X(ω)=∫−∞∞x(t)e−jωtdt
The intuition for this transform comes from the Fourier Series. Only periodic signals can be represented by the Fourier Series. If we start with a finite signal x(t), then we can just make it periodic by copying the domain over which it is nonzero so it repeats over a period T. Call this signal x~(t). Since x~ is periodic, we can find its fourier series coefficients.
ak=T1∫Tx~(t)e−jnT2πt=T1∫Tx(t)e−jnT2πt=T1∫−∞∞x(t)e−jnT2πt
These steps are possible because x~(t)=x(t) over a single period, and x(t) is zero outside that period.
Tak=∫−∞∞x(t)e−jnT2πt
Notice that if we let T approach infinity, then ω0=T2π becomes very small, so the Tak can almost be thought of as samples of some continuous time function. What this means is for a general aperiodic signal, regardless of if it is finite or not, we can think of it as having "infinite period" and thus made up of a continuous set of frequencies. This is what motivates the CTFT.
The Inverse Continuous Time Fourier Transform takes us from the frequency domain reprsentation of a function X(ω) to its time domain representation x(t)
x(t)=2π1∫−∞∞X(ω)ejωtdω
We can arrive at this equation by starting from the Fourier series again. Our faux signal x~(t) which was the periodic function we constructed out of our aperiodic one is represented by its Fourier Series
x~(t)=∑k−∞∞akejkω0t=∑k=−∞∞(T1X(ω))ejωt∣w=kω0.
Notice this is just rewriting ak as the samples of the Fourier Transform X(ω). T=ω02π so
x~(t)=2π1∑k=−∞∞ω0X(ω)ejωt∣w=kω0
x(t)=limT→∞x~(t)=limω0→0x~(t)=2π1∫−∞∞X(ω)ejωtdω.
For all these properties, assume that x(t)↔X(ω) and y(t)↔Y(ω) Linearity:
ax(t)+by(t)↔aX(ω)+bY(ω)
x(t−t0)↔e−jωt0X(ω)
x(at)↔∣a∣1X(aω)
x∗(t)↔X∗(−ω)
dtdx(t)↔jωX(ω),dωdX(ω)↔−jtx(t)
(x∗y)(t)↔X(ω)Y(ω),x(t)y(t)↔2π1(X∗Y)(ω)
ejω0tx(t)↔X(ω−ω0)
∫−∞∞∣x(t)∣2dt=2π1∫−∞∞∣X(ω)∣2dω
A big question that arises when thinking about the Fourier Transform is whether or not the integral ∫x(t)e−jωt actually converges.
If ∫−∞∞∣x(t)∣dt converges, then X(ω) exists and is continuous. In addition, X(ω) approaches 0 as ∣ω∣ approaches ∞.
∣x(t)e−jωt∣=∣x(t)∣∣e−jωt∣=∣x(t)∣.
So if one converges, the other must converge. However, this means that x(t)=1, x(t)=sin(ωt), x(t)=cos(ωt) don’t have a "strict" Fourier Series because the integral doesn’t converge for these periodic signals. In order to get around this, we can define a "generalized" Fourier Transform which operates on periodic signals.
Starting with x(t)=1, we know that in the frequency domain, the only consituent frequency is ω=0. This means that X(ω)=kδ(ω) where k is some scalar. Using the Inverse Fourier Transform,
x(t)=2π1∫kδ(ω)ejωtdω=2πk.
That means k=2π, so
x(t)=1↔X(ω)=2πδ(ω).
x(t)=ejω0t=2πδ(ω−ω0).
The generalized Fourier Transform for a periodic signal x(t) is
X(ω)=∑−∞∞ak⋅2πδ(ω−ω0)
where ak are the coefficients of the Fourier Series of x(t).
