Application of the time-shifting property in case of Fourier-Transform of cosine

Question

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?

Answer 1

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

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

Answer 2

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

Answer 3

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: $$e^{-j\frac{\pi}{2}}=-j$$ $$e^{j\frac{\pi}{2}}=j$$