Z-Transform and Difference Equation

Contents

**Evaluate the Z-Transform of Standard Functions** 2

**State and prove Initial and Final value Theorems** 7

**State and prove Convolution Theorem** 9

**Calculate Z-Transform for the Function multiplied by k** 11

**Calculate Z-Transform for the Function divided by k** 13

**Describe Inverse Z-Transform** 14

**Calculate Inverse Z-Transform using Partial Fraction Method** 16

**Calculate Inverse Z-Transform using Residue Theorem** 18

**Recall the Difference Equation and its Formation** 18

**Find the solution of First Order Linear Difference Equation using Z-Transform** 18

**Find the solution of Second Order Linear Difference Equation using Z-Transform** 18

**Find the solution of Higher Order Linear Difference Equation using Z-Transform** 18

**Recall Z-Transform**

The z-transform is a mathematical technique used to transform a discrete-time signal or sequence into a function of a complex variable, z. It is an important tool in digital signal processing and is used to analyze and design digital filters, control systems, and other systems in which discrete-time signals are used.

The z-transform of a discrete-time signal x[n] is defined as:

X(z) = Z{x[n]} = x[n] z^{(-n)}

where z is a complex variable, and the sum is taken over all values of n from 0 to ∞.

The z-transform can be viewed as a generalization of the Fourier transform, which is used to transform continuous-time signals into a function of a complex variable, jω.

Properties of the z-transform include linearity, time shifting, scaling, and convolution, which are analogous to the properties of the Fourier transform.

Example: Consider the sequence x[n] = {1, 2, 3, 4, …}. The z-transform of x[n] is:

X(z) = Z{x[n]} = x[n] z^{(-n)} = 1 + 2z^{(-1)} + 3z^{(-2) }+ 4z^{(-3)} + …

This is an example of a power series, which converges for certain values of z. For example, if |z| > 1, the series diverges, and if |z| < 1, the series converges absolutely. The region of convergence (ROC) is the set of values of z for which the series converges.

The ROC is an important concept in the z-transform, as it determines the range of values for which the z-transform is valid. In the example above, the ROC is |z| > 0, which means that the z-transform is valid for all values of z except z = 0.

The z-transform has several applications in digital signal processing, such as in the design of digital filters, which are used to modify the frequency content of a signal. By applying the z-transform to a discrete-time signal, it is possible to analyze its frequency content and design filters that modify the signal in a desired way. The z-transform is also used in the analysis and design of digital control systems, which are used to control the behavior of physical systems using discrete-time signals.

**Evaluate the Z-Transform of Standard Functions**

The z-transform is a mathematical tool used to analyze and process digital signals. It is defined as the summation of a discrete-time signal x[n] multiplied by the complex exponential function z^{(-n)}, where z is a complex variable. The z-transform can be used to transform a discrete-time signal from the time domain to the frequency domain, where it can be analyzed and manipulated using algebraic operations.

In order to evaluate the z-transform of a standard function, it is necessary to substitute the function into the z-transform equation and perform the summation over the appropriate range of n values. Some common standard functions and their z-transforms are as follows:

- Unit impulse function:

The unit impulse function δ[n] is defined as:

δ[n] = 1, n=0

δ[n] = 0, n≠0

The z-transform of δ[n] is:

Z{δ[n]} = δ[n] z^{(-n)} = 1

- Unit step function:

The unit step function u[n] is defined as:

u[n] = 1, n≥0

u[n] = 0, n<0

The z-transform of u[n] is:

Z{u[n]} = u[n] z^{(-n)} = 1/(1-z^{(-1)})

- Exponential function:

The exponential function a^{n} u[n] is defined as:

a^{n }u[n], n≥0

0, n<0

The z-transform of a^{n} u[n] is:

Z{a^{n} u[n]} = a^{n} u[n] z^{(-n)} = (a z^{(-1)})^{n}

= 1/(1-a z^{(-1)}), |a z^{(-1)}| < 1

- Sinusoidal function:

The sinusoidal function sin(ωn) is defined as:

sin(ωn), n≥0

0, n<0

The z-transform of sin(ωn) is:

Z{sin(ωn)} = sin(ωn) z^{(-n)}

= (z^{(-1)} sin(ω))/(1 – 2z^{(-1)} cos(ω) + z^{(-2)})

These are some of the common standard functions and their z-transforms. By evaluating the z-transforms of these functions, it is possible to analyze and manipulate digital signals in the frequency domain.

**Recall the Properties of Z-Transform: Linear property, Scale change property, and Shifting property**

The z-transform is a mathematical tool used to analyze and process digital signals. It is defined as the summation of a discrete-time signal x[n] multiplied by the complex exponential function z^{(-n)}, where z is a complex variable. The z-transform can be used to transform a discrete-time signal from the time domain to the frequency domain, where it can be analyzed and manipulated using algebraic operations.

There are several properties of the z-transform that make it a powerful tool for analyzing digital signals. Some of the most important properties are the linear property, the scale change property, and the shifting property.

