ECE 352 Systems II

Manish K. Gupta, PhD Office: Caldwell Lab 278

Email: guptam @ ece. osu. edu Home Page: http://www.ece.osu.edu/~guptam

TA: Zengshi Chen Email: chen.905 @ osu. edu Office Hours for TA : in CL 391:  Tu & Th 1:00-2:30 pm

Home Page: http://www.ece.osu.edu/~chenz/

Acknowledgements

• Various graphics used here has been taken from public resources instead of redrawing it. Thanks to those who have created it.

• Thanks to Brian L. Evans and Mr. Dogu Arifler

• Thanks to Randy Moses and Bradley Clymer

ECE 352

Slides edited from: • Prof. Brian L. Evans and Mr. Dogu Arifler

Dept. of Electrical and Computer Engineering The University of Texas at Austin course:

EE 313 Linear Systems and Signals Fall 2003

tfbtfdt

dbtf

dt

dbtf

dt

db

tyatydt

daty

dt

daty

dt

d

m

m

mm

m

m

n

n

nn

n

011

011

1

1

k

kk

dt

dD

dt

dD

dt

dD

2

22

tfbDbDbDbtyaDaDaDDP

mm

mm

DQ

nn

n

)(

011

1

)(

011

1

Continuous-Time Domain Analysis• Example: Differential systems

– There are n derivatives of y(t) and m derivatives of f(t)– Constants a0, a1, …, an-1 and b0, b1, …, bm

– Linear constant-coefficient differential equation

• Using short-hand notation,above equation becomes

m

l

ll

n

k

kk

DbbDP

DaaDQ

10

10

)(

)(

Continuous-Time Domain Analysis• Polynomials

Q(D) and P(D)where an = 1:

– Recall n derivatives of y(t) and m derivatives of f(t)– We will see that this differential system behaves as an

(m-n)th-order differentiator of high frequency signals if m > n

– Noise occupies both low and high frequencies– We will see that a differentiator amplifies high

frequencies. To avoid amplification of noise in high frequencies, we assume that m n

tyDQctfDPctfcDP

tfDPtyDQ

111111

tyDQctfcDP 2222

tyDQctyDQc

tfcDPtfcDPtfctfcDP

2211

22112211

tytyDQtftfDP 2121

Continuous-Time Domain Analysis• Linearity: for any complex constants c1 and c2,

)(on dependent

zero are conditions initial all

)( zero-non toresponse

)( oft independen

only conditions system internal from results

0)(when

response state-zero responseinput -zeroresponse Total

tf

tf

tf

tf

Continuous-Time Domain Analysis• For a linear system,

– The two components are independent of each other

– Each component can be computed independently of the other

T[·] y(t)f(t)

Continuous-Time Domain Analysis• Zero-input response

– Response when f(t)=0

– Results from internal system conditions only

– Independent of f(t)

– For most filtering applications (e.g. your stereo system), we want no zero-input response.

• Zero-state response– Response to non-zero f(t)

when system is relaxed

– A system in zero state cannot generate any response for zero input.

– Zero state corresponds to initial conditions being zero.

Zero-Input Response• Simplest case

• Solution:

• For arbitrary constant C– Could C be complex?

– How is C determined?

0000 tyatydt

d tyatydt

d000

taeCty 0 0

dt

dDtyDQ where00

00011

1 tyaDaDaD n

nn

tnn

nn

t

t

eCdt

ydtyD

eCdt

ydtyD

eCdt

dytDy

00

22

02

02

00

Zero-Input Response• General case:

• The linear combination of y0(t) and its n successive derivatives are zero.

• Assume that y0(t) = C e t

Zero-Input Response• Substituting into the differential equation

• y0(t) = C et is a solution provided that Q() = 0.

• Factor Q() to obtain n solutions:

Assuming that no two i terms are equal

0011

1

zeronon

t

Q

nn

n

zeronon

eaaaC

0)( 21 nQ

tn

tt neCeCeCty 21210

Zero-Input Response• Could i be complex? If complex,

• For conjugate symmetric roots, and conjugate symmetric constants,

iii j

termgoscillatin

termdampening

sincos tjte

eeee

iit

tjttjt

i

iiiii

termgoscillatin

1

termdampening

111 cos2 CteCeCeC ittt iii

Zero-Input Response• For repeated roots, the solution changes.• Simplest case of a root repeated twice:

• With r repeated roots

002 tyD

tetCCty 210

0 0 tyD r

trr etCtCCty 1

210

System Response• Characteristic equation

– Q(D)[y(t)] = 0

– The polynomial Q()• Characteristic of system

• Independent of the input

– Q() roots 1, 2, …, n

• Characteristic roots a.k.a. characteristic values, eigenvalues, natural frequencies

• Characteristic modes (or natural modes) are the time-domain responses corresponding to the characteristic roots– Determine zero-input

response

– Influence zero-state response

RLC Circuit• Component values

L = 1 H, R = 4 , C = 1/40 FRealistic breadboard components?

