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10.03 Initial & Final Value Theorems in the Z-Domain
Welcome back! Today, we are going to master one of the most reliable scoring sections of Chapter 10 under Instructor 1: Initial & Final Value Theorems (IVT & FVT) in the Z-Domain.
Just like their continuous-time Laplace counterparts, these discrete boundary theorems allow us to calculate the immediate initial sample and the long-term steady-state final value directly from the -domain algebraic expression [8.66, 8.132]. This completely bypasses the tedious process of executing partial fractions or long division to invert the function [8.90]!
In KUET examinations, this topic routinely commands 5 to 15 marks. Let’s walk through the strict mathematical proofs, map out the s-plane/z-plane stability boundaries, and solve the classic exam-killer problems step-by-step to lock in maximum marks!
1. The Initial Value Theorem (IVT)
The Initial Value Theorem determines the first term of a causal discrete sequence directly from its -domain representation [8.88].
1.1 Theorem Statement
If a sequence is causal {meaning for all }, and its Z-transform converges for [8.88], then:
1.2 Mathematically Rigorous Proof [5 Marks]
To prove this theorem, we start with the fundamental definition of the unilateral Z-transform for a causal sequence [8.128, 8.88]:
Expanding this power series term-by-term [8.88]: X(z) = x[0] + rac{x[1]}{z} + rac{x[2]}{z^2} + rac{x[3]}{z^3} + \dots
Now, let’s evaluate the limit of both sides as [8.88]:
ight]$$ Since the sequence terms $x[n]$ are finite values, the limit of each rational term with $z$ in the denominator approaches zero: $$\lim_{z o \infty} rac{x[n]}{z^n} = 0 \quad ext{for all } n \ge 1$$ Substituting these zero limits back into the expanded series: $$\lim_{z o \infty} X(z) = x[0] + 0 + 0 + 0 + \dots$$ $$\mathbf{x[0] = \lim_{z o \infty} X(z)} \quad \mathbf{[ ext{Q.E.D.}]}$$ --- ## 2. The Final Value Theorem (FVT) The **Final Value Theorem** evaluates the steady-state, long-term behavior of a causal sequence as discrete time approaches infinity ($n o \infty$) [8.132, 151]. ### 2.1 Theorem Statement If a sequence $x[n]$ is causal, and all poles of its scaled transform $(z-1)X(z)$ lie strictly **inside the unit circle** in the $z$-plane [8.132], then the steady-state value $x[\infty]$ is given by: $$\mathbf{x[\infty] = \lim_{n o \infty} x[n] = \lim_{z o 1} \left(1 - z^{-1} ight) X(z) = \lim_{z o 1} rac{z-1}{z} X(z)}$$ --- ### 2.2 Mathematically Rigorous Proof [8-10 Marks] To prove FVT, we analyze the Z-transform of a first-difference sequence $d[n] = x[n] - x[n-1]$ [8.131]: $$\mathcal{Z}\left\{ x[n] - x[n-1] ight\} = \sum_{n=0}^{\infty} \left( x[n] - x[n-1] ight) z^{-n}$$ Using the **Linearity** [8.21] and **Time-Shifting (Delay)** [8.27] properties of unilateral transforms: $$\mathcal{Z}\left\{ x[n] - x[n-1] ight\} = X(z) - z^{-1} X(z) = \left( 1 - z^{-1} ight) X(z)$$ Now, let's equate this algebraic expression back to the expanded summation definition: $$\sum_{n=0}^{\infty} \left( x[n] - x[n-1] ight) z^{-n} = \left( 1 - z^{-1} ight) X(z)$$ Let's evaluate the limit of both sides as $z o 1$: $$\lim_{z o 1} \sum_{n=0}^{\infty} \left( x[n] - x[n-1] ight) z^{-n} = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)$$ Because $\lim_{z o 1} z^{-n} = 1^{-n} = 1$ for all finite $n$, we can substitute $z = 1$ directly into the summation