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Properties of the Laplace Transform

Related calculators: Laplace Transform Calculator , Inverse Laplace Transform Calculator

The Laplace transform has a number of interesting properties.

Property 1. Linearity of the Laplace transform: L(af(t)+bg(t))=aL(f(t))+L(g(t)){L}{\left({a}{f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}\right)}={a}{L}{\left({f{{\left({t}\right)}}}\right)}+{L}{\left({g{{\left({t}\right)}}}\right)}, where a{a} and b{b} are some constants.

The proof is straightforward through the definition:

L(a(f(t)+bg(t)))=∫0∞e−st(af(t)+bg(t))dt=∫0∞(af(t)e−st+bg(t)e−st)dt{L}{\left({a}{\left({f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}\right)}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{\left({a}{f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}\right)}{d}{t}={\int_{{0}}^{\infty}}{\left({a}{f{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}+{b}{g{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}\right)}{d}{t}.

Now, using linearity, we can write that ∫0∞(af(t)e−st+bg(t)e−st)dt=a∫0∞e−stf(t)dt+b∫e−stg(t)dt=aL(f(t))+bL(g(t)){\int_{{0}}^{\infty}}{\left({a}{f{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}+{b}{g{{\left({t}\right)}}}{{e}}^{{-{s}{t}}}\right)}{d}{t}={a}{\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}+{b}\int{{e}}^{{-{s}{t}}}{g{{\left({t}\right)}}}{d}{t}={a}{L}{\left({f{{\left({t}\right)}}}\right)}+{b}{L}{\left({g{{\left({t}\right)}}}\right)}.

This completes the proof.

Note that there can be more than 2 terms, and the property will still hold.

Example 1 . Calculate L(5e3t−2sin⁡(7t)+cos⁡(t)){L}{\left({5}{{e}}^{{{3}{t}}}-{2}{\sin{{\left({7}{t}\right)}}}+{\cos{{\left({t}\right)}}}\right)}.

From the table of Laplace transforms, it is known that L(e3t)=1s−3{L}{\left({{e}}^{{{3}{t}}}\right)}=\frac{{1}}{{{s}-{3}}}, L(sin⁡(7t))=7s2+72{L}{\left({\sin{{\left({7}{t}\right)}}}\right)}=\frac{{7}}{{{{s}}^{{2}}+{{7}}^{{2}}}}, and L(cos⁡(t))=ss2+12{L}{\left({\cos{{\left({t}\right)}}}\right)}=\frac{{s}}{{{{s}}^{{2}}+{{1}}^{{2}}}}; so, using Property 1, we obtain the following:

L(5e3t−2sin⁡(7t)+cos⁡(t))=5L(e3t)−2L(sin⁡(7t))+L(cos⁡(t))={L}{\left({5}{{e}}^{{{3}{t}}}-{2}{\sin{{\left({7}{t}\right)}}}+{\cos{{\left({t}\right)}}}\right)}={5}{L}{\left({{e}}^{{{3}{t}}}\right)}-{2}{L}{\left({\sin{{\left({7}{t}\right)}}}\right)}+{L}{\left({\cos{{\left({t}\right)}}}\right)}=

=5s−3−27s2+49+ss2+1=5s−3−14s2+49+ss2+1=\frac{{5}}{{{s}-{3}}}-{2}\frac{{7}}{{{{s}}^{{2}}+{49}}}+\frac{{s}}{{{{s}}^{{2}}+{1}}}=\frac{{5}}{{{s}-{3}}}-\frac{{14}}{{{{s}}^{{2}}+{49}}}+\frac{{s}}{{{{s}}^{{2}}+{1}}}.

Property 2. Scaling by constant: L(f(at))=1sF(sa){L}{\left({f{{\left({a}{t}\right)}}}\right)}=\frac{{1}}{{s}}{F}{\left(\frac{{s}}{{a}}\right)}, where a{a} is a positive constant.

The proof is straightforward:

L(f(at))=∫0∞e−stf(at)dt{L}{\left({f{{\left({a}{t}\right)}}}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({a}{t}\right)}}}{d}{t}.

Using the change of variable technique, we make the substituion u=at{u}={a}{t}; then, du=adt{d}{u}={a}{d}{t}, which yields dt=dua{d}{t}=\frac{{{d}{u}}}{{a}}. Since t{t} is changing from 0{0} to ∞\infty, we have that u{u} is changing from a⋅0=0{a}\cdot{0}={0} to a⋅∞=∞{a}\cdot\infty=\infty.

So,

∫0∞e−stf(at)dt=∫0∞e−suaf(u)dua=1a∫0∞e−sauf(u)du=1aF(sa){\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({a}{t}\right)}}}{d}{t}={\int_{{0}}^{\infty}}{{e}}^{{-{s}\frac{{u}}{{a}}}}{f{{\left({u}\right)}}}\frac{{{d}{u}}}{{a}}=\frac{{1}}{{a}}{\int_{{0}}^{\infty}}{{e}}^{{-\frac{{s}}{{a}}{u}}}{f{{\left({u}\right)}}}{d}{u}=\frac{{1}}{{a}}{F}{\left(\frac{{s}}{{a}}\right)}.

This completes the proof.

Property 3. L(tf(t))=−ddsF(s){L}{\left({t}{f{{\left({t}\right)}}}\right)}=-\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}, where L(f(t))=F(s){L}{\left({f{{\left({t}\right)}}}\right)}={F}{\left({s}\right)}.

