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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/

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

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-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 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 Convergenceh[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 PairsStability

- 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

- 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

- 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)

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

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

- Linearity
- Shifting

Z-transform Properties

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

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

- 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]

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

- 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 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

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]

- 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 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 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

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.

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)

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

- 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?

- 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

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

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]

AliasingAliasing

- 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

- 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

- 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

- 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 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

- 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

- For r repeated roots of Q(g) = 0

y0[k] = (C1 + C2k + … + Crkr-1) gk

- Similar to the continuous-time case

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)

Im l

Marginally Stable

Marginally Stable

Unstable

Re g

Re l

-1

1

Stable

Stable

Unstable

Stability in Both DomainsContinuous-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

- 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

- 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

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

Example

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

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

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

- 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

- 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.

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.

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.

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