- Linear Property:

The linear property of the z-transform states that if two discrete-time signals x_{1}[n] and x_{2}[n] have z-transforms X1(z) and X2(z), respectively, and if a and b are constants, then the z-transform of the linear combination of x_{1}[n] and x_{2}[n] is given by:

Z{a x_{1}[n] + b x_{2}[n]} = a X1(z) + b X_{2}(z)

In other words, the z-transform is a linear operator that obeys the superposition principle. This property makes it easy to analyze and manipulate signals that are composed of multiple components.

- Scale Change Property:

The scale change property of the z-transform states that if a discrete-time signal x[n] has a z-transform X(z), then the z-transform of the scaled version of x[n] is given by:

Z{k x[n]} = K X(kz)

where k is a constant. In other words, scaling a discrete-time signal by a constant k simply scales its z-transform by a factor of k.

- Shifting Property:

The shifting property of the z-transform states that if a discrete-time signal x[n] has a z-transform X(z), then the z-transform of the shifted version of x[n] is given by:

Z{x[n-m]} = z^{(-m)} X(z)

where m is a non-negative integer. In other words, shifting a discrete-time signal x[n] to the right by m samples in the time domain corresponds to multiplying its z-transform by z^{(-m)} in the frequency domain.

Examples:

- Linear Property:

Suppose we have two discrete-time signals:

x_{1}[n] = {1, 2, 3, 4} and x_{2}[n] = {2, 3, 4, 5}

The z-transforms of these signals are:

X_{1}(z) = 1 + 2z^{(-1)} + 3z^{(-2)} + 4z^{(-3)}

X_{2}(z) = 2 + 3z^{(-1)} + 4z^{(-2)} + 5z^{(-3)}

Using the linear property of the z-transform, we can find the z-transform of the linear combination of x1[n] and x2[n]:

Z{3 x1[n] – 2 x2[n]} = 3 X1(z) – 2 X2(z)

= 3(1 + 2z^{(-1)} + 3z^{(-2)} + 4z^{(-3)}) – 2(2 + 3z^{(-1)} + 4z^{(-2)} + 5z^{(-3)})

= -1 + 3z^{(-1)} – z^{(-2) }– z^{(-3) }

**State and prove Initial and Final value Theorems**

The z-transform is a mathematical tool used to analyze and process discrete-time signals. It is defined as the summation of a discrete-time signal x[n] multiplied by the complex exponential function z^{(-n)}, where z is a complex variable. The z-transform can be used to transform a discrete-time signal from the time domain to the frequency domain, where it can be analyzed and manipulated using algebraic operations.

The initial and final value theorems are two important results related to the behavior of a discrete-time signal in the time domain as n goes to negative ∞ or positive ∞, respectively. These theorems can be used to determine the steady-state behavior of a system or to evaluate the behavior of a system at the beginning or end of a sequence.

- Initial Value Theorem:

The initial value theorem states that the initial value of a discrete-time signal x[n], i.e., the value of x[n] when n goes to negative ∞, is given by:

lim _{n->-∞} x[n] = lim _{z->∞} [zX(z)]

where X(z) is the z-transform of x[n].

Proof:

To prove the initial value theorem, we can start by expressing the z-transform of x[n] as:

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

We can rewrite this expression as:

X(z) = [x[0] + x[1] z^{(-1)} + x[2] z^{(-2)} + …] + [x[1] z^{(-1)} + x[2] z^{(-2)} + x[3] z^{(-3)} + …] + …

The first term inside the square brackets corresponds to the initial value of x[n], while the second and subsequent terms correspond to the steady-state behavior of x[n]. As n goes to negative ∞, the terms inside the square brackets become negligible, and the z-transform of x[n] reduces to the first term only:

lim _{n->-} X(z) = x[0]

Taking the inverse z-transform of both sides gives:

lim _{n->-} x[n] = lim _{z->} [zX(z)]

which is the initial value theorem.

- Final Value Theorem:

The final value theorem states that the final value of a discrete-time signal x[n], i.e., the value of x[n] when n goes to positive ∞, is given by:

lim _{n->} x[n] = lim _{z->1} [(z-1)X(z)]

where X(z) is the z-transform of x[n].

Proof:

To prove the final value theorem, we can start by expressing the z-transform of x[n] as:

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

We can rewrite this expression as:

X(z) = [x[0] + x[1] z^{(-1)} + x[2] z^{(-2)} + …] + [x[1] z^{(-1)} + x[2] z^{(-2)} + x[3] z^{(-3)} + …] + …

The first term inside the square brackets corresponds to the initial value of x[n], while the second and subsequent terms correspond to the steady-state behavior of x[n]. As n goes to positive ∞, the terms inside the square brackets become negligible, and the z-transform of x[n] reduces to the second and subsequent terms only:

lim _{n->} X(z) = [x[1] z^{(-1)} + x[2] z^{(-2)} + x[3]

**State and prove Convolution Theorem**

The convolution theorem is an important property of the z-transform that relates the convolution of two discrete-time signals to their z-transforms. It states that the z-transform of the convolution of two signals is equal to the product of their individual z-transforms. The convolution theorem can be used to simplify the analysis of linear time-invariant (LTI) systems, which are systems whose response to an input signal is linear and time-invariant.

