Archive for March, 2009

Linear Systems – Point of View

March 20, 2009 at 5:51 pm · Filed under Signal Processing

This post is going to be much shorter than the previous one, and might be very obvious to the most of you. However, I still found it worth writing.

A continuous time linear system can be described using a function of 2 variables, i.e.

h(t,\tau),

which is the impulse response (a function of t) for an impulse signal input arriving at time \tau.

The figure below describes a general linear system:

LinearSystem

The output is calculated by convolving the input and the impulse response of the system, with respect to \tau:

y(t)=\overset{+\infty}{\underset{-\infty}{\int}}h(t,\tau)\cdot x(\tau)d\tau\equiv A,

which is actually a superposition of all impulse responses.
However, a linear system can be observed in a slightly different manner (yet equivalent).

Let me re-write the integral by changing variables:

A\underset{\tau=t-\alpha}{=}\overset{-\infty}{\underset{+\infty}{\int}}h(t,t-\alpha)\cdot x(t-\alpha)d(-\alpha)=\overset{+\infty}{\underset{-\infty}{\int}}h(t,t-\alpha)\cdot x(t-\alpha)d\alpha\equiv B

Now lets define a new function, \overset{\sim}{h}(t,\tau), as the impulse response at time t for an impulse arriving at time t-\tau.

We can say that \overset{\sim}{h}(t,\tau) is a different representation of the system, yet fully describes it.

Note that the relation between \overset{\sim}{h}(t,\tau) and h(t,\tau) is given by:

\overset{\sim}{h}(t,\tau)=h(t,t-\tau)

It follows that:

B=\overset{+\infty}{\underset{-\infty}{\int}}\overset{\sim}{h}(t,\alpha)\cdot x(t-\alpha)d\alpha

Replacing \alpha with \tau, we get that:

y(t)=\overset{+\infty}{\underset{-\infty}{\int}}\overset{\sim}{h}(t,\tau)\cdot x(t-\tau)d\tau.

The latter shows the output of a linear system as a superposition of functions,each is a delayed variation of the input signal, x(t), multiplied by some function of t, \overset{\sim}{h}(t,\tau).

This way is sometimes preferable, for example in wireless communication systems involving multipath channel.

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Sampling – Projection Onto a Sinc(.) Basis ?

March 13, 2009 at 10:34 pm · Filed under Signal Processing

The idea for my first post – the one you’re reading right now – came to my mind after I attended a guest lecture at Tel-Aviv University. The latter was given by Prof. Alan V. Oppenheim from MIT. His lecture carried the title “sampling, sampling”, so you can guess what it was about. What I would like to talk about this time is a brief note, made by Prof. Oppenheim during his lecture, which yet caught my attention. Let me quickly refresh your memory with the basics of uniform sampling model, and I promise to get back to that point right afterwards.

Lets assume a continuous-time signal, X(t); sampling it yields a discrete-time signal, X[n] = X(nT), where T is the sampling period. The mathematical model commonly used to describe the sampler is multiplication of X(t) by an impulse train, I(t)=\underset{n}{\sum}\delta(t-nT), and then converting every Dirac delta function (continuous time impulse) to a Kronecker delta function (discrete time impulse).

Under the assumption that X(t) satisfies the Nyquist–Shannon criterion, i.e. band-limited (-\frac{\pi}{T},\frac{\pi}{T}), it can be fully recovered from its samples X[n] by using an ideal interpolator. The latter first converts every Kronecker delta function of X[n] to a Dirac delta function, and then applies an ideal LPF with a cut-off frequency of \frac{\pi}{T}.

The figure below describes this model:

Sampler and Interpolator Figure

Sampler and Interpolator Figure

The formula for the recovered signal,X_{r}(t) , is given by:

X_{r}(t)=\underset{n=-\infty}{\overset{\infty}{\sum}}X[n]\cdot Sinc(\frac{\pi(t-nT)}{T}),

which is sometimes referred to as Whittaker–Shannon interpolation formula.

