Title: Default for 20.3 Further Laplace Transforms
Course: HELM Workbooks 2023-2024

The second shift theorem

The second shift theorem is similar to the first except that, in this case, it is the time-variable that is shifted not the \(s\)-variable. Consider a causal function \(f(t)u(t)\) which is shifted to the right by amount \(a\), that is, the function \(f(t-a)u(t-a)\) where \(a>0\). The following figure illustrates the two causal functions.

[Image: Two graphs, showing a casual function and the same function, shifted to the right by a.]

The Laplace transform of the shifted function is easily obtained: \[\begin{align}{\cal L}\{f(t-a)u(t-a)\}&= \int_0^{\infty}\text{e}^{-st}f(t-a)u(t-a)\,dt \\ &=\int_a^{\infty}\text{e}^{-st}f(t-a)\,dt\end{align}\]

(Note the change in the lower limit from \(0\) to \(a\) resulting from the step function switching on at \(t=a\)). We can re-organise this integral by making the substitution \(x=t-a\). Then \(\displaystyle \mathrm{d}t=\mathrm{d}x\) and when \(t=a, x=0\) and when \(t=\infty\) then \(x=\infty\).

Therefore \[\begin{align} \int_a^{\infty}\text{e}^{-st}f(t-a)\,dt &=\int_0^{\infty}\text{e}^{-s(x+a)}f(x)\,dx\\ &=\text{e}^{-sa}\int_0^{\infty}\text{e}^{-sx}f(x)\,dx\end{align}\]

The final integral is simply the Laplace transform of \(f(x)\), which we know is \(F(s)\) and so, finally, we have the statement of the second shift theorem:

Key Point

Second Shift Theorem

If \({\cal L}\{f(t)\}=F(s)\) then \({\cal L}\{f(t-a)u(t-a)\}=\text{e}^{-sa}F(s)\)

Obviously, this theorem has its uses in finding the Laplace transform of time-shifted causal functions but it is also of considerable use in finding inverse Laplace transforms since, using the inverse formulation of the theorem of the previous Key Point we get:

Key Point

Inverse Second Shift Theorem

If \({\cal L}^{-1}\{F(s)\}=f(t)\) then \({\cal L}^{-1}\{\text{e}^{-sa}F(s)\}=f(t-a)u(t-a)\)

Example

Find the inverse Laplace transform of \(\displaystyle {\text{e}^{-3s}\over {s^2}}\).

Solution

You should obtain \((t-3)u(t-3)\) for the following reasons. We know that the inverse Laplace transform of \(1/s^2\) is \(t.u(t)\) (from the table of common Laplace transforms) and so, using the second shift theorem (with \(a=3\)), we have \[ {\cal L}^{-1}\biggl\{\text{e}^{-3s}{1\over {s^2}}\biggr\}=(t-3)u(t-3).\]

This function is graphed in the following figure:

[Image]

Exercise

Find the inverse Laplace transform of \(\displaystyle {{s\over {s^2-2s+2}}}\):
Feedback: You should obtain \(\text{e}^t(\cos t+\sin t)\). To obtain this, complete the square in the denominator: \(s^2-2s+2=(s-1)^2+1\) and so \[\begin{align} {s\over {s^2-2s+2}} &={s\over {(s-1)^2+1}} \\ &={(s-1)+1\over {(s-1)^2+1}} \\ &= {s-1\over {(s-1)^2+1}}+{1\over {(s-1)^2+1}}. \end{align}\] Now, using the first shift theorem \[ {\cal L}^{-1}\biggl\{{s-1\over {(s-1)^2+1}}\biggr\}=\text{e}^t\cos t.u(t)\quad \text{ since}\quad {\cal L}^{-1}\biggl\{{s\over {s^2+1}}\biggr\}=\cos t.u(t)\] and \[ {\cal L}^{-1}\biggl\{{1\over {(s-1)^2+1}}\biggr\}=\text{e}^t\sin t.u(t)\quad \text{ since}\quad {\cal L}^{-1}\biggl\{{1\over {s^2+1}}\biggr\}=\sin t.u(t)\] Thus \[ {\cal L}^{-1}\biggl\{{s\over {s^2-2s+2}}\biggr\}=\text{e}^t(\cos t+\sin t)u(t).\]

The Laplace transform of a derivative

Here we consider not a causal function \(f(t)\) directly but its derivatives \(\displaystyle{df\over {dt}}\), \(\displaystyle{d^2f\over {dt^2}}\), (which are also causal.) The Laplace transform of derivatives will be invaluable when we apply the Laplace transform to the solution of constant coefficient ordinary differential equations.

If \({\cal L}\{f(t)\}\) is \(F(s)\) then we shall seek an expression for \(\displaystyle{\cal L}\{{df\over {dt}}\}\) in terms of the function \(F(s)\).

