ECE 352 Systems II

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ECE 352 Systems II

• Manish K. Gupta, PhD
• Office Caldwell Lab 278
• Email guptam _at_ ece. osu. edu
• Home Page http//www.ece.osu.edu/guptam
• TA Zengshi Chen Email chen.905 _at_ osu. edu
• Office Hours for TA in CL 391  Tu Th
  100-230 pm
• Home Page http//www.ece.osu.edu/chenz/

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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

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• 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

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Z-transforms
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Z-transforms

• For discrete-time systems, z-transforms play the
  same role of Laplace transforms do in
  continuous-time systems
• As with the Laplace transform, we compute forward
  and inverse z-transforms by use of transforms
  pairs and properties

Bilateral Forward z-transform
Bilateral Inverse z-transform
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Region of Convergence

• Region of the complex z-plane for which forward
  z-transform converges

• Four possibilities (z0 is a special case and may
  or may not be included)

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Z-transform Pairs

• hk dk
• Region of convergence entire z-plane
• hk dk-1
• Region of convergence entire z-plane
• hn-1 ? z-1 H(z)

• hk ak uk
• Region of convergence z gt a which is the
  complement of a disk

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Stability

• Rule 1 For a causal sequence, poles are inside
  the unit circle (applies to z-transform functions
  that are ratios of two polynomials)
• Rule 2 More generally, unit circle is included
  in region of convergence. (In continuous-time,
  the imaginary axis would be in the region of
  convergence of the Laplace transform.)
• This is stable if a lt 1 by rule 1.
• It is stable if z gt 1 gt a by rule 2.

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Inverse z-transform

• Yuk! Using the definition requires a contour
  integration in the complex z-plane.
• Fortunately, we tend to be interested in only a
  few basic signals (pulse, step, etc.)
• Virtually all of the signals well see can be
  built up from these basic signals.
• For these common signals, the z-transform pairs
  have been tabulated (see Tables)

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Example

• Ratio of polynomial z-domain functions
• Divide through by the highest power of z
• Factor denominator into first-order factors
• Use partial fraction decomposition to get
  first-order terms

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Example (cont)

• Find B0 by polynomial division
• Express in terms of B0
• Solve for A1 and A2

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Example (cont)

• Express Xz in terms of B0, A1, and A2
• Use table to obtain inverse z-transform

• With the unilateral z-transform, or the bilateral
  z-transform with region of convergence, the
  inverse z-transform is unique.

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Z-transform Properties

• Linearity
• Shifting

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Z-transform Properties

• Convolution definition
• Take z-transform
• Z-transform definition
• Interchange summation
• r k - m
• Z-transform definition

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Example
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Difference Equations
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Linear Difference Equations

• Discrete-timeLTI systemscan becharacterizedby
  differenceequations
• yk (1/2) yk-1 (1/8) yk-2 fk
• Taking z-transform of the difference equation
  gives description of the system in the z-domain

?
fk
yk

UnitDelay

1/2
yk-1
UnitDelay
1/8
yk-2
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Advances and Delays

• Sometimes differential equations will be
  presented as unit advances rather than delays
• yk2 5 yk1 6 yk 3 fk1 5 fk
• One can make a substitution that reindexes the
  equation so that it is in terms of delays
• Substitute k with k-2 to yield
• yk 5 yk-1 6 yk-2 3 fk-1 5 fk-2
• Before taking the z-transform, recognize that we
  work with time k ? 0 so uk is often implied
• yk-1 yk-1 uk ? yk-1 uk-1

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Example

• System described by a difference equation
• yk 5 yk-1 6 yk-2 3 fk-1 5 fk-2
• y-1 11/6, y-2 37/36
• fk 2-k uk

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Transfer Functions

• Previous example describes output in time domain
  for specific input and initial conditions
• It is not a general solution, which motivates us
  to look at system transfer functions.
• In order to derive the transfer function, one
  must separate
• Zero state response of the system to a given
  input with zero initial conditions
• Zero input response to initial conditions only

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Transfer Functions

• Consider the zero-state response
• No initial conditions y-k 0 for all k gt 0
• Only causal inputs f-k 0 for all k gt 0
• Write general nth order difference equation

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Stability

• Knowing Hz, we can compute the output given any
  input
• Since Hz is a ratio of two polynomials, the
  roots of the denominator polynomial (called
  poles) control where Hz may blow up
• Hz can be represented as a series
• Series converges when poles lie inside (not on)
  unit circle
• Corresponds to magnitudes of all poles being less
  than 1
• System is said to be stable

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Relation between hk and Hz

• Either can be used to describe the system
• Having one is equivalent to having the other
  since they are a z-transform pair
• By definition, the impulse response, hk, is
• yk hk when fk dk
• Zhk Hz Zdk ? Hz Hz 1
• hk ? Hz
• Since discrete-time signals can be built up from
  unit impulses, knowing the impulse response
  completely characterizes the LTI system

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Complex Exponentials

• Complex exponentials havespecial property when
  theyare input into LTI systems.
• Output will be same complexexponential weighted
  by Hz
• When we specialize the z-domain to frequency
  domain, the magnitude of Hz will control which
  frequencies are attenuated or passed.