Statement of the Convolution Theorem:

Let x[n] and h[n] be two discrete-time signals with z-transforms X(z) and H(z), respectively. Then, the z-transform of their convolution y[n] = x[n] * h[n] is given by:

Y(z) = X(z) * H(z)

where * denotes the convolution operation and * denotes the multiplication operation.

Proof:

To prove the convolution theorem, we start with the definition of the z-transform:

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

H(z) = h[n] z^{(-n)}

The convolution of x[n] and h[n] is defined as:

y[n] = x[n] * h[n] = x[k] h[n-k]

The z-transform of y[n] is therefore given by:

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

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

We can rearrange the terms inside the inner sum as follows:

x[k] h[n-k] = x[k] h[n-k] + x[k] h[n-k]

The first term inside the square brackets corresponds to the convolution between x[n] and h[n] over the range k = -∞ to k = n, while the second term corresponds to the convolution over the range k = n+1 to k = ∞. Using this expression, we can rewrite the z-transform of y[n] as:

Y(z) = [ x[k] h[n-k]] z^{(-n)} + x[k] h[n-k]] z^{(-n)}

We can now evaluate each of these sums separately. For the first sum, we can use the change of variables m = n-k to obtain:

sum from x[k] h[n-k] = x[n-m] h[m]

Substituting this expression back into the z-transform of y[n], we obtain:

Y(z) = [ x[n-m] h[m]] z^{(-n)} + [ x[k] h[n-k]] z^{(-n)}

We can now exchange the order of the two sums in the first term to obtain:

[ x[n-m] h[m]] z^{(-n) }= h[m] [ x[n-m] z^{(-n)}]

= H(z) X(z)

**Calculate Z-Transform for the Function multiplied by k**

This Learning Outcome states that you should be able to calculate the Z-transform of a function multiplied by a scalar constant k. The Z-transform is a mathematical tool used in signal processing and system analysis to convert a discrete-time signal or system function into a complex frequency domain representation.

The Z-transform of a function multiplied by a scalar constant k can be calculated using the following formula:

X(z) = Z {k*x[n]} = k*Z{x[n]}

where X(z) is the Z-transform of the function k*x[n], x[n] is the original function, and Z{} denotes the Z-transform operation.

For example, consider the discrete-time function x[n] = {1, 2, 3, 4}. Let k = 2. To calculate the Z-transform of k*x[n], we can use the formula above:

X(z) = Z {2*x[n]} = 2*Z{x[n]}

Using the standard Z-transform table, we can find that the Z-transform of x[n] is:

Z{x[n]} = 1 + 2z^{(-1)} + 3z^{(-2)} + 4z^{(-3)}

Therefore, the Z-transform of k*x[n] is:

X(z) = 2*Z{x[n]} = 2*(1 + 2z^{(-1) }+ 3z(^{-2)} + 4z^{(-3)})

X(z) = 2 + 4z^{(-1)} + 6z^{(-2)} + 8z^{(-3)}

Another example is the function x[n] = {-1, 2, -3, 4, -5}. Let k = -3. To calculate the Z-transform of k*x[n], we can use the formula above:

X(z) = Z {-3*x[n]} = -3*Z{x[n]}

Using the standard Z-transform table, we can find that the Z-transform of x[n] is:

Z{x[n]} = -1 + 2z^{(-1)} – 3z^{(-2)} + 4z^{(-3)} – 5z^{(-4)}

Therefore, the Z-transform of k*x[n] is:

X(z) = -3*Z{x[n]} = 3 – 6z^{(-1)} + 9z^{(-2)} – 12z^{(-3)} + 15z^{(-4)}

In conclusion, calculating the Z-transform of a function multiplied by a scalar constant k involves multiplying the Z-transform of the original function by the scalar constant k. This concept is useful in signal processing and system analysis to analyze the effect of scaling on the Z-transform of a function.

**Calculate Z-Transform for the Function divided by k**

This Learning Outcome states that you should be able to calculate the Z-transform of a function divided by a scalar constant k. The Z-transform is a mathematical tool used in signal processing and system analysis to convert a discrete-time signal or system function into a complex frequency domain representation.

The Z-transform of a function divided by a scalar constant k can be calculated using the following formula:

X(z) = Z {x[n]/k} = (1/k)*Z{x[n]}

where X(z) is the Z-transform of the function x[n]/k, x[n] is the original function, and Z{} denotes the Z-transform operation.

