Topic: Frequency domain view of filtering.

## Question

(Note: There is a very high chance of a question like this on the final.) Define a signal x(t) and take samples every T (using a specific value of T). Store the samples in a discrete-time signal z[n]. Obtain a mathematical expression for the Fourier transform of x(t) and sketch it. Obtain a mathematical expression for the Fourier transform of y[n] and sketch it.

Oops! Actually, y[n] should be z[n] above. -pm

Let's hope we get a lot of different signals from different students! Post Your answer/questions below.

• Instructor's note: if you make a mistake and somebody else comments on it, please do not remove it when you make corrections later. Instead, just start over below your previous answer. The idea is to learn from our mistakes! -pm

$x(t) = \delta(t), T=1$

\begin{align} z[n] &= x_T[n] \\ &= \delta(t+T) \end{align}

Fourier Transform of x(t) = 1

$y[n] = x(t)*z[n]$ <-- is this correct?

• No, there is a serious problem with your signal y[n]: it's defined as a function of both t and n. It should NOT depend on n.

I only solved the general form for this problem.

\begin{align} \mathcal{F}(x(t)) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} \end{align}

\begin{align} z[n] = comb_T(x(t)) \end{align}

• Oops! Your z[n] is actually a function of t (because a comb is the multiplication of two functions of t, namely the signal x(t) and an impulse train p(t) .

I'm not sure what y[n] is equal to. I'm assuming that y[n] is the same as z[n]. Then the FT of y[n] is

\begin{align} Y(e^{j\omega}) = \frac{1}{T}rep_\frac{1}{T}(X(e^{j\omega})) \end{align}

- Mike Wolfer

T=3

\begin{align} x(t)=cos(2t) \end{align}

Sampling consists mathematically of a substitution of t->3n Therefore,

\begin{align} z {\color{red} [}(n){\color{red} ]} = cos(2(3n)) \end{align}

To make calculations easier Euler Form is used:

\begin{align} cos(6n) -> {\color{red} =} 1/2(e^{i6n}+e^{-i6n}) \end{align}

Inverse CTFT: (You should use the formula in terms of f in hertz.)

$x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \! X(\omega)e^{j \omega t} dw$

$1/2e^{i2t}+1/2e^{-i2t} = \frac{1}{2\pi} \int_{-\infty}^{\infty} \! X(\omega)e^{j \omega t} dw$

Table -> (Here you could just say "The function X(f) that satisfies this equation is")

\begin{align} x(f)=1/2(\delta(f - 2/(2\pi)) +\delta(f + 2/(2\pi))) \end{align}

Sketch consists of two positive spikes, at 1/pi and -1/pi on the X axis(f).

IDFT:

$x[n] = \frac{1}{2\pi} \int_{-\infty {\color{red}-\pi} }^{\infty {\color{red} \pi} } X(\omega)e^{j \omega n} d\omega$

$1/2e^{i6n}+1/2e^{-i6n}= \frac{1}{2\pi} \int_{-\infty {\color{red}-\pi}}^{\infty {\color{red} \pi}} X(\omega)e^{j \omega n} d\omega$

Table -> (You could just say "The function z(w) that satisfies this equation is the following, where k is an integer chosen so that the spikes lie between -pi and pi.")

\begin{align} z(w)=\pi( \delta(\omega -6{\color{red}+2\pi k}) +\delta(\omega+6{\color{red}+2\pi k}) ){\color{red}, -\pi<w<\pi } \end{align}

Sketch consists of a pair of spikes repeated every 2 pi. Given that the original is not in the initial +/-pi range, it is shifted by 2 pi intervals till the spikes are within.

6-2pi-> spikes at +/- .283

Instructor's note: The method used above directly relies on the FT formulas. This is a good way to attack such problems if you don't remember the relationship between the sampling and the signal (although I do hope you will remember it for the test). -pm

Can you please elaborate on how to handle sampling mathematically? The only representation we have is in a graphical manner, and that makes it really difficult to demonstrate the relationship between sampling and the signal in a concrete, non-picture form. ~Ajfunche

Instructor's answer. Sure. Let me first give you a single mathematical expression to describe this relationship:

$\text{If } x_d[n]=x(nT), \text{ then } X_d(\omega) = \frac{1}{T} \sum_{k=-\infty}^\infty X \left( \frac{\omega}{2\pi T}+\frac{k}{T}\right).$

Now to check that this expression is correct, it is a good idea to check that $X(\omega)$ is periodic with period $2\pi$ We have

\begin{align} X_d(\omega+2\pi) &=\frac{1}{T} \sum_{k=-\infty}^\infty X \left( \frac{\omega+2\pi}{2\pi T}+\frac{k}{T}\right) \\ &=\frac{1}{T} \sum_{k=-\infty}^\infty X \left( \frac{\omega}{2\pi T}+\frac{1}{T}+\frac{k}{T}\right) \\ &=\frac{1}{T} \sum_{k=-\infty}^\infty X \left( \frac{\omega}{2\pi T}+\frac{(1+k)}{T}\right) \\ (\text{letting }k'=k+1) &=\frac{1}{T} \sum_{k'=-\infty}^\infty X \left( \frac{\omega}{2\pi T}+\frac{(k')}{T}\right) \\ (\text{replacing }k' \text{ by } k) &= \frac{1}{T} \sum_{k=-\infty}^\infty X \left( \frac{\omega}{2\pi T}+\frac{k}{T}\right) \\ &= X_d(\omega) \end{align}

So yes, the expression defines a function that is periodic with period $2\pi$.

Now, to understand where this expression comes from, one first needs to look at the Fourier transform of the comb of the signal, i.e. the "analog sampling"

\begin{align} x_s (t) &= \text{comb}_T \left\{ x(t) \right\}, \\ &= x(t) p_T(t), \\ &= x(t) \sum_{k=\infty}^\infty \delta (t-kT),\\ &= \sum_{k=\infty}^\infty x(kT) \delta (t-kT). \end{align}

As we have shown in class, "the fourier transform of a comb is a rep". More precisely

\begin{align} X_s (f) &= \frac{1}{T} \text{rep}_{\frac{1}{T}} \left\{ X(f) \right\}, \\ &= \frac{1}{T} X(f) * p_{\frac{1}{T}}(f) \\ &=\frac{1}{T} X(f)* \sum_{k=\infty}^\infty \delta (f-\frac{k}{T}),\\ &= \frac{1}{T} \sum_{k=\infty}^\infty X (f-\frac{k}{T}). (*) \end{align}

One can show (in a similar way as above) that this function is periodic with period 1/T.

Now there is a very simple relationship between $X_s (f)$ and $X_d(\omega)$, namely

$X_d(\omega) =X_s \left(\frac{\omega}{2\pi T} \right).$

I do not have time to write the proof now, but it is in your lecture notes.

The expression we are concerned about comes from this relationship combined with Equation $*$ above. -pm