Now, by the definition of the Laplace transform \[ {\cal L}\biggl\{{df\over {dt}}\biggr\}=\int_0^{\infty}\text{e}^{-st}{df\over {dt}}\,dt\] This integral can be simplified using integration by parts:

\[\begin{eqnarray*} \int_0^{\infty}\text{e}^{-st}{df\over {dt}}\,dt&=&\biggl[\text{e}^{-st}f(t)\biggr]_0^{\infty}-\int_0^{\infty}(-s)\text{e}^{-st}f(t)\,dt\\ &=&-f(0)+s\int_0^{\infty}\text{e}^{-st}f(t)\,dt \end{eqnarray*}\] (As usual, we assume that contributions arising from the upper limit, \(t=\infty\), are zero.) The integral on the right-hand side is precisely the Laplace transform of \(f(t)\) which we naturally replace by \(F(s)\). Thus \[ {\cal L}\biggl\{{df\over {dt}}\biggr\}=-f(0)+sF(s)\]

As an example, we know that if \(f(t)=\sin t\ u(t)\) then \[{\cal L}\{f(t)\}={1\over {s^2+1}}= F(s)\] and so, according to the result just obtained,

\[\begin{eqnarray*} {\cal L}\biggl\{ \frac{df}{dt} \biggr\}={\cal L}\{\cos t\ u(t)\}&=&-f(0)+sF(s)\\ &=&0+s\biggl({1\over {s^2+1}}\biggr)\\ &=&{s\over {s^2+1}} \end{eqnarray*}\] a result we know to be true.

We can find the Laplace transform of the second derivative in a similar way to find:

\[\displaystyle{\cal L}\biggl\{{d^2f\over {dt^2}}\biggr\}=-f'(0)-sf(0)+s^2F(s)\] (The reader might wish to derive this result.) Here \(f'(0)\) is the derivative of \(f(t)\) evaluated at \(t=0\).

Key Point

Laplace Transforms of Derivatives

If \({\cal L}\{f(t)\}=F(s)\) then\[\begin{eqnarray*}{\cal L}\biggl\{{df\over {dt}}\biggr\}&=&-f(0)+sF(s)\\ {\cal L}\biggl\{{d^2f\over {dt^2}}\biggr\}&=&-f'(0)-sf(0)+s^2F(s)\end{eqnarray*}\]

Example

If \(\displaystyle{{\cal L}\{f(t)\}=F(s)\quad \text{ and}\quad {d^2f\over {dt^2}}-{df\over {dt}}=3t}\) with initial conditions \(f(0)=1,\quad f'(0)=0\), find the explicit expression for \(F(s)\).

1. Begin by finding \(\displaystyle{{\cal L}\biggl\{{d^2f\over {dt^2}}\biggr\},\ \ {\cal L}\biggl\{{df\over {dt}}\biggr\}\ \text{ and}\ {\cal L}\{3t\}}\)
Solution
\[\begin{eqnarray*} {\cal L}\{3t\}&=&3/s^2\\ {\cal L}\biggl\{{df\over {dt}}\biggr\}&=&-f(0)+sF(s)=-1+sF(s)\\ {\cal L}\biggl\{{d^2f\over {dt^2}}\biggr\}&=&-f'(0)-sf(0)+s^2F(s)=-s+s^2F(s) \end{eqnarray*}\]
2. Now complete the calculation to find \(F(s)\):
Solution
You should find \(\quad\displaystyle {F(s)={s^3-s^2+3\over s^3(s-1)}}\)
since, using the transforms we have found: \[\begin{eqnarray*} -s+s^2F(s)-(-1+sF(s))&=&{3\over {s^2}}\\\text{ so}\qquad F(s)[s^2-s]&=&{3\over {s^2}}+s-1={s^3-s^2+3\over {s^2}} \end{eqnarray*}\] leading to \( \displaystyle F(s)={s^3-s^2+3\over s^3(s-1)}\)

Exercise

Find the Laplace transforms of

1. \(t^3\text{e}^{-2t}u(t)\)
2. \(\text{e}^t\sinh 3t.u(t)\)
3. \(\sin (t-3). u(t-3)\)

Exercise

If \(F(s)={\cal L}\{f(t)\}\) find expressions for \(F(s)\) if

1. \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} -3\frac{\mathrm{d}y}{\mathrm{d}t}+4y=\sin t\qquad y(0)=1, y'(0)=0\)
2. \(\displaystyle 7\frac{\mathrm{d}y}{\mathrm{d}t}-6y=3u(t)\qquad y(0)=0,\)

Exercise

Find the inverse Laplace transforms of

1. \(\displaystyle{{6\over {(s+3)^4}}}\)
2. \(\displaystyle{{15\over {s^2-2s+10}}}\)
3. \(\displaystyle{{3s^2+11s+14\over {s^3+2s^2-11s-52}}}\)
4. \(\displaystyle{{\text{e}^{-3s}\over {s^4}}}\)
5. \(\displaystyle{{\text{e}^{-2s-2}(s+1)\over {s^2+2s+5}}}\)

The delta function (or impulse function)

There is often a need for considering the effect on a system (modelled by a differential equation) by a forcing function which acts for a very short time interval. For example, how does the current in a circuit behave if the voltage is switched on and then very shortly afterwards switched off? How does a cantilevered beam vibrate if it is hit with a hammer (providing a force which acts over a very short time interval)? Both of these engineering ‘systems’ can be modelled by a differential equation. There are many ways the ‘kick’ or ‘impulse’ to the system can be modelled. The function we have in mind could have the graphical representation (when \(a\) is small) shown in Figure 14.