terms: $$\sum_{n=0}^{\infty} \left( x[n] - x[n-1] ight) = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)$$ The infinite summation on the left is a **telescoping series**. Let's write it as the limit of its partial sums as $N o \infty$: $$\lim_{N o \infty} \sum_{n=0}^{N} \left( x[n] - x[n-1] ight) = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)$$ Let's expand this partial sum explicitly: $$\sum_{n=0}^{N} \left( x[n] - x[n-1] ight) = (x[0] - x[-1]) + (x[1] - x[0]) + (x[2] - x[1]) + \dots + (x[N] - x[N-1])$$ Notice how all intermediate terms cancel out perfectly {telescoping behavior}: * $x[0]$ cancels with $-x[0]$ * $x[1]$ cancels with $-x[1]$ * $x[N-1]$ cancels with $-x[N-1]$ This leaves only the boundary elements: $$\sum_{n=0}^{N} \left( x[n] - x[n-1] ight) = x[N] - x[-1]$$ Since the sequence $x[n]$ is causal, the initial boundary condition is strictly zero: $x[-1] = 0$. Substituting this back: $$\lim_{N o \infty} x[N] = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)$$ By definition, $\lim_{N o \infty} x[N] = x[\infty]$. Therefore: $$\mathbf{x[\infty] = \lim_{z o 1} \left( 1 - z^{-1} ight) X(z)} \quad \mathbf{[ ext{Q.E.D.}]}$$ --- ## 3. The Crucial Pole-Stability Boundary Criterion In continuous-time Laplace analysis, the Final Value Theorem is bounded by the imaginary $j\Omega$-axis. In the discrete $z$-domain, **FVT is strictly bounded by the Unit Circle ($|z| = 1$)**. ``` Complex z-Plane Stability Boundaries Im(z) ▲ │ Unstable Region ┌───┴───┐ (|z| > 1) ┌──┘ │││ └──┐ ┌─┘ │││ └─┐ │ Stable Region │ ──────┼───────┼─────────┼──────► Re(z) │ (|z| < 1) │ Pole at z = 1 └─┐ │││ ┌─┘ Allowed for FVT └──┐ │││ ┌──┘ └───┼───┘ │ ``` > [!CAUTION] > > **Strict Mathematical Prerequisites for FVT:** > The Final Value Theorem is valid **if and only if** the limit $x[\infty]$ exists as a single, finite steady-state value [8.132]. This imposes three absolute rules on the poles of $X(z)$: > 1. **Stable Open Region:** All poles of $X(z)$ must lie strictly **inside** the unit circle ($|p_i| < 1$). > 2. **Stable Step Offset:** A single, non-repeated pole is permitted exactly **on** the unit circle at $z = 1$ {representing a constant step DC level like $u[n]$} [8.131]. > 3. **The Instability Trap:** If any poles of $X(z)$ lie **on** the unit circle other than $z = 1$ {e.g., oscillatory poles at $z = e^{\pm j\omega_0}$ like sines/cosines} or **outside** the unit circle ($|p_i| > 1$ {representing growing exponentials}), **FVT fails completely**! The algebraic limit will yield a meaningless number. --- ### Proof of FVT Failure on Oscillatory Signals Let $x[n] = \cos\left(n rac{\pi}{2} ight) u[n] = \{1, 0, -1, 0, 1, 0, -1, 0, \dots\}$. The sequence oscillates forever and has no steady-state limit. Its Z-transform is: $$X(z) = rac{z^2}{z^2 + 1} \quad ext{with poles at } p_{1,2} = \pm j \quad (|p_i| = 1)$$ If we blindly apply FVT without checking poles: $$x[\infty] = \lim_{z o 1} \left(1 - z^{-1} ight) rac{z^2}{z^2 + 1} = \lim_{z o 1} \left( rac{z-1}{z} ight) rac{z^2}{z^2 + 1} = rac{(0) \cdot 1}{1 + 1} = 0$$ This algebraic result ($0$) is completely false because the physical signal never decays to zero! This is why conducting a **Pole Stability Audit** is your mandatory first step on exam questions to avoid losing marks. --- ## 4. High-Yield Solved "Exam Killers" Let's solve the most heavily tested, recurring KUET past year questions with complete