Time Convolution and Multiplication Properties of Laplace Transform

---

---

Laplace Transform

The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or s-domain.

Mathematically, if x(t) is a time domain function, then its Laplace transform is defined as −

$$\mathrm{L[x(t)] \:=\: X(s) \:=\: \int_{-\infty}^{\infty}\: x(t) e^{-st} \: dt \quad \dots\: (1)}$$

Equation (1) gives the bilateral Laplace transform of the function x(t). But for the causal signals, the unilateral Laplace transform is applied, which is defined as −

$$\mathrm{L[x(t)] \:=\: X(s) \:=\: \int_0^{\infty}\: x(t) e^{-st} \: dt \quad \dots\: (2)}$$

Also, the inverse Laplace transform is defined as −

$$\mathrm{L^{-1}[X(s)] \:=\: x(t) \:=\: \frac{1}{2\pi j} \int_{\sigma - j\infty}^{\sigma + j\infty}\: X(s) e^{st} \: ds \quad \dots\: (3)}$$

Time Convolution Property of Laplace Transform

Statement – The time convolution property of the Laplace transform states that the Laplace transform of convolution of two signals in time domain is equivalent to the product of their respective Laplace transforms. Therefore, if

$$\mathrm{x_1(t) \:\overset{LT}\longleftrightarrow\: X_1(s) \quad \text{and} \quad x_2(t) \:\overset{LT}\longleftrightarrow\: X_2(s)}$$

Then,

$$\mathrm{x_1(t) \:\cdot\: x_2(t)\:\overset{LT}\longleftrightarrow\: X_1(s)X_2(s)}$$

Proof

If x₁(t) and x₂(t) are two time domain causal signals, then their convolution is defined as,

$$\mathrm{x_1(t) \:\cdot\: x_2(t) \:=\: \int_0^t x_1(t \:-\: \tau) x_2(\tau) \: d\tau}$$

Now, from the definition of Laplace transform, we have,

$$\mathrm{L[x_1(t)\:\cdot\:x_2(t)]\:=\:\int_0^{\infty} \left[ \int_0^t x_1(t\:-\:\tau) x_2(\tau)\:d\tau \right] e^{-st}\:dt}$$

$$\mathrm{\Rightarrow\:L[x_1(t)\:\cdot\:x_2(t)]\:=\:\int_0^{\infty} \left[ \int_0^{\infty} x_1(t\:-\:\tau) x_2(\tau)\:d\tau \right] e^{-st} \:dt}$$

Let (t − τ) = u, then,

$$\mathrm{t\:=\:(u\:+\:\tau) \quad \text{and} \quad dt \:=\: du}$$

$$\mathrm{\therefore\:L[x_1(t)\:\cdot\:x_2(t)]\:=\:\int_0^{\infty} \left[ \int_0^{\infty}\: x_1(u) x_2(\tau)\:d\tau \right] e^{-s(u + \tau)}\:du}$$

Rearranging the integrations, we have,

$$\mathrm{L[x_1(t)\:\cdot\:x_2(t)]\:=\:\int_0^{\infty}\:x_1(u) e^{-su}\:du\: \int_0^{\infty}\: x_2(\tau) e^{-s\tau}\:d\tau}$$

$$\mathrm{\therefore\:L[x_1(t)\:\cdot\:x_2(t)]\:=\:X_1(s) X_2(s)}$$

Or it can be represented as,

$$\mathrm{x_1(t)\:\cdot\:x_2(t) \:\overset{LT}\longleftrightarrow\: X_1(s) X_2(s)}$$

Therefore, it proves the time convolution property of the Laplace transform.

Convolution in s-Domain Property of Laplace Transform

Statement – The convolution in s-domain property, also known as multiplication property or modulation property of Laplace transform. The frequency convolution property of Laplace transform states that the Laplace transform of product of two time domain signals is equivalent to the convolution of their respective Laplace transforms. Therefore, if

$$\mathrm{x_1(t)\:\overset{LT}\longleftrightarrow\:X_1(s)\quad \text{and} \quad x_2(t)\:\overset{LT}\longleftrightarrow\:X_2(s)}$$

Then,

$$\mathrm{x_1(t) x_2(t) \:\overset{LT}\longleftrightarrow\: \frac{1}{2\pi j} [X_1(s)\: \cdot\: X_2(s)]}$$

