equivalent circuit)

BE VERY CAREFUL ABOUT THE POLARITY OF VOLTAGE SOURCE AND THE DIRECTION OF CURRENT SOURCE!!

ZC

Cv(0-) I(s)

+ -

V(s)



Page 6-8

2) Inductor

Alternatively, you can also represent it as an impedance and a parallel current source (Norton equivalent circuit)

3) Resistor V(s) = RI(s) 4) Voltage and Current Sources (Don’t forget to apply Laplace Transform on

them)

5) Op-Amp : same ideal model assumption

ZL

i(0-)/s I(s)

+ -

V(s)



Page 6-9

6) Mutual Inductance (Transformers)

⇓ Laplace transform model: Obtain it by using inductance model



Page 6-10

Example: Find Complex Norton Equivalent circuit given 0)0( =−cv

Solution 1) Compute the Short-Circuit Current scs IsI =)(

Straightforward to see: )(2)( sIsI S −= To compute I(s), apply mesh analysis on the left loop:

3

2)(2)(

3

1)(

)(3

)()3

1(1

)(3 1)(1

+−=−=⇒

+=⇒

+=+=+Ω×=

ssIsI

ssI

sIs

ssI

sssIsI

s

s

No need to do inverse Laplace transform as the equivalent circuit is in the s-domain.



Page 6-11

2) Find the equivalent impedance sZ

Normally, we can just kill all the independent sources and combine the impedances (using resistive combination rules). However, as there is a dependent source, we need to drive it with a test voltage:

)(2

)(

)(

)(

sI

sV

sI

sVZ test

test

tests ==

Mesh analysis on the left loop:

0)(

)(3

)() 1(

0)(3

1)(

=⇒

−=Ω⇒

=+Ω×

sI

sIs

sI

s

sIsI

So we got an interesting result: CIRCUITOPEN 0

)(⇒∞== sV

Z tests

Vtest(s)

a

ZL

V test(s) +

-

Itest(s)

Zs



Page 6-12

Example: Find the transfer function H(s) = Vo(s)/Vi(s) of the following circuit. Assume all initial conditions are zero. This is called the Sallen-Key circuit, which we will see again in filter design.

Rewrite everything in Laplace domain, we have

We recognize the op-amp configuration as a non-inverting amplifier, so we have

a

b

R

RK += 1



Page 6-13

To find Vo, we need Vb which depends on Va. All other nodal voltages are known. Thus, we need two nodal equations:

Applying KCL at node a, we have:

01

121

=−+−+−

sC

KVV

R

VV

R

VV babaia

11

12

121

1111V

RVsKC

RVsC

RR ba =

+−

++⇒ (1)

Applying KCL at node b, we have:

022

=+−b

ab sVCR

VV

011

122

=

++−⇒ ba VsKC

RV

R (2)

Combining equations (1) and (2) by eliminating Va, we get:

[ ] 1)1(

1

1121222

2121 +−+++=

sKCRCRCRsCCRRV

V

i

b

Since bo KVV = , we have

[ ] 1)1(1121222

2121 +−+++=

sKCRCRCRsCCRR

K

V

V

i

o

where a

b

R

RK += 1


Applications of the Laplace Transform - Vis Centercheung/courses/ee422g/lecture6.pdf · EE 422G...

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Transcript of Applications of the Laplace Transform - Vis Centercheung/courses/ee422g/lecture6.pdf · EE 422G...