• Loop equations(D2 + 4 D + 40) [y0(t)] = 0

• Characteristic polynomial 2 + 4 + 40 =

( + 2 - j 6)( + 2 + j 6)

• Initial conditionsy(0) = 2 Aý(0) = 16.78 A/s

L R

C

y(t)

f(t)

Envelope

y0(t) = 4 e-2t cos(6t - /3) A

ECE 352

Linear Time-Invariant System

• Any linear time-invariant system (LTI) system, continuous-time or discrete-time, can be uniquely characterized by its– Impulse response: response of system to an impulse

– Frequency response: response of system to a complex exponential e j 2 f for all possible frequencies f

– Transfer function: Laplace transform of impulse response

• Given one of the three, we can find other two provided that they exist

ECE 352

Impulse response

Impulse response of a system is response of the system to an input that is a unit

impulse (i.e., a Dirac delta functional in continuous time)

ECE 352

Example Frequency Response

• System response to complex exponential e j for all possible frequencies where = 2 f

• Passes low frequencies, a.k.a. lowpass filter

|H()|

p ss p

passband

stopband stopband

H()

ECE 352

Kronecker Impulse

• Let [k] be a discrete-time impulse function, a.k.a. the Kronecker delta function:

• Impulse response h[k]: response of a discrete-time LTI system to a discrete impulse function

00

01

k

kk

k

[k]

1

Transfer Functions

Zero-State Response• Q(D) y(t) = P(D) f(t)• All initial conditions are 0 in zero-state response

• Laplace transform of differential equation, zero-state component

sYty

sFstfdt

dtfD

sYstydt

dtyD

kk

kk

rr

rr

input

response state zero

L

L

sQ

sP

sF

sYsH

sFsPsYsQ

sFtf

Transfer Function• H(s) is called the transfer function because it

describes how input is transferred to the output in a transform domain (s-domain in this case)Y(s) = H(s) F(s)

y(t) = L-1{H(s) F(s)} = h(t) * f(t) H(s) = L{h(t)}

• The transfer function is the Laplace transform of the impulse response

Transfer Function• Stability conditions for an LTIC system

– Asymptotically stable if and only if all the poles of H(s) are in left-hand plane (LHP). The poles may be repeated or non-repeated.

– Unstable if and only if either one or both of these conditions hold: (i) at least one pole of H(s) is in right-hand plane (RHP); (ii) repeated poles of H(s) are on the imaginary axis.

– Marginally stable if and only if there are no poles of H(s) in RHP, and some non-repeated poles are on the imaginary axis.

Ts

Ts

esF

sYsH

esFsY

Ttfty

Examples• Laplace transform

• Assume input f(t) & output y(t) are causal

• Ideal delay of T seconds

0

dtetfsF ts

ssF

sYsH

sFsfsFssY

dt

dfty

)(0

s

sH

ysFs

sY

dftyt

1

010

Examples• Ideal integrator with

y(0-) = 0

• Ideal differentiator with f(0-) = 0

Cascaded Systems• Assume input f(t) and output y(t) are causal

• Integrator first,then differentiator

• Differentiator first,then integrator

• Common transfer functions– A constant (finite impulse response)– A polynomial (finite impulse response)– Ratio of two polynomials (infinite impulse response)

1/s sF(s)/s

tdf

0f(t)

F(s)

f(t)

F(s)

s 1/ss F(s)

dt

dff(t)

F(s)

f(t)

F(s)

Frequency-Domain Interpretation

• y(t) = H(s) e s t

for a particular value of s

• Recall definition offrequency response:

sH

sts

ts

ts

dehe

deh

ethty

h(t) y(t)est

h(t) y(t)ej 2f t

fH

fjtfj

tfj

tfj

dehe

deh

ethty

2 2

2

2

Frequency-Domain Interpretation• s is generalized frequency: s = + j 2 f• We may convert transfer function into

frequency response by if and only if region of convergence of H(s) includes the imaginary axis

• What about h(t) = u(t)?We cannot convert this to a frequency response

However, this system has a frequency response

• What about h(t) = (t)?

