#14 Circuit Theory-2 #14 | Laplace Transform Properties - Shift, Differentiation, Integration

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

1. 1. Manipulation rules

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

   Combine basic Laplace pairs with time shift, s shift, differentiation and integration.

   Keep the one-sided initial-time convention and convergence conditions consistent.

   Narration transcript

   Last lesson we defined the Laplace transform and built three fundamental pairs: 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. Today we add the manipulation tools — shifting in time, shifting in frequency, differentiation, and integration. With these in hand, any signal you can build from the table can be transformed by inspection.

2. 2. Correct delay magnitude

   For delay a≥0, f(t−a)u(t−a) transforms to exp(−as)F(s).

   With s=σ+jω, the delay factor has magnitude exp(−aσ) and phase −aω.

   Magnitude is unchanged on the imaginary axis σ=0, not at arbitrary complex s.

   Narration transcript

   First, the time-shift theorem. Suppose we have a signal f(t) with transform F(s), and we delay it by a seconds: f of t minus a, multiplied by u of t minus a so it stays causal. Set up the Laplace integral. The unit step kills everything below t equals a, so the lower limit becomes a. Substitute tau equals t minus a, and the integral splits into e to the minus a s, times the ordinary Laplace integral of f. The conclusion: a delay of a seconds in time becomes multiplication by e to the minus a s in the s-domain. The magnitude is unchanged, only the phase shifts.

3. 3. Exponential weighting

   Multiplying f(t) by exp(−at) replaces F(s) by F(s+a).

   Shift the convergence region accordingly; distinguish this from delaying the signal.

   Narration transcript

   Now the frequency-shift theorem, also called s-shift. If we multiply f(t) by e to the minus a t in the time domain, what happens in the s-domain? Plug it into the integral and combine the two exponentials: e to the minus a t, times e to the minus s t, equals e to the minus s plus a, times t. So the integral has the same shape as the original transform, with s replaced by s plus a. The result: multiplying by an exponential in time shifts the s variable by a. This is the symmetric mirror of the time-shift theorem.

4. 4. Initial conditions in derivatives

   The one-sided first derivative gives sF(s)−f(0−), under the full-origin convention.

   The second derivative gives s²F(s)−s f(0−)−f′(0−). Use a matching ordinary 0+ convention when no origin impulse is included.

   Narration transcript

   Differentiation in time. This is the property that turns differential equations into algebra. Take the Laplace integral of f prime of t, and apply integration by parts: u equals e to the minus s t, dv equals f prime of t dt. Then du equals minus s e to the minus s t dt, and v equals f of t. The boundary term evaluates to minus f of zero plus zero at infinity (because Re s greater than the abscissa of convergence). The remaining integral is just s times F of s. Putting it together: the Laplace transform of f prime equals s F of s minus f of zero. For the second derivative, apply the rule twice: s squared F of s, minus s f of zero, minus f prime of zero. Each derivative becomes a power of s, plus the initial conditions.

5. 5. Integration

   Integrating f from zero to t with zero integration constant gives F(s)/s in its convergence domain.

   A nonzero integration constant contributes a separate constant/s term.

   Narration transcript

   Integration in time. If we integrate f from zero to t, what happens in the s-domain? The mirror of differentiation: instead of multiplying by s, we divide by s. The Laplace transform of the integral of f from zero to t equals F of s over s. The proof swaps the order of integration in the two-variable integral and splits naturally. The takeaway is symmetric: derivatives correspond to multiplying by s, integrals correspond to dividing by s. This is exactly why circuits with capacitors and inductors collapse into algebraic networks in the s-domain.

6. 6. Multiplication by time

   Multiplying by t gives −dF/ds where differentiation under the integral is justified.

   Applying it to 1/s yields the ramp transform 1/s².

   Narration transcript

   One more useful property. Multiplying f of t by t in the time domain corresponds to taking minus the derivative of F with respect to s in the s-domain. As a quick application: t times u of t goes to minus the derivative of one over s, which is one over s squared. We just derived another transform pair without doing any integral.

7. 7. Property summary

   Delay: exp(−as)F(s); exponential weighting: F(s+a); derivatives include initial values; integration divides by s.

   For positive a, time scaling f(at) gives F(s/a)/a. Higher derivatives and repeated integration require the corresponding complete conditions.

   Narration transcript

   Here is the running properties table for quick reference. Time shift adds e to the minus a s. Frequency shift replaces s with s plus a. The first derivative becomes s F of s minus f of zero. The second derivative adds an s f of zero and an f prime of zero correction. Integration divides by s. Multiplication by t differentiates with respect to s. Time scaling f of a t maps to one over a, F of s over a. These eight rules plus the six pairs from last lesson cover essentially every signal in linear circuit analysis.

8. 8. Worked initial-value problem

   For y′+2y=0 and y(0)=1, transformation gives (s+2)Y=1.

   Thus Y=1/(s+2) and y(t)=exp(−2t) for t≥0. Its initial value and differential equation both check directly.

   Narration transcript

   Time for a small but complete example showing why this matters. Solve y prime of t plus two y of t equals zero, with y of zero equals one. Take the Laplace transform of both sides. The differentiation rule gives s Y of s minus one for the y prime term, plus two Y of s. Group: s plus two, times Y of s, equals one. Algebra: Y of s equals one over s plus two. Recognize this as the exponential pair from last lesson, so y of t equals e to the minus two t. We just solved a differential equation by doing pure algebra in the s-domain. Next lesson we will handle more complex right-hand sides using partial fractions.

9. 9. Review

   A delay changes Laplace magnitude away from the imaginary axis.

   Preserve signs of initial-condition terms and distinguish delay, exponential weighting and time scaling.

   Narration transcript

   Three takeaways. Time-shift maps to e to the minus a s, and frequency-shift replaces s with s plus a — the symmetric pair. Differentiation in time multiplies by s and pulls in initial conditions, integration in time divides by s. With these manipulation rules plus the six basic pairs, you can transform almost any signal in your head. Next lesson we go in the other direction: inverse Laplace and partial fractions.

Source video: Circuit Theory-2 #14 | Laplace Transform Properties - Shift, Differentiation, Integration (6:46)