If X(t) satisfies the Nyquist–Shannon criterion, we get X_{r}(t)=X(t). The common proof is achieved in the frequency domain.

Let’s now go back to the point. Prof. Oppenheim wanted us to think of X[n] as a projection of X(t) onto a Sinc(.) basis – \{Sinc(\frac{\pi(t-nT)}{T})\}_{n=-\infty}^{\infty}. This seems very logical when one’s looking at the interpolation formula; \{Sinc(\frac{\pi(t-nT)}{T})\}_{n=-\infty}^{\infty}is the basis, and X[n] are the coefficients. The equality X_{r}(t)=X(t) is proven (for X(t) functions that satisfy the Nyquist–Shannon criterion), and everything seems to fall into place.

However, I was wondering to myself, how could it be that sampling a continuous-time signal is the same as projecting it onto a Sinc(.) basis ?

How would a “straight-forward” proof of this claim look like ? How would the given data of X(t) (band limitation) be used in that proof ?

Here are my notes, hope you find them interesting:

Let’s define:

\varphi_{n}(t)\equiv Sinc(\frac{\pi(t-nT)}{T})

Its Fourier transform is:

\phi_{n}(w)\equiv F\{\varphi_{n}(t)\}=T\cdot e^{-jwnT}\cdot\left\{ \begin{array}{ccc} 1 & , & |w|<\frac{\pi}{T} \\ 0 & , & o.w \end{array}\right.

Let’s also define:

<\varphi_{n}(t),\varphi_{m}(t)>\equiv\int\varphi_{n}(t)\varphi_{m}^{*}(t)dt

(who has just said “inner product” ?)

By Parseval’s/Plancherel’s theorem:

<\varphi_{n}(t),\varphi_{m}(t)>=A\cdot<\phi_{n}(w),\phi_{m}(w)>=B\cdot\overset{+\frac{\pi}{T}}{\underset{-\frac{\pi}{T}}{\int}}e^{-jw(n-m)T}dw=

=\left\{ \begin{array}{ccc} -B\cdot\frac{1}{j(n-m)T}e^{-jw(n-m)T}\mid_{-\frac{\pi}{T}}^{+\frac{\pi}{T}}=0 & , & n\neq m \\ C & , & n=m\end{array}\right.=C\cdot\delta[n-m]

This shows that the functions \{Sinc(\frac{\pi(t-nT)}{T})\}_{n=-\infty}^{\infty} are orthogonal to one another. Now we would like to examine the projection of X(t) on Sinc(\frac{\pi(t-nT)}{T}):

<X(t),\varphi_{n}(t)>=\alpha\cdot<X(w),\phi_{n}(w)>=\alpha\cdot T\overset{+\frac{\pi}{T}}{\underset{-\frac{\pi}{T}}{\int}}X(w)\cdot e^{+jwnT}dw\equiv R,

where X(w)=F\{X(t)\} is the Fourier transform of X(t).

Thus, if X(t) is band-limited (-\frac{\pi}{T},\frac{\pi}{T}), i.e. X(w)=0,w\notin(-\frac{\pi}{T},\frac{\pi}{T}), R can be re-written as:

R=\alpha\cdot T\overset{+\infty}{\underset{-\infty}{\int}}X(w)\cdot e^{+jw(nT)}dw=\beta\cdot F^{-1}\{X(w)\}\mid_{t=nT}=\beta\cdot X(nT),

meaning that the projection of X(t) (which satisfies the Nyquist–Shannon criterion) on Sinc(\frac{\pi(t-nT)}{T}) is equivalent to sampling X(t) at t=nT !

Nice, isn’t it?

Only one comment: What I have shown here is not a proof. It is merely an insight. A rigorous proof will have to show that the set of functions \{Sinc(\frac{\pi(t-nT)}{T})\}_{n=-\infty}^{\infty} is a complete orthogonal system, with an inner product as (not) defined above.

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