exam-scoring layout. ### 4.1 The Finite-Duration Sequence [5 Marks] > [!question] **KUET 2025 Fig 16(c) / 2022 / 2020 / 2019** > > If the Z-transform of a causal sequence $x[n]$ is given by: > $$X(z) = 2 + 3z^{-1} + 4z^{-2}$$ > Find the initial value $x[0]$ and the final value $x[\infty]$ of the corresponding sequence. #### Step 1: Calculate the Initial Value $x[0]$ Apply the Z-domain Initial Value Theorem [8.132]: $$x[0] = \lim_{z o \infty} X(z)$$ $$x[0] = \lim_{z o \infty} \left[ 2 + 3z^{-1} + 4z^{-2} ight] = \lim_{z o \infty} \left[ 2 + rac{3}{z} + rac{4}{z^2} ight]$$ Since $ rac{3}{\infty} o 0$ and $ rac{4}{\infty^2} o 0$: $$\mathbf{x[0] = 2} \quad \mathbf{[ ext{Verified}]}$$ --- #### Step 2: Conduct the Pole Stability Audit To apply FVT, we must verify pole stability [8.132]. Let's rewrite $X(z)$ in positive powers of $z$: $$X(z) = rac{2z^2 + 3z + 4}{z^2}$$ The denominator roots yield a **double pole at the origin**: $$z^2 = 0 \implies p_{1,2} = 0$$ Since the pole magnitudes $|0| = 0 < 1$ lie strictly inside the unit circle, the system is absolutely stable. The steady-state value converges, making FVT strictly valid! --- #### Step 3: Calculate the Final Value $x[\infty]$ Apply the Z-domain Final Value Theorem [8.132]: $$x[\infty] = \lim_{z o 1} \left(1 - z^{-1} ight) X(z)$$ $$x[\infty] = \lim_{z o 1} \left(1 - rac{1}{z} ight) \left[ 2 + rac{3}{z} + rac{4}{z^2} ight]$$ Substitute $z = 1$: $$x[\infty] = \left(1 - rac{1}{1} ight) \left[ 2 + rac{3}{1} + rac{4}{1} ight] = (0) \cdot (9)$$ $$\mathbf{x[\infty] = 0} \quad \mathbf{[ ext{Verified}]}$$ --- #### Step 4: Double-Domain Inspection Verification Let's verify our boundary limits by directly expanding $X(z)$ to obtain the discrete sequence $x[n]$ [8.106]: $$X(z) = 2z^0 + 3z^{-1} + 4z^{-2} \leftrightarrow x[n] = \{ \mathbf{2}, 3, 4, 0, 0, 0, \dots \}$$ * Initial element at $n = 0$ is $x[0] = 2$ {Matches IVT perfectly!}. * As $n o \infty$, the finite sequence has settled to $0$, so $x[\infty] = 0$ {Matches FVT perfectly!}. This complete cross-verification proves our analytical methods are 100% correct! --- ### 4.2 The Rational Fractional Proof [10 Marks] > [!question] **KUET 2024 / 2019 / 2015 (10-Mark Classic - Heavily Tested)** > > Prove that the final value of $x[n]$ for: > $$X(z) = rac{z^2}{(z-1)(z-0.2)}$$ > is $1.25$ and its initial value is unity. #### Step 1: Prove the Initial Value is Unity Apply the Z-domain Initial Value Theorem [8.132]: $$x[0] = \lim_{z o \infty} X(z) = \lim_{z o \infty} rac{z^2}{(z-1)(z-0.2)}$$ To evaluate this high-order limit, divide both the numerator and the denominator by the highest power of $z$, which is $z^2$: $$X(z) = rac{z^2}{z^2 - 1.2z + 0.2} = rac{1}{1 - 1.2z^{-1} + 0.2z^{-2}}$$ Now, apply the limit as $z o \infty$: $$x[0] = \lim_{z o \infty} rac{1}{1 - rac{1.2}{z} + rac{0.2}{z^2}}$$ $$x[0] = rac{1}{1 - 0 + 0} = 1$$ $$\mathbf{x[0] = 1 \quad [ ext{PROVED}]}$$ --- #### Step 2: Conduct the Pole Stability Audit We must isolate the roots of the denominator of $X(z)$ [8.132]: $$(z-1)(z-0.2) = 0 \implies p_1 = 1, \ p_2 = 0.2$$ * The pole $p_2 = 0.2$ lies strictly inside the unit circle ($|0.2| = 0.2 < 1$). * The pole $p_1 = 1$ is a simple, non-repeated pole located exactly on the unit circle. * Since all poles are within $|z| \le 1$ with no multiple or oscillatory poles on the boundary, the sequence is stable and settles to a constant DC value. FVT is strictly