**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, 0_{1} is selected so that x(s) is analytic (no singularities) for s > 0_{1}. 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} = e^{0} 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 = e^{0} = 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 = e^{0} > 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 = e^{0} = 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 45^{0}

**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)z^{0} = 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

x_{1}(m) x_{1}(z) …………….(20)

x_{2}(m) x_{2}(z) ……………………(21)

Then the linearity implies that for any linear combination of x_{1}(m) and x_{2}(m) are have

a_{1}x_{1}(m) + a_{2}x_{2}(m) a_{1}x_{1}(z) + a_{1}x_{2}(z) ……………(22)

- (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^{m}_{0} is

z_{o}^{m}x(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 x_{1}(m) and x_{2}(m)

x_{1}(m) x_{1}(z)

x_{2}(m) x_{2}(z)

the convolutional property states that

x_{1} (m)^{*} x_{2} (m) x_{1}(z) x_{2}(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_{k}x(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_{k}z^{-k} y(z) + b_{k}z^{-k} x(z) ………………….(32)

- (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} + b_{1}z^{-1} +……….+ b_{m} z^{-m}/1 – a_{1}z^{-1} – ………a_{n}z^{-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**

- 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

- 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

- 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

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

(a) z^{5} – 0.25^{5}/z^{4} (z – 0.25), z > 0.25

(b) z^{5} – 0.25^{5}/z^{4}(z – 0.25), z > 0

(c) z^{5} – 0.25^{5}/z^{3} (z – 0.25), z < 0.25

(d) z^{5} – 0.25^{5}/z^{4} (z – 0.25) ,all z

- 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

- The z-transform of 3
^{n}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

- 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

- 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

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

(a) z/2 (2z – 1)/(z^{2} – z + 1), 0 < |z| < 1

(b) z/2 (2z – 1)/(z^{2} – z + 1), |z| > 1

(c) z/2 (1 – 2z)/(z^{2} – z + 1), 0 < |z| < 1

(d) z/2 (1 – 2z)/(z^{2} – z + 1), |z|> 1

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

(a) 3z^{5} + 6 + z^{-1} – 4z^{-2}, 0 < |z| <

(b) 3z^{5} + 6 + z^{-1 }– 4z^{-2}, 0 < |z| <

(c) 3z^{-5} + 6 + z – 4z^{2}, 0 < |z| <

(d) 3z^{-5} + 6 + z – 4z^{2}, 0 < |z| <

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

(a) 2z^{2} + 4z + 5 + 7z + z^{3}, z = _{oo}

(b) 2z^{-2} + 4z^{-1} + 5 + 7z + z^{3}, z = _{00}

(c) 2z^{-2 }+ 4z^{-1} + 5 + 7z + z^{3}, 0 < |z| < _{00}

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

- 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 – 2z^{2} + 4z^{4} – 5z^{5}

(d) 1 – z^{2} + z^{4} – z^{5}

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

(a) z^{4}/z^{2} – 16

(b) (z + 2)^{2}/(z + 2)^{2} – 16

(c) 1/z^{2} – 16

(d) (z – 2)^{2}/(z – 2)^{2} – 16

- The z-transform of the signal y[n] = 1/2
^{n}x[n] is

(a) (z + 2)^{2}/(z + 2)^{2} – 16

(b) z^{2}/z^{2} – 4

(c) (z – 2)^{2}/(z – 2)^{2} – 16

(d) z^{2}/z^{2} – 64

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

(a) z^{2}/16z^{2} – 257z^{4} – 16

(b) -16 z^{2}/(z^{2} – 16)^{2}

(c) z^{2}/257 z^{2} – 16 z^{4} – 16

(d) 16 z^{2}/(z^{2} – 16)^{2}

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

(a) 32 z^{2}/(z^{2} – 16)^{2}

(b) -32 z^{2}/(z^{2} – 16)^{2}

(c) 32 z/(z^{2} – 16)^{2}

(d) -32 z/(z^{2} – 16)^{2}

- 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) z^{2} (1 + z)/z^{2} – 16

(c) z^{2}(-1 + z)/z^{2} – 16

(d) none of the above

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

(a) z^{-3}/(z^{2} – 16)^{2}

(b) z^{7}/(z^{2} – 16)^{2}

(c) z^{5}/(z^{2} – 16)^{2}

(d) z/(z^{2} – 16)^{2}

- The time signal corresponding to x(2z) is

(a) n^{2}3^{n} u[2n]

(b) (-3/2)^{n} n^{2}u[n]

(c) (3/2)^{n} n^{2}u[n]

(d) 6^{n}n^{2} u[n]

- The time signal corresponding to x(z
^{-1}) is

(a) n^{2}3^{-n}[-n]

(b) n^{2}m^{-u} u[-n]

(c) 1/n^{2} 3^{1/n} u[n]

(d) 1/n^{2} 3^{1/n} u[-n]

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

(a) (n – 1)^{3} 3^{n-1} u [n-1]

(b) n^{3}3^{n} u [ n – 1]

(c) (1 – n)^{3} 3^{n-1} u [n – 1]

(d) (n – 1)^{3} 3^{n-1} u [n]

- The time signal corresponding to z
^{2}– 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]

- 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]

- 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

- 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) z^{n} g(n) z^{-n}

(c) f(k) g(n – k) z^{-n}

(d) f(n – k) g(n) z^{-n}s

**Answers with Solutions**

- (d)

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

- (d)

x(z) = x[n]z^{2} = z^{k}, all z

- (a)

x(z) = z^{-n} = 1/1 – z^{-1}, |z| > 1

- (d)

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

= z^{5} – 0.25^{5}/z^{4}(z – 0.5), all z

- (d)

x(z) = (1/4 z^{-1})^{n}

= (4z)^{n} = 1/1 – 4z, |z| < 1/4

- (b)

x(z) = (3z^{-1})^{n} = (1/3 z)^{n}

= 1/3 z/1 – 1/3 z, |z| < 3 = z/3 – z

- (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

- (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

- (b)

a^{n} 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 – e^{3} z^{-1} + 1/1 – e^{3} z^{-1}]

= 1/2 [2 – z^{-1} [e^{3} + e^{3}] + z^{-2}

= z/2 (2z – 1)/(z^{2} – z + 1), |z| > 1

- (b)

x(n + n_{o}) x(z)

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

x(z) = 3z^{5} + 6 + z^{-1} – 4z^{-2}, 0 < |z| < _{oo}

- (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) 2z^{2} + 4z + 5 + 7z^{-1} + z^{-3}, 0 < |z| < _{oo}

- (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

- (c)

x(n – n_{o}) z^{-no} x(z)

y[n] = x [n- 2] y(z) = z^{-2 }x(z)

= 1/z^{2} – 16

- (b)

y[n] = 1/2^{n} x[n] y(z) = x(2z)

= z^{2}/z^{2} – 4

- (c)

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

x [-n] x(1/z)

- (a)

y[n] = nx [n] y(z) = – z/dx(z)/dz

= 32z^{2}/(z^{2} – 16)^{2}

- (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/(z^{2} – 16)^{2}

- (c)

y(z) = x(2z) y[n] = 1/2^{n} x[n]

y[n] = 1/2^{n} n^{2} 3^{n} u[n]

- (b)

y(z) = x(1/2) y(n) = x [-n]

y(n) = n^{2}3^{-n} u[-n]

- (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]

- (a)

y(z) = z^{2} – z^{-2}/2 x(z) y[n]

= 1/2 [x[n + 2] – x[ n – 2]]

- (b)

y(z) = x(z) h(z)

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

- (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)

- (a)

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