relationship between laplace fourier and z transform | what is The Relationship between the Laplace, the Fourier and the z-Transform

The Relationship between the Laplace, the Fourier and the z-Transform difference , relationship between laplace fourier and z transform ? 

Z-Transform

z-transform   The z-transform is the discrete time counter-part of the laplace transform. infact, it can be shown to be the laplace transform of the discretiesd version of a continuous time signal x(t). in the process, we obtain the s-plane mapping required by z = e^st . the z-transform converts a difference equation into an algebric equation, and hence, is useful in solving difference equation.

z-transform, like the laplace transform, is an indispensable mathematical tool for the design analysis and monitoring of systems. the z-transform is the discrete time counter part of the laplace transform and a generalisation of the fourier transform of a sampled signal. Like laplace transform the z-transform of a sampled insight into the transient behavior, the steady state behaviour and the stability of discrete time systems. A working knowledge of the z-transform is essential to the study of digital filters and systems.

Derivation of the z-Transform

The z-transform is the discrete-time counterpart of the laplace transform. in this section we derive the z-transfrom from the laplace transform a discrete time signal. the laplace transform x(t), is given by the integral

x(s)  = x(t)e^-st  dt ………………….(1)

where, the complex variable s = 0 + j_oo, and the lower limit of t = 0 allow the possibility that the signal x(t) may include an impulse. the inverse laplace transform is defined by

x(t) = x(s) e^st  ds ………………….(2)

where, 0₁ is selected so that x(s) is analytic (no singularities) for s > 0₁. the z-transform can be derived from eq. (1) by sampling the continuous time input signal x(t). for a sampled signal x(mT_s), normally denoted as x(m) assuming the sampling period T_S = 1, the laplace transform eq. (1) becomes

x(e^s) = x(m) e^-sm  ……………….(3)

substituting the variable e^s in eq. (3) with the variable z, we obtain the one-sided z-transform equation

x(z) = x(m) z^-m ………………………..(4)

the two-sided z-transform is defined as

x(z) = x(m) z^-m ………………….(5)

note that for a one-sided signal, x(m) = 0 for m < 0. eqs. (4) and (5) are equivalent.

The Relationship between the Laplace, the Fourier and the z-Transform

The laplace transform, the fourier transform and the z-transform are closely related in that they all employ complex exponential as their basis function. for right-sided signals (zero-valued for negative time index) the laplace transform is a generalisation of the fourier transform of a continuous time signal and the z-transform is a generalisation of the fourier transform of a discrete time signal. in the previous section, we have shown that the z-transform can be derived as the laplace transform of a discrete time signal. in the following, we explore the relation between the z-transform and the fourier transform. using the relationship,

z = e^s = e⁰ e^joo = re^j2f ……..(6)

where, s = 0 + j_oo and _oo = 2f, we can rewrite the z-transform eq. (4) in the following form

x(z) = x(m) r^-m e^-j2mf …………………..(7)

note that when r = e⁰ = 1 the z-tranform becomes the fourier transform of a sampled of a sampled signal given by

x(z = e^-j2f) = x(m)e^-j2fm …………………(8)

Therefore, the z-transform is a generalization of the fourier transform of a sampled signal. like the laplace transform, the basis functions for the z-transform are damped or growing sinusoids of the form z^-m = r^m e^-2jfm as shown in fig. 1. these signals are particularly suitable for tansient signal analysis. the fourier basis functions are steady complex exponential,e^-j2fm of time-invariant amplitudes and phase, suitable for steady state or time-invariant signal analysis.

A similar relationship exists between the laplace transform and the fourier transform of a continuous time signal. the laplace transform is a one-sided transform with the lower limit of integration at t = 0^–, whereas the fourier transform is a two-sided transform with the lower limit of integration at t = – _oo. however, for a one-sided signal which is zero-valued for t < 0, the limits of integration for the laplace and the fourier transforms are identical. in that case if the variable s in the laplace transform is replaced with the frequency variable j2f, then the laplace integral becomes the fourier integral. hence, for a one-sided signal, the fourier transform is a special case of the laplace transform corresponding to s = 2jf and 0 .

Example 1.  Shown that the laplace transform of a sampled signal is periodic with respect to the frequency axis j_oo of the complex frequency variable s = 0 + j_oo.

