Z-Transform. x (n) Sampler

Size: px

Start display at page:

Download "Z-Transform. x (n) Sampler"

Hugo Palmer
5 years ago
Views:

1 Chapter Two A- Discrete Time Signals: The discrete time signal x(n) is obtained by taking samples of the analog signal xa (t) every Ts seconds as shown in Figure below. Analog signal Discrete time signal xa(t) x(n)=xa(nts) fs=/ts xa(t) x (n) Sampler t 0 n Ts Ts 3Ts..5Ts ( )= ( ) = < = < = where fs is sampling frequency and Ts is sampling period. ( )= Example : Consider an analog sinusoidal signal ( + ) which, when sampled periodically at a rate fs = /Ts samples per second, yields x( ) = where = ( )= ( + )= +, = π is analog frequency and T = π is digital frequency

2 Chapter Two Sampler Mathematical Model: The sampler can be defined as an impulse separator with intervals of Ts and ( T ). amplitude of ( ) ( ) () () ( )= ( )= ( ( ) ( )= 0 Ts 3Ts ) ( )= ( Ts (0) ( ) + ) ( ( ) ( )+ ( ) ( ) + ) Sampling Theorem To recover the analog signal from its sampled values the sampling frequency must be greater or equal Nyquist rate. maximum frequency of sampled signal, and = original signal,where is the is the minimum rate to cover the is at the Nyquist rate. Example : Consider the analog signal: x (t) = 3 cos(50πt) + 0 sin(300πt) cos(00πt), what is the Nyquist rate for this signal? = 5, = 50, = 50, Therefore the maximum frequency = 50, and 300, then, the Nyquist rate is 300Hz. ( )= ( ) = 3 cos(0.67 ) + 0 sin( ) cos(0.33 )

3 Some Elementary Discrete Time Systems:-. Unit Sample Sequence () = = ( ) = = 0. Unit Step Sequence () = 0 0, 0,,,,, () = 0, 0 < 0 ( ) = 0 < 3. Real valued exponential sequence () = (), (Positive time sequence), for example, () = 0.9 () 0 0. () = ( ), 4. Complex valued exponential sequence (Negative time sequence). () = ( ), for example,() = () =. 5. Sinusoidal Sequence () = sin( + )() () = cos( + ) (). 6. Unit Ramp Sequence () = n (),. 7. Unit Sample Synthesis () = ()( ); 8. The geometric series = = ()() = (0), < () n ( ) () n o n n 3

4 =, < =, < () Response of LTI Systems to Arbitrary Inputs: (The Convolution Sum) The input is the arbitrary signal x(n) that is expressed as a sum of weighted impulses: x(n) = ()( ) The response of the system to x (n) is the corresponding sum of weighted outputs y(n)=t[x(n)]=[ ()( ) ] = ()[( )]] Since the impulse response h(n)=t[()] then by the time-invariance property, the response of the system to the delayed unit sample sequence ( ) is h(n-k)=t[( )] Then the output response of the LTI system as a function of the input sign al x(n) and the impulse response h(n) can be expressed as () = () h() = ()h( ) = h()( ) The equation above is called convolution sum. To summarize, the process of computing the convolution between x(k) and h(k) involves the following four steps:. Folding. Fold h(k) about k = 0 to obtain h( k).. Shifting. Shift h( k) by no to the right (left) if no is positive (negative), to obtain h(n 0 k). x(n) T[x(n)] h(n) y(n) 4

5 3. Multiplication. Multiply x(k) by h(n 0 k) to obtain the product sequence v no (k) =x(k)h(n 0 k). 4. Summation. Sum all the values of the product sequence v no (k) to obtain the value of the output at time n = n 0. Example 3: Find () h()by using a graphical method? h() () () = () h() = n ()h( ) n h( ) h( ) k n-3 n- n- n k. () = 0. = 0-3 h( ) k () k (0) 0.5 (0) = ()h( ) =.5 5

6 3. = h( ) () = ()h( ) =.5 3 k () k () = - 0 h( ) () = ()h( ) = 3 3 k () k () = 3 0 h(3 ) 3 (3) = ()h(3 ) =.5 3 k () k (3) = 4 0 h(4 ) k () k (4)

