Signals and Systems

9.3 The Inverse Laplace Transform

Imron Rosyadi

The Inverse Laplace Transform

Recovering Signals from the s-Domain

Recap: From Time to s-Domain

The Laplace Transform \(X(s)\) of a signal \(x(t)\) is defined as:

\[ X(s) = \int_{-\infty}^{+\infty} x(t) e^{-st} dt \]

Where \(s = \sigma + j\omega\). This means \(X(s)\) can be interpreted as the Fourier Transform of an exponentially weighted signal:

\[ X(\sigma+j \omega)=\mathcal{F}\left\{x(t) e^{-\sigma t}\right\}=\int_{-\infty}^{+\infty} x(t) e^{-\sigma t} e^{-j \omega t} d t \tag{9.53} \]

Note

The Region of Convergence (ROC) for \(X(s)\) is crucial. It defines the range of \(\sigma\) values for which the integral converges.

Remind students that the Laplace transform extends the Fourier transform by allowing for exponentially growing or decaying signals. The real part of ‘s’, sigma, acts as a damping factor, ensuring the integral converges. Emphasize that knowing just \(X(s)\) algebraically is not enough; the ROC is equally important for defining the transform.

The Inverse Laplace Transform Formula

We can recover \(x(t)\) from \(X(s)\) using the inverse Fourier transform relationship:

\[ x(t) e^{-\sigma t}=\mathcal{F}^{-1}\{X(\sigma+j \omega)\}=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} X(\sigma+j \omega) e^{j \omega t} d \omega \tag{9.54} \]

Multiplying by \(e^{\sigma t}\), we get:

\[ x(t)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} X(\sigma+j \omega) e^{(\sigma+j \omega) t} d \omega \tag{9.55} \]

By changing the variable of integration from \(\omega\) to \(s\) (\(ds = j d\omega\) since \(\sigma\) is constant), we obtain the fundamental inverse Laplace transform equation:

\[ x(t)=\frac{1}{2 \pi j} \int_{\sigma-j \infty}^{\sigma+j \infty} X(s) e^{s t} d s \tag{9.56} \]

Warning

The integral in Eq. (9.56) is a contour integral in the complex \(s\)-plane. For most ECE undergraduate courses, direct evaluation is often complex and solved using alternative methods.

Explain that the integration path is a vertical line in the s-plane, where the real part \(\sigma\) is constant and within the ROC. This formula is powerful but computationally intensive. We will focus on a more practical approach for rational functions.

Practical Inversion: Partial Fraction Expansion (PFE)

For rational Laplace Transforms, \(X(s) = P(s)/Q(s)\), we typically use Partial Fraction Expansion (PFE). This method avoids direct evaluation of the complex integral.

The procedure involves:

1. Factorizing the denominator polynomial \(Q(s)\).
2. Expanding \(X(s)\) into a sum of simpler terms (e.g., first-order terms).
3. Identifying the inverse Laplace transform for each term, considering its Region of Convergence (ROC).

For distinct poles, \(X(s)\) can be expanded as:

\[ X(s)=\sum_{i=1}^{m} \frac{A_{i}}{s+a_{i}} \tag{9.57} \]

Important

The ROC of the original \(X(s)\) is vital for correctly determining the inverse transform of each partial fraction term!

Practical Inversion: Partial Fraction Expansion (PFE)

Example Structure:

Given \(X(s) = \frac{s+3}{(s+1)(s+2)}\)

1. Factorized (already done)

2. PFE: \(X(s) = \frac{A}{s+1} + \frac{B}{s+2}\)

3. Solve for \(A\) and \(B\). \(A = \left.[(s+1)X(s)]\right|_{s=-1}\) \(B = \left.[(s+2)X(s)]\right|_{s=-2}\)

   Then, based on ROC, find \(x(t)\).

Highlight that PFE simplifies the complex \(X(s)\) into a sum of basic forms whose inverse transforms are known. The key challenge for students is understanding how the overall ROC applies to each individual term to determine its time-domain characteristic (right-sided, left-sided).

Inverse Transform of Basic Terms

The inverse Laplace transform of a first-order term \(\frac{1}{s+a}\) depends entirely on its ROC.

