chapter 2chapter 2 review of signal processing basics

Chapter 2Chapter 2Review of Signal Processing Basics

Marc T. Thompson, Ph.D.Thompson Consulting, Inc.

9 Jacob Gates RoadHarvard MA 01451Harvard, MA 01451

Phone: (978) 456-7722 Fax: (240) 414-2655

E il @ h dEmail: [email protected]: http://www.thompsonrd.com

Slides to accompany Intuitive Analog Circuit Design by Marc T. Thompson© 2006-2008, M. Thompson

Summary• Review of 1st-order systems• Relationship between bandwidth and risetime• Relationship between bandwidth and risetime• 2nd order systems• Resonance, damping and quality factor

E th d• Energy methods• Transfer functions, pole/zero plots and Bode plots• Calculating risetime for systems in cascadeg y• Comments on PSPICE

10/30/2008 Signal Processing Basics 2

Laplace Notation• Basic idea: Laplace transform converts differential

equation to algebraic equation

• Laplace method is used in sinusoidal steady state after all startup transients have died out

Circuit domain Laplace (s) domain Resistance, R R Inductance L LsCapacitance C 1Capac ta ce C

Cs1

Circuit Laplace transformed circuit


System Function H(s)• Find “transfer function” H(s) (also called “transfer

function”) by solving Laplace transformed circuitfunction ) by solving Laplace transformed circuit

1)(1

1

1

)()()(

21

22

+++

=+

==CsRR

CsR

RRCs

R

svsvsH o

1)()( 2121

++++ CsRRCs

RRsvi


Does this System Function Make Sense Intuitively?

+R1

+

-vo(s)+vi(s) -

1/(Cs)R2

At very low frequencies (s 0), the capacitor is an open-circuit:

At very high frequencies (s ∞), the capacitor is an short circuit:short-circuit:


Does this System Function Make Sense Intuitively?

+R1

+

-vo(s)+vi(s) -

1/(Cs)R2


First-Order Systemy

• Voltage-driven RC lowpass filter

Vtt−τ )1()(

eRVti

eVtvt

r

o

=

−=−τ

τ

)(

)1()(

RCRr

=τ

)(

τω 1

h =Time constant Bandwidth [Hz]

ττ 2.2=R

πω2

hhf =

35.0=τ

RisetimeBandwidth [rad/s]

10/30/2008 Signal Processing Basics 7h

R f=τ

Another First-Order SystemAnother First Order System• Current-driven RC

eIRtvt

t

o −=−

−τ )1()(

RCeIti

t

r

==

τ

τ)(

ττ 2.2=Rτ

ω 1h =

R f35.0

=τR

πω2

hhf = hf


First-Order Systems --- Some Details

• Frequency response:

• Phase response:Phase response:

• -3 dB bandwidth:


10 - 90% Risetime• Defined as the time it takes a step response to transition

from 10% of final value to 90% of final valueThi l t i f fi t d t ith h t• This plot is for a first-order system with no overshoot or ringing


Relationship Between Risetime and Bandwidthp

• Exact for a first-order system:

• Approximate for higher-order systems


First-Order System Step and Frequency Response

Step Response

1

Am

plitu

de

0.4

0.6

0.8

Time (sec.)0 1 2 3 4 5 6

0

0.2

BodeDiagrams

Mag

nitu

de (d

B)

Bode Diagrams

-20

-10

0

-3 dB point= 1 rad/sec

F ( d/ )

Pha

se (d

eg);

10-1

100

101

-80-60-40-20 Angle = -45 deg.

at -3dB point


Frequency (rad/sec)

“Group Delay”

11)(+

=s

sHτ 1

11)(

ωτω

+=

jjH

( )1

2

tan)(

1)(1)(

ωτω

ωτω

∠

+=

−jH

jH

( )

( )21)()(

tan)(

ωττ

ωωω

ωτω

+=

∠−=

−=∠

djHdjG

jH

( )Group delay is a measure of how much time delay the frequency components in a signal undergo. Mathematically, the group delay of a

t i th ti d i ti f th h ith t t Tsystem is the negative derivative of the phase with respect to omega. To find group delay for the first-order system, we make use of the identity:

( )dduu

dd

21

11tan+

=−


( )dxudx 21+

First-Order System --- Low and High Frequency Behavior

•• ClosedClosed--form solution for frequency response:form solution for frequency response:

11)(+

=s

sHτ

1)( jH

ω

ωτω

2 1)(1)(

1)(

=

+=

jH

jjH

( )ωτω

ωτ1

2

tan)(

1)(−−=∠

+

jH

• How does this response behave at frequencies well above and well below the pole frequency ?


y

First-Order System --- High Frequency Behavior

For high frequencies, where ωτ >> 1

tan-1(x) = π/2 – 1/x + 1/(3x3) - … for x > 1

ωτω

ωτ

1)(1≈

>>jH

ωτπω ωτ

12

)( 1 +−≈∠ >>jH


First-Order System --- Low Frequency Behavior

For low frequencies, where ωτ << 1

tan-1(x) = x – x3/3 + x5/5 - … for x < 1 and 21

11 x

x−≈

+for x << 1

21

ωτω

ωτω

ωτ

ωτ

−≈∠

≈−≈

<<

<<

1

21

)(

1)(211)(

jH

jH

Well below the pole frequency, the magnitude is approximately 1.The phase is approximately linear phase, behaving like an approximate time delay “Group delay” is negative derivative of angle

ωτ <<1)( j

approximate time delay. Group delay is negative derivative of angle with respect to frequency, or:

τω ωτ ≈<<1)( jG


ωτ <<1)( j

Step Response of Single Pole

Step ResponseStepresponseofsinglepoleat -1r/s

• Single pole at -1 rad/sec.• Note that risetime = 2.2 sec

0.8

0.9

1

Step response of single pole at 1 r/s

plitu

de

0.5

0.6

0.7

Am

02

0.3

0.4

0 1 2 3 4 5 6

0

0.1

0.210-90% risetime= 2.2 sec


Time (sec.)

Step Response of Single Pole with High Frequency Pole Added

• Poles at -1 rad/sec. and -10 rad/sec.• Note the time delay of approximately 0 1 secNote the time delay of approximately 0.1 sec.

Step Response

0.9

1

Compare to system with real-axis poles at -1 and -10 r/s

e

0.6

0.7

0.8

Am

plitu

de

0.3

0.4

0.5

System with extra pole added

0 1 2 3 4 5 6

0

0.1

0.2


Time (sec.)

Bode Plot Single Pole with High Frequency Pole Added

• Poles at -1 rad/sec. and -10 rad/sec.Compare to system with real-axis poles at -1 and -10 r/s

-20

-10

0

Single pole Single pole + additional high freq. pole

); M

agni

tude

(dB)

-60

-50

-40

-30p

Pha

se (d

eg)

-100-80-60-40-20

Frequency (rad/sec)

10-1

100

101

102

-160-140-120


q y( )

Second-Order Mechanical Systemy

Spring force:

Newton’s law for moving mass:

Differential equation for mass motion:

Guess a solution of the form:

Differential equation for mass motion:


Second-Order Mechanical Systemy

Put this proposed solution into the differential equation:

This solution works if:


Second-Order Electrical Systemy

1

22

2

2

22 2121

11

1

1

)()()(

nn

n

i

o

ssssRCsLCsC

LsR

CssvsvsH

ωζωω

ζ ++=

++=

++=

++==

n LC1

=ω

2nnCs ωω

Natural frequency

o

n

ZR

CL

RRC21

21

2===

ωςDamping ratio


Second-Order System Frequency Response

⎟⎟⎞

⎜⎜⎛−+

=2

121)(

ωςωω

jjH

=

⎟⎟⎠

⎜⎜⎝

+ 2

1)(

1

ω

ωω nn

jH

⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛2

2

22

12)(

ωω

ωςω

nn

j

⎟⎟⎞

⎜⎜⎛

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

−=∠ −− 11 2tan

2

tan)( ςωωωςω

ω nnjH ⎟⎟⎠

⎜⎜⎝ −

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=∠ 22

2

2tan

1tan)(

ωωωω

ωn

n

jH


Second-Order System Frequency Response

Frequency response for natural frequency = 1 and various damping ratios

dB)

20

-10

0

10

20

30

Resonance

ase

(deg

); M

agni

tude

(d

-40

-30

-20

0

Pha

-150

-100

-50

Frequency (rad/sec)

10-1

100

101

2 70701<= ςforM


707.021 2 <−= ςςωω fornp707.0

12 2<

−= ς

ςςforM p

Resonant Circuit --- Pole/Zero Plotseso a C cu o e/ e o o sωj

x njω+

ωj ωj

σ σxnω−

2 poles

σxx

x njω−

Zero damping

2 poles

C i i l d i OverdampedZero dampingζ=0

Critical dampingζ=1

Overdampedζ>>1

Signal Processing Basics

Second-Order System Frequency Response at Natural Frequency

• Now, what happens if we excite this system exactly at the t l f ? Th inatural frequency, or ω = ωn. ? The response is:

121)(ςωω

== n

sH

2)( π

ωω −=∠ = nsH


Quality Factor, or “Q”Quality factor is defined as:

diss

stored

PE

ω

where Estored is the peak stored energy in the system and Pdiss is the average power dissipationis the average power dissipation.


