0
$\begingroup$
  1. Time-shifting property: $x[n-n_d] \xrightarrow{\mathscr{F}} e^{-j\omega n_d} X(e^{j\omega}) $
  2. Fourier-Transform of cosine-signal: $\cos(\omega_0n) \xrightarrow{\mathscr{F}} \frac{1}{2}(\delta(\omega - \omega_0) + \delta(\omega+\omega_0)) $

Combining 1. & 2. together, I am getting: $\cos(\omega_0n - \frac{\pi}{2}) \xrightarrow{\mathscr{F}} e^{-j\frac{\pi}{2}}\frac{1}{2}(\delta(\omega - \omega_0) + \delta(\omega+\omega_0)) $, but instead the Fourier-Transform of

$$\cos(\omega_0n - \tfrac{\pi}{2})$$

is

$$\cos(\omega_0n - \tfrac{\pi}{2}) \xrightarrow{\mathscr{F}} \tfrac{1}{2}\delta(\omega - \omega_0)e^{-j\frac{\pi}{2}} + \tfrac{1}{2}\delta(\omega+\omega_0)e^{j\frac{\pi}{2}}$$

Can anyone tell what I'm doing wrong here?

$\endgroup$
1
$\begingroup$

You have done a wrong calculation. First, you need to write the cosine as

$$ \cos(\omega_0n-\pi/2)=\cos\left(\omega_0(n-\tfrac{\pi}{2\omega_0})\right) $$

i.e. the time-shift needs to be performed on the non-scaled version of the time variable $n$. Then, you apply the Fourier Transform:

$$ \mathscr{F}\left\{\cos\left(\omega_0(n-\tfrac{\pi}{2\omega_0})\right)\right\}=\exp\left(-j\tfrac{\pi\omega}{2\omega_0}\right)\tfrac{1}{2}\big(\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\big) $$

And now, with the filtering property of the Dirac impulse you end up with the correct result

$$ \mathscr{F}\left\{\cos\left(\omega_0(n-\tfrac{\pi}{2\omega_0})\right)\right\}=\tfrac{1}{2}e^{-j\pi/2}\delta(\omega-\omega_0)+\tfrac{1}{2}e^{j\pi/2}\delta(\omega+\omega_0) $$

$\endgroup$
1
$\begingroup$

Two things

  1. Time shift equation is wrong. Should be $e^{-j \omega n_d}$.
  2. You need to express the cosine as a time shift $cos(\omega (n - \pi /2 /\omega))$ Time shift is not the same as phase shift and the time shift is actually pi/2 divided by the frequency
$\endgroup$
1
$\begingroup$

To find the Fourier transform of $\cos(\omega_0n - \frac{\pi}{2})$, you can use

$$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$ That is: $$\begin{align} \cos(\omega_0n - \tfrac{\pi}{2})&=\cos(\omega_0n)\cos\left(\frac{\pi}{2}\right)+\sin(\omega_0n)\sin\left(\frac{\pi}{2}\right)\\ &=\sin(\omega_0n)\end{align}$$

Which gives you $$\mathcal{F}\{\sin(\omega_0n)\}=\frac{j}{2}\left(\delta(\omega + \omega_0) - \delta(\omega-\omega_0)\right)$$

which in fact is equal to $$\tfrac{1}{2}\delta(\omega - \omega_0)e^{-j\frac{\pi}{2}} + \tfrac{1}{2}\delta(\omega+\omega_0)e^{j\frac{\pi}{2}}$$ when you expand the two complex exponentials: $$e^{-j\frac{\pi}{2}}=-j$$ $$e^{j\frac{\pi}{2}}=j$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.