7 (4) = ()h(4 ) = () = 0 () = {0, 0,.5,. 5, 3,.5, 0.5, 0, 0, } () n HW : the impulse response of a linear time invariant system is h(n)={,,,-} Determine the response of the system to the input signal x(n)={,,3,} Example 4: Suppose a linear shift invariant system with input x(n) and output y(n) is characterized by its unit sample response h(n)=a n u(n) for 0 < a <. Find the () response y(n) of such system to the input signal x(n)=u(n). () = h()( ) = ()( ) = = ().... ( ) ()( ) n o n o.... n n.... n n o 7

8 System Properties -Linear and nonlinear: A system is said to be linear if T[a x (n)+a x (n)]=a T[x (n)]+a T[x (n)] For any two input x (n) and x (n) and for any complex constant a and a. - Shift invariance (Time invariance) : T[x(n-n 0 )]=y(n-n 0 ) 3- Causality: If has zero values for n<0. 4-Stability: the system is stable if for any bounded input x(n) <,the output will be bounded y(n) < (BIBO) For LTI system, stability is S= () < Example 5: a given system is represented by the following input output formula y(n)=t[x(n)]=n x(n). Show whether the system is linear or nonlinear and time varying or not. Linearity: let x(n)= a x (n)+a x (n) y(n)=n[a x (n)+a x (n)]= a n x (n)+a n x (n)= a T[x (n)]+a T[x(n)] system is linear Shift invariance: let x(n)=u(n) y(n)=nu(n) If x(n)=u(n-) y(n)=n u(n-) y(n-) system is shift varying Example 6 : a LTI system is characterized by h(n)=a n u(n). Classify this system if causal or non-causal an stable or unstable. Causality: the system is causal since h(n) is zero for n<0 Stability: S= () = = < 0 If a < the system is BIBO and hence it is stable since S is finite. If a is unstable since S diverges 8

9 B- The Direct From Fourier Transform to Z- Transform: The time representation of the discrete signal is: () = ( )( ) By taking the Fourier transform for both sides F{()} = ( ) F(( )) ( ) F{()} = () Where = The Z-transform is defined by replacing with z to yield X() = {()} = () {()} = = () () The z in the above definition is a complex variable and the set of z values for which the summation converges is called the region of convergence (ROC) for the transform. In general this region is bounded by < <, where is lower (minimum) limit of this region may be zero, and is upper (maximum) limit of it may be infinity. 9

10 Example 7: Find the Z-transform including the region of convergence of () = 0 0 < 0 X() = () = { ()} = ( ) By using the geometric series = =, < X() = ( ) = = The above result converges if < > Most useful z-transforms can be expressed in the form,x() = (),where P(z) and Q(z) are polynomials in z represent the numerator and denominator of X(z) respectively. The values of z for which P(z) and X(z) equal to zero are called the zeros of X(z), and the values with Q(z) = 0 and X(z) goes to infinity are called the poles. Imaginary () Zero at origin Pole at Z=a Real > Boundary of region of convergence (ROC) Example 8: Find the Z-transform including the region of convergence of () = ( ) 0

X() = () = { ( )} = Imagenary ( ) By letting n= m in the above summation, that is changed the variables, and since interchanging the order of summation has not changed the sign, X(z) becomes X() = =

11 X() = () = { ( )} = Imagenary ( ) By letting n= m in the above summation, that is changed the variables, and since interchanging the order of summation has not changed the sign, X(z) becomes X() = = ( ) The last step was obtained by adding the m=0 term to the sum and subtracting it from the total. By using the geometric progression formula to evaluate the infinite sum, X(z) becomes. X() = = and hence the ROC of this transform is / < < Zero at origion Pole at Z=b Real < Boundary of region of convergence (ROC) = Example 9: Find the Z-transform including the region of convergence of () = () ( ) X() = () = { () ( )}

12 X() = { ()} { ( )} = + From previous two examples the regions of convergence of them are > and < respectively. And the final region of convergence of the result in this example is the intersection between these two regions. { > } { < } For drawing this region, it depends upon the values of the two variables a and b. If <, the above intersection is the empty set, i.e. the transform doesn't converge. If >, the transform converges in the annular region as shown in Figure below. Imagenary Imagenary ROC b a Real a b Real Example 0: Find the Z-transform including the region of convergence of ) () = () ) () = ( ) > 0 ) X() = () = {()} = =,

