Applying dead time (pure time delay) after inverse Laplace

Question

If I have the transfer function: $P(S)=\frac{5e^{-3.4s}}{s+1}$

And I perform the inverse Laplace transform on: $\frac{5}{s+1}$

Which gives me: $5e^{-5t}$

How do I now apply the dead time (pure time delay) of: $e^{-3.4s}$ ?

UPDATE - So if I can plot the response in MATLAB using the step() function, and the code below, I get this:

num = 5;
den = [1 1];
P = tf(num,den,'InputDelay',3.4)
figure()
step(P)

What I'm trying to do; is to apply a step input to the transfer function, perform the inverse Laplace transform, apply the time delay, and then plot the response with respect to time (t), and get the same looking MATLAB plot.

Answer 1

The inverse transform of $e^{-cs}\cdot F(s)$ is $u_c(t)\cdot f(t-c)$.

So... $$F(s)=e^{-3.4s}\cdot \frac5{s+1}$$ $$f(t)=5u_{3.4}(t)\cdot e^{-(t-3.4)}$$ $$f(t)=5u_{3.4}(t)\cdot e^{\ 3.4-t}$$ Remember that $u_c(t)$ is defined as $0$ for $t<c$, and equal to $1$ for $t \ge c$. Basically, it's a function that's $0$ until it "turns on" and becomes $1$ at $t=c$.