0Refor 1

ss

sH

1 allfor 1 fHssH

fjs

sHfH2

Unilateral Laplace Transform• Differentiation in time property

f(t) = u(t)

What is f ’(0)? f’(t) = (t).f ’(0) is undefined.

By definition of differentiation

Right-hand limit, h = h = 0, f ’(0+) = 0

Left-hand limit, h = - h = 0, f ’(0-) does not exist

t

f(t)

1

h

tfhtftf

h

0lim

Block DiagramsH(s)F(s) Y(s)

H1(s) + H2(s)F(s) Y(s)

H1(s)

F(s) Y(s)

H2(s)

=

H1(s)F(s) Y(s)H2(s) H1(s)H2(s)F(s) Y(s)=W(s)

G(s) 1 + G(s)H(s)

F(s) Y(s)G(s)F(s) Y(s)

H(s)

-

=E(s)

Derivations• Cascade

W(s) = H1(s)F(s) Y(s) = H2(s)W(s)

Y(s) = H1(s)H2(s)F(s) Y(s)/F(s) = H1(s)H2(s)

One can switch the order of the cascade of two LTI systems if both LTI systems compute to exact precision

• Parallel CombinationY(s) = H1(s)F(s) + H2(s)F(s)

Y(s)/F(s) = H1(s) + H2(s)

H1(s)F(s) Y(s)H2(s) H2(s)F(s) Y(s)H1(s)

Derivations• Feedback System

• Combining these two equations

sEsGsY

sYsHsFsE

sF

sHsG

sGsY

sFsGsYsHsGsY

sYsHsFsGsY

1

• What happens if H(s) is a constant K?– Choice of K controls all

poles in the transfer function

– This will be a common LTI system in Intro. to Automatic Control Class (required for EE majors)

Stability

Stability• Many possible

definitions

• Two key issues for practical systems– System response to zero

input

– System response to non-zero but finite amplitude (bounded) input

• For zero-input response– If a system remains in a particular

state (or condition) indefinitely, then state is an equilibrium state of system

– System’s output due to nonzero initial conditions should approach 0 as t

– System’s output generated by initial conditions is made up of characteristic modes

Stability• Three cases for zero-input response

– A system is stable if and only if all characteristic modes 0 as t

– A system is unstable if and only if at least one of the characteristic modes grows without bound as t

– A system is marginally stable if and only if the zero-input response remains bounded (e.g. oscillates between lower and upper bounds) as t

Characteristic Modes• Distinct characteristic roots 1, 2, …, n

0Re if

0Re if

0Re if0

lim

10

λ

λe

λ

e

ecty

tjt

t

n

j

tj

j

– Where = + j in Cartesian form

– Units of are in radians/second

Stable Unstable

Im{}

Re{}

MarginallyStable

Left-handplane (LHP)

Right-handplane (RHP)

Characteristic Modes• Repeated roots

– For r repeated roots of value .

– For positive k

r

i

tii etcty

1

10

0Reif

0Reif

0Reif0

lim

tk

tet

• Decaying exponential decays faster thantk increases for any value of k– One can see this by using

the Taylor Series approximation for et about t = 0:

...6

1

2

11 3322 ttt

Stability Conditions• An LTIC system is asymptotically stable if and

only if all characteristic roots are in LHP. The roots may be simple (not repeated) or repeated.

• An LTIC system is unstable if and only if either one or both of the following conditions exist:(i) at least one root is in the right-hand plane (RHP)(ii) there are repeated roots on the imaginary axis.

• An LTIC system is marginally stable if and only if there are no roots in the RHP, and there are no repeated roots on imaginary axis.

dtfhdtfhty

dtfh

tfthty

dττhCty Ctf

tCtftf

and ,

then, i.e.bounded, is )( If

Response to Bounded Inputs• Stable system: a bounded input (in amplitude)

should give a bounded response (in amplitude)• Linear-time-invariant (LTI) system

• Bounded-Input Bounded-Output (BIBO) stable

h(t) y(t)f(t)

Impact of Characteristic Modes• Zero-input response consists of the system’s

characteristic modes• Stable system characteristic modes decay

exponentially and eventually vanish• If input has the form of a characteristic mode,

then the system will respond strongly• If input is very different from the characteristic

modes, then the response will be weak

tuee

AtuetuAe

tfthty

tt

tt

responseweak amplitude small

response strongamplitude large

resonance

tuet

ty

t

Impact of Characteristic Modes• Example: First-order system with characteristic

mode e t

• Three cases

tueRC

th RC

t

1

e-1/RC

1/RC

t

h(t)