Sol. in eq. (3) substitute s + jk2, where k is an integer variable for the frequency variable to obtain

x(e^{s + j2k}) = x(m) e^{-(s + jk2)m}

= x(m) e^-sm e^-jk2m/= 1 …………..(9)

hence, the laplace transform of a sample signal is periodic with a period of 2 as shown in fig. 2(a).

The z-plane and the Unit Circle

The frequency variables of the laplace transform s = 0 + j_oo and the z-transform z = re^j are complex variables with real and imaginary parts and can be visualized in a two dimensional plane. fig. 2(a) and 2(b) show the s-plane of the laplace transform and the z-plane of z-transform. in the s-plane the vertical j-axis is the frequency axis and the horizontal 0-axis gives the exponential rate of decay or the rate of growth of the amplitude of the complex sinusoid as also shown in fig. 1. As shown in example 1 when a signal is sampled in the time domain its laplace transform, and hence, the s-plane, becomes periodic with respect to the j-axis. this is illustrated by the periodic horizontal dashed lines in fig. 2(a). periodic processes can be conveniently represented using a circular polar diagram such as the z-plane and its associated unit circle. now, imagine bending the j-axis of the s-plane of the sampled signal of fig. 2(a) in the direction of the left hand side half side half of the s-plane to form a circle such that the points and meet. the resulting circule is called the unit circle and the resulting diagram is called the z-plane. the area to the left of the s-plane, i.e., for < 0 or r = e < 1, is mapped into the area inside the unit circle, this is the region of stable causal signals and systems. the area to the right of the s-plane, < 0 or r = e⁰ > 1, is mapped onto the outside of the unit circle this is the region of unstable signals and systems. the j-axis, with 0 or r = e⁰ = 1, is itself mapped onto the unit circle line. hence, the cartesian co-ordinates used in s-plane for continuous time signals, fig. 2(a) is mapped into a polar representation in the z-plane for discrete-time signals fig. 2(b). fig. 3 illustrates that an angle of 2 i.e., one round the unit circle, corresponds to a frequency of F_S Hz where, F_S is the sampling frequency. hence, a frequency of f Hz corresponds to an angle 0 given by

0 = 2/f_s f radian ……………….(10)

for example at a sampling rate of F_s = 40 kHz, a frequency of 5 kHz corresponds to an angle of 2 x 5/40 = 0.05 rad of 45⁰

The Region of Convergence (ROC)

Since, the z-transform is an infinite power series, it exists only for those values of the variable z for which the series converges to a finite sum. the region of Convergence (ROC) of x(z) is the set of all the values of z for which x(z) attains a finite computable value.

Example 2.  Determine the z-transform, the region of convergence and the fourier transform of the following signal.

Sol. substituting for x(m) in the z-transform eq. (4) we obtain

x(z) = x(m) z^-m = (0)z⁰ = 1 …………………(12)

for all values of the variable z we have x(z) = 1, hence as shown in the shaded area of below figure, the region of convergence is the entire z-plane.

The fourier transform of x(m) may be obtained by evaluating x(z) in eq. (12) at z = e^j as

x(e^j) = 1 …………………………..(13)

Example 3. determine the z-transform, the region of convergence and the fourier transform of the following signal

x(m) = (m- k) = {1,   m = k   0,    m = k    …………………….(14)

Sol.  substituting for x(m) in the z-transform eq. (4) we obtain

x(z) = (m – k) z^-m = z^-k …………………..(15)

The z-transform is x(z) = – z^-k = 1/z^k . hence, x(z) is finite-valued for all the values of z except for z = 0. as shown by the shaded area of the above figure, the region of convergence is the entire z-plane except the point z = 0. the fourier transform is obtained by evaluating x(z) in eq. (15) at z = e^j  is

x(e^j) = e^-jk …………………….(16)

Example 4.  Determine the z-transform, the region of convergence and the fourier transform of the following signal.

x(m) = (m + k) + (m – k) = {1,  m = + k   0, m = + k      ……………….(17)