Q of Series Resonant RLC

LC=

1ω

LIE

LC

pkstored =2

21

RIP pkdiss =2

21

2

ZCL

LIQ o

pk==⎟

⎟⎞

⎜⎜⎛

⎟⎞

⎜⎛=

2

21

1RRRILC

Qpk

==⎟⎟⎟

⎠⎜⎜⎜

⎝

⎟⎠

⎜⎝

=2

21


Q of Parallel Resonant RLC=

1ω

pkstored CVE

LC

= 2

21

pkdiss R

VP =

2

21

2

pk

ZRR

V

CV

CQ ==

⎟⎟⎟⎞

⎜⎜⎜⎛

⎟⎠

⎞⎜⎝

⎛= 2

2

21

1

opk ZCL

RVLC

Q⎟⎟

⎠⎜⎜

⎝⎠⎝

2

21


Relationship Between Damping Ratio and “Quality Factor” Q

• A second order system can also be characterized by it’s “Q lit F t ” Q“Quality Factor” or Q.

QsH ==1)( QsH

n==

= ςωω 2)(

11)(H

• Use Q in transfer function of series resonant circuit:

112)(

2

2

2

2

++=

++=

Qssss

sH

nnnn ωωωζ

ω


Qnnnn

Series Resonant Circuit at ResonanceSeries Resonant Circuit at ResonanceThe magnitude of this transfer function is:

222

1)(

⎟⎞

⎜⎛

⎟⎞

⎜⎛

=sHωω

21 ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

Qnn ωω

ωω

Exactly at resonance (ω = ωn), the magnitude of the transfer function is:

QsHn=

=ωω)(


Second-Order System Step Response• Shown for varying values of damping ratio.

Step Response

16

1.8

2Step response for natural frequency = 1 and various damping ratios

Damping ratio = 0.1

tude

1

1.2

1.4

1.6

Am

plit

0.6

0.8

1

0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4


Time (sec.)

Second-Order System Step Responsey p pStep Response

1.6

1.8

2Step response for natural frequency = 1 and various damping ratios

Am

plitu

de

0.8

1

1.2

1.4

0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

Time (sec.)

0 1 2 3 4 5 6 7 8 9 10

21 1

tansin1)(ς

ωςω

⎟⎟⎞

⎜⎜⎛

⎟⎟⎞

⎜⎜⎛ −

+−= −− t

tetvn LCn

1=ω

2

2

1

tansin1

1)(

ςωω

ςω

ς ⎟⎠

⎜⎝

⎟⎟

⎠⎜⎜

⎝+

−−= do ttv

LR

2=ς

Damped natural frequency


1 ςωω −= nd C2Damped natural frequency

Logarithmic Decrement

• Method of estimating damping ratio based on measurementmeasurement

)sin( 1111 φωςω +

=− tCex d

tn

)sin( 1222 φωςω +− tCex d

tn


Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• Simplify:

11ntex −ςω

• TD = oscillation period

)(2

11 Dn Ttex +−= ςω

TD oscillation period = (2π)/ωd

• Simplify againx

DnTexx ςω=

2

1

⎞⎛⎞⎛⎞⎛• Take log of both sides: ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛

−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−==≡⎟⎟

⎠

⎞⎜⎜⎝

⎛22

2

1

12

1ln

ςπς

ςωςςωδ D

dDn TT

xx


Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• The logarithmic decrement δ is the natural log of the ratio

of any two successive oscillation amplitudes.


Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• Comment: If we

measure how the amplitude decreased cycle-by-cycle, we can find dampingcan find damping ratio

• Example: In a free vibration test thevibration test, the ratio of amplitudes is 2.5 to 1 on

⎞⎛successive oscillations... 145.0

12916.0

15.2ln

2=⇒

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−==⎟

⎠⎞

⎜⎝⎛= ς

ςπςδ


Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• If damping ratio is very low,

πςδ 22l 1 ⎟⎞

⎜⎛

⎟⎞

⎜⎛ x

• In the case of low damping, it may not be easy to make

πςς

πςδ 212ln

22

1 ≈⎟⎟

⎠⎜⎜

⎝ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

xx

In the case of low damping, it may not be easy to make measurement of successive oscillations. We can make measurements at time t1 and at N cycles later. Note that:

⎞⎛⎞⎛⎞⎛⎞⎛⎟⎟⎠

⎞⎜⎜⎝

⎛•••⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

++ 14

3

3

2

2

1

1

1

N

N

N xx

xx

xx

xx

xx

• Take log of both sides

δNxxxxx N =⎟⎟⎠

⎞⎜⎜⎝

⎛+•••⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ 3211 lnlnlnlnln


xxxxx NN⎟⎠

⎜⎝

⎟⎠

⎜⎝

⎟⎠

⎜⎝

⎟⎠

⎜⎝

⎟⎠

⎜⎝ ++ 14321

Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• Take log of both sides

xxxxx ⎟⎞

⎜⎛

⎟⎞

⎜⎛

⎟⎞

⎜⎛

⎟⎞

⎜⎛

⎟⎞

⎜⎛

δNxx

xx

xx

xx

xx

N

N

N

=⎟⎟⎠

⎞⎜⎜⎝

⎛+•••⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

++ 14

3

3

2

2

1

1

1 lnlnlnlnln

• This means we can find the logarithmic decrement as:

⎟⎞

⎜⎛

⎟⎞

⎜⎛ 1l1 xδ ⎟⎟

⎠⎜⎜⎝

⎟⎠⎞

⎜⎝⎛=

+1

1lnNxN

δ


A Potpourri of Resonant CircuitsA Potpourri of Resonant Circuits• Both damped and undamped• All circuits simulated with PSIM


Example 1: Undamped Resonant CircuitExample 1: Undamped Resonant Circuit• L = 1/(2π); C = 1/(2π). Initial conditions: capacitor voltage

= 0 and inductor current = 1 Amp


Example 1: What Do We Know About This Ci it?Circuit?

• L = 1/(2π); C = 1/(2π)• Zo = 1 Ohmo• ωo = 2π rad/sec; fo = 1 Hz


Example 1: Undamped Resonant Circuit RResponse

• L = 1/(2π); C = 1/(2π)

• Note ratio of voltage/current = 1 ohm


Example 1: Series Resonant Step ResponseExample 1: Series Resonant Step Response

• L = 1/(2π); C = 1/(2π). Assume zero initial conditions( ); ( )


Example 1: Series Resonant Step ResponseExample 1: Series Resonant Step Response• Zo = 1 Ohm• Natural frequency: ω = 2π rad/sec; f = 1 HzNatural frequency: ωo 2π rad/sec; fo 1 Hz


Example 1: Series Resonant Step ResponseExample 1: Series Resonant Step Response• Zo = 1 Ohm• ωo = 2π rad/sec; fo = 1 Hz


Example 1: Series Resonant Step ResponseExample 1: Series Resonant Step Response• What is inductor current? Remember Zo = 1 Ohm


Example 1: Series Resonant Step RResponse

• What is inductor current? Remember Zo = 1 Ohm

Peak-peak = 20V

Peak-peak = 20A


Example 2: Now Add 0.1 Ω ResistorExample 2: Now Add 0.1 Ω Resistor


Example 2: Now Add 0.1 Ω ResistorExample 2: Now Add 0.1 Ω Resistor• R << Zo, so damping is small


Example 2: Now Add 0.1 Ω ResistorExample 2: Now Add 0.1 Ω Resistor


Example 3: Change the 0.1 Ω to 1 ΩExample 3: Change the 0.1 Ω to 1 Ω• R = Zo in this case


Example 3: Change the 0.1 Ω to 1 ΩExample 3: Change the 0.1 Ω to 1 Ω


Example 4: Change the 1 Ω to 10 ΩExample 4: Change the 1 Ω to 10 Ω• R >> Zo in this case


Example 4: Series Resonant Step ResponseExample 4: Series Resonant Step Response• What’s the initial inductor current, approximately (and

why)?why)?


Example 4: Series Resonant Step ResponseExample 4: Series Resonant Step Response• What’s the initial inductor current, approximately?