System Time Constant• When an input is applied to a system, a certain

amount of time elapses before the system fully responds to that input– Time lag or response time is the system time constant

– No single mathematical definition for all cases

• Special case: RC filter– Time constant is = RC

– Instant of time at whichh(t) decays to e-1 0.367 of its maximum value

t0 tht

h(t)

h(t0) h(t)

ĥ(t)

RCt

dteRCRC

t

h

RC

t

h

0

11

System Time Constant• General case:

– Effective duration is th seconds where area under ĥ(t)

– C is an arbitrary constant between 0 and 1

– Choose th to satisfy this inequality

• General case appliedto RC time constant:

000

)( )( )(ˆ dtthCthtdtth h

th

h(t) y(t)u(t)

t

u(t)

1

t

h(t)

A

t

y(t)

A th

tr

Step Response• y(t) = h(t) * u(t)

• Here, tr is the rise time of the system• How does the rise time tr relate to the system

time constant of the impulse response?• A system generally does not respond to an input

instantaneously

tr

Filtering• A system cannot effectively respond to periodic

signals with periods shorter than th

• This is equivalent to a filter that passes frequencies from 0 to 1/th Hz and attenuates frequencies greater than 1/th Hz (lowpass filter)– 1/th is called the cutoff frequency– 1/tr is called the system’s bandwidth (tr = th)

• Bandwidth is the width of the band of positive frequencies that are passed “unchanged” from input to output

Transmission of Pulses• Transmission of pulses through a system (e.g.

communication channel) increases the pulse duration (a.k.a. spreading or dispersion)

• If the impulse response of the system has duration th and pulse had duration tp seconds, then the output will have duration th + tp

System Realization

Passive Circuit Elements• Laplace transforms with

zero-valued initial conditions

• Capacitor

• Inductor

• Resistor

sLsI

sVsH

sIsLsVdt

diLtv

sCsI

sVsH

sIsC

sV

sVsCsIdt

dvCti

1

1

RsI

sVsH

sIRsV

tiRtv

+

–

v(t)

+

–

v(t)

+

–

v(t)

First-Order RC Lowpass Filter

x(t) y(t)

++

C

R

X(s) Y(s)

++R

sC

1

Time domain

Laplace domainCR

s

CRsX

sY

sX

sCR

sCsY

sIsC

sY

sCR

sXsI

1

1

)(

)(

)(

1

1

)(

)(

1)(

1)(

)(

i(t)

I(s)

Passive Circuit Elements• Laplace transforms with

non-zero initial conditions

• Capacitor

s

isIsL

iLsIsL

isIsLsV

dt

diLtV

0

0

0

• Inductor

0

1

0

1

0

vCsIsC

s

vsI

sCsV

vsVsCsI

dt

dvCti

Operational Amplifier

• Ideal case: model this nonlinear circuit as linear and time-invariantInput impedance is extremely high (considered infinite)

vx(t) is very small (considered zero)

+

_

y(t)

+

_

+_vx(t)

Operational Amplifier Circuit

• Assuming that Vx(s) = 0,

• How to realize gain of –1?• How to realize gain of 10?

+

_

Y(s)

+

_

+_Vx(s)

Zf(s)

Z(s)

+_

F(s)

I(s)

sZ

sZ

sF

sYsH

sFsZ

sZsY

sZ

sFsI

sZsIsY

f

f

f

H(s)

Differentiator• A differentiator amplifies high frequencies, e.g.

high-frequency components of noise:H(s) = s where s = + j 2fFrequency response is H(f) = j 2 f | H( f ) |= 2 f |

• Noise has equal amounts of low and high frequencies up to a physical limit

• A differentiator may amplify noise to drown out a signal of interest

• In analog circuit design, one would use integrators instead of differentiators

Initial and Final Values• Values of f(t) as t 0 and t may be

computed from its Laplace transform F(s) • Initial value theorem

If f(t) and its derivative df/dt have Laplace transforms, then provided that the limit on the right-hand side of the equation exists.

• Final value theoremIf both f(t) and df/dt have Laplace transforms, then

provided that s F(s) has no poles in the RHP or on the imaginary axis.

ssFfs

lim0

ssFtfst 0limlim

52

32102

sss

ssY

0

521

32

10lim

52

3210lim

lim0

2

2

2

ss

ss

ss

s

ssYy

s

s

s

65

30

52

3210lim

limlim

20

0

ss

s

ssYty

s

st

Final and Initial Values Example

• Transfer functionPoles at s = 0, s = -1 j2

Zero at s = -3/2