Sol.  substituting for x(m) in the z-transform eq. (4) we obtain

x(z) = x(m) z^-m = z^k + z^-k ………………………(18)

hence, x(z) is finite-valued for all the values of z except for z = 0 and z = _oo. as shown by the shaded area of the below figure, the region of convergence is the entire z-plane except the points z = 0 and z = ₀₀ not shown. the fourier transform is obtained by evaluating eq. (18) at z = e^j as

x(e^j) = e^-jk + e^{+ jk} = 2cos (k) …………….(19)

Properties of the z-transform

As z-transform is a generalisation of the fourier transform of a sampled signal it has similar properties to the fourier transform as descrided in the following

Linearity

given two signals

x₁(m)  x₁(z) …………….(20)

x₂(m)  x₂(z) ……………………(21)

Then the linearity implies that for any linear combination of x₁(m) and x₂(m) are have

a₁x₁(m) + a₂x₂(m) a₁x₁(z) + a₁x₂(z) ……………(22)

1. (22) is known as the superposition principle.

Time Shifting

the variable z has a useful interpretation in terms of time delay. if

x(m) x(z)

x(m – k)  z^k x(z) …………………….(23)

this property can be proved by taking the z-transform of x(m – k)

x(z) = x(m – k) z^-m  x(n) z^{-(n + k)}

= z^-k  x(n) z^-n = z^-k x(z) ………………(24)

where, we have made a variable substitution n = m – k. that is the effect of a time shif by k sampling-interval-time units is equivalent to multiplication of the z-transform by z^-k. note that z^-1 delays the signal by 1 unit and z^-k by  k units and z⁺¹ is a non-causal unit time advance and z^+k advances a signal in time by k units.

Multiplication by an Exponential Sequence (Frequency Modulation)

The z-transfrom relation for the product of a signal x(m) and the exponential sequence z^m₀ is

z_o^mx(m)  x(z/z_o) ………………(25)

this property can be shown by substituting z_o^m x(m)  x(z/z_o) in the z-transform equation

x(m) z_o^m z^-m  x(m) (z/z_o)^-m = x(z/z_o) …………….(26)

note that for the case when z = e^j and z_o = e^j then we have the frequency modulation equation

e^j0m x(m)   x(e^j) ……………….(27)

Convolution

For two signals x₁(m) and x₂(m)

x₁(m)   x₁(z)

x₂(m)  x₂(z)

the convolutional property states that

x₁ (m)^* x₂ (m)  x₁(z) x₂(z) ……………(28)

where, the asterisk sing^* denotes the convolution operation. that is the convolutio of two signals in the time domain is equivalent to multiplication of their z-transforms and vice-versa.

Differentiation in the z-Domain

x(m) x(z)

mx(m)  -zdx (z)/dz ………………..(29)

this property can be proved by taking the derivative of the z-transform equation w.r.t the variable z as

dx (z)/dz = d/dz  x(m) z^{-m -1}  ……………………(30)

= – z^-1  mx(m) z^-m

Transfer Function

consider the general linear time-invariant difference equation describing the input-output relationship of a discrete-time system

y(m) = a_k y(m – k) + b_kx(m – k) …………………(31)

in eq. (31), the signal x(m) is the system input, y(m) is the system output and a_k and b_k are the system coefficients. taking the z-transform of eq. (31), we obtain

y(z) = a_kz^-k y(z) + b_kz^-k x(z) ………………….(32)

1. (32) can be re-arranged and expressed in terms of the ratio of a numerator polynomial y(z) and a denominator polynomial x(z) as

h(z) = y(z)/x(z) = b_o + b₁z^-1 +……….+ b_m z^-m/1 – a₁z^-1 – ………a_nz^-n

= _{k  = 0} bz^-k/1 – az^-k ………………(33)

h(z) is known as the system transfer function. the frequency response of a system h(0) may be obtained by substituting z = e^j in eq. (33).

Inverse z-transform by Inspection

in this method the discrete-time equation for a signal or a system is obtained from its z-transform by recognising simple z-transform pairs and substituting the time domain terms for their corresponding z-domain terms.

