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: $$\frac{1}{a+i2\pi*f}=exp(-at)*u(t)=\frac{10}{5+i2\pi*f}=10*exp(-5t)*u(t)$$ $$G(f)*exp(-i*2*\pi*f*t_0)=g(t-t_0)$$ $$exp(-i*\pi*f*8)=exp(-i*\pi*2*f*4) =>t_0=4$$ $$exp(-i*2*\pi*f*4)=g(t-4)$$ 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?
1 Answer
$\begingroup$
$\endgroup$
1
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?
answered Nov 18, 2019 at 16:48
\expfor the exponential, removing a few*that could be mistaken for a convolution, and some\cdot. Classicaly, you can avoid some of them if you write expressions in different ways, from the simpler to the more complex, typically: sign/integer/i/pi/variable. For instance : $e^{(-i)\cdot \pi \cdot f\cdot 8}$ rewrites $e^{-8i\pi f } $. $\endgroup$