Properties of the Laplace Transform

Posted on Edited on In Views:

Sources:

  1. B. P. Lathi & Roger Green. (2018). Chapter 4: The Laplace Transform. Signal Processing and Linear Systems (3rd ed., pp. 350-360). Oxford University Press.

For a quick reference table, see Wikipedia page on the Laplace transform.

Time shifting

The time-shifting property states that if \[ x(t) \Longleftrightarrow X(s) \] then for \(t_0 \geq 0\) \[ x\left(t-t_0\right) \Longleftrightarrow X(s) e^{-s t_0} \]

Observe that \(x(t)\) starts at \(t=0\)1, and, therefore, \(x\left(t-t_0\right)\) starts at \(t=t_0\). This fact is implicit, but is not explicitly indicated in Eq. (4.12). This often leads to inadvertent errors.

To avoid such a pitfall, we should restate the property as follows. If \[ \color{teal} {x(t) u(t) \Longleftrightarrow X(s)} \] then \[ \color{salmon} {x\left(t-t_0\right) u\left(t-t_0\right) \Longleftrightarrow X(s) e^{-s t_0} \quad t_0 \geq 0} \]

Proof

\[ \mathcal{L}\left[x\left(t-t_0\right) u\left(t-t_0\right)\right]=\int_0^{\infty} x\left(t-t_0\right) u\left(t-t_0\right) e^{-s t} d t \]

Setting \(t-t_0=\tau\), we obtain \[ \mathcal{L}\left[x\left(t-t_0\right) u\left(t-t_0\right)\right]=\int_{-t_0}^{\infty} x(\tau) u(\tau) e^{-s\left(\tau+t_0\right)} d \tau \]

Because \(u(\tau)=0\) for \(\tau<0\) and \(u(\tau)=1\) for \(\tau \geq 0\), the limits of integration can be taken from 0 to \(\infty\). Thus, \[ \begin{aligned} \mathcal{L}\left[x\left(t-t_0\right) u\left(t-t_0\right)\right] & =\int_0^{\infty} x(\tau) e^{-s\left(\tau+t_0\right)} d \tau \\ & =e^{-s t_0} \int_0^{\infty} x(\tau) e^{-s \tau} d \tau \\ & =X(s) e^{-s t_0} \end{aligned} \]

Note that \(x\left(t-t_0\right) u\left(t-t_0\right)\) is the signal \(x(t) u(t)\) delayed by \(t_0\) seconds. The time-shifting property states that delaying a signal by \(t_0\) seconds amounts to multiplying its transform \(e^{-s t_0}\).

This property of the unilateral Laplace transform holds only for positive \(t_0\) because if \(t_0\) were negative, the signal \(x\left(t-t_0\right) u\left(t-t_0\right)\) may not be causal.

Frequency shifting

The frequency-shifting property states that if \[ \color{teal} {x(t) \Longleftrightarrow X(s)} \] then \[ \color{salmon} {x(t) e^{s_0 t} \Longleftrightarrow X\left(s-s_0\right)} \]

Proof

\[ \mathcal{L}\left[x(t) e^{s_0 t}\right]=\int_{0^{-}}^{\infty} x(t) e^{s_0 t} e^{-s t} d t=\int_{0^{-}}^{\infty} x(t) e^{-\left(s-s_0\right) t} d t=X\left(s-s_0\right) \]

The time-differentiation property

If \[ \color{teal} {x(t) \Longleftrightarrow X(s)} \] then \[ \frac{d x(t)}{d t} \Longleftrightarrow s X(s)-x\left(0^{-}\right) \]

Repeating this property a second time (differentiating twice) yields \[ \frac{d^2 x(t)}{d t^2} \Longleftrightarrow s^2 X(s)-s x\left(0^{-}\right)-\dot{x}\left(0^{-}\right) \]

Repeated differentiation yields \[ \begin{aligned} \frac{d^n x(t)}{d t^n} & \Longleftrightarrow s^n X(s)-s^{n-1} x\left(0^{-}\right)-s^{n-2} \dot{x}\left(0^{-}\right)-\cdots-x^{(n-1)}\left(0^{-}\right) \\ & =s^n X(s)-\sum_{k=1}^n s^{n-k} x^{(k-1)}\left(0^{-}\right) \end{aligned} \] where \(x^{(r)}\left(0^{-}\right)\)is \(d^r x / d t^r\) at \(t=0^{-}\).

