Inverse Laplace transform - MATLAB ilaplace (2025)

Compute the inverse Laplace transform of 1/s^2. By default, the inverse transform is in terms of t.

syms sF = 1/s^2;f = ilaplace(F)

f =$$t$$

Compute the inverse Laplace transform of 1/(s-a)^2. By default, the independent and transformation variables are s and t, respectively.

syms a sF = 1/(s-a)^2;f = ilaplace(F)

f =$$t\hspace{0.17em}{\mathrm{e}}^{a\hspace{0.17em}t}$$

Specify the transformation variable as x. If you specify only one variable, that variable is the transformation variable. The independent variable is still s.

syms xf = ilaplace(F,x)

f =$$x\hspace{0.17em}{\mathrm{e}}^{a\hspace{0.17em}x}$$

Specify both the independent and transformation variables as a and x in the second and third arguments, respectively.

f = ilaplace(F,a,x)

f =$$x\hspace{0.17em}{\mathrm{e}}^{s\hspace{0.17em}x}$$

Compute the following inverse Laplace transforms that involve the Dirac and Heaviside functions.

syms s tf1 = ilaplace(1,s,t)

f1 =$$\delta \text{<span>dirac</span>}\left(t\right)$$

F = exp(-2*s)/(s^2+1);f2 = ilaplace(F,s,t)

f2 =$$\mathrm{heaviside}\left(t-2\right)\hspace{0.17em}\sin \left(t-2\right)$$

Create two functions $\mathit{f}\left(\mathit{t}\right)=\mathrm{heaviside}\left(\mathit{t}\right)$ and $\mathit{g}\left(\mathit{t}\right)=\exp \left(-\mathit{t}\right)$. Find the Laplace transforms of the two functions by using laplace. Because the Laplace transform is defined as a unilateral or one-sided transform, it only applies to the signals in the region $\mathit{t}\ge 0$.

syms t positivef(t) = heaviside(t);g(t) = exp(-t);F = laplace(f);G = laplace(g);

Find the inverse Laplace transform of the product of the Laplace transforms of the two functions.

h = ilaplace(F*G)

h =$$1-{\mathrm{e}}^{-t}$$

According to the convolution theorem for causal signals, the inverse Laplace transform of this product is equal to the convolution of the two functions, which is the integral ${\int }_0^{\mathit{t}}\mathit{f}\left(\tau \right)\mathit{g}\left(\mathit{t}-\tau \right)\mathit{d}\tau$ with $\mathit{t}\ge \mathrm{0}$. Find this integral.

syms tauconv_fg = int(f(tau)*g(t-tau),tau,0,t)

conv_fg =$$1-{\mathrm{e}}^{-t}$$

Show that the inverse Laplace transform of the product of the Laplace transforms is equal to the convolution, where h is equal to conv_fg.

isAlways(h == conv_fg)

ans = logical 1

Find the inverse Laplace transform of the matrix M. Specify the independent and transformation variables for each matrix entry by using matrices of the same size. When the arguments are nonscalars, ilaplace acts on them element-wise.

syms a b c d w x y zM = [exp(x) 1; sin(y) 1i*z];vars = [w x; y z];transVars = [a b; c d];f = ilaplace(M,vars,transVars)

f =
$$\left(\begin{array}{cc}{\mathrm{e}}^x\hspace{0.17em}\delta \text{<span>dirac</span>}\left(a\right)& \delta \text{<span>dirac</span>}\left(b\right)\\ \mathrm{ilaplace}\left(\sin \left(y\right),y,c\right)& {\delta \text{<span>dirac</span>}}^{\prime }\left(d\right)\hspace{0.17em}\mathrm{i}\end{array}\right)$$

If ilaplace is called with both scalar and nonscalar arguments, then it expands the scalars to match the nonscalars by using scalar expansion. Nonscalar arguments must be the same size.

syms w x y z a b c df = ilaplace(x,vars,transVars)

f =
$$\left(\begin{array}{cc}x\hspace{0.17em}\delta \text{<span>dirac</span>}\left(a\right)& {\delta \text{<span>dirac</span>}}^{\prime }\left(b\right)\\ x\hspace{0.17em}\delta \text{<span>dirac</span>}\left(c\right)& x\hspace{0.17em}\delta \text{<span>dirac</span>}\left(d\right)\end{array}\right)$$

Compute the Inverse Laplace transform of symbolic functions. When the first argument contains symbolic functions, then the second argument must be a scalar.

syms F1(x) F2(x) a bF1(x) = exp(x);F2(x) = x;f = ilaplace([F1 F2],x,[a b])

f =$$\left(\begin{array}{cc}\mathrm{ilaplace}\left({\mathrm{e}}^x,x,a\right)& {\delta \text{<span>dirac</span>}}^{\prime }\left(b\right)\end{array}\right)$$

If ilaplace cannot compute the inverse transform, then it returns an unevaluated call to ilaplace.

syms F(s) tF(s) = exp(s);f(t) = ilaplace(F,s,t)

f(t) =$$\mathrm{ilaplace}\left({\mathrm{e}}^s,s,t\right)$$

Return the original expression by using laplace.

F(s) = laplace(f,t,s)

F(s) =$${\mathrm{e}}^s$$