Example 4: Response for R >> ZoExample 4: Response for R Zo

• At low frequencies, this behaves like an RC lowpass filter• At high frequencies when capacitor is almost a short thisAt high frequencies, when capacitor is almost a short, this

behaves as a LPF with L/R time constant• Why is this?


Second Order System --- Pole Location Variation with Damping

Critically Very underdamped

ωj

j

dampedOverdamped

ωjωj

σ

x njω+

σxnω−

2 polesσxx

x njω−

2 poles


Overdamped Case1)( 2=sH

Overdampedωj• At low frequencies, for RCs >> LCs2,

R L th

1)( 2 ++ RCsLCs

sH

σxx

or R >> Ls, then …

11)(+

≈RCs

sHσxx

• At high frequencies, with RCs>>1, then

1+RCs

then

⎟⎞

⎜⎛ +

≈+

≈1

11)( 2

sLRCsRCsLCssH


⎟⎠

⎜⎝

+1sR

RCs

Example 5: Series Resonant Circuit with Di dDiode

• Q: With a positive voltage step, what does the diode do in this circuit?this circuit?



• A: Nothing…it’s always reverse biased


Example 6: Now, Flip the DiodeExample 6: Now, Flip the Diode• Q: What does the diode do in this circuit?



• A: It shorts out the capacitor; the circuit behaves like a series LR circuit with L/R time constant 159 secseries LR circuit with L/R time constant 159 sec.


Example 7: Second-Order System Step Response with Extra PoleResponse, with Extra Pole

• Second order system with ωn = 1 r/s, damping ratio = 0.5• Extra pole added at -10 r/sp

Step ResponseCompare step response of wn=1, damp=0.5 2nd order system with pole added at -10 r/s

e

0.8

1

Effects of extra pole

Am

plitu

de

0.4

0.6Effects of extra pole

0 2 4 6 8 10 12

0

0.2


Time (sec.)

Example 8: Second-Order System Step Response, with Extra Pole

• Second orderStep Response

Compare step response of wn=1, damp=0.5 2nd order system with pole added at -2 r/s

Second order system with ωn = 1 r/s, damping

ti 0 5 08

1

ratio = 0.5• Extra pole added at -2 r/s A

mpl

itude

0.6

0.8

0.2

0.4

Time (sec.)

0 2 4 6 8 10 12

0


Widely-Spaced Pole ApproximationWidely-Spaced Pole Approximation

For a transfer function of the form:

BA2

1 BAss ++2

If the poles are on the real axis and are widely spaced, we can approximate them by:

As fast −≈

ABsslow

fast

/−≈


Widely-Spaced Pole Approximation• Assume that we have 3 real-axis polesp

11)(sH ( ) ( ) 1)1)(1)(1()(

3212

3231213

321321 +++++++=

+++=

sssssssH

τττττττττττττττ


Widely-Spaced Pole Approximation• We can re-write this transfer function:

23

1)( =sH

3213

12

23

3 1)(

τττ=+++

asasasa

3231212

τττττττττ

++=++=

aa

3211 τττ ++=a


Widely-Spaced Pole Approximation• Pole locations can be approximated if the poles are widely spaced:widely spaced:

111−≈−≈−≈plow

213231212

13211

1ττττττττττττ

−≈−≈++

−≈−≈

++ap

aplow

13211

33213213

1τττττττττττ

++

≈≈≈≈

aa

phigh

221

1

323121

321

2

1

τττττττττ−≈−≈

++≈≈

apmedium


Widely-Spaced Pole Approximation• In the general case with k poles, if they are widely spaced and on the real axis:spaced and on the real axis:

1ka 1−≈ −

k

kk a

p

1=oa


Widely-Spaced Pole Approximation ---Sanity Check

• Examine system with poles at -1, -10, -100, -1000 and 10000 r/sand -10000 r/s

1111.110122.110122.110111.1101)( 213346510 ++×+×+×+

= −−−− ssssssH

srp

10122.1

/110,1110

10111.1

3

10

6

5

×

−≈×

−≈

−

−

−

srp

srp

/10010122.110122.1

/010,110111.110122.1

3

1

3

64

−≈××

−≈

−≈××

−≈

−

−

−

• The approximation does

srp

srp

/901

/9.910122.1

111.112

−≈−≈

−≈×

−≈ −

approximation does a decent job


srp /9.0111.12 ≈≈

Undamped Resonant Circuitp

Lv

dtdi cL = C

idtdv Lc −

=

2

LCv

dtdi

Cdtvd cLc −=−=

12

2

Guess that the voltage v(t) is sinusoidal with v(t) = Vosinωt. Putting this into the equation for capacitor voltage results in:

