Inverse Z-Transform: Decoding E(z) = Z/((z-2)(z-1)^2)

by Aria Freeman 54 views

Hey guys! Ever stumbled upon a Z-transform and felt like you're deciphering an alien language? Well, you're not alone! Z-transforms are super useful in digital signal processing and control systems, but cracking the inverse can sometimes feel like a puzzle. Today, we're going to break down a specific example, E(z) = z/((z-2)(z-1)^2), and walk through the process of finding its inverse Z-transform. Buckle up, and let's get started!

Understanding the Z-Transform

Before we dive into the nitty-gritty, let's quickly recap what the Z-transform is all about. In simple terms, the Z-transform converts a discrete-time signal, which is a sequence of numbers, into a complex-frequency representation. Think of it as a way to analyze signals in a different domain, much like the Laplace transform for continuous-time signals. The inverse Z-transform does the opposite: it takes us from the complex-frequency domain back to the discrete-time domain, giving us the original sequence. Understanding Z-transform is crucial before diving in this inverse Z-transform. This transformation is essential for analyzing the stability and behavior of discrete-time systems.

The Challenge: E(z) = z/((z-2)(z-1)^2)

Our mission today is to find the inverse Z-transform of the function E(z) = z/((z-2)(z-1)^2). This particular function presents a common challenge: the presence of repeated factors in the denominator. The (z-1)^2 term means we'll need to employ a specific technique called partial fraction decomposition with repeating factors. Don't worry; it sounds more complicated than it is! We'll break it down step by step so even if you're new to this, you'll be able to follow along. Our journey starts with recognizing the structure of E(z) and understanding why partial fraction decomposition is our go-to strategy here.

Why Partial Fractions?

The key to tackling this problem lies in partial fraction decomposition. Why? Because we have a complex fraction, and we want to break it down into simpler fractions that we know how to invert using standard Z-transform pairs. Think of it like simplifying a complicated recipe into individual ingredients. By expressing E(z) as a sum of simpler fractions, we can then use a Z-transform table (or our knowledge of common transforms) to find the inverse transform of each part. This approach transforms a daunting task into a series of manageable steps. Essentially, partial fraction decomposition allows us to leverage known inverse transforms of simpler functions.

Step 1: Setting up the Partial Fraction Decomposition

Okay, let's get our hands dirty! The first step is to set up the partial fraction decomposition. Because we have a repeated factor, (z-1)^2, our decomposition will look like this:

z/((z-2)(z-1)^2) = A/(z-2) + B/(z-1) + C/(z-1)^2

Notice how we have a term for each power of the repeated factor, up to the highest power. This is crucial for handling repeated roots correctly. Now, our goal is to find the values of A, B, and C. These constants will determine the weights of our simpler fractions and, ultimately, the shape of our inverse Z-transform. The setup is the foundation, and accurate determination of A, B, and C is paramount for a correct solution. The structure accounts for the repeated root at z=1, which is critical in the overall solution.

Step 2: Solving for the Coefficients (A, B, and C)

Alright, time to put on our algebra hats and solve for A, B, and C. The most common way to do this is to multiply both sides of the equation by the original denominator, (z-2)(z-1)^2. This clears out the fractions and gives us a polynomial equation:

z = A(z-1)^2 + B(z-2)(z-1) + C(z-2)

Now, we can use different strategies to solve for A, B, and C. One popular method is to substitute specific values of z that will eliminate some of the unknowns. For example:

  • Let z = 1: This makes the terms with (z-1) equal to zero, leaving us with 1 = C(1-2), so C = -1.
  • Let z = 2: This makes the terms with (z-2) equal to zero, giving us 2 = A(2-1)^2, so A = 2.

Now that we have A and C, we can substitute another convenient value for z or expand the equation and equate coefficients. Let's expand and equate coefficients of the z term:

z = A(z^2 - 2z + 1) + B(z^2 - 3z + 2) + C(z - 2) z = 2(z^2 - 2z + 1) + B(z^2 - 3z + 2) - 1(z - 2)

Expanding and collecting terms, we get:

0z^2 + 1z + 0 = (2+B)z^2 + (-4-3B-1)z + (2+2B+2)

Equating coefficients of z^2 we get, 0 = 2 + B, so B = -2. This method, known as equating coefficients, is an alternative approach to the substitution method, often providing a robust check on our calculations.

Double-Checking Our Work

Before moving on, it's always a good idea to double-check our values. Let's plug A = 2, B = -2, and C = -1 back into our partial fraction decomposition:

z/((z-2)(z-1)^2) = 2/(z-2) - 2/(z-1) - 1/(z-1)^2

Does this look right? A quick mental check can save us from propagating errors. The accuracy of these coefficients is the backbone of the entire inverse transform process; therefore, verification at this stage is a smart move.