13 ) X() = () = {( )} = {()} = {()} =, > 0 Example : Find the Z-transform including the region of convergence of () = () X() = () = {()} = = {()} =, > Properties of Z-transform. Linearity If X () = { ()} h < < X () = { ()} h < < Then { () + ()} = X () + X () = {}. Translation {Shifting} {()} = X() h < < Then {( )} = X()h h Proof: {( )} = Let n-n o =m ( ) 3

14 {()} = () = X() Example : Find the Z-transform including the region of convergence of () = ( 4) X() = {( 4)} = {()} = 3. Multiplication by an exponential = { ()} = X() = X h < < { ± ()} = X() = X Proof:, > h ± < < ± ) { ()} = () = ()( ) = ( ) ) { ()} = () = ()( ) = ( ) Example 3: Find the Z-transform including the region of convergence of {()} = () = { ()} = ( ) { ()} = ( ) 4. Multiplication by n (Ramp) {()} = X() () = () =, > ( ) = ( ) < >, h < < ( h) 4

15 Example 4: Find the Z-transform including the region of convergence of { ()} { ()} = () {()} = { ()} = = ( ).. ( ) = ( ) 5. Initial value theorem (0) = lim () = lim X() 6. Final value theorem ( ) = lim () = lim ( )X() 7. Discrete convolution () () = () ( ) { () ()} = X ()X () Proof: { () ()} =, h, () ( ) = () ( ) = X () () = = X ()X () HW: Find {() ()} 8. Time reversal () X(), < < ( ) X( ), < < HW: Find Z{-n u(-n)} Example 5: Find the Z-transform including the region of convergence of () = cos( )() 5

16 + = + = + = + ( )( ) = ( + ) + = cos( ) cos ( = + ) + cos +, > ± > Example 6: Find the Z-transform including the region of convergence of Use {cos( )()} = () = (n ) () cos( ( ))( ) and any necessary properties. {()} = () cos( )() h = () cos( )() = {cos( )()} {cos( )()} = {cos( )()} cos cos + >, = cos cos + = cos cos + {()} = cos cos + = H.W:. {cos[0.( )] ()}. e sin( ) () 3. {sinh( ) ()} 6

17 = lim ( Chapter Two C- The Inverse of Z-transform {()}:- Example 7: Determine the signal () for X() = ln( + ), > using {()} = X(), ln( ) = X() = + ( ) X() = ( ) + Take the inverse of Z-transform for both sides to get: () = ( ) + () = ( ) ( ) < > () = ( ) ( ) There are many methods to find the inverse of Z-transform - Partial fraction method i) Distinct poles X() = Example 8: Find () for X() = for the following ROC's:.. ) >, b) < 0.5 and c) 0.5 < < X() = X() = ( )( 0.5) = = lim ( 0.5) =, = lim. + ( ) = 0.5 7

18 = ( Chapter Two () =., X() =. Im ROC a- > () = () (0.5) () Real b- < 0.5 () = ( ) + (0.5) ( ) Im ROC 0.5 Real Im ROC c- 0.5 < < () = ( ) (0.5) () 0.5 Real Then ii) Multiple order poles If X() = = ( ) = ( Example 9: Find ()) for X() = X() = ( )( ) = ( )( ) ()() X() = ( + 8

19 = lim ( + ) =, ( + ) = lim ( + ) = 3 4 = lim X() = ( ( () = Notes: ()is a right hand sequence (causal or positive time sequence). ( )is a left hand sequence (anti-causal or negative time sequence). Example 0: Find the inverse Z-transform of X() = + ( 0.5) ( 0.5) h > X() + = ( 0.5) ( 0.5) = = lim ( 0.5) +. =! lim ( 0.5) + ( 0.5) + ( 0.5) + ( 0.5) = 6, ( 0.5) = lim. ( 0.5) = 0. + ( 0.5) =! lim = lim...5( 0.5) = 80 ( 0.5).5 ( 0.5) 9

20 = X() = () = 6 lim. + = 80 ( 0.5) 6 ( 0.5) 0 ( 0.5) + 80 ( 0.5) 80 ( 0.5) ( ) () 0 () + 80 () 80 4 () Notes: At > = (). () = () () = ( ) In general! () () = ( )( ) ( +) ( )! + () HW: Using the partial fraction method to find () for - X() = ( ) ()()() () < < 3, () > 3, () < - X() = (.) ( ) > 3- X() = (.)( ) < > - Power series inverse method (division) Example : Find the inverse Z-transform by division method for X() = h () >, () < 3 a) For ROC >, we must divide to obtain negative power of z since > indicates a right hand sequence. 0