valid! --- #### Step 3: Prove the Final Value is 1.25 Apply the Z-domain Final Value Theorem [8.132]: $$x[\infty] = \lim_{z o 1} \left(1 - z^{-1} ight) X(z)$$ $$x[\infty] = \lim_{z o 1} \left( rac{z-1}{z} ight) \left[ rac{z^2}{(z-1)(z-0.2)} ight]$$ Let's simplify this algebraic expression before taking the limit: * Cancel out the $(z-1)$ term in both the numerator and denominator: $$x[\infty] = \lim_{z o 1} \left( rac{1}{z} ight) \left[ rac{z^2}{z-0.2} ight]$$ * Simplify the $z$ terms: $$x[\infty] = \lim_{z o 1} rac{z}{z - 0.2}$$ Now substitute the boundary limit $z = 1$ directly: $$x[\infty] = rac{1}{1 - 0.2} = rac{1}{0.8}$$ $$x[\infty] = rac{10}{8} = 1.25$$ $$\mathbf{x[\infty] = 1.25 \quad [ ext{PROVED}]}$$ Both boundaries have been rigorously proven, securing a flawless 10/10 mark allocation! --- ## 5. Common Mistakes That Cost Marks > [!warning] **Critical Exam Pitfalls** > > 1. **Omitting the Pole Stability Check:** Applying FVT directly to an unstable transform. For example, if $X(z) = rac{z}{z-2}$, applying FVT blindly yields $\lim_{z o 1} (1-z^{-1}) rac{z}{z-2} = 0$. But the actual sequence is $x[n] = 2^n u[n]$ which grows exponentially to $\infty$. Omitting this stability audit will result in a **3 to 4 mark deduction**! > 2. **Confusing IVT Limits:** Attempting to evaluate the initial value of $x[n]$ by setting $z o 0$ instead of $z o \infty$. Remember, the high-frequency limit in the time-continuous domain ($s o \infty$) maps to the infinite radial exterior of the $z$-plane ($z o \infty$) due to $z = e^{s T_s}$ [8.131]. > 3. **Dropping the $(1-z^{-1})$ Multiplier in FVT:** Evaluating $\lim_{z o 1} X(z)$ instead of $\lim_{z o 1} (1-z^{-1}) X(z)$. If $X(z)$ has a pole at $z = 1$, evaluating $X(1)$ will divide by zero and explode, leaving you stuck in the middle of your exam calculus. --- ## 6. Verbatim PYQ Reference Bank * **KUET 2025 Q6(b):** If $X(z) = 2 + 3z^{-1} + 4z^{-2}$, find the initial and final values of the corresponding sequence, $x[n]$. **(15 Marks)** * **KUET 2024 Q7(b):** Prove that the final value of $x[n]$ for $X(z) = rac{z^2}{(z-1)(z-0.2)}$ is 1.25 and its initial value is unity. **(10 Marks)** * **KUET 2022 Q6(b):** If $X(z) = 2 + 3z^{-1} + 4z^{-2}$, find the initial and final values of the corresponding sequence $x[n]$. **(05 Marks)** * **KUET 2020 Q6(b):** Find the initial and final values of the sequence $x[n]$ if its Z-transform is $X(z) = 2 + 3z^{-1} + 4z^{-2}$. **(05 Marks)** * **KUET 2019 Q7(b):** Prove that the final value of $x[n]$ for $X(z) = rac{z^2}{(z-1)(z-0.2)}$ is 1.25 and its initial value is unity. **(07 Marks)** * **KUET 2015 Q7(b):** Prove that the final value of $x[n]$ for $X(z) = rac{z^2}{(z-1)(z-0.2)}$ is 1.25 and its initial value is unity. **(10 Marks)** --- ## 7. Interactive Self-Check Checklist - [ ] Can you state and write down the term-by-term proof of the Z-domain Initial Value Theorem? - [ ] Can you prove the discrete telescoping series cancellation that establishes the Final Value Theorem? - [ ] Do you know how to perform a Pole Stability Audit to check if FVT is mathematically valid? - [ ] Why does FVT fail for oscillatory sequences (e.g., unit circle poles other than $z = 1$)? - [ ] Can you solve the classic 10-mark $1.25$ proof without checking partial fractions? --- *Source: signals-and-systems textbook (K. Deergha Rao) Ch 8 [8.131] / Rabiul sir class notes (Lec-16) / KUET QB 24-15.*