For example, consider the discrete-time function x[n] = {2, 4, 6, 8}. Let k = 2. To calculate the Z-transform of x[n]/k, we can use the formula above:

X(z) = Z {x[n]/2} = (1/2)*Z{x[n]}

Using the standard Z-transform table, we can find that the Z-transform of x[n] is:

Z{x[n]} = 2 + 4z^{(-1)} + 6z^{(-2)} + 8z^{(-3)}

Therefore, the Z-transform of x[n]/k is:

X(z) = (1/2)*Z{x[n]} = 1 + 2z^{(-1) }+ 3z^{(-2)} + 4z^{(-3)}

Another example is the function x[n] = {1, -3, 5, -7}. Let k = -1/3. To calculate the Z-transform of x[n]/k, we can use the formula above:

X(z) = Z {x[n]/(-1/3)} = (-3)*Z{x[n]}

Using the standard Z-transform table, we can find that the Z-transform of x[n] is:

Z{x[n]} = 1 – 3z^{(-1) }+ 5z^{(-2)} – 7z^{(-3)}

Therefore, the Z-transform of x[n]/k is:

X(z) = (-3)*Z{x[n]} = -3 + 9z^{(-1)} – 15z^{(-2) }+ 21z^{(-3)}

In conclusion, calculating the Z-transform of a function divided by a scalar constant k involves dividing the Z-transform of the original function by the scalar constant k. This concept is useful in signal processing and system analysis to analyze the effect of scaling on the Z-transform of a function.

**Describe Inverse Z-Transform**

This Learning Outcome states that you should be able to describe the inverse Z-transform. The Z-transform is a mathematical tool used in signal processing and system analysis to convert a discrete-time signal or system function into a complex frequency domain representation. The inverse Z-transform is the mathematical operation used to convert a function in the Z-domain back to its corresponding discrete-time function in the time domain.

The inverse Z-transform can be represented using the following formula:

x[n] = (1/2πj) * ∮ X(z)z^{(n-1)} dz

where x[n] is the inverse Z-transform of the function X(z), and ∮ represents the complex contour integral in the counterclockwise direction around the ROC (region of convergence) of X(z).

In simple terms, the inverse Z-transform formula calculates the coefficients of the discrete-time function x[n] by integrating the Z-transform X(z) around a closed path in the Z-plane. The inverse Z-transform formula has a similarity with the inverse Fourier transform formula, where the Fourier transform is used to convert a signal from the time domain to the frequency domain, and the inverse Fourier transform is used to convert the signal back to the time domain.

Let’s take an example to understand the inverse Z-transform. Consider the Z-transform of a discrete-time function:

X(z) = 1/(1 – 0.5z^{(-1)})

The ROC for this function is |z| > 0.5. To find the inverse Z-transform, we need to use the formula mentioned above:

x[n] = (1/2πj) * ∮ X(z)z^{(n-1}) dz

where the contour integral is taken around a closed path that encloses the ROC. We can use the partial fraction expansion to simplify the X(z) function:

X(z) = 1/(1 – 0.5z^{(-1)}) = (2/(z-0.5)) = (1/0.5)*1/(1-(z^{(-1)}/0.5))

Using the Z-transform table, we know that the inverse Z-transform of 1/(1-a.z^{(-1)}) is a^{n}.u[n]. Therefore, we can rewrite X(z) as:

X(z) = 2/(z-0.5) = 4z^{(-1)}/(1-0.5z^{(-1)}) = 4z^{(-1)}*∑ (0.5) ^{n}*u[n]

where u[n] is the unit step function.

Now, we can use the inverse Z-transform formula to find x[n]:

x[n] = (1/2πj) * ∮ X(z)z^{(n-1)} dz

Taking the contour integral around a circle with a radius greater than 0.5 in the counterclockwise direction, we can evaluate the integral and obtain the inverse Z-transform:

x[n] = 2*(0.5)^{n}*u[n-1]

Therefore, the original discrete-time function is:

x[n] = 2*(0.5)^{(n-1)}*u[n-1]

In conclusion, the inverse Z-transform is a mathematical tool used to convert a function in the Z-domain back to its corresponding discrete-time function in the time domain. The formula for the inverse Z-transform involves integrating the Z-transform around a closed path in the Z-plane. The inverse Z-transform is useful in signal processing and system analysis to analyze the behavior of discrete-time systems in the time domain.

**Calculate Inverse Z-Transform using Partial Fraction Method**

This Learning Outcome states that you should be able to calculate the inverse Z-transform using the partial fraction method. The partial fraction method is a technique used to decompose a rational function in the Z-domain into simpler terms that can be more easily inverted using the Z-transform tables or other techniques.

The partial fraction method involves the following steps:

- Factorise the polynomial in the denominator of the rational function into linear and quadratic terms.
- Express the rational function as a sum of simpler terms with constant coefficients, where each term has a denominator corresponding to one of the factors in step 1.
- Use the Z-transform tables or other techniques to calculate the inverse Z-transform of each simpler term.
- Combine the inverse Z-transforms of the simpler terms to obtain the inverse Z-transform of the original function.

