- #1
sam.green
- 6
- 0
Hello,
I am working through Signals and Systems Demystified, but I am at an impasse.
I would like to take the inverse Fourier transform of
[tex]
H(f)=\begin{cases}
-j&\text{if } f > 0\\
j&\text{if } f<0\end{cases}
[/tex]
So
[tex]
h(t) = \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{0}^{\infty} -je^{j2\pi ft}df
= \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{-\infty}^{0} je^{j2\pi f t}df
= 2\int_{-\infty}^{0} je^{j2\pi f t}df
= \frac{j2e^{j2\pi ft}}{j2\pi t}|_{-\infty}^{0}
[/tex]
So
[tex]
h(t) = \frac{1}{\pi t} - \lim_{f \rightarrow -\infty} \frac{e^{j2\pi ft}}{\pi t}
[/tex]
The book says that [tex]h(t) = \frac{1}{\pi t}[/tex] is the answer, but I don't understand how. Isn't [tex]\lim_{f \rightarrow -\infty} \frac{e^{j2\pi ft}}{\pi t}[/tex] periodic and nonconverging?
I am working through Signals and Systems Demystified, but I am at an impasse.
I would like to take the inverse Fourier transform of
[tex]
H(f)=\begin{cases}
-j&\text{if } f > 0\\
j&\text{if } f<0\end{cases}
[/tex]
So
[tex]
h(t) = \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{0}^{\infty} -je^{j2\pi ft}df
= \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{-\infty}^{0} je^{j2\pi f t}df
= 2\int_{-\infty}^{0} je^{j2\pi f t}df
= \frac{j2e^{j2\pi ft}}{j2\pi t}|_{-\infty}^{0}
[/tex]
So
[tex]
h(t) = \frac{1}{\pi t} - \lim_{f \rightarrow -\infty} \frac{e^{j2\pi ft}}{\pi t}
[/tex]
The book says that [tex]h(t) = \frac{1}{\pi t}[/tex] is the answer, but I don't understand how. Isn't [tex]\lim_{f \rightarrow -\infty} \frac{e^{j2\pi ft}}{\pi t}[/tex] periodic and nonconverging?