Example 15. find the inverse z-transform of

h(z) = y(z)/x(z) = 1/1 – az^-1 ……………..(34)

Sol.  from eq. (35) we have

y(z) = az^-1 y(z) + x(z) ……………..(35)

by inspection and through the substituting of z^-k y(z) for y(m – k) and z^-k x(z) for x(m – k), we obtain the discrete-time equivalent of eq. (36)

y(m) = ay(m – k) + x(m) ………………(36)

now, if the input x(m) is a discrete-time impulse (m), given by

(m) = {1,  m = 0   0,    m = 0 ………………..(37)

then the output of the feedbacj system of eq. (38) will be

y(m) {a^m ,  m > 0   0 ,   m < 0 ………………..(38)

Intro Exercise – 4

1. The z-transform of [n – k], k > 0 is

(a) z^k, z > 0

(b) z^-k , z < 0

(c) z^k, z = 0

(d) z^-k, z = 0

2. The z-transform of u[n + k], k > 0 is

(a) z^-k, z = 0

(b) z^k, z = 0

(c) z^-k, all z

(d) z^k, all z

3. The z-transform of u[n] is

(a) 1/1 – z^-1, |z| > 1

(b) 1/1 – z^-1 , |z| < 1

(c) z/1 – z^-1, |z| < 1

(d) z/1 – z^-1, |z| > 1

4. The z-transform of (1/4)ⁿ (u[n] – u[n -5] is

(a) z⁵ – 0.25⁵/z⁴ (z – 0.25), z > 0.25

(b) z⁵ – 0.25⁵/z⁴(z – 0.25), z > 0

(c) z⁵ – 0.25⁵/z³ (z – 0.25), z < 0.25

(d) z⁵ – 0.25⁵/z⁴ (z – 0.25) ,all z

5. The z-transform of (1/4)⁴ u[-n] is

(a) 4z/4z – 1, |z| > 1/4

(b) 4z/4z – 1, |z| < 1/4

(c) 1/1 – 4z, |z| > 1/4

(d) 1/1 – 4z, |z| < 1/4

6. The z-transform of 3ⁿ u[-n – 1] is

(a) z/3 – z, |z| > 3

(b) z/3 – z, |z| < 3

(c) 3/3 – z, |z| > 3

(d) 3/3 – z, |z| < 3

7. The z-transform of (2/3)^|n| is

(a) -5z/(2z – 3)(3z – 2), -3/2 < z < -2/3

(b) -5z/(2z – 3)(3z – 2), 2/3 < |z|<3/2

(c) 5z/(2z – 3)(3z – 2), 2/3 <|z|<2/3

(d) 5z/(2z – 3)(3z – 2), -3/2 <z <-2/3

8. The z-transform of (1/2)ⁿ u[n] + (1/4)ⁿ u[-n – 1] is

(a) 1/1-1/2 z^-1 – 1/1 – 1/4 z^-1, 1/4 <|z|<1/2

(b) 1/1 – 1/2 z^-1 + 1/1- 1/4 z^-1, 1/4 < |z| <1/2

(c) 1/1 – 1/2 z^-1 – 1/1- 1/4 z^-1, |z| > 1/2

(d) none of the above

9. The z-transform of cos (3n) u[n] is

(a) z/2 (2z – 1)/(z² – z + 1), 0 < |z| < 1

(b) z/2 (2z – 1)/(z² – z + 1), |z| > 1

(c) z/2 (1 – 2z)/(z² – z + 1), 0 < |z| < 1

(d) z/2 (1 – 2z)/(z² – z + 1), |z|> 1

10. The z-transform of (3, 0, 0, 0, 0, 6, 1, – 4) is

(a) 3z⁵ + 6 + z^-1 – 4z^-2, 0 < |z| <

(b) 3z⁵ + 6 + z^-1 – 4z^-2, 0 < |z| <

(c) 3z^-5 + 6 + z – 4z², 0 < |z| <

(d) 3z^-5 + 6 + z – 4z², 0 < |z| <

11. The z-transform of x[n] = {2,4,5,7,0,1) is

(a) 2z² + 4z + 5 + 7z + z³, z = _oo

(b) 2z^-2 + 4z^-1 + 5 + 7z + z³, z = ₀₀

(c) 2z^-2 + 4z^-1 + 5 + 7z + z³, 0 < |z| < ₀₀