21 ± ± X() = as compare with etc X() = + ( ) + ( ) + (0) + () + () + We can recognize that () =, () = () = 3 ( ) + 4 ( ) + 9 b) For ROC <, must divide to get positive power of z since < indicates a left hand sequence, for negative n ± ± 6 3 X() = + 4 ( ) =, ( ) = 4 () = ( + ) + 4( + ) +

22 3- Discrete convolution Example 3: Find the inverse Z-transform by using discrete convolution method for X() =. ()(.) X() = ( () = () = () = () ()=0.63() (0.63) ( ) D-The one-sided Z-transform The one-sided or unilateral Z-transform of a signal () is defined by: X () = () () X () Example 4: determine the one-sided Z-transform of the signals. () = {, 4, 5, 7, 0, } X () = () = ( ), > 0 X () = 3. () = ( + ), > 0 0 Shifting property: - Shift to right with Initial Condition (I. C.) {Time delay} {( )}

23 Proof: X () = () X () = ( ) let = = +, h = 0 =, = = X () = () = () = () + () = X () + () Example 5: determine the one-sided Z-transform of the signals - () = () - () = ( ) h () = - X () = = - X () = + () = = + ( ) + ( ) Shift to left with Initial Condition (I. C.) {Time advance} {( + )} = X () (), > 0 Proof: X () = () X () = ( + ) let = + =, h = 0 =, = = 3

24 X () = () = () = () () = X () () Example 6: find Z + transform for ( + ) X () = () X () = ( + ) = X () () = {X () (0)} E. Discrete Time System Realization and Difference Equation Direct Form Realizations X(z) The transfer function of an arbitrary K th order is given by: H(z) Y(z) b z + b z +. +b z + b H(z) = = l b lz l a z + a z + a z +. +a z + a a z For causal systems, we must have L K. To implement the equation above we need to divide the numerator and denominator by a 0 z K and rewrite the equation in the form: H(z) = b z a + b z a +. + b z a + b z a + a a z + a a z +. + a a z + a a z 4

25 Chapter Two Example 7 : Second Order system Realization: The transfer function of second order system is given by: H(z) = b +b z +a z +b z +a z Keep in mind that we express H(z) in terms of z to produce a practical realization that uses time-delay elements rather than time-advance elements. The linear constant coefficient difference equation (LCCDE) is written as: Y(z) ( + a z +a z ) = X(z)(b + b z +b z ) Taking inverse Z-transform, the LCCDE is given by: y(n) + a y(n ) + a y(n ) = b x(n) + b x(n ) + b x(n ) The Direct form implementation of H(z) is shown b0 x(n) z- a b ( ) z- y(n) z- ( ) a b z- ( ) ( ) Example 8: Given that H(z) = represents a causal system. Find the difference equation and draw the realization structure. H(z) = Y(z) z+ = = X(z) z z + 3 ( ) + ( +3 ) 5

26 Y(z) X(z) = ()( + 3 ) = ()( + ) The LCCDE, () ( ) + 3( ) = ( ) + ( ) () = ( ) 3( ) + ( ) + ( ) The direct form realization of the discrete time system is shown, x(n) y(n) z - z - ( ) ( ) z - z - ( ) ( ) HW: Given the impulse response of a system h(n)=a n u(n). Find the LCCDE and draw the realization structure. Inverse Systems When connected in cascade, a system H(z) and its inverse Hi(z) form an identity system, where the final output exactly equals the original input. For this to occur, the transfer function of the composite system must be unity, which is to say that H(z)Hi(z) =. Example 9: A discrete time system is realized by the structure shown below. a) Determine the impulse response. b) Determine a realization for its inverse system, that is the system which produces x(n) as an output when y(n) is used as input. 6

27 x(n) w(n) y(n) z - a) () = () ( ) () = () + 3 ( ) () () Taking Z-transform for () and () get: W(z)=X(z)+0.8z - W(z) W(z)= (). Y(z)= W(z)+3z - W(z) () = ( + 3z )() (3) (4) Substitute (3) in (4) get: Y(z) = ( + 3z ) (). () () =. =H(z) H(z)=. =. +. The impulse response, h(n)=3(0.8) ( ) + (0.8) () b) Hi(z)= =. =.. = () (). () The difference equation for the inverse system is determined as: X(z)(+.5z - )=Y(z)( z - ) x(n)=.5 x(n-)+0.5 y(n)-0.4 y(n-) The direct form realization of Hi(z) is shown, 0.5 y(n) x(n) z z - 7

