#13 Circuit Theory-2 #13 | The Laplace Transform - Definition, Step, Impulse

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. From phasors to transients

   Corrected solution notes. The audio is the original recording and may contain errors or incomplete phrases. Where they differ, use the corrected written solution.

   Single-frequency phasors describe sinusoidal steady state. Laplace methods also represent switching transients and initial-state contributions for suitable linear models.

   Narration transcript

   For twelve lessons we worked in AC steady state -- single frequency, sinusoidal sources, balanced phasors. We solved circuits algebraically using impedance, but only when nothing was switching, only when transients had died out. Today we step into a more powerful tool that handles transients and steady state at the same time. The Laplace transform.

2. 2. Transform differential equations

   Derivatives become algebraic factors in s together with initial-condition terms.

   Solve the transformed equation and invert it; do not drop stored-energy terms unless the problem states zero initial conditions.

   Narration transcript

   Differential equations are hard. Algebraic equations are easy. The Laplace transform takes a function of time and produces a function of a complex variable s, in such a way that differentiation in time becomes multiplication by s in the new domain. A second-order RLC circuit equation that you would solve with characteristic roots and integrating factors, becomes a polynomial equation in s. You solve it, then transform back. The whole hard problem is replaced by algebra.

3. 3. One-sided definition

   Use the one-sided transform with initial-time convention stated explicitly. For ordinary causal functions, integrate f(t)exp(−st) from zero to infinity.

   When including an impulse at the origin and pre-switch initial values, use the 0− convention so its full weight is included.

   Narration transcript

   Here is the definition. The one-sided Laplace transform of f(t) is the integral, from zero to infinity, of f(t) times e to the minus s t, dt. Three things to notice. The lower limit is zero, so we are looking at signals that start at t equals zero. The kernel is e to the minus s t, an exponential probe. And s itself is a complex number -- which is what makes Laplace strictly more powerful than Fourier.

4. 4. Complex variable and convergence

   s=σ+jω, giving kernel exp(−σt)exp(−jωt).

   Positive σ adds decay, while negative σ adds growth. Setting σ=0 gives a Fourier transform only when the imaginary axis lies in the appropriate convergence domain or a distributional interpretation is supplied.

   Narration transcript

   Let us split s. Write s as sigma plus j omega. Then e to the minus s t equals e to the minus sigma t, times the complex exponential e to the minus j omega t, which is cosine omega t minus j sine omega t. So the Laplace kernel is a damped sinusoid -- sigma controls how fast it decays, omega controls how fast it oscillates. When sigma is zero, e to the minus s t reduces to the Fourier kernel. So Fourier is the special case sigma equals zero. Laplace contains Fourier and goes further.

5. 5. Unit step

   The causal unit step transforms to 1/s for Re(s)>0.

   Evaluate the upper limit only within this region of convergence.

   Narration transcript

   Time for our first transform pair. Take f(t) equals u(t), the unit step. The integral becomes integral from zero to infinity of one times e to the minus s t, dt. The antiderivative is minus one over s, e to the minus s t. Evaluated from zero to infinity: at infinity, the term goes to zero whenever the real part of s is greater than zero. At zero, it equals minus one over s. The result: u(t) transforms to one over s. The condition real part of s greater than zero is called the region of convergence -- the values of s for which the integral actually exists.

6. 6. Impulse

   With the full-origin-impulse convention, δ(t) transforms to one.

   Hence the zero-state impulse output of an LTI system has transform equal to its transfer function.

   Narration transcript

   The impulse pair is even simpler. Take f(t) equals delta of t, the unit impulse at the origin. By the sifting property, the integral of delta of t times any function evaluates that function at t equals zero. So the Laplace integral collapses to e to the minus s times zero, which is one. The transform of the impulse is just the constant one. This explains why impulse responses are so important -- in the s-domain, they are the transfer function itself.

7. 7. Correct decay parameter

   exp(−at)u(t) transforms to 1/(s+a), with Re(s)>−a for real a.

   For positive a, a is a decay rate in inverse seconds. The time constant is τ=1/a, not a.

   Narration transcript

   One more building block. Take f(t) equals e to the minus a t, u(t), a decaying exponential. Plug it into the integral. We get integral from zero to infinity of e to the minus s plus a, times t, dt. This is the same shape as the unit step integral, with s replaced by s plus a. The result: one over s plus a. Notice how the time constant a in the time domain becomes a shift of the s variable in the s-domain. The region of convergence is real part of s greater than minus a.

8. 8. Sinusoidal pairs

   cos(ωt)u(t) transforms to s/(s²+ω²); sin(ωt)u(t) transforms to ω/(s²+ω²).

   For nonzero real frequency, the causal integral formulas have region Re(s)>0.

   Narration transcript

   Two more pairs that we will use constantly, derived in one step with Euler's formula. Cosine omega t equals one half of e to the j omega t plus e to the minus j omega t. Each piece is just an exponential, so each transforms with the previous rule. Add them up: cosine omega t u(t) transforms to s over s squared plus omega squared. Sine omega t transforms to omega over s squared plus omega squared. Memorize this pattern -- it shows up everywhere.

9. 9. Basic table

   Impulse: one; step: 1/s; ramp: 1/s²; causal decaying exponential: 1/(s+a).

   Use the stated causal convention and convergence region with the sine and cosine pairs as well.

   Narration transcript

   Here is the running table. Delta of t goes to one. u(t) goes to one over s. e to the minus a t goes to one over s plus a. t u(t) goes to one over s squared. Cosine omega t goes to s over s squared plus omega squared. Sine omega t goes to omega over s squared plus omega squared. Six pairs. Combined with the shift, scaling, and differentiation rules from the next lessons, these cover almost every signal you will meet in circuit analysis.

10. 10. Kernel example

    For f(t)=exp(−0.3t)u(t) at s=0.5+j2, the integrand is exp(−0.8t)exp(−j2t).

    Its integral is 1/(0.8+j2). One complex sample evaluates the transform at that s; it does not alone identify the full signal.

    Narration transcript

    Watch this. Here is f(t) equals e to the minus zero point three t. As we multiply by the kernel e to the minus s t with s equal to zero point five plus j two, the curve unwinds out of the page into a 3D spiral. The x axis is time. The y axis is the real part of the integrand. The z axis is the imaginary part. The combined decay sigma plus zero point three pulls the spiral toward zero, and omega twists it around the axis. The Laplace transform F(s) is just the integral of this spiral, projected down to a single complex number. That number captures the entire signal at this one point in the s-plane.

11. 11. Review

    Keep initial conditions, convergence and the initial-time convention visible.

    A positive exponential rate a has time constant 1/a. The transform converts the specified differential problem into algebra without changing its physical assumptions.

    Narration transcript

    Three takeaways. The Laplace transform converts differential equations into algebra by mapping time domain functions to functions of a complex variable s. The kernel e to the minus s t is a damped sinusoid, generalizing the Fourier kernel by adding decay. The fundamental pairs we need are u(t) goes to one over s, delta of t goes to one, and e to the minus a t goes to one over s plus a. Next lesson we add the time-shift, differentiation, and integration properties.

Source video: Circuit Theory-2 #13 | The Laplace Transform - Definition, Step, Impulse (7:22)