# Fourier Transform of a Modified Impulse Train

1. Jul 20, 2007

### WolfOfTheSteps

I hope this is OK to post here. I thought it would be better here than in the math questions forum, since you are EEs, and probably have more experience dealing with things related to the delta function.

Problem

Let

$$\hat{x}(t) = \sum_{k=-\infty}^{\infty}\delta(t-2k).$$

Now let

$$x(t) = 2\hat{x}(t) + \hat{x}(t-1).$$

Find the Fourier Transform of $x(t)$.

Given Solution

Here is the official solution given by Oppenheim:

From Table 4.2,

$$\hat{X}(j\omega) = \pi \sum_{k=-\infty}^{\infty}\delta(\omega - \pi k)$$

Therefore,

$$X(j\omega)=\hat{X}(j\omega)[2+e^{-\omega}] = \pi \sum_{k=-\infty}^{\infty}\delta(\omega-\pi k)[2 + (-1)^k].$$

My Two Questions

Question 1. When applying the time-shifting property of Fourier series, why did they multiply by $e^{-\omega}$ and not $e^{-j\omega}$??

Question 2. How did the $e^{-\omega}$ become $(-1)^k$ in the final answer?

Thank you!

Last edited: Jul 20, 2007
2. Jul 21, 2007

### rbj

it's a mistake. i don't have the book, but if you represented this correctly, they are wrong and you are right. delaying by one sample multiplies the continuous spectrum by $e^{-j\omega}$ . this will fix your Q2 also.

3. Jul 21, 2007

### WolfOfTheSteps

Maybe not! How does $e^{-j\omega}$ become $(-1)^k$ in the final answer? I know $e^{-j k \pi} = (-1)^k$ but how could that come from $e^{-j\omega}$ when omega is variable and there is no k? Is the final answer also a typo?

Thanks!

4. Jul 24, 2007

### mdelisio

Remember, you are multiplying the $e^{-j \omega}$ by a chain of delta functions (in $\omega$-space). The only thing that matters is where the delta function is nonzero.

Remember $f(x)\delta(x-a) = f(a)\delta(x-a)$

5. Jul 24, 2007

### WolfOfTheSteps

Of course! Which happens exactly when $\omega = \pi k$. I should have realized that!

Thanks!