)i (1)i (2 tt )sin()sin(2 tLC

t ωωω −=−

This means that the resonant frequency is the standard (as expected) q y ( p )resonance:

LCr12 =ω


LC

Energy Methodsgy

Storage Mode Relationship Comments Capacitor/electric field 21 CVE

pstorage

2

2CVEelec =

Inductor/magnetic field storage dVBLIE

omag ∫==

μ221 2

2

Kinetic energy 2

21 MvEk =

Rotary energy 2

21 ωIEr =

I ≡ mass moment of inertia (kg m2)2r (kg- m )

Spring 2

21 kxEspring = k ≡ spring constant (N/m)

Potential energy hMgE Δ=Δ Δh ≡ height changeote t a e e gy hMgEp ΔΔ Δh height changeThermal energy TCE THT Δ=Δ CTH ≡ thermal capacitance

(J/K)


Energy Methodsgy

By using energy methods we can find the ratio of maximum capacitor voltage to maximum inductor current. Assuming that the capacitor is initially charged t V lt d b i th t it t d E ½CV2 dto Vo volts, and remembering that capacitor stored energy Ec = ½CV2 and inductor stored energy is EL = ½LI2, we can write the following:

22 11 22

21

21

oo LICV =


Energy Methods

What does this mean about the magnitude of the inductor current ? Well, we can solve for the ratio of Vo/Io resulting in:

o ZLV≡= o

o

ZCI

≡

The term “Zo” is defined as the characteristic impedance of a resonant circuit. Let’s

th t h i d t it i it ith C 1 i F d d L 1assume that we have an inductor-capacitor circuit with C = 1 microFarad and L = 1 microHenry. This means that the resonant frequency is 106 radians/second (or 166.7 kHz) and that the characteristic impedance is 1 Ohm.


Example 9: Simulation

• Initial conditions at t = 0: capacitor voltage = 1; inductor current = 0.


Transfer Functions, Pole-Zero and Bode Plots


Risetime for Systems in Cascade

For multiple systems in cascade, the risetimes do not simply add; for instance, for N systems wired in series, each with its own risetime τR1, τR2, … τRN, the overall y , R1, R2, RN,risetime of the cascade τR is:

223

22

21 RNRRRR τττττ ⋅⋅⋅+++≈

Note that this equation works if each system is buffered/isolated from the next sytem; i.e. the systems don’t load each other down.


Example 10: Risetime for Systems in Cascade

For vo1: sec220)10)(1000)(2.2(2.2 7

1 μτ === −RCR For vo2: Consider TR2, which is risetime of 2nd RC circuit by itself:

7

( )( ) sec311sec2202

sec220)10)(1000)(2.2(2.22

22

1,

72

μμτττ

μτ

≈≈+≈

=== −

RRsystemR

R RC


Example 10: Risetime for Systems in Cascade


Example 13: Risetime for 3 Systems in CascadeExample 13: Risetime for 3 Systems in Cascade

FFor vo3: ( )( ) sec381sec22032

32

22

1, μμττττ ==++≈ RRRsystemR


Example 13: Risetime for 3 Systems in CascadeExample 13: Risetime for 3 Systems in Cascade


Comments on PSPICE• PSPICE is a useful tool but beware of some pitfallsPSPICE is a useful tool, but beware of some pitfalls

– GIGO problem– Models

S ti th f t ’ d l i t• Sometimes the manufacturers’ models are incorrect – Tolerances

• Set tolerances sufficiently small. However, this will increase simulation time

– Convergence issues• Transient response is not guaranteed to convergeTransient response is not guaranteed to converge

– Maximum time step• Choose a time step small enough to capture the

detail that you need If you make the time step toodetail that you need. If you make the time step too small, simulation time can be very large

– Parasitics


Comments on PSPICE (cont.)• Continued ...

– Parasitic elements can cause your PSPICE– Parasitic elements can cause your PSPICE simulation not to match protoboard results. Parasitic capacitance and inductance; resistive h t th tshunt paths, etc.

– You should always do a sanity check to verify that your PSPICE result is reasonabley

– PSPICE simulation is not a replacement for good design and hand calculations.


References


chapter 2chapter 2 review of signal processing basics

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