# 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 = est . 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 + joo, 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) est  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(mTs), normally denoted as x(m) assuming the sampling period TS = 1, the laplace transform eq. (1) becomes

x(es) = x(m) e-sm  ……………….(3)

substituting the variable es 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 = es = e0 ejoo = rej2f ……..(6)

where, s = 0 + joo 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 = rm 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 joo of the complex frequency variable s = 0 + joo.

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

x(es + 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 + joo and the z-transform z = rej 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 FS Hz where, FS is the sampling frequency. hence, a frequency of f Hz corresponds to an angle 0 given by

0 = 2/fs f radian ……………….(10)

for example at a sampling rate of Fs = 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 = ej as

x(ej) = 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/zk . 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 = ej  is

x(ej) = 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 = zk + 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 = ej as

x(ej) = 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)  zk 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 zm0 is

zomx(m)  x(z/zo) ………………(25)

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

x(m) zom z-m  x(m) (z/zo)-m = x(z/zo) …………….(26)

note that for the case when z = ej and zo = ej then we have the frequency modulation equation

ej0m x(m)   x(ej) ……………….(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) = ak y(m – k) + bkx(m – k) …………………(31)

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

y(z) = akz-k y(z) + bkz-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) = bo + b1z-1 +……….+ bm z-m/1 – a1z-1 – ………anz-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 = ej 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) {am ,  m > 0   0 ,   m < 0 ………………..(38)

Intro Exercise – 4

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

(a) zk, z > 0

(b) z-k , z < 0

(c) zk, z = 0

(d) z-k, z = 0

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

(a) z-k, z = 0

(b) zk, z = 0

(c) z-k, all z

(d) zk, all z

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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| <

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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]

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

(a) n23-n[-n]

(b) n2m-u u[-n]

(c) 1/n2 31/n u[n]

(d) 1/n2 31/n u[-n]

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

(a) (n – 1)3 3n-1 u [n-1]

(b) n33n u [ n – 1]

(c) (1 – n)3 3n-1 u [n – 1]

(d) (n – 1)3 3n-1 u [n]

1. 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]

1. 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]

1. 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

1. 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

1. (d)

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

1. (d)

x(z) = x[n]z2 = zk, all z

1. (a)

x(z) = z-n = 1/1 – z-1, |z| > 1

1. (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

1. (d)

x(z) = (1/4 z-1)n

= (4z)n = 1/1 – 4z, |z| < 1/4

1. (b)

x(z) = (3z-1)n = (1/3 z)n

= 1/3 z/1 – 1/3 z, |z| < 3 = z/3 – z

1. (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

1. (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

1. (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

1. (b)

x(n + no)    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

1. (d)

x(n + no)   zno  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

1. (b)

x(n – no)    x(z)

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

x(z) = 1 – z-2 + z-4 – z-5, z = 0

1. (c)

x(n – no) z-no  x(z)

y[n] = x [n- 2]  y(z) = z-2 x(z)

= 1/z2 – 16

1. (b)

y[n] = 1/2n x[n]   y(z) = x(2z)

= z2/z2 – 4

1. (c)

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

x [-n]   x(1/z)

1. (a)

y[n] = nx [n]   y(z) = – z/dx(z)/dz

= 32z2/(z2 – 16)2

1. (b)

x(n – no)   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

1. (c)

y(z) = x(2z)   y[n] = 1/2n x[n]

y[n] = 1/2n n2 3n u[n]

1. (b)

y(z) = x(1/2)   y(n) = x [-n]

y(n) = n23-n u[-n]

1. (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 3n-1 u[n – 1]

1. (a)

y(z) = z2 – z-2/2 x(z)   y[n]

= 1/2 [x[n + 2] – x[ n – 2]]

1. (b)

y(z) = x(z)  h(z)

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

1. (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)

1. (a)

f(n) g(n) z-n