Proof

\[ \mathcal{L}\left[\frac{d x(t)}{d t}\right]=\int_{0^{-}}^{\infty} \frac{d x(t)}{d t} e^{-s t} d t \]

Integrating by parts, we obtain \[ \mathcal{L}\left[\frac{d x(t)}{d t}\right]=\left.x(t) e^{-s t}\right|_{0^{-}} ^{\infty}+s \int_{0^{-}}^{\infty} x(t) e^{-s t} d t \]

For the Laplace integral to converge [i.e., for \(X(s)\) to exist], it is necessary that \(x(t) e^{-s t} \rightarrow 0\) as \(t \rightarrow \infty\) for the values of \(s\) in the ROC for \(X(s)\). Thus, \[ \mathcal{L}\left[\frac{d x(t)}{d t}\right]=-x\left(0^{-}\right)+s X(s) \]

Repeated application of this procedure yields the equation for \(\frac{d^n x(t)}{d t^n}\).

The frequency-differentiation property

The dual of the time-integration property is the frequency-integration property, which states that \[ \frac{x(t)}{t} \Longleftrightarrow \int_s^{\infty} X(z) d z \]

The time-integration property

The time-integration property states that if \[ x(t) \Longleftrightarrow X(s) \] then \[ \int_{0^{-}}^t x(\tau) d \tau \Longleftrightarrow \frac{X(s)}{s} \quad \text { and } \quad \int_{-\infty}^t x(\tau) d \tau \Longleftrightarrow \frac{X(s)}{s}+\frac{\int_{-\infty}^{0^{-}} x(\tau) d \tau}{s} \]

Proof. To prove the first part, we define \[ g(t)=\int_{0^{-}}^t x(\tau) d \tau \] so that \[ \frac{d}{d t} g(t)=x(t) \quad \text { and } \quad g\left(0^{-}\right)=0 \]

Now, if \[ g(t) \Longleftrightarrow G(s) \] then \[ X(s)=\mathcal{L}\left[\frac{d}{d t} g(t)\right]=s G(s)-g\left(0^{-}\right)=s G(s) \] Therefore, \[ G(s)=\frac{X(s)}{s} \] or \[ \int_{0^{-}}^t x(\tau) d \tau \Longleftrightarrow \frac{X(s)}{s} \]

To prove the second part, observe that \[ \int_{-\infty}^t x(\tau) d \tau=\int_{-\infty}^{0^{-}} x(\tau) d \tau+\int_{0^{-}}^t x(\tau) d \tau \]

Note that the first term on the right-hand side is a constant for \(t \geq 0\). Taking the Laplace transform of the foregoing equation and using the first part of Eq. (4.16), we obtain \[ \int_{-\infty}^t x(\tau) d \tau \Longleftrightarrow \frac{\int_{-\infty}^{0^{-}} x(\tau) d \tau}{s}+\frac{X(s)}{s} \]

The frequency-integration property

The dual of the time-integration property is the frequency-integration property, which states that \[ \frac{x(t)}{t} \Longleftrightarrow \int_s^{\infty} X(z) d z \]

The scaling property

The scaling property states that if \[ x(t) \Longleftrightarrow X(s) \] then for \(a>0\) \[ x(a t) \Longleftrightarrow \frac{1}{a} X\left(\frac{s}{a}\right) \]

The proof is given in Ch. 7.

Note that \(a\) is restricted to positive values because if \(x(t)\) is causal, then \(x(a t)\) is anticausal (is zero for \(t \geq 0\) ) for negative \(a\), and anticausal signals are not permitted in the (unilateral) Laplace transform.

