Homework Three

From Class Wiki
Revision as of 18:50, 14 October 2009 by Nicholas.Christman (talk | contribs)
Jump to navigation Jump to search

October 5th, 2009, class notes (as interpreted by Nick Christman)


The topic covered in class on October 5th was about how to deal with signals that are not periodic.

Given the following Fourier series (Equation 1), what if the signal is not periodic?

x(t)=x(t+T)=n=αnej2πntT where αn=1TT2T2x(t)ej2πntTdt

To investigate this potential disaster, let's look at what happens as the period increases (i.e. not periodic). Essentially, as T we can say the following:

1T

df

nT

f

n=1T

( )df

With this, we get the following (Equation 2):

x(t)=limT[n=(1TT2T2x(t)ej2πntTdt)ej2πntT](x(t)ej2πftdt)ej2πtfdf

Given the above equivalence, we say the following:

X(f)=x(t)ej2πftdt

Therefore, we have obtained an equation to relate the Fourier analysis of a function in the time-domain to the frequency-domain:

X(f)=x(t)ej2πftdt x(t)|ej2πtf x(t) projected onto ej2πtf
x(t)=X(f)ej2πftdf X(f)|ej2πtf x(f) projected onto ej2πtf


From this we can see that x(t) is the inverse Laplace transform of X(f). Similarly, X(f) is the Laplace transform of x(t)


The next thing we did was rearranged some limits within Equation 2 (given the above similarities) to give us the following:

(x(t)ej2πftdt)ej2πtfdf

x(t)(ej2πf(tt)df)dt

Notice that ej2πf(tt)δ(tt)

Similarly,

(X(f)ej2πftdt)ej2πtfdf

X(f)(ej2πt(ff)df)dt

Again, notice that ej2πf(ff)δ(ff)=δ(ff)


Back