[Image: Graph of a step function that 'switches' on at x=d and off at x=d+a, with amplitude b.]

This can be represented formally using step functions; it switches on at \(t=d\) and switches off at \(t=d+a\) and has amplitude \(b\): \[\qquad\displaystyle f(t)=b[u(t-d)-u(t-\{d+a\})]\]

The effect on the system is related to the area under the curve rather than just the amplitude \(b\). Our aim is to reduce the time interval over which the forcing function acts (i.e. reduce \(a\)) whilst at the same time keeping the total effect (i.e. the area under the curve) a constant. To do this we shall take \(b=1/a\) so that the area is always equal to 1. Reducing the value of \(a\) then gives the sequence of inputs shown in Figure 15.

[Image]

As the value of \(a\) decreases the height of the rectangle increases (to ensure the value of the area under the curve is fixed at value 1) until, in the limit as \(a\rightarrow 0\), the 'function' becomes a 'spike’ at \(t=d\). The resulting function is called a delta function (or impulse function) and denoted by \(\delta (t-d)\). This notation is used because, in a very obvious sense, the delta function described here is ‘located’ at \(t=d\). Thus the delta function \(\delta (t-1)\) is ‘located’ at \(t=1\) whilst the delta function \(\delta (t)\) is 'located’ at \(t=0\).

If we were defining an ordinary function we would write \(\qquad\displaystyle \delta (t-d)=\lim_{a\rightarrow 0}{1\over a}[u(t-d)-u(t-\{d+a\})]\)

However, this limit does not exist. The important property of the delta function relates to its integral: \[\begin{align*} \int_{-\infty}^{\infty}\delta (t-d)\,dt &=\lim_{a\rightarrow0}\int_{-\infty}^{\infty}{1\over a}[u(t-d)-u(t-\{d+a\})]\,dt \\ &=\lim_{a\rightarrow0}\int_d^{d+a}{1\over a}\,dt\\ &=\lim_{a\rightarrow0}\biggl[{d+a\over a}-{d\over a}\biggr]=1\qquad \end{align*}\] which is what we expect since the area under each of the limiting curves is equal to \(1\).

A more technical discussion obtains the more general result:

Key Point

Sifting Property of the Delta Function

\[\int_{-\infty}^{\infty}f(t)\delta (t-d)\,dt=f(d)\]

This is called the sifting property of the delta function as it sifts out the value \(f(d)\) from the function \(f(t)\). Although the integral here ranges from \(t=-\infty\) to \(t=+\infty\) in fact the same result is obtained for any range if the range of the integral includes the point \(t=d\). That is, if \(\alpha\le d\le\beta\) then \[\qquad\int_{\alpha}^{\beta}f(t)\delta (t-d)\,dt=f(d)\]

Thus, as long as the delta function is 'located' within the range of the integral the sifting property holds. For example, \[\qquad \int_1^2\sin t\ \delta (t-1.1)\,dt=\sin 1.1=0.8912\qquad\qquad \int_0^{\infty}\text{e}^{-t}\delta (t-1)\,dt=\text{e}^{-1}=0.3679 \]

Exercise

Write expressions for delta functions located at \(t=-[value]\) and at \(t=[value]\)

Hint: use the syntax delta(t) to input the delta function \(\delta(t)\).

At \(t=-[value]\):

At \(t=[value]\):
Feedback: The correct answers are \(\delta (t+[value])\) and \(\delta (t-[value])\).

Exercise

Evaluate the integral \(I = \displaystyle{\int_{-1}^3(\sin t\ \delta (t+2)-\cos t\ \delta (t))\,dt}\).

\(I =\)
Feedback: You should obtain the value \(-1\) since the first delta function, \(\delta (t+2)\), is located outside the range of integration and thus \[ \int_{-1}^3(\sin t \ \delta (t+2)-\cos t\ \delta (t))\,dt = \int_{-1}^3-\cos t\ \delta (t)\,dt =-\cos 0=-1.\]

The Laplace transform of the delta function

Here we consider \({\cal L}\{\delta(t-d)\}\). From the definition of the Laplace transform: \[ {\cal L}\{\delta (t-d)\}=\int_0^{\infty}\text{e}^{-st}\delta (t-d)\,dt=\text{e}^{-sd} \] by the sifting property of the delta function. Thus

Key Point

Laplace Transform of the Delta Function

\[{\cal L}\{\delta (t-d)\}=\text{e}^{-sd}\quad \text{ and, putting } \ d=0,\quad {\cal L}\{\delta (t)\}=\text{e}^0=1\]

Example

Find the Laplace transform of \(3\delta (t-3).\)

Solution

\(3e^{-3s}\)