Let’s take an example to illustrate the partial fraction method for calculating the inverse Z-transform. Consider the following rational function in the Z-domain:

X(z) = (3z^{(-1)} + 2)/(1 – 2z^{(-1)} + z^{(-2)})

The denominator of the function can be factorised as:

1 – 2z^{(-1)} + z^{(-2)} = (1 – z^{(-1)})^{2}

Therefore, we can express X(z) as a sum of two simpler terms:

X(z) = (3z^{(-1)} + 2)/((1 – z^{(-1)})^{2}) = A/(1 – z^{(-1)}) + B/(1 – z^{(-1)})^{2}

where A and B are constants that we need to determine. We can find the values of A and B by multiplying both sides of the equation by the denominators of the simpler terms and simplifying:

(3z^{(-1)} + 2) = A*(1 – z^{(-1)})^{(-1)} + B*(1 – z^{(-1)})^{(-2)}

Multiplying both sides by (1 – z^{(-1)})^{2} and substituting z = 1, we get:

5 = A + B

Differentiating both sides with respect to z and multiplying by z, we get:

-3z^{(-2)} = -A*(1 – z^{(-1)})^{(-2)} – 2B*(1 – z^{(-1)})^{(-3)}

Substituting z = 1, we get:

-3 = -A – 2B

Solving the equations simultaneously, we get A = 2 and B = 3. Therefore, we can express X(z) as:

X(z) = 2/(1 – z^{(-1)}) + 3/(1 – z^{(-1)})^{2}

Using the Z-transform tables, we can calculate the inverse Z-transform of each simpler term:

The inverse Z-transform of 2/(1 – z^{(-1)}) is 2u[n-1].

The inverse Z-transform of 3/(1 – z^{(-1)})^{2} is 3n*u[n-1].

Therefore, the inverse Z-transform of X(z) is:

x[n] = 2u[n-1] + 3n*u[n-1]

In conclusion, the partial fraction method is a technique used to calculate the inverse Z-transform of a rational function in the Z-domain. The method involves factoring the denominator of the function into simpler terms and expressing the function as a sum of simpler terms. The coefficients of the simpler terms can be found by solving a system of equations. The inverse Z-transform of each simpler term can be calculated using the Z-transform tables or other techniques.

**Calculate Inverse Z-Transform using Residue Theorem**

This Learning Outcome states that you should be able to calculate the inverse Z-transform using the residue theorem. The residue theorem is a powerful technique used to calculate the inverse Z-transform of a function in the Z-domain that has poles.

The residue theorem states that if a function in the Z-domain has a pole of order k at z = a, then the inverse Z-transform of the function is given by:

x[n] = Residue{X(z)*z^{n}, z=a}

where X(z) is the function in the Z-domain, and Residue{} denotes the residue of the function at z = a. The residue is a complex number that can be calculated using the formula:

Residue{X(z), z=a} = (1/(k-1)!)*lim(z-a) ^{(k-1)} * d^{(k-1)}/dz^{(k-1)} [X(z)*(z-a)

^{k}]

where k is the order of the pole at z = a.

Let’s take an example to illustrate the use of the residue theorem to calculate the inverse Z-transform. Consider the following function in the Z-domain:

X(z) = (3z – 2)/(z^{2} – 4z + 3)

The denominator of the function can be factorized as^{:}

z^{2}– 4z + 3 = (z – 1)*(z – 3)

Therefore, the function has poles at z = 1 and z = 3. We can write X(z) as a sum of two terms:

X(z) = A/(z – 1) + B/(z – 3)

where A and B are constants that we need to determine. To find the values of A and B, we can multiply both sides of the equation by the denominators of the terms and simplify:

3z – 2 = A*(z – 3) + B*(z – 1)

Setting z = 1 and z = 3, we get:

A = ½

B = ½

Therefore, we can express X(z) as:

X(z) = 1/(2*(z – 1)) + 1/(2*(z – 3))

Using the residue theorem, we can calculate the inverse Z-transform of each term. The first term has a pole of order 1 at z = 1, and the second term has a pole of order 1 at z = 3. Therefore, the inverse Z-transform of each term is given by:

The inverse Z-transform of 1/(2*(z – 1)) is (1/2)*u[n-1].

The inverse Z-transform of 1/(2*(z – 3)) is (1/2)*3 ^{n}*u[n-1].

Therefore, the inverse Z-transform of X(z) is:

x[n] = (1/2)*u[n-1] + (3/2)*3 ^{n}*u[n-1]

In conclusion, the residue theorem is a powerful technique used to calculate the inverse Z-transform of a function in the Z-domain that has poles. The method involves factorizing the denominator of the function into simpler terms and expressing the function as a sum of simpler terms. The coefficients of the simpler terms can be found by solving a system of equations. The inverse Z-transform of each simpler term can be calculated using the residue theorem. Finally, the inverse Z-transform of the original function is obtained by combining the inverse Z-transform of the simpler terms.

