Laplace Transform of Periodic Functions (Time Periodicity Property of Laplace Transform)

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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,

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

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,

$$\mathrm{L[x(t)]\:=\:X(s)\:=\:\int_{0}^{\infty}\:x(t)e^{-st}\:dt\:\:\dotso\:(2)}$$

Laplace Transform of Periodic Functions

The Laplace transform of periodic functions can be determined by using the time shifting property $\mathrm{[\text{i.e, }x(t \:-\: T)\:=\:e^{sT}\:X(s)]}$. Consider a causal periodic function x(t) which satisfies the condition $\mathrm{x(t)\:=\:x(t\:+\:nT)}$ or all t > 0, where T is the time period of x(t) and n = 0, 1, 2,....

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

$$\mathrm{L[x(t)] \:=\: X(s) \:=\: \int_{0}^{\infty}\: x(t) e^{-st} \: dt \quad \dotso\: (3)}$$

The above expression can also be written as,

$$\mathrm{X(s) \:=\: \int_{0}^{T}\: x(t)\: e^{-st} \: dt \:+\: \int_{T}^{2T}\: x(t) e^{-st} \: dt \:+\: \int_{2T}^{3T}\: x(t) e^{-st} \: dt \:+\: \dotso \:+\:\int_{nT}^{(n+1)T}\: x(t)\:e^{-st}\:dt\:+\:\dotso}$$

$$\mathrm{\Rightarrow\: X(s) \:=\: \int_{0}^{T}\: x(t) e^{-st} \: dt \:+\: e^{-sT} \int_{0}^{T}\: x(t\:+\:T) e^{-st} \: dt\:+\: e^{-2sT} \int_{0}^{T}\: x(t\:+\:2T) e^{-st} \: dt \:+\: \dotso\:+\: e^{-nsT} \int_{0}^{T}\: x(t\:+\:nT) e^{-st} \: dt \:+\: \dotso \quad \dots (4)}$$

As the function x(t) is a periodic function, therefore,

$$\mathrm{x(t) \:=\:x(t\:+\:T)\:=\:x(t\:+\:2T)\:=\:\dotso}$$

Hence, the equation (4) can be written as

$$\mathrm{X(s) \:=\: \int_{0}^{T}\: x(t) e^{-st} \: dt \:+\: e^{-sT} \int_{0}^{T}\: x(t) e^{-st} \: dt \:+\: e^{-2sT} \int_{0}^{T}\: x(t) e^{-st} \: dt \:+\: \dotso \:+\: e^{-nsT} \int_{0}^{T}\: x(t) e^{-st} \: dt \:+\: \dotso}$$

$$\mathrm{\Rightarrow\: X(s) \:=\: \left[ 1 \:+\: e^{-sT} \:+\: e^{-2sT} \:+\: \dotso \:+\: e^{-nsT} \:+\: \dotso \right]\: \int_{0}^{T}\: x(t) e^{-st} \: dt \quad \dotso\: (5)}$$

Using the binomial series expansion, we can write,

$$\mathrm{X(s) \:=\: \left[1 \:-\: e^{-sT} \right]^{-1}\: \int_{0}^{T}\: x(t) e^{-st} \: dt}$$

$$\mathrm{\therefore \: X(s) \:=\: \frac{1}{1 \:-\: e^{-sT}}\: X_1(s) \quad \dotso\: (6)}$$

Where,

$$\mathrm{X_1(s)\:=\:\int_{0}^{T}\:x(t)\:e^{-st}\:dt\:\:\dotso\:(7)}$$

The $\mathrm{X_1(s)}$ is the Laplace transform of the first period of the time function. The eqn. (6) represents the periodicity property of the Laplace transform.

Numerical Example

Find the Laplace transform of function $\mathrm{x(t)\:=\:sin\pi t\:u(t)}$ using periodicity property of Laplace transform.

Solution

The given function is,

$$\mathrm{x(t)\:=\:sin\:\pi t\:u(t)}$$

The given signal is a periodic signal with a time period of T which is given by,

$$\mathrm{T \:=\: \frac{2\pi}{\omega} \:=\: \frac{2\pi}{\pi}\:=\:2sec}$$

Now, by using the periodicity property of Laplace transform, we have,

$$\mathrm{L[x(t)] \:=\: X(s) \:=\: \frac{1}{[1−e^{−2s}]}\:\int^{2}_{0}\:sin(\pi t)e^{−st}\:dt}$$

$$\mathrm{\Rightarrow\: X(s) \:=\: \frac{1}{1 \:-\: e^{-2s}}\: \int_{0}^{2}\: \left( \frac{e^{j\pi t} \:-\: e^{-j\pi t}}{2j} \right) e^{-st} \: dt}$$

$$\mathrm{\Rightarrow\: X(s) \:=\: \frac{1}{1 \:-\: e^{-2s}}\: \left[ \frac{1}{2j} \left( \int_{0}^{2}\: e^{j\pi t} e^{-st} \: dt \:-\: \int_{0}^{2}\: e^{-j\pi t} e^{-st} \: dt \right) \right]}$$

$$\mathrm{\Rightarrow\: X(s) \:=\: \frac{1}{1 \:-\: e^{-2s}}\: \left[ \frac{1}{2j} \left( \int_{0}^{2}\: e^{-(s \:-\: j\pi) t} \: dt \:-\: \int_{0}^{2}\: e^{-(s \:+\: j\pi) t} \: dt \right) \right]}$$

$$\mathrm{\Rightarrow \:X(s) \:=\: \frac{1}{1 \:-\: e^{-2s}}\: \left\{ \frac{1}{2j} \left[ \frac{e^{-(s \:-\: j\pi) t}}{-(s \:-\: j\pi)} \:-\: \frac{e^{-(s \:+\: j\pi) t}}{-(s \:+\: j\pi)} \right]_{0}^{2} \right\}}$$

On solving the limits, we get,

$$\mathrm{X(s) \:=\: \frac{1}{1 \:-\: e^{-2s}} \left\{ \frac{1}{2j} \left[ \frac{-(s \:+\: j\pi) \left( e^{-2(s \:-\: j\pi)t} \:-\: 1 \right) \:+\: (s \:-\: j\pi) \left( e^{-2(s \:+\: j\pi)t} \:-\: 1 \right)}{s^2 \:+\: \pi^2} \right] \right\}}$$

$$\mathrm{\Rightarrow\: X(s) \:=\: \frac{1}{1 \:-\: e^{-2s}} \left[ \frac{\pi (1 \:-\: e^{-2s})}{s^2 \:+\: \pi^2} \right]}$$

$$\mathrm{\therefore\: X(s) \:=\: \frac{\pi}{s^2 \:+\: \pi^2}}$$