Step 3: Rewriting in Standard Z-Transform Form

Now that we have our partial fraction decomposition, we need to rewrite each term in a form that matches standard Z-transform pairs. Remember, our goal is to use a Z-transform table (or our knowledge of common transforms) to find the inverse. The key here is to have 'z' in the numerator:

Our decomposition is:

E(z) = 2/(z-2) - 2/(z-1) - 1/(z-1)^2

We want it to look like:

E(z) = 2 * [z/z(z-2)] - 2 * [z/z(z-1)] - [z/z(z-1)^2]

Let's multiply and divide each term by z:

E(z) = 2 * (z/z) * [1/(z-2)] - 2 * (z/z) * [1/(z-1)] - (z/z) * [1/(z-1)^2] E(z) = 2 * [z/(z(z-2))] - 2 * [z/(z(z-1))] - [z/(z(z-1)^2)]

But instead, to simplify the inverse transform directly we rewrite them as:

E(z) = 2 * [1/(z-2)] - 2 * [1/(z-1)] - [1/(z-1)^2]

This might seem like a small step, but it's crucial for matching the forms in our Z-transform table. By manipulating the expression to match known transform pairs, we set the stage for a straightforward application of inverse Z-transform properties. This manipulation is a technique to align our decomposed fractions with standard transform pairs.

Step 4: Applying the Inverse Z-Transform

Here comes the exciting part! We're finally ready to apply the inverse Z-transform. We'll use our Z-transform table (or our knowledge of common transforms) to find the inverse of each term. Remember, the Z-transform is a linear operator, so we can take the inverse of each term separately and then add them up.

Let's recall our rewritten E(z):

E(z) = 2/(z-2) - 2/(z-1) - 1/(z-1)^2

Now, we'll use the following standard Z-transform pairs:

  • Z^-1[z/(z-a)] = a^n u[n] where u[n] is the unit step function.
  • Z^-1[1/(z-a)] = a^(n-1) u[n-1]
  • Z-1[1/(z-1)2] = n(1)^(n-1) u[n-1] = nu[n-1]

Applying these transforms, we get:

  • Z^-1[2/(z-2)] = 2 * 2^(n-1) u[n-1] = 2^n u[n-1]
  • Z^-1[-2/(z-1)] = -2 * (1)^(n-1) u[n-1] = -2 u[n-1]
  • Z-1[-1/(z-1)2] = -1 * n(1)^(n-1) u[n-1] = -n u[n-1]

Putting it all together, the inverse Z-transform of E(z) is:

e[n] = 2^n u[n-1] - 2 u[n-1] - n u[n-1]

And there you have it! We've successfully found the inverse Z-transform. The use of standard pairs, along with linearity property, allows us to systematically tackle each fraction.

Step 5: Simplifying the Result (Optional)

We have our answer, but we can often simplify it further. In this case, we can factor out the unit step function:

e[n] = (2^n - n - 2) u[n-1]

This is a more compact and readable form of our result. The unit step function, u[n-1], indicates that this sequence starts at n = 1 (since u[n-1] = 0 for n < 1 and u[n-1] = 1 for n >= 1). Simplification is not always necessary, but it can often provide a clearer picture of the behavior of the discrete-time signal. The simplified expression gives insight into the signal's behavior as a function of n.

Final Answer and Conclusion

So, the inverse Z-transform of E(z) = z/((z-2)(z-1)^2) is:

e[n] = (2^n - n - 2) u[n-1]

Woohoo! We made it! We successfully navigated the world of partial fractions, repeated factors, and Z-transform pairs. This problem highlights the power of partial fraction decomposition in simplifying complex expressions and making them amenable to inverse Z-transformation. Remember, the key is to break down the problem into manageable steps, use the right techniques, and double-check your work along the way. Keep practicing, and you'll become a Z-transform master in no time! Remember, guys, the more you practice, the easier these transformations will become!

Key Takeaways

  • Partial fraction decomposition is a powerful tool for inverting Z-transforms of rational functions.
  • Repeated factors in the denominator require special attention in the partial fraction setup.
  • Matching terms to standard Z-transform pairs is crucial for successful inversion.
  • Always double-check your coefficients and final result.

This journey through the inverse Z-transform illustrates the step-by-step approach necessary for complex signal processing problems. The process embodies a general strategy of breaking problems down into simpler parts, addressing each part methodically, and then reassembling the solution. Remember that practice makes perfect in mastering these signal transformation techniques.