Proof.

From the definition, it follows that

F(s)=∫0∞e−stf(t)dt{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}.

Differentiating this equality with respect to s{s} gives (note that the limits of integration are constant):

ddsF(s)=∫0∞ddse−stf(t)dt\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}={\int_{{0}}^{\infty}}\frac{{d}}{{{d}{s}}}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t},

or

ddsF(s)=∫0∞−te−stf(t)dt\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}={\int_{{0}}^{\infty}}-{t}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t},

which finally gives

−ddsF(s)=∫0∞te−stf(t)dt=L(tf(t))-\frac{{d}}{{{d}{s}}}{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{t}{{e}}^{{-{s}{t}}}{f{{\left({t}\right)}}}{d}{t}={L}{\left({t}{f{{\left({t}\right)}}}\right)}.

This completes the proof.

Example 2. Calculate L(tsin⁡(t)){L}{\left({t}{\sin{{\left({t}\right)}}}\right)}.

Since L(sin⁡(t))=1s2+1{L}{\left({\sin{{\left({t}\right)}}}\right)}=\frac{{1}}{{{{s}}^{{2}}+{1}}}, using property 3, we obtain that

L(tsin⁡(t))=−dds1s2+1=1(s2+1)2⋅d(s2+1)ds=2s(s2+1)2{L}{\left({t}{\sin{{\left({t}\right)}}}\right)}=-\frac{{d}}{{{d}{s}}}\frac{{1}}{{{{s}}^{{2}}+{1}}}=\frac{{1}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}\cdot\frac{{{d}{\left({{s}}^{{2}}+{1}\right)}}}{{{d}{s}}}=\frac{{{2}{s}}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}.

Property 4. L(f(t)t)=∫s∞F(u)du{L}{\left(\frac{{f{{\left({t}\right)}}}}{{t}}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}, where F(s)=L(f(t)){F}{\left({s}\right)}={L}{\left({f{{\left({t}\right)}}}\right)}.

Proof.

This property follows from property 3.

Let g(t)=f(t)t{g{{\left({t}\right)}}}=\frac{{f{{\left({t}\right)}}}}{{t}} and L(g(t))=G(s){L}{\left({g{{\left({t}\right)}}}\right)}={G}{\left({s}\right)}. We need to prove that G(s)=∫s∞F(u)du{G}{\left({s}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}.

So, f(t)=tg(t){f{{\left({t}\right)}}}={t}{g{{\left({t}\right)}}}.

Taking the Laplace transform of both sides gives:

L(f(t))=L(tg(t)){L}{\left({f{{\left({t}\right)}}}\right)}={L}{\left({t}{g{{\left({t}\right)}}}\right)}.

Or:

F(s)=−ddsG(s){F}{\left({s}\right)}=-\frac{{d}}{{{d}{s}}}{G}{\left({s}\right)} (using property 3).

Integrating this equality with respect to s{s} and taking the limits s{s} and ∞\infty yields:

∫s∞−ddsG(s)ds=∫s∞F(u)du{\int_{{s}}^{\infty}}-\frac{{d}}{{{d}{s}}}{G}{\left({s}\right)}{d}{s}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}.

−(lim⁡a→∞G(a)−G(s))=∫s∞F(u)du-{\left(\lim_{{{a}\to\infty}}{G}{\left({a}\right)}-{G}{\left({s}\right)}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}.

Since G(s)=∫0∞e−stg(t)dt{G}{\left({s}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{g{{\left({t}\right)}}}{d}{t}, we have that lim⁡s→∞G(s)=0\lim_{{{s}\to\infty}}{G}{\left({s}\right)}={0}.

So,

−(0−G(s))=∫s∞F(u)du-{\left({0}-{G}{\left({s}\right)}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}.

G(s)=∫s∞F(u)du{G}{\left({s}\right)}={\int_{{s}}^{\infty}}{F}{\left({u}\right)}{d}{u}.

This completes the proof.

Example 3. Calculate L(sin⁡(t)t){L}{\left(\frac{{{\sin{{\left({t}\right)}}}}}{{t}}\right)}.

Since L(sin⁡(t))=1s2+1{L}{\left({\sin{{\left({t}\right)}}}\right)}=\frac{{1}}{{{{s}}^{{2}}+{1}}}, using property 4, we obtain that

L(sin⁡(t)t)=∫s∞1p2+1dp=lim⁡m→∞∫sm1p2+1dp=lim⁡m→∞atan⁡(p)∣sm={L}{\left(\frac{{{\sin{{\left({t}\right)}}}}}{{t}}\right)}={\int_{{s}}^{\infty}}\frac{{1}}{{{{p}}^{{2}}+{1}}}{d}{p}=\lim_{{{m}\to\infty}}{\int_{{s}}^{{m}}}\frac{{1}}{{{{p}}^{{2}}+{1}}}{d}{p}=\lim_{{{m}\to\infty}}{\operatorname{atan}{{\left({p}\right)}}}{{\mid}_{{s}}^{{m}}}=

=lim⁡m→∞atan⁡(m)−atan⁡(s)=π2−atan⁡(s)=acot⁡(s)=\lim_{{{m}\to\infty}}{\operatorname{atan}{{\left({m}\right)}}}-{\operatorname{atan}{{\left({s}\right)}}}=\frac{\pi}{{2}}-{\operatorname{atan}{{\left({s}\right)}}}={\operatorname{acot}{{\left({s}\right)}}}.