28 HW: Consider the discrete time system shown below. a) Determine the analytical expression for the impulse response of the system. b) Determine a realization for its inverse system. x(n) w(n) y(n) 0.9 z - z - 3 F. Solution of Linear Constant Coefficient Difference Equations Example 30: Find the solution to the following linear constant coefficient difference equation y(n)-.5 y(n-)+0.5 y(n-)=(0.5) n for n 0 with initial condition y(-)=4 and y(-)=0. Solution : Taking Z-transform for both sides gives: Y(z)-.5[z - Y(z)+y(-)]+0.5[z - Y(z)+z - y(-)+y(-)]= Y(z)-.5z - Y(z)-6+0.5z - Y(z)+z - +5=. Y(z)[-.5z z - ] = +. () = ( 0.5) () = ( 0.5)( 0.5)( ) = / /3 8

29 Y(z)= y(n)= (0.5) u(n) + (0.5) u(n) + u(n) HW: Determine the response y(n), n 0 of the system described by the second order difference equation: y(n)-4y(n-)+4y(n-)=x(n)-x(n-) when the input is x(n)=( ) () and the initial conditions are y(-)=y(-)=0. 9

30 Tutorial Sheet No. Z- Transform. Determine Z-transform including region of convergence a) x(n)=n a n u(-n-) b) x(n) = {, 0, -,0,, -l) c) () = ( ) ( ) < 0. Compute the convolution of the following signals () = ( ) ( ) < 0 and () = () 3. Using long division, determine the inverse z-transform of + () = + 4. Determine the causal signal x(n) if its z-transform X(z) is given by: a) () = b) () = c) () = Given () = () find x(n) for the following ROC using partial fraction method: a) ROC z > b) ROC z < c) ROC < z < 30

31 Chapter Two 6. Determine the response of the system with impulse response h(n)=an u(n) to the input signal x(n)=u(n)-u(n-0). 7. Determine the impulse response of the following systems i) x(n) z- z- z- y(n) z- ii) x(n) 0.8 y(n) z- 0.6 z- 8. Implementation of LTI systems Consider the recursive discrete-time system described by the difference equation: y(n)=0.8y(n-)-0.64 y(n-) x(n) Assume zero initial conditions. 9. Let x(n) be a causal sequence with z-transform X(z) whose pole-zero plot is shown below. Sketch the pole-zero plots and the ROC of the following sequence: Im(z) a) x(n)=x(-n) b)x(n)=e 0.5 x(n) Re(z)

32 0. We want to design a causal discrete-time LTI system with the property that if the input is () = () ( ) Then the output is () = () a) Determine the impulse response h(n) and the system function H(z). b) Find the difference equation that characterizes this system. c) Determine a realization of the system using direct form implementation.. Consider the causal system () = ( ) + () + ( ) Determine: (a) The impulse response (b) The zero-state step response (c) The step response if y(-) = A 0 (d) The response to the input () = ( ) 0 <. Realize an FIR ilter with impulse response h(n) given by: a) () = [() ( )] b) () = () ( ) + ( ) + ( ) ( ) + ( ) 3. Use the one-sided z-transform to determine y(n), n 0 in the following cases: a) () + ( ) ( ) = ; ( ) = ( ) = b) (). ( ) +. ( ) = ; ( ) =, ( ) = c) ( + ) ( + ) + () = (); () =, () = 3

# FIR. [ ] = b k. # [ ]x[ n " k] [ ] = h k. x[ n] = Ae j" e j# ˆ n Complex exponential input. [ ]Ae j" e j ˆ. ˆ )Ae j# e j ˆ. y n. y n.

# FIR. [ ] = b k. # [ ]x[ n k] [ ] = h k. x[ n] = Ae j e j# ˆ n Complex exponential input. [ ]Ae j e j ˆ. ˆ )Ae j# e j ˆ. y n. y n. [ ] = h k M [ ] = b k x[ n " k] FIR k= M [ ]x[ n " k] convolution k= x[ n] = Ae j" e j ˆ n Complex exponential input [ ] = h k M % k= [ ]Ae j" e j ˆ % M = ' h[ k]e " j ˆ & k= k = H (" ˆ )Ae j e j ˆ ( )