**Recall the Difference Equation and its Formation**

This learning outcome focuses on recalling the difference equation and its formation. A difference equation is a mathematical expression that describes the relationship between the input and output of a discrete-time system. It is used to model discrete-time systems, which are systems that process signals that are defined only at discrete time instants.

The general form of a linear difference equation is given by:

a[n]y[n] + a[n-1]y[n-1] + … + a[0]y[n-m] = b[n]x[n] + b[n-1]x[n-1] + … + b[0]x[n-k]

where y[n] is the output of the system at time instant n, x[n] is the input of the system at time instant n, and a[n], a[n-1], …, a[0] and b[n], b[n-1], …, b[0] are the coefficients of the difference equation. The terms m and k are the orders of the system and represent the number of past output values and past input values used in the equation, respectively.

The formation of a difference equation involves deriving an expression that relates the output of a discrete-time system to its input and past output values. The general approach is to use the input-output relationship of the system and express it in terms of the current and past values of the input and output. This can be done using the following steps:

- Identify the input and output of the system.
- Derive an expression that relates the current output to the current input and past output values.
- Express the current input and past output values in terms of the input and output at previous time instants.
- Simplify the resulting expression to obtain the difference equation.

Let’s take an example to illustrate the formation of a difference equation. Consider a discrete-time system that receives an input x[n] and produces an output y[n] according to the following input-output relationship:

y[n] = 0.5y[n-1] + 0.2x[n] + 0.3x[n-1]

To obtain a difference equation that describes this system, we can use the following steps:

- Identify the input and output of the system: In this case, the input is x[n] and the output is y[n].
- Derive an expression that relates the current output to the current input and past output values: From the input-output relationship, we have:

y[n] = 0.5y[n-1] + 0.2x[n] + 0.3x[n-1]

- Express the current input and past output values in terms of the input and output at previous time instants: We can express y[n-1] and x[n-1] in terms of y[n-2] and x[n-2], respectively:

y[n-1] = 0.5y[n-2] + 0.2x[n-1] + 0.3x[n-2]

- Simplify the resulting expression to obtain the difference equation: Substituting the expressions for y[n-1] and x[n-1] in the input-output relationship, we get:

y[n] = 0.5y[n-1] + 0.2x[n] + 0.3(0.5y[n-2] + 0.2x[n-1] + 0.3x[n-2])

**Find the solution of First Order Linear Difference Equation using Z-Transform**

This learning outcome focuses on finding the solution of a first-order linear difference equation using Z-transform. A first-order linear difference equation is an equation of the form:

y[n] + a1y[n-1] = b0x[n]

where y[n] is the output at time instant n, x[n] is the input at time instant n, a1 and b0 are constants, and n is a discrete-time index.

To find the solution of this equation using Z-transform, we can apply the Z-transform to both sides of the equation. The Z-transform of a sequence y[n] is defined as:

Y(z) = Z{y[n]} = ∑ y[n]z^{(-n)}

where z is the Z-transform variable. The Z-transform of the input sequence x[n] is X(z). The Z-transform of the left-hand side of the difference equation can be expressed as:

Y(z) + a1z^{(-1)}Y(z) = (1 + a1z^{(-1)})Y(z)

where we have used the linearity and time-shift properties of the Z-transform.

The Z-transform of the right-hand side of the difference equation can be expressed as:

b0X(z)

Substituting these expressions in the difference equation, we get:

(1 + a1z^{(-1)})Y(z) = b0X(z)

Solving for Y(z), we get:

Y(z) = (b0 / (1 + a1z^{(-1)}))X(z)

This is the Z-transform solution of the first-order linear difference equation. To obtain the time-domain solution, we can use the inverse Z-transform, which is defined as:

y[n] = Z^{(-1)}{Y(z)} = (1/2πj) ∫[γ-i∞, γ+i∞] Y(z)z^{n} dz

where γ is a contour in the complex plane that encloses all the poles of Y(z).

Let’s take an example to illustrate the solution of a first-order linear difference equation using Z-transform. Consider the difference equation:

y[n] + 0.5y[n-1] = x[n]

We can apply the Z-transform to both sides of this equation to obtain:

Y(z) + 0.5z^{(-1)}Y(z) = X(z)

Solving for Y(z), we get:

Y(z) = (1 / (1 + 0.5z^{(-1)}))X(z)

To obtain the time-domain solution, we need to find the inverse Z-transform of Y(z). The Z-transform has a pole at z = 0.5, so we need to choose a contour in the complex plane that encloses this pole. Let’s choose a contour that is a circle with radius 0.6 centered at the origin. Applying the inverse Z-transform, we get:

y[n] = (1/2πj) ∫[C] Y(z)z^{n} dz

= (1/2πj) ∫[C] (1 / (1 + 0.5z^{(-1)}))X(z)z^{n} dz

= Residue at z=0.5 of Y(z)z^{n}X(z) + Residue at z=0 of Y(z)z^{n}X(z)

= (0.5^{n})u[n] + (-0.5^{n})u[n-1]

where u[n] is the unit step function.