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Z and Laplace Transforms
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Z and Laplace Transforms

• Are complex-valued functions of a complex
  frequency variable
• Laplace s ? j 2 ? f
• Z z e j W
• Transform difference/differential equations into
  algebraic equations that are easier to solve

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Z and Laplace Transforms

• No unique mapping from Z to Laplace domain or
  vice-versa
• Mapping one complex domain to another is not
  unique
• One possible mapping is impulse invariance.
• The impulse response of a discrete-time LTI
  system is a sampled version of a continuous-time
  LTI system.

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Z and Laplace Transforms
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Impulse Invariance Mapping

• Impulse invariance mapping is z e s T

s -1 ? j ? z 0.198 ? j 0.31 (T 1) s 1 ?
j ? z 1.469 ? j 2.287 (T 1)
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Sampling Theorem
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Sampling

• Many signals originate as continuous-time
  signals, e.g. conventional music or voice.
• By sampling a continuous-time signal at isolated,
  equally-spaced points in time, we obtain a
  sequence of numbers
• k ? , -2, -1, 0, 1, 2,
• Ts is the sampling period

Sampled analog waveform
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Shannon Sampling Theorem

• A continuous-time signal x(t) with frequencies no
  higher than fmax can be reconstructed from its
  samples xk x(k Ts) if the samples are taken
  at a rate fs which is greater than 2 fmax.
• Nyquist rate 2 fmax
• Nyquist frequency fs/2.
• What happens if fs 2fmax?
• Consider a sinusoid sin(2 p fmax t)
• Use a sampling period of Ts 1/fs 1/2fmax.
• Sketch sinusoid with zeros at t 0, 1/2fmax,
  1/fmax,

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Sampling Theorem Assumptions

• The continuous-time signal has no frequency
  content above the frequency fmax
• The sampling time is exactly the same between any
  two samples
• The sequence of numbers obtained by sampling is
  represented in exact precision
• The conversion of the sequence of numbers to
  continuous-time is ideal

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Why 44.1 kHz for Audio CDs?

• Sound is audible in 20 Hz to 20 kHz range
• fmax 20 kHz and the Nyquist rate 2 fmax 40
  kHz
• What is the extra 10 of the bandwidth used?
• Rolloff from passband to stopband in the
  magnitude response of the anti-aliasing filter
• Okay, 44 kHz makes sense. Why 44.1 kHz?
• At the time the choice was made, only recorders
  capable of storing such high rates were VCRs.
• NTSC 490 lines/frame, 3 samples/line, 30
  frames/s 44100 samples/s
• PAL 588 lines/frame, 3 samples/line, 25 frames/s
  44100 samples/s

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Sampling

• As sampling rate increases, sampled waveform
  looks more and more like the original
• Many applications (e.g. communication systems)
  care more about frequency content in the waveform
  and not its shape
• Zero crossings frequency content of a sinusoid
• Distance between two zero crossings one half
  period.
• With the sampling theorem satisfied, sampled
  sinusoid crosses zero at the right times even
  though its waveform shape may be difficult to
  recognize

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Aliasing

• Analog sinusoid
• x(t) A cos(2pf0t f)
• Sample at Ts 1/fs
• xk x(Ts k) A cos(2p f0 Ts k f)
• Keeping the sampling period same, sample
• y(t) A cos(2p(f0 lfs)t f)
• where l is an integer

• yk y(Ts k) A cos(2p(f0 lfs)Tsk f) A
  cos(2pf0Tsk 2p lfsTsk f) A cos(2pf0Tsk
  2p l k f) A cos(2pf0Tsk f) xk
• Here, fsTs 1
• Since l is an integer,cos(x 2pl) cos(x)
• yk indistinguishable from xk

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Aliasing

• Since l is any integer, an infinite number of
  sinusoids will give same sequence of samples
• The frequencies f0 l fs for l ? 0 are called
  aliases of frequency f0 with respect fs to
  because all of the aliased frequencies appear to
  be the same as f0 when sampled by fs

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Generalized Sampling Theorem

• Sampling rate must be greater than twice the
  bandwidth
• Bandwidth is defined as non-zero extent of
  spectrum of continuous-time signal in positive
  frequencies
• For lowpass signal with maximum frequency fmax,
  bandwidth is fmax
• For a bandpass signal with frequency content on
  the interval f1, f2, bandwidth is f2 - f1

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Difference Equations and Stability
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Example Second-Order Equation