(d) 2z² + 4z + 5 + 7z^-1 + z^-3, 0 < |z| < ₀₀

12. The z-transform of x[n] = (1,0 – 1,0, 1 – 1) is

(a) 1 + 2z^-2 – 4z^-4 + 5z^-5

(b) 1 – z^-2 + z^-4 – z^-5

(c) 1 – 2z² + 4z⁴ – 5z⁵

(d) 1 – z² + z⁴ – z⁵

13. The z-transform of the signal x[n – 2] is

(a) z⁴/z² – 16

(b) (z + 2)²/(z + 2)² – 16

(c) 1/z² – 16

(d) (z – 2)²/(z – 2)² – 16

14. The z-transform of the signal y[n] = 1/2ⁿ x[n] is

(a) (z + 2)²/(z + 2)² – 16

(b) z²/z² – 4

(c) (z – 2)²/(z – 2)² – 16

(d) z²/z² – 64

15. The z-transform of the signal x[-n]* x[n] is

(a) z²/16z² – 257z⁴ – 16

(b) -16 z²/(z² – 16)²

(c) z²/257 z² – 16 z⁴ – 16

(d) 16 z²/(z² – 16)²

16. The z-transform of the signal nx[n] is

(a) 32 z²/(z² – 16)²

(b) -32 z²/(z² – 16)²

(c) 32 z/(z² – 16)²

(d) -32 z/(z² – 16)²

17. The z-transform of the signal x[n +1] + x[n – 1] is

(a) (z + 1)²/(z + 1)² – 16 + (z – 1)²/(z – 1)² – 16

(b) z² (1 + z)/z² – 16

(c) z²(-1 + z)/z² – 16

(d) none of the above

18. The z-transform the signal x[n]* x[n – 3] is

(a) z^-3/(z² – 16)²

(b) z⁷/(z² – 16)²

(c) z⁵/(z² – 16)²

(d) z/(z² – 16)²

19. The time signal corresponding to x(2z) is

(a) n²3ⁿ u[2n]

(b) (-3/2)ⁿ n²u[n]

(c) (3/2)ⁿ n²u[n]

(d) 6ⁿn² u[n]

20. The time signal corresponding to x(z^-1) is

(a) n²3^-n[-n]

(b) n²m^-u u[-n]

(c) 1/n² 3^1/n u[n]

(d) 1/n² 3^1/n u[-n]

21. The time signal corresponding to d/dz x(z) is

(a) (n – 1)³ 3^n-1 u [n-1]

(b) n³3ⁿ u [ n – 1]

(c) (1 – n)³ 3^n-1 u [n – 1]

(d) (n – 1)³ 3^n-1 u [n]

22. The time signal corresponding to z² – z^-2 x(z) is

(a) 1/2 (x[n + 2] – x[n – 2])

(b) x[n + 2] – x[n – 2]

(c) 1/2 (x[n – 2] – x[n + 2])

(d) d x[n – 2] – x [n + 2]

23. The time signal corresponding to {x[z]}² is

(a) [x[n]]²

(b) x[n]* x[n]

(c) x[n]*x[-n]

(d) x[-n]* x[-n]

24. The system described by the difference equation y[n] -2y(n – 1)+ y (n – 2) = x(n) – x (n – 1) has y(n) = 0 and n < 0

if x(n) = (n), the y(z) will be

(a) 2

(b) 1

(c) zero

(d) -1

25. Given that, f and G are the one-sided z-transforms of discrete time functions f(n) and g(n), the z-transform of

f(k) g(n – k) is given by

(a) f(n) g(n) z^-n

(b) (n) zⁿ g(n) z^-n

(c) f(k) g(n – k) z^-n

(d) f(n – k) g(n) z^-ns

Answers with Solutions

1. (d)

x(z) = x[n] z^-n = z^-k, z = 0

2. (d)

x(z) = x[n]z² = z^k, all z

3. (a)

x(z) = z^-n = 1/1 – z^-1, |z| > 1

4. (d)

x(z) = (1/4 z^-1)⁴ = 1- (1/4 z^-1)/1 – (1/4 z^-1)