Therefore, the solution of the first-order linear difference equation is:

y[n] = (0.5^{n})u[n] + (-0.5^{n})u[n-1]

**Find the solution of Second Order Linear Difference Equation using Z-Transform**

This learning outcome focuses on finding the solution of a second-order linear difference equation using Z-transform. A second-order linear difference equation is an equation of the form:

y[n] + a1y[n-1] + a2y[n-2] = b0x[n]

where y[n] is the output at time instant n, x[n] is the input at time instant n, a1, a2, and b0 are constants, and n is a discrete-time index.

To find the solution of this equation using Z-transform, we can apply the Z-transform to both sides of the equation. The Z-transform of a sequence y[n] is defined as:

Y(z) = Z{y[n]} = ∑ y[n]z^{(-n)}

where z is the Z-transform variable. The Z-transform of the input sequence x[n] is X(z). The Z-transform of the left-hand side of the difference equation can be expressed as:

Y(z) + a1z^{(-1)}Y(z) + a2z^{(-2)}Y(z) = (1 + a1z^{(-1)} + a2z^{(-2)})Y(z)

where we have used the linearity and time-shift properties of the Z-transform.

The Z-transform of the right-hand side of the difference equation can be expressed as:

b0X(z)

Substituting these expressions in the difference equation, we get:

(1 + a1z^{(-1)} + a2z^{(-2)})Y(z) = b0X(z)

Solving for Y(z), we get:

Y(z) = (b0 / (1 + a1z^{(-1)} + a2z^{(-2)}))X(z)

This is the Z-transform solution of the second-order linear difference equation. To obtain the time-domain solution, we can use the inverse Z-transform.

There are three cases that we may encounter while solving for the inverse Z-transform of the above expression:

- The roots of the denominator polynomial 1 + a1z
^{(-1)}+ a2z^{(-2)}are real and distinct. In this case, we can use partial fraction decomposition to express the Z-transform as a sum of simpler fractions, and then use the inverse Z-transform to obtain the time-domain solution. - The roots of the denominator polynomial 1 + a1z
^{(-1)}+ a2z^{(-2)}are real and repeated. In this case, we can use partial fraction decomposition with repeated factors to express the Z-transform as a sum of simpler fractions, and then use the inverse Z-transform to obtain the time-domain solution. - The roots of the denominator polynomial 1 + a1z
^{(-1)}+ a2z^{(-2)}are complex conjugates. In this case, we can use the residue theorem to find the inverse Z-transform of the Z-transform solution.

Let’s take an example to illustrate the solution of a second-order linear difference equation using Z-transform. Consider the difference equation:

y[n] + 1.2y[n-1] + 0.36y[n-2] = 0.8x[n]

To solve this second-order linear difference equation using the Z-transform, we first need to take the Z-transform of both sides of the equation. Using the linearity property of the Z-transform and the shifting property, we obtain:

Y(z) + 1.2z^{(-1)}Y(z) + 0.36z^{(-2)}Y(z) = 0.8X(z)

Simplifying the equation, we get:

Y(z) = (0.8 / (1 + 1.2z^{(-1) }+ 0.36z^{(-2)})) X(z)

Now, we can factor the denominator of the right-hand side of the equation to obtain:

Y(z) = (0.8 / ((z^{(-1)} + 0.6)(z^{(-1)} + 0.6))) X(z)

Using partial fraction decomposition, we can express this equation in terms of simple fractions:

Y(z) = (A / (z^{(-1)} + 0.6)) + (B / (z^{(-1)} + 0.6)^{2}) X(z)

where A and B are constants that can be found by equating the coefficients of the terms in the right-hand side of the equation. Solving for A and B, we get:

A = 0.8 / (1 – 0.6) = 2

B = 0.8 / ((1 – 0.6)^{2}) = 5

Substituting these values into the equation, we get:

Y(z) = (2 / (z^{(-1)} + 0.6)) + (5 / (z^{(-1)} + 0.6)^{2}) X(z)

Now, we can take the inverse Z-transform of both sides of the equation to obtain the solution in the time domain. Using the table of Z-transform pairs, we can find that the inverse Z-transform of 1 / (z^{(-1)} + a) is a^{n} u[n], where u[n] is the unit step function. Using this, we can write:

y[n] = 2(-0.6)^{n} u[n] + 5n(-0.6)^{n} u[n]

where u[n] is the unit step function, which is 1 for n >= 0 and 0 otherwise.

