Impulse response of a transfer function Lets say we have a transfer function from an LTI-system that goes as follows, and we want the impulse response for it:
$$\frac{10\cdot e^{\left(-\mathrm{i}\right)\cdot \pi\cdot f\cdot 8}}{5+\mathrm{i}\cdot 2\cdot \pi\cdot f}$$
The way I would approach solving this, would be to use the following fourier transform formulas:
$$\frac{1}{a+i2\pi*f}=exp(-at)*u(t)=\frac{10}{5+i2\pi*f}=10*exp(-5t)*u(t)$$
$$G(f)*exp(-i*2*\pi*f*t_0)=g(t-t_0)$$
$$exp(-i*\pi*f*8)=exp(-i*\pi*2*f*4) =&gt;t_0=4$$
$$exp(-i*2*\pi*f*4)=g(t-4)$$
Which would end up with the following result:
$$\frac{10\cdot e^{\left(-\mathrm{i}\right)\cdot \pi\cdot f\cdot 8}}{5+\mathrm{i}\cdot 2\cdot \pi\cdot f}=10*exp(-5t)*u(t)*g(t-4)$$
However, this seems to be incorrect. Did I do an error with the fourier transforms, or have I misunderstood some of the formulas?
You're almost there; you just need to connect a few dots.  Let your $\frac{10}{5 + i 2 \pi f} = G(f)$.  Then $g(t) = 10 u(t) e^{-5 t}$.

Now we get into that exponent part.  Your $h(t) = g(t - t_0)$ is correct, but you're applying it incorrectly.  You need to apply it to the part that I've labeled as $g(t)$: $h(t) = g(t - t_0) = 10 u(t - t_0) e^{-5 (t - t_0)} = 10 u(t - 4) e^{-5 (t - 4)}$.

So, basically, it's the output you'd expect from the "pure" transfer function, just delayed by 4 time units (presumably seconds). 

Does that help?