Determine the inverse $z$-transform of $\frac{1+z^{-1}}{1-z^{-1}+0.5 z^{-2}}$. Is this a proper fraction?
Answer (1)
Multiply the numerator and denominator by z 2 to obtain X ( z ) = z 2 − z + 0.5 z ( z + 1 ) .
Complete the square in the denominator: z 2 − z + 0.5 = ( z − 0.5 ) 2 + ( 0.5 ) 2 .
Decompose X ( z ) into X ( z ) = 3 ( z − 0.5 ) 2 + ( 0.5 ) 2 0.5 z + ( z − 0.5 ) 2 + ( 0.5 ) 2 ( z − 0.5 ) z .
Find the inverse z -transform: x [ n ] = ( 2 1 ) n ( 3 sin ( 4 π n ) + cos ( 4 π n ) ) u [ n ] .
x [ n ] = ( 2 1 ) n ( 3 sin ( 4 π n ) + cos ( 4 π n ) ) u [ n ]
Explanation
Problem Setup We are asked to find the inverse z -transform of the function X ( z ) = 1 − z − 1 + 0.5 z − 2 1 + z − 1 . This is a standard problem in signal processing, where we want to find the time-domain sequence x [ n ] corresponding to the given z -transform.
Eliminating Negative Powers of z First, let's multiply the numerator and the denominator by z 2 to eliminate the negative powers of z . This gives us X ( z ) = z 2 ( 1 − z − 1 + 0.5 z − 2 ) z 2 ( 1 + z − 1 ) = z 2 − z + 0.5 z ( z + 1 ) = z 2 − z + 0.5 z 2 + z
Completing the Square Now, we want to express X ( z ) in a form that allows us to easily find the inverse z -transform. We can rewrite the denominator by completing the square: z 2 − z + 0.5 = z 2 − z + 4 1 + 4 1 = ( z − 0.5 ) 2 + ( 0.5 ) 2 So, we have X ( z ) = ( z − 0.5 ) 2 + ( 0.5 ) 2 z 2 + z
Decomposing X(z) We want to decompose X ( z ) into a form that resembles known inverse z -transforms. Specifically, we want to express it as X ( z ) = A ( z − 0.5 ) 2 + ( 0.5 ) 2 0.5 z + B ( z − 0.5 ) 2 + ( 0.5 ) 2 ( z − 0.5 ) z Multiplying by z in the numerator is a trick that allows us to use standard inverse z-transforms.
Finding A and B To find A and B , we equate the numerators: z 2 + z = A ( 0.5 z ) + B ( z ( z − 0.5 )) = 0.5 A z + B z 2 − 0.5 B z z 2 + z = B z 2 + ( 0.5 A − 0.5 B ) z Equating the coefficients of z 2 and z , we get: B = 1 0.5 A − 0.5 B = 1 ⟹ 0.5 A − 0.5 = 1 ⟹ 0.5 A = 1.5 ⟹ A = 3 So, A = 3 and B = 1 .
Rewriting X(z) Thus, we have X ( z ) = 3 ( z − 0.5 ) 2 + ( 0.5 ) 2 0.5 z + ( z − 0.5 ) 2 + ( 0.5 ) 2 ( z − 0.5 ) z
Identifying r and omega Recall that the inverse z -transform of z 2 − 2 rz c o s ( ω ) + r 2 r n s i n ( ωn ) u [ n ] ⋅ z is z 2 − 2 rz c o s ( ω ) + r 2 rz s i n ( ω ) . In our case, we have ( z − 0.5 ) 2 + ( 0.5 ) 2 in the denominator, which can be written as z 2 − z + 0.5 . Comparing this with z 2 − 2 rz cos ( ω ) + r 2 , we have r 2 = 0.5 and 2 r cos ( ω ) = 1 . Thus, r = 0.5 = 2 1 and cos ( ω ) = 2 r 1 = 2 ⋅ 2 1 1 = 2 2 . This means ω = 4 π .
Finding the Inverse z-transform Also, the inverse z -transform of z 2 − 2 rz c o s ( ω ) + r 2 r n c o s ( ωn ) u [ n ] ⋅ z is z 2 − 2 rz c o s ( ω ) + r 2 z ( z − r c o s ( ω )) .
Therefore, the inverse z -transform of X ( z ) is x [ n ] = 3 ( 0.5 ) n sin ( 4 π n ) u [ n ] + ( 0.5 ) n cos ( 4 π n ) u [ n ] x [ n ] = ( 0.5 ) n ( 3 sin ( 4 π n ) + cos ( 4 π n ) ) u [ n ] Since 0.5 = 2 1 , we can write x [ n ] = ( 2 1 ) n ( 3 sin ( 4 π n ) + cos ( 4 π n ) ) u [ n ]
Final Answer The inverse z -transform of the given function is x [ n ] = ( 2 1 ) n ( 3 sin ( 4 π n ) + cos ( 4 π n ) ) u [ n ]
Examples
In signal processing, the z-transform is used to analyze discrete-time signals and systems. Determining the inverse z-transform allows us to convert a system's transfer function from the z-domain back to the time domain, which is crucial for understanding the system's behavior. For example, in control systems, we can analyze the stability and response of a digital filter by examining its inverse z-transform. This helps engineers design and implement effective signal processing algorithms.
