Q14Advanced Engineering Mathematics

Question

Use convolution theorem to show that Z^{- 1} {\frac{z ( z + 1 )}{( z - 1 ) ^{3}}} = n^{2}

Answer 1

The problem requires calculating the inverse Z-transform of a complex rational function, $X (z) = \frac{z ( z + 1 )}{( z - 1 ) ^{3}}$ . While partial fraction expansion is a common technique, this specific problem is more elegantly and rapidly solved by utilizing the fundamental properties of the Z-transform, specifically the property that relates multiplication by $n$ in the discrete time domain to differentiation in the complex Z-domain.

1. Establishing Base Transforms

We begin with the most fundamental building block in discrete time analysis: the unit step sequence $u_{n} = 1$ (for $n \geq 0$ ). Its well-known Z-transform is $\frac{z}{z - 1}$ .

The fundamental differentiation property states that if $Z {f (n)} = F (z)$ , then multiplying the sequence by $n$ results in the derivative of the transform scaled by $- z$ : $Z {n f (n)} = - z \frac{d}{d z} F (z)$ . We apply this to the unit step sequence to find the transform of the linear ramp sequence $f (n) = n$ :

Z {n} = - z \frac{d}{d z} (\frac{z}{z - 1}) = - z (\frac{( z - 1 ) ( 1 ) - z ( 1 )}{( z - 1 ) ^{2}}) = - z (\frac{- 1}{( z - 1 ) ^{2}}) = \frac{z}{( z - 1 ) ^{2}}

2. Applying the Property Sequentially

Our target transform has a cubic term $(z - 1)^{3}$ in the denominator, suggesting that the property has been applied twice. We apply the differentiation property a second time to the sequence $f (n) = n^{2}$ . We can view this as $Z {n \cdot (n)}$ , where we are applying the operator to the transform we just derived for $n$ .

Z {n^{2}} = - z \frac{d}{d z} (Z {n}) = - z \frac{d}{d z} (\frac{z}{( z - 1 ) ^{2}})

We meticulously apply the quotient rule for derivatives $(\frac{u}{v})^{'} = \frac{v u ^{'} - u v ^{'}}{v ^{2}}$ to perform the differentiation:

Let numerator $u = z$ , so $u^{'} = 1$ .
Let denominator $v = (z - 1)^{2}$ , so $v^{'} = 2 (z - 1)$ .
The denominator squared is $v^{2} = (z - 1)^{4}$ .

\frac{d}{d z} (\frac{z}{( z - 1 ) ^{2}}) = \frac{( z - 1 ) ^{2} ( 1 ) - z ( 2 ( z - 1 ))}{( z - 1 ) ^{4}}

We factor out the common term $(z - 1)$ from the numerator to simplify the algebraic expression:

= \frac{( z - 1 ) [( z - 1 ) - 2 z ]}{( z - 1 ) ^{4}} = \frac{- z - 1}{( z - 1 ) ^{3}} = - \frac{z + 1}{( z - 1 ) ^{3}}

Finally, we multiply by the leading $- z$ factor dictated by the differentiation property:

Z {n^{2}} = - z (- \frac{z + 1}{( z - 1 ) ^{3}}) = \frac{z ( z + 1 )}{( z - 1 ) ^{3}}

This resulting expression perfectly matches the target function provided in the problem statement. Therefore, by mathematical logical equivalence, the inverse Z-transform is the discrete time sequence $n^{2}$ .

Q14Advanced Engineering Mathematics

Question

Answer

1. Establishing Base Transforms

2. Applying the Property Sequentially