Case 1: Right-Sided Signal

If the ROC is to the right of the pole \(s=-a\) (\(\operatorname{Re}\{s\} > -a\)):

\[ \frac{1}{s+a} \stackrel{\mathcal{L}^{-1}}{\longleftrightarrow} e^{-at}u(t) \]

This corresponds to a causal or right-sided exponential.

Case 2: Left-Sided Signal

If the ROC is to the left of the pole \(s=-a\) (\(\operatorname{Re}\{s\} < -a\)):

\[ \frac{1}{s+a} \stackrel{\mathcal{L}^{-1}}{\longleftrightarrow} -e^{-at}u(-t) \]

This corresponds to an anti-causal or left-sided exponential.

Tip

Remember: The ROC for a right-sided signal is always to the right of its rightmost pole. The ROC for a left-sided signal is always to the left of its leftmost pole. For a two-sided signal, the ROC is a strip between poles.

Visually differentiate between the two cases. Emphasize the sign change for the left-sided signal and the role of \(u(t)\) and \(u(-t)\). This is a fundamental concept for inverse Laplace transforms.

Example 9.9: Right-Sided Signal

Let’s find the inverse Laplace transform of:

\[ X(s)=\frac{1}{(s+1)(s+2)}, \quad \operatorname{Re}\{s\}>-1 \tag{9.58} \]

1. Partial-Fraction Expansion: \(X(s)=\frac{A}{s+1}+\frac{B}{s+2}\)

   Using the cover-up method: \(A = \left.[(s+1)X(s)]\right|_{s=-1} = \frac{1}{-1+2} = 1\) \(B = \left.[(s+2)X(s)]\right|_{s=-2} = \frac{1}{-2+1} = -1\)

   So, \(X(s)=\frac{1}{s+1}-\frac{1}{s+2} \tag{9.62}\)

2. Apply ROC to terms: The overall ROC is \(\operatorname{Re}\{s\}>-1\). This is to the right of both poles (\(s=-1\) and \(s=-2\)). Therefore, both terms correspond to right-sided signals.

   • For \(\frac{1}{s+1}\): ROC is \(\operatorname{Re}\{s\}>-1 \implies e^{-t}u(t)\)
   • For \(\frac{1}{s+2}\): ROC is \(\operatorname{Re}\{s\}>-2 \implies -e^{-2t}u(t)\)

Example 9.9: Right-Sided Signal (cont.)

3. Combine the terms:

   \[ x(t) = (e^{-t} - e^{-2t})u(t) \tag{9.65} \]

Walk through each step clearly. Explain why both terms are right-sided based on the overall ROC extending to the right of all poles.

Example 9.9: Pole-Zero Plot and ROC

The pole-zero plot for \(X(s)=\frac{1}{(s+1)(s+2)}\) with ROC \(\operatorname{Re}\{s\}>-1\):

Interpretation of ROC:

The overall ROC, \(\operatorname{Re}\{s\}>-1\), is to the right of both poles.

• For the term \(\frac{1}{s+1}\): its ROC is \(\operatorname{Re}\{s\}>-1\). This implies a right-sided signal: \(e^{-t}u(t)\).

• For the term \(\frac{1}{s+2}\): its ROC is \(\operatorname{Re}\{s\}>-1\). Since this is to the right of its pole at \(-2\), this also implies a right-sided signal: \(-e^{-2t}u(t)\).

Thus, both terms contribute right-sided signals.

Explain that for the individual terms, their ROCs are determined by the overall ROC. If the overall ROC is to the right of all poles, then each term’s ROC must also be to the right of its respective pole.

Example 9.10: Left-Sided Signal

Consider the same \(X(s)\), but with a different ROC:

\[ X(s)=\frac{1}{(s+1)(s+2)}, \quad \operatorname{Re}\{s\}<-2 \]

1. Partial-Fraction Expansion: Still \(X(s)=\frac{1}{s+1}-\frac{1}{s+2}\)

2. Apply ROC to terms: The overall ROC is \(\operatorname{Re}\{s\}<-2\). This is to the left of both poles (\(s=-1\) and \(s=-2\)). Therefore, both terms correspond to left-sided signals.

