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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/.

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ece 352 systems ii

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
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
slide3
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

z transforms5
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

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

Four possibilities (z=0 is a special case and may or may not be included)

Im{z}

Im{z}

Entire plane

Disk

Re{z}

Re{z}

Im{z}

Im{z}

Intersection of a disk and complement of a disk

Complement of a disk

Re{z}

Re{z}

Region of Convergence
z transform pairs
h[k] = d[k]

Region of convergence: entire z-plane

h[k] = d[k-1]

Region of convergence: entire z-plane

h[n-1]  z-1 H(z)

h[k] = ak u[k]

Region of convergence: |z| > |a| which is the complement of a disk

Z-transform Pairs
stability
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| < 1 by rule #1.
    • It is stable if |z| > 1 > |a| by rule #2.
inverse z transform
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 we’ll see can be built up from these basic signals.
    • For these common signals, the z-transform pairs have been tabulated (see Tables)
example
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
example con t
Example (con’t)
  • Find B0 by polynomial division
  • Express in terms of B0
  • Solve for A1 and A2
example con t12
Express X[z] 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.

Example (con’t)
z transform properties
Z-transform Properties
  • Linearity
  • Shifting
z transform properties14
Z-transform Properties
  • Convolution definition
  • Take z-transform
  • Z-transform definition
  • Interchange summation
  • r = k - m
  • Z-transform definition
linear difference equations
Linear Difference Equations
  • Discrete-timeLTI systemscan becharacterizedby differenceequations

y[k] = (1/2) y[k-1] + (1/8) y[k-2] + f[k]

  • Taking z-transform of the difference equation gives description of the system in the z-domain

+

f[k]

y[k]

+

UnitDelay

+

1/2

y[k-1]

UnitDelay

1/8

y[k-2]

advances and delays
Advances and Delays
  • Sometimes differential equations will be presented as unit advances rather than delays

y[k+2] – 5 y[k+1] + 6 y[k] = 3 f[k+1] + 5 f[k]

  • One can make a substitution that reindexes the equation so that it is in terms of delays

Substitute k with k-2 to yield

y[k] – 5 y[k-1] + 6 y[k-2] = 3 f[k-1] + 5 f[k-2]

  • Before taking the z-transform, recognize that we work with time k 0 so u[k] is often implied

y[k-1] = y[k-1] u[k]  y[k-1] u[k-1]

example19
Example
  • System described by a difference equation

y[k] – 5 y[k-1] + 6 y[k-2] = 3 f[k-1] + 5 f[k-2]

y[-1] = 11/6, y[-2] = 37/36

f[k] = 2-k u[k]

transfer functions
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
transfer functions21
Transfer Functions
  • Consider the zero-state response
    • No initial conditions: y[-k] = 0 for all k > 0
    • Only causal inputs: f[-k] = 0 for all k > 0
  • Write general nth order difference equation
stability22

F[z]

H[z]

Y[z]

Stability
  • Knowing H[z], we can compute the output given any input
  • Since H[z] is a ratio of two polynomials, the roots of the denominator polynomial (called poles) control where H[z] may blow up
  • H[z] 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
relation between h k and h z
Relation between h[k] and H[z]
  • 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, h[k], is

y[k] = h[k] when f[k] = d[k]

Z{h[k]} = H[z] Z{d[k]} H[z] = H[z] · 1

h[k]  H[z]

  • Since discrete-time signals can be built up from unit impulses, knowing the impulse response completely characterizes the LTI system
complex exponentials
Complex Exponentials
  • Complex exponentials havespecial property when theyare input into LTI systems.
  • Output will be same complexexponential weighted by H[z]
  • When we specialize the z-domain to frequency domain, the magnitude of H[z] will control which frequencies are attenuated or passed.
z and laplace transforms26
Z and Laplace Transforms
  • Are complex-valued functions of a complex frequency variable

Laplace: s =  + j 2 f

Z: z = ejW

  • Transform difference/differential equations into algebraic equations that are easier to solve
z and laplace transforms27

f[k]

H[z]

y[k]

Z

Laplace

H[esT]

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.
z and laplace transforms28

f[k]

H[z]

y[k]

Z

Laplace

H[esT]

Z and Laplace Transforms
impulse invariance mapping

Im{s}

Im{z}

1

Re{s}

Re{z}

-1

1

1

-1

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)

sampling

s(t)