• yk2 - 0.6 yk1 - 0.16 yk 5 fk2
  withy-1 0 and y-2 6.25 and fk 4-k
  uk
• Zero-input response
• Characteristic polynomial g2 - 0.6 g - 0.16 (g
  0.2) (g - 0.8)
• Characteristic equation
  (g 0.2) (g - 0.8) 0
• Characteristic roots
  g1 -0.2 and g2 0.8
• Solution
  y0k C1 (-0.2)k C2 (0.8)k
• Zero-state response

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Example Impulse Response

• hk2 - 0.6 hk1 - 0.16 hk 5 dk2with
  h-1 h-2 0 because of causality
• In general, from Lathi (3.41),
• hk (b0/a0) dk y0k uk
• Since a0 -0.16 and b0 0,
• hk y0k uk C1 (-0.2)k C2 (0.8)k uk
• Lathi (3.41) is similar to Lathi (2.41)

Lathi (3.41) balances impulsive events at origin
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Example Impulse Response

• Need two values of hk to solve for C1 and C2
• h0 - 0.6 h-1 - 0.16 h-2 5 d0 ? h0 5
• h1 - 0.6 h0 - 0.16 h-1 5 d1 ? h1 3
• Solving for C1 and C2
• h0 C1 C2 5
• h1 -0.2 C1 0.8 C2 3
• Unique solution ? C1 1, C2 4
• hk (-0.2)k 4 (0.8)k uk

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Example Solution

• Zero-state response solution (Lathi, Ex. 3.13)
• ysk hk fk (-0.2)k 4(0.8)k uk
  (4-k uk)
• ysk -1.26 (4)-k 0.444 (-0.2)k 5.81
  (0.8)k uk
• Total response yk y0k ysk
• yk C1(-0.2)k C2(0.8)k -1.26
  (4)-k 0.444 (-0.2)k 5.81 (0.8)k uk
• With y-1 0 and y-2 6.25
• y-1 C1 (-5) C2(1.25) 0
• y-2 C1(25) C2(25/16) 6.25
• Solution C1 0.2, C2 0.8

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Repeated Roots

• For r repeated roots of Q(g) 0
• y0k (C1 C2 k Cr kr-1) gk
• Similar to the continuous-time case

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Stability for an LTID System

• Asymptotically stable if andonly if all
  characteristic rootsare inside unit circle.
• Unstable if and only if one orboth of these
  conditions exist
• At least one root outside unit circle
• Repeated roots on unit circle
• Marginally stable if and only if no roots are
  outside unit circle and no repeated roots are on
  unit circle (see Figs. 3.17 and 3.18 in Lathi)

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Stability in Both Domains
Continuous-Time Systems
Discrete-Time Systems
Marginally stable non-repeated characteristic
roots on the unit circle (discrete-time systems)
or imaginary axis (continuous-time systems)
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Frequency Response ofDiscrete-Time Systems
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Frequency Response

• For continuous-time systems the response to
  sinusoids are
• For discrete-time systems in z-domain
• For discrete-time systems in discrete-time
  frequency

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Response to Sampled Sinusoids

• Start with a continuous-time sinusoid
• Sample it every T seconds (substitute t k T)
• We show discrete-time sinusoid with
• Resulting in
• Discrete-time frequency is equal to
  continuous-time frequency multiplied by sampling
  period

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Example

• Calculate the frequency response of the system
  given as a difference equation as
• Assuming zero initial conditions we can take the
  z-transform of this difference equation
• Since

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Example

• Group real and imaginary parts
• The absolute value (magnitude response) is

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Example

• The angle (phase response) is
• where 0 comes from the angle of the nominator and
  the term after comes from the denominator of
• Reminder Given a complex number a j b the
  absolute value and angle is given as

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Example

• We can calculate the output of this system for a
  sinusoid at any frequency by substituting ? with
  the frequency of the input sinusoid.

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Discrete-time Frequency Response

• As in previous example, frequency response of a
  discrete-time system is periodic with 2?
• Why? Frequency response is function of the
  complex exponential which is periodic with 2?
• Absolute value of discrete-time frequency
  response is even and angle is odd symmetric.
• Discrete-time sinusoid is symmetric around ?

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Aliasing and Sampling Rate

• Continuous-time sinusoid can have a frequency
  from 0 to infinity
• By sampling a continuous-time sinusoid,
• Discrete-time frequency ? unique from 0 to ?
• We only can represent frequencies up to half of
  the sampling frequency.
• Higher frequencies exist would be wrapped to
  some other frequency in the range.

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Effect of Poles and Zeros of Hz

• The z-transform of a difference equation can be
  written in a general form as
• We can think of complex number as a vector in the
  complex plane.
• Since z and zi are both complexnumbers the
  difference is againa complex number thus a
  vectorin the complex plane.

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Effect of Poles and Zeros of Hz

• Each difference term in Hz may be represented
  as a complex number in polar form
• Magnitude is the distance ofthe pole/zero to the
  chosenpoint (frequency) on unit circle.
• Angle is the angle of vectorwith the horizontal
  axis.

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Effect of Poles/Zeros (Lathi)