= z⁵ – 0.25⁵/z⁴(z – 0.5), all z

5. (d)

x(z) = (1/4 z^-1)ⁿ

= (4z)ⁿ = 1/1 – 4z, |z| < 1/4

6. (b)

x(z) = (3z^-1)ⁿ = (1/3 z)ⁿ

= 1/3 z/1 – 1/3 z, |z| < 3 = z/3 – z

7. (b)

x(z) = (3/2 z^-1)ⁿ + (2/3 z^-1)ⁿ

= -1/1 – 3/2 z^-1) + 1/(1 – 2/3 z^-1)

-5/6 z/(z – 3/2) (z – 2/3), 2/3 < |z| < 3/2

8. (d)

x(z) = 1/1 – 1/2 z^-1 – 1/1 – 1/4 z^-1, |z| > and |z| < 1/4,

no region of convergence exists

9. (b)

aⁿ u(n)  z/z – a

x[n] = 1/2 e (3)ⁿ u[n] + 1/2 e(3)ⁿ u[n]

x(z) = 1/2 [1/1 – e³ z^-1 + 1/1 – e³ z^-1]

= 1/2 [2 – z^-1 [e³ + e³] + z^-2

= z/2 (2z – 1)/(z² – z + 1), |z| > 1

10. (b)

x(n + n_o)    x(z)

x[n] = 3 [n + 5] + 6 [n] + [n – 1] – 4[n – 2]

x(z) = 3z⁵ + 6 + z^-1 – 4z^-2, 0 < |z| < _oo

11. (d)

x(n + n_o)   z^no  x(z)

x[n] = 2 [n + 2] + 4 [n + 1] + 5[n] + 7[n – 1] + [n – 3]

x(z) 2z² + 4z + 5 + 7z^-1 + z^-3, 0 < |z| < _oo

12. (b)

x(n – n_o)    x(z)

x[n] = [n] – [n – 2] + [n – 4] – [n – 5]

x(z) = 1 – z^-2 + z^-4 – z^-5, z = 0

13. (c)

x(n – n_o) z^-no  x(z)

y[n] = x [n- 2]  y(z) = z^-2 x(z)

= 1/z² – 16

14. (b)

y[n] = 1/2ⁿ x[n]   y(z) = x(2z)

= z²/z² – 4

15. (c)

y[n] = x [-n] * x[n]    y[z] = x (1/z) x(z)

x [-n]   x(1/z)

16. (a)

y[n] = nx [n]   y(z) = – z/dx(z)/dz

= 32z²/(z² – 16)²

17. (b)

x(n – n_o)   z^-no x(z)

y[n] = x [n + 1] + x [n – 1]  y(z) = (z + z^-1) x(z)

y(z) = z/(z² – 16)²

19. (c)

y(z) = x(2z)   y[n] = 1/2ⁿ x[n]

y[n] = 1/2ⁿ n² 3ⁿ u[n]

20. (b)

y(z) = x(1/2)   y(n) = x [-n]

y(n) = n²3^-n u[-n]

21. (c)

y(z) = d/dz x(z) = – z^-1 [-z d/dz x (z)

y(z)   y(n) = – (n – 1) x [-n – 1]

y(n) = – (n – 1)³ 3^n-1 u[n – 1]

22. (a)

y(z) = z² – z^-2/2 x(z)   y[n]

= 1/2 [x[n + 2] – x[ n – 2]]

23. (b)

y(z) = x(z)  h(z)

y(z) = x(z)  x(z)   y([n] = x[n] * x[n]

24. (c)

y[n] – 2y [n – 1] + y[n – 2] = x[n] – x[ n – 1]

for n = 0,

y(0) 2y (-1) + y(-2) = x(0) – x(-1)

 

y(0) = x(0) -x (-1)

y(n) = 0 for n < 0

for n = 1,

y(1) = – 2y(0) + y(-1) = x(1) – x(0)

y(1) = x(1) – x(0) + 2x(0) -2x(-1)

y(1) = x(1) + x(0) – 2x(-1)

for n = 2,

y(2) = x(2) -x(1) + 2y(1) – y(0)

y(2) = x(2) – x(1) + 2x(1) + 2x(0) – 4x(-1) -x(0) + x(-1)

y(2) = x(2) + x(1) + x(0) – 3x(-1)

y(2) = d(2) + d(1) + d(0) – 3d(-1)

25. (a)

f(n) g(n) z^-n