X(ω)=∑−∞∞x[n]e−jωn
x[n]=2π1∫<2π>X(ω)ejωndω
For all these properties, assume that x[n]↔X(ω) and y[n]↔(ω) Time Shift:
x[n−n0]↔e−jωn0X(ω)
X(ω−ω0)↔ejω0nx[n]
x[−n]↔X(−ω)
x∗[n]=X∗(−ω)
xM[n]↔X(Mω),xM[n]={x[Mn]0when nmodM=0else.
nx[n]↔jdωdX(ω)
x[n]y[n]↔2π1∫2πX(θ)Y(ω−θ)dθ
(x∗y)[n]=X(ω)Y(ω)
If ∑−∞∞∣x[n]∣ converges, then X(ω)exists and is continuous.
Just like in continuous time, periodic signals like x[n]=1,x[n]=sin(ω0t),x[n]=cos(ω0t)... are problematic because they don’t converge under the "strict" transform, so they require a generalized transform. In the frequency domain, a constant signal like x[n]=1 will be the sum of all frequencies. This will look like a sum of Dirac Deltas
X(ω)=k∑l=−∞∞δ(ω−2πl).
x(t)=2π1∫2πk∑l=−∞∞δ(ω−2πl)=2πk∑l=−∞∞∫2πδ(ω−2πl)=2πk∫2πδ(ω−2π⋅0)=2πk.
Therefore k=2π, so
x[n]=1↔X(ω)=2π∑l=−∞∞δ(ω−2πl)
x[n]=ejω0n↔X(ω)=2π∑l=−∞∞δ(ω−ω0−2πl).
Once again using the Fourier Series representation of x[n], we can define the generalized Discrete Time Fourier Transform.
For a perioidic signal x[n], the generalized Discrete Time Fourier Transform is
x[n]↔2π∑−∞∞akδ(ω−N2πk)
where ak are the Fourier Series coefficients of x[n]
For a length N finite sequence {x[n]}0n−1, the Discrete Fourier Transform of the signal is a length N finite sequence {X[k]}0n−1 where
X[k]=∑n=0N−1x[n]e−jN2πkn.
One way to interpret the DFT is in terms of the Fourier series for a disrete periodic signal x~[n]=x[n mod N]. Recall that the coefficient of the kth term of the Fourier Series is
ak=N1∑n=0N−1x[n]e−jN2πkn.
Notice that the ak of the Fourier Series are the DFT values except scaled by a factor of N. This gives an intuitive inverse DFT.
For a length N finite sequence {X[k]}0N−1 representing the DFT of a finite perioidc signal {x[n]}0N−1, the inverse DFT is given by
x[n]=N1∑k=0N−1X[k]ejN2πkn.
One important property of the DFT is its complex conjugacy. When x[n] is a real valued signal, then X[n−k]=X[k]∗. This can easily be shown by substituting N−k into the DFT formula. Further intuition for the DFT comes from relating it to the DTFT. Suppose we have a finite signal x[n] which is 0 for n<0 and n>N−1. The DTFT of this signal is
X(ω)=∑n=−∞∞x[n]e−jωn=∑n=0N−1x[n]e−jωn.
Suppose we sample the DTFT at intervals of N2πk, then
X[k]=X(N2πk)=∑n=0N−1x[n]e−jN2πkn.
Thus we can think of the DFT as a N point sample of the DTFT.
2D CTFT: X(ω1,ω2)=∫−∞∞∫−∞∞x(t1,t2)e−jω0te−jω1tdt1dt2
Inverse 2D CTFT: x(t1,t2)=(2π)21∫−∞∞∫−∞∞X(ω1,ω2)ejω1t1ejω2t2dω1dω2
2D DTFT: X(ω1,ω2)=∑n2=−∞∞∑n1=−∞∞x[n1,n2]e−jω1n1e−jω2n2
Inverse 2D DTFT: x[n1,n2]=∑ω2=−∞∞∑ω1=−∞∞X(ω1,ω2)ejω1n1ejω2n2
2D DFT: X[k1,k2]=∑n2=0N2−1∑n1=0N1−1x[n1,n2]e−jN12πk1n1e−jN22πk2n2
2D DFT: x[n1,n2]=∑k2=0N2−1∑k1=0N1−1X[k1,k2]ejN12πk1n1ejN22πk2n2
If x(t1,t2)=x(t1)x(t2), then X(ω1,ω2)=X(ω1)X(ω2)