Ts

t

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

shannon sampling theorem
Shannon Sampling Theorem
  • A continuous-time signal x(t) with frequencies no higher than fmax can be reconstructed from its samples x[k] = x(k Ts) if the samples are taken at a rate fswhich 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, …
sampling theorem assumptions
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
why 44 1 khz for audio cds
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

sampling35
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
aliasing
Analog sinusoid

x(t) = A cos(2pf0t+ f)

Sample at Ts = 1/fs

x[k] = 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

y[k] = 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) = x[k]

Here, fsTs = 1

Since l is an integer,cos(x + 2pl) = cos(x)

y[k] indistinguishable from x[k]

Aliasing
aliasing37
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 f0with respect fsto because all of the aliased frequencies appear to be the same as f0when sampled by fs
generalized sampling theorem
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
example second order equation
Example: Second-Order Equation
  • y[k+2] - 0.6 y[k+1] - 0.16 y[k] = 5 f[k+2] withy[-1] = 0 and y[-2] = 6.25 and f[k] = 4-k u[k]
  • 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 y0[k] = C1 (-0.2)k + C2 (0.8)k

  • Zero-state response
example impulse response
Example: Impulse Response
  • h[k+2] - 0.6 h[k+1] - 0.16 h[k] = 5 d[k+2]with h[-1] = h[-2] = 0 because of causality
  • In general, from Lathi (3.41),

h[k] = (b0/a0) d[k] + y0[k] u[k]

  • Since a0 = -0.16 and b0 = 0,

h[k] = y0[k] u[k] = [C1 (-0.2)k + C2 (0.8)k] u[k]

  • Lathi (3.41) is similar to Lathi (2.41):

Lathi (3.41) balances impulsive events at origin

example impulse response42
Example: Impulse Response
  • Need two values of h[k] to solve for C1 and C2

h[0] - 0.6 h[-1] - 0.16 h[-2] = 5 d[0] h[0] = 5

h[1] - 0.6 h[0] - 0.16 h[-1] = 5 d[1] h[1] = 3

  • Solving for C1 and C2

h[0] = C1 + C2 = 5

h[1] = -0.2 C1 + 0.8 C2 = 3

Unique solution C1 = 1, C2 = 4

  • h[k] = [(-0.2)k + 4 (0.8)k] u[k]
example solution
Example: Solution
  • Zero-state response solution (Lathi, Ex. 3.13)

ys[k] = h[k] * f[k] = {[(-0.2)k + 4(0.8)k] u[k]} * (4-ku[k])

ys[k] = [-1.26 (4)-k + 0.444 (-0.2)k + 5.81 (0.8)k] u[k]

  • Total response: y[k] = y0[k] + ys[k]

y[k] = [C1(-0.2)k + C2(0.8)k] + [-1.26 (4)-k + 0.444 (-0.2)k + 5.81 (0.8)k] u[k]

  • 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

repeated roots
Repeated Roots
  • For r repeated roots of Q(g) = 0

y0[k] = (C1 + C2k + … + Crkr-1) gk

  • Similar to the continuous-time case
stability for an ltid system

Im g

Unstable

Re g

Marginally Stable

g

Stable

|g|

b

-1

1

Lathi, Fig. 3.16

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)
stability in both domains

Im g

Im l

Marginally Stable

Marginally Stable

Unstable

Re g

Re l

-1

1

Stable

Stable

Unstable

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)

frequency response
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
response to sampled sinusoids
Response to Sampled Sinusoids
  • Start with a continuous-time sinusoid
  • Sample it every Tseconds (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
example50
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
example51
Example
  • Group real and imaginary parts
  • The absolute value (magnitude response) is
example52
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
example53
Example
  • We can calculate the output of this system for a sinusoid at any frequency by substituting  with the frequency of the input sinusoid.
discrete time frequency response
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 
aliasing and sampling rate
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.
effect of poles and zeros of h z

Im

Re

Effect of Poles and Zeros of H[z]
  • 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.
effect of poles and zeros of h z57

Im

x

o

o

Re

x

Effect of Poles and Zeros of H[z]
  • Each difference term in H[z] 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.
effect of poles zeros lathi

wT

wT

o

x

x

-

-/2

wT

wT

x

o

x

x

o

x

-

-

Effect of Poles/Zeros (Lathi)