   • For \(\frac{1}{s+1}\): The ROC \(\operatorname{Re}\{s\}<-1\) (consistent with overall ROC) implies \(-e^{-t}u(-t)\).
   • For \(\frac{1}{s+2}\): The ROC \(\operatorname{Re}\{s\}<-2\) implies \(-(-e^{-2t}u(-t)) = e^{-2t}u(-t)\) (due to the negative sign in PFE).

3. Combine the terms:

   \[ x(t) = [-e^{-t} + e^{-2t}]u(-t) \tag{9.68} \]

Emphasize how the change in ROC drastically changes the time-domain signal. Point out the sign manipulation for the second term due to the PFE and the left-sided rule.

Example 9.11: Two-Sided Signal

Once again, the same \(X(s)\), but with a ROC between the poles:

\[ X(s)=\frac{1}{(s+1)(s+2)}, \quad -2 < \operatorname{Re}\{s\} < -1 \]

1. Partial-Fraction Expansion: Still \(X(s)=\frac{1}{s+1}-\frac{1}{s+2}\)

2. Apply ROC to terms: The overall ROC is \(-2 < \operatorname{Re}\{s\} < -1\).

   • For the pole at \(s=-1\): The ROC \(\operatorname{Re}\{s\} < -1\) (consistent with overall ROC) means it’s to the left of this pole. \(\implies \frac{1}{s+1} \longleftrightarrow -e^{-t}u(-t)\)

   • For the pole at \(s=-2\): The ROC \(\operatorname{Re}\{s\} > -2\) (consistent with overall ROC) means it’s to the right of this pole. \(\implies \frac{1}{s+2} \longleftrightarrow e^{-2t}u(t)\)

3. Combine the terms:

   \[ x(t) = -e^{-t}u(-t) - e^{-2t}u(t) \tag{9.69} \]

This example shows a two-sided signal, which results when the ROC lies between poles. Explain how each term’s ROC is determined by its relation to the overall ROC and its own pole location.

Interactive Demo: Partial Fraction Coefficients

Let’s calculate the coefficients for a simple rational function: \(X(s) = \frac{1}{(s+a)(s+b)} = \frac{A}{s+a} + \frac{B}{s+b}\)

Input values for \(a\) and \(b\) (assume \(a \neq b\)).

viewof pole_a = Inputs.number({value: 1, min: -5, max: 5, step: 0.1, label: "Pole a (for s+a)"});
viewof pole_b = Inputs.number({value: 2, min: -5, max: 5, step: 0.1, label: "Pole b (for s+b)"});

Explain the cover-up method for finding A and B. Encourage students to try different values and observe the coefficients. This provides immediate feedback and reinforces the PFE calculation. Point out the potential for errors if poles are not distinct.

Interactive Demo: Visualizing Inverse Laplace Transforms

Explore how ROC changes the time-domain signal. For \(X(s) = \frac{1}{(s+1)(s+2)}\), the PFE is \(\frac{1}{s+1} - \frac{1}{s+2}\).

Adjust the ROC boundaries below and observe the resulting \(x(t)\).

viewof sigma_low = Inputs.number({value: -2.5, min: -3.5, max: 0.5, step: 0.1, label: "ROC Lower Bound (e.g., -inf = -3.5)"});
viewof sigma_high = Inputs.number({value: -0.5, min: -3.5, max: 0.5, step: 0.1, label: "ROC Upper Bound (e.g., +inf = 0.5)"});

Tip

Poles at: \(s_1 = -1\), \(s_2 = -2\). - Try setting ROC: (-0.5 < Re{s}) for Example 9.9 (Right-sided). - Try setting ROC: (Re{s} < -2.5) for Example 9.10 (Left-sided). - Try setting ROC: (-2.0 < Re{s} < -1.0) for Example 9.11 (Two-sided).

Explain the logic in the Python code: how sigma_low and sigma_high determine if each pole’s contribution is right-sided or left-sided. Encourage students to set the sliders to match the ROCs from Examples 9.9, 9.10, and 9.11 to see the corresponding signals. This is a powerful visualization of the ROC’s importance. Mention that the sigma_low and sigma_high inputs simulate the boundaries of the ROC. For an “infinite” ROC, students should set it to the max/min of the slider range.