Time convolution and frequency convolution

Another pair of properties states that if \[ x_1(t) \Longleftrightarrow X_1(s) \quad \text { and } \quad x_2(t) \Longleftrightarrow X_2(s) \] then (time-convolution property) \[ x_1(t) * x_2(t) \Longleftrightarrow X_1(s) X_2(s) \] and (frequency-convolution property) \[ x_1(t) x_2(t) \Longleftrightarrow \frac{1}{2 \pi j}\left[X_1(s) * X_2(s)\right] \]

Observe the symmetry (or duality) between the two properties. Proofs of these properties are postponed to Ch. 7.

\(h(t) \Longleftrightarrow H(s)\)

In the previous post, we have proved that \(H(s)\), the transfer function of an LTIC system, is the bilateral Laplace transform of the system's impulse response \(h(t)\), that is, \[ \color{orange} {h(t) \Longleftrightarrow H(s)} . \]

If \(h(s)\) starts from \(t=0\), \(H(s)\) is the unilateral Laplace transform of \(h(t)\).

Recall that in an LTIC system we have \(y(t)=x(t) * h(t)\), we apply the time-convolution property to obtain \[ \begin{equation} \label{eq4_18} Y(s)=X(s) H(s) . \end{equation} \]

The response \(y(t)\) is the zero-state response of the LTIC system to the input \(x(t)\). From Eq. (4.18), it follows that \[ H(s)=\frac{Y(s)}{X(s)}=\frac{\mathcal{L}[\text { zero-state response }]}{\mathcal{L}[\text { input }]} \]

This may be considered an alternate definition of the LTIC system transfer function \(H(s)\). It is the ratio of the transform of zero-state response to the transform of the input.

Initial and final values

In certain applications, it is desirable to know the values of \(x(t)\) as \(t \rightarrow 0\) and \(t \rightarrow \infty\) [initial and final values of \(x(t)\) ] from the knowledge of its Laplace transform \(X(s)\). Initial and final value theorems provide such information.

The initial value theorem states that if \(x(t)\) and its derivative \(d x / d t\) are both Laplace transformable, then \[ x\left(0^{+}\right)=\lim _{s \rightarrow \infty} s X(s) \] provided the limit on the right-hand side of Eq. (4.20) exists. The final value theorem states that if both \(x(t)\) and \(d x / d t\) are Laplace transformable, then \[ \lim _{t \rightarrow \infty} x(t)=\lim _{s \rightarrow 0} s X(s) \] provided \(s X(s)\) has no poles in the RHP or on the imaginary axis. To prove these theorems, we begin by the time-differentiation property as mentioned before \[ \frac{d x(t)}{d t} \Longleftrightarrow s X(s)-x\left(0^{-}\right) . \] We calculate: \[ \begin{aligned} s X(s)-x\left(0^{-}\right) & =\int_{0^{-}}^{\infty} \frac{d x(t)}{d t} e^{-s t} d t \\ & =\int_{0^{-}}^{0^{+}} \frac{d x(t)}{d t} e^{-s t} d t+\int_{0^{+}}^{\infty} \frac{d x(t)}{d t} e^{-s t} d t \\ & =\left.x(t)\right|_{0^{-}} ^{0^{+}}+\int_{0^{+}}^{\infty} \frac{d x(t)}{d t} e^{-s t} d t \\ & =x\left(0^{+}\right)-x\left(0^{-}\right)+\int_{0^{+}}^{\infty} \frac{d x(t)}{d t} e^{-s t} d t \end{aligned} \]

Therefore, \[ s X(s)=x\left(0^{+}\right)+\int_{0^{+}}^{\infty} \frac{d x(t)}{d t} e^{-s t} d t \] and \[ \begin{aligned} \lim _{s \rightarrow \infty} s X(s) & =x\left(0^{+}\right)+\lim _{s \rightarrow \infty} \int_{0^{+}}^{\infty} \frac{d x(t)}{d t} e^{-s t} d t \\ & =x\left(0^{+}\right)+\int_{0^{+}}^{\infty} \frac{d x(t)}{d t}\left(\lim _{s \rightarrow \infty} e^{-s t}\right) d t \\ & =x\left(0^{+}\right) \end{aligned} \]

  1. Well, the unilaterate Laplace transform essentially regulates the signal to transform, namely \(x(t)\), to be "unilaterate", which means it must start at t=0. The "causality" is not true.↩︎