Therefore, the solution of the given difference equation is:

y[n] = 2(-0.6)^{n} u[n] + 5n(-0.6)^{n} u[n]

**Find the solution of Higher Order Linear Difference Equation using Z-Transform**

This learning outcome focuses on finding the solution of a higher-order linear difference equation using Z-transform. A higher-order linear difference equation is an equation of the form:

y[n] + a1y[n-1] + a2y[n-2] + … + aNy[n-N] = b0x[n]

where y[n] is the output at time instant n, x[n] is the input at time instant n, a1, a2, …, aN, and b0 are constants, N is the order of the equation, and n is a discrete-time index.

To find the solution of this equation using Z-transform, we can apply the Z-transform to both sides of the equation. The Z-transform of a sequence y[n] is defined as:

Y(z) = Z{y[n]} = ∑ y[n]z^{(-n)}

where z is the Z-transform variable. The Z-transform of the input sequence x[n] is X(z). The Z-transform of the left-hand side of the difference equation can be expressed as:

Y(z) + a1z^{(-1)}Y(z) + a2z^{(-2)}Y(z) + … + aNz^{(-N)}Y(z) = (1 + a1z^{(-1)} + a2z^{(-2)} + … + aNz^{(-N)})Y(z)

where we have used the linearity and time-shift properties of the Z-transform.

The Z-transform of the right-hand side of the difference equation can be expressed as:

b0X(z)

Substituting these expressions in the difference equation, we get:

(1 + a1z^{(-1)} + a2z^{(-2)} + … + aNz^{(-N)})Y(z) = b0X(z)

Solving for Y(z), we get:

Y(z) = (b0 / (1 + a1z^{(-1)} + a2z^{(-2)} + … + aNz^{(-N)}))X(z)

This is the Z-transform solution of the higher-order linear difference equation. To obtain the time-domain solution, we can use the inverse Z-transform.

There are three cases that we may encounter while solving for the inverse Z-transform of the above expression:

- The roots of the denominator polynomial 1 + a1z
^{(-1)}+ a2z^{(-2)}+ … + aNz^{(-N)}are real and distinct. In this case, we can use partial fraction decomposition to express the Z-transform as a sum of simpler fractions, and then use the inverse Z-transform to obtain the time-domain solution. - The roots of the denominator polynomial 1 + a1z
^{(-1)}+ a2z^{(-2)}+ … + aNz^{(-N)}are real and repeated. In this case, we can use partial fraction decomposition with repeated factors to express the Z-transform as a sum of simpler fractions, and then use the inverse Z-transform to obtain the time-domain solution. - The roots of the denominator polynomial 1 + a1z
^{(-1)}+ a2z^{(-2)}+ … + aNz^{(-N) }are complex conjugates. In this case, we can use the residue theorem to find the inverse Z-transform of the Z-transform solution.

Let’s take an example to illustrate the solution of a higher-order linear difference equation using Z-transform. Consider the difference equation:

y[n] + 2y[n-1] + 2y[n-2] + y[n-3] = 4x[n]

To solve this higher-order linear difference equation using the Z-transform, we first need to take the Z-transform of both sides of the equation. Using the linearity property of the Z-transform and the shifting property, we obtain:

Y(z) + 2z^{(-1)}Y(z) + 2z^{(-2)}Y(z) + z^{(-3)}Y(z) = 4X(z)

Simplifying the equation, we get:

Y(z) = (4 / (1 + 2z^{(-1)} + 2z^{(-2)} + z^{(-3)})) X(z)

Now, we can factor the denominator of the right-hand side of the equation to obtain:

Y(z) = (4 / ((z^{(-1)} + 1)(z^{(-1)} + 1 + j)(z^{(-1)} + 1 – j))) X(z)

where j is the imaginary unit. Using partial fraction decomposition, we can express this equation in terms of simple fractions:

Y(z) = (A / (z^{(-1)} + 1)) + (B / (z^{(-1)} + 1 + j)) + (C / (z^{(-1)} + 1 – j)) X(z)

where A, B, and C are constants that can be found by equating the coefficients of the terms in the right-hand side of the equation. Solving for A, B, and C, we get:

A = 4 / ((1 – j)(1 + j)) = 1

B = 4j / ((1 + j – j^{2})(1 + j)) = -1 – j

C = 4(-j) / ((1 – j – j^{2})(1 – j)) = -1 + j

Substituting these values into the equation, we get:

Y(z) = (1 / (z^{(-1)} + 1)) – (1 + j) / (z^{(-1)} + 1 + j) – (1 – j) / (z^{(-1)} + 1 – j) X(z)

Now, we can take the inverse Z-transform of both sides of the equation to obtain the solution in the time domain. Using the table of Z-transform pairs, we can find that the inverse Z-transform of 1 / (z^{(-1)} + a) is a^{n} u[n], where u[n] is the unit step function. Using this, we can write:

y[n] = (-1)^{n} u[n] – (n+1)cos(nθ)u[n] + nsin(nθ)u[n]

where θ is the angle whose tangent is 1/2, that is, θ = arctan(1/2).

Therefore, the solution of the given difference equation is:

y[n] = (-1)^{n} u[n] – (n+1)cos(nθ)u[n] + nsin(nθ)u[n]