Proof

The inverse Laplace transform of x₁(t) is,

$$\mathrm{x_1(t)\:=\:\frac{1}{2\pi j} \int_{\sigma - j\infty}^{\sigma+j\infty}X_1(p) e^{pt}\: dp}$$

Thus, from the definition of the Laplace transform, we have,

$$\mathrm{L[x_1(t) x_2(t)] \:=\: \int_{-\infty}^{\infty} [x_1(t) x_2(t)] e^{-st}\: dt}$$

$$\mathrm{\Rightarrow\: L[x_1(t) x_2(t)] \:=\: \int_{-\infty}^{\infty} \left[ \frac{1}{2\pi j} \int_{\sigma - j\infty}^{\sigma + j\infty} X_1(p) e^{pt}\: dp \right]\: x_2(t) e^{-st}\: dt}$$

Rearranging the order of integration, we have,

$$\mathrm{L[x_1(t) x_2(t)] \:=\: \frac{1}{2\pi j} \int_{\sigma - j\infty}^{\sigma + j\infty}\: X_1(p) \left[ \int_{-\infty}^{\infty}\: x_2(t) e^{-(s \:-\: p)t}\: dt \right]\: dp}$$

$$\mathrm{\Rightarrow\: L[x_1(t) x_2(t)] \:=\: \frac{1}{2\pi j} \int_{\sigma - j\infty}^{\sigma + j\infty}\: X_1(p) X_2(s \:-\: p)\: dp}$$

$$\mathrm{\therefore\: L[x_1(t) x_2(t)] \:=\: \frac{1}{2\pi j} [X_1(s) \:\cdot\: X_2(s)]}$$

Or it can also be represented as,

$$\mathrm{x_1(t) x_2(t) \:\overset{LT}\longleftrightarrow\: \frac{1}{2\pi j} [X_1(s) \:\cdot \:X_2(s)]}$$

Thus, it proves the multiplication property or convolution in s-domain property of the Laplace transform.

Numerical Example 1

A single x(t) has Laplace transform as,

$$\mathrm{X(s) \:=\: \frac{s \:+\: 3}{s^2 \:+\: 2s \:+\: 1}}$$

Then, find the Laplace transform of $\mathrm{y(t)\:=\:x(t)\:\cdot\:x(t)}$.

Solution

The given Laplace transform of x(t) is,

$$\mathrm{X(s) \:=\: \frac{s \:+\: 3}{s^2 \:+\: 2s \:+\: 1}}$$

Then, the Laplace transform of y(t), i.e., Y(s) is,

$$\mathrm{L[y(t)] \:=\: Y(s) \:=\: L[x(t) \:\cdot\: x(t)]}$$

Now, using time convolution property $\mathrm{\left[\text{i.e., }\: x_1(t) \:\cdot\: x_2(t) \:\overset{LT}\longleftrightarrow\: X_1(s) X_2(s)\right]}$ of Laplace transform, we have,

$$\mathrm{Y(s) \:=\: X(s) \:\cdot\: X(s) \:=\: [X(s)]^2}$$

$$\mathrm{\Rightarrow\: Y(s) \:=\: \left[ \frac{s \:+\: 3}{s^2 \:+\: 2s \:+\: 1} \right]^2}$$

Numerical Example 2

Using the convolution in s-domain property of the Laplace transform, find the Laplace transform of the function

$$\mathrm{x(t) \:=\: \delta(t) \sin(\omega t)}$$

Solution

The given function is,

$$\mathrm{x(t) \:=\: \delta(t) \sin(\omega t)}$$

Since we know,

$$\mathrm{L[\delta(t)] \:=\: 1 \quad \text{and} \quad L[\sin(\omega t)] \:=\: \frac{\omega}{s^2 \:+\: \omega^2}}$$

Hence, by using the convolution in s-domain property $\mathrm{\left\{\text{i.e., } \:x_1(t) x_2(t) \:\overset{LT}\longleftrightarrow\: \frac{1}{2\pi j} [X_1(s) \:\cdot\: X_2(s)]\right\}}$ of Laplace transform, we have,

$$\mathrm{L[\delta(t)\: \sin(\omega t)] \:=\: \frac{1}{2\pi j} \left( \frac{\omega}{s^2 \:+\: \omega^2} \right)}$$