Minor changes

source/signal_processing/code/dft.R

@@ -11,7 +11,7 @@ X3=Xkn(3,0)+Xkn(3,1)+Xkn(3,2)+Xkn(3,3)
 
 X=c(X0,X1,X2,X3) # This is your final fft
 
-f0=0/N*fs # k/N*fs
+f0=0/N*fs # f=k/N*fs
 f1=1/N*fs
 f2=2/N*fs
 f3=3/N*fs

source/signal_processing/code/idft.R

@@ -0,0 +1,12 @@
+X=fft(c(1,2,3,4)) # Get the DFT of our signal
+N=4 # Signal size
+xkn=function(k,n){X[k+1]*exp(2i*pi*(k/N)*n)}
+
+# Perform the inverse DFT:
+x0=1/N*(xkn(0,0)+xkn(1,0)+xkn(2,0)+xkn(3,0))
+x1=1/N*(xkn(0,1)+xkn(1,1)+xkn(2,1)+xkn(3,1))
+x2=1/N*(xkn(0,2)+xkn(1,2)+xkn(2,2)+xkn(3,2))
+x3=1/N*(xkn(0,3)+xkn(1,3)+xkn(2,3)+xkn(3,3))
+
+x=Re(c(x0,x1,x2,x3)) # Take the real
+stopifnot(any(Im(x)==0)) # Also see how all imaginary parts are 0!

source/signal_processing/dft.rst

@@ -49,3 +49,63 @@ In depth example
 Let's take a discrete signal :math:`\vec{x}=\{1,2,3,4\}`, sampled at a frequency :math:`f_s=10 Hz`.
 
 .. literalinclude:: code/dft.R
+   :language: R
+
+Now let's do its inverse DFT:
+
+.. literalinclude:: code/idft.R
+   :language: R
+
+Understanding how imaginary parts cancel out in during the inverse DFT is important.
+The periodicity of the DFT plays a key role for that.
+For our signal :math:`\vec{x}` we have:
+
+.. math::
+   \vec{X}=\{10+0j, -2+2j, -2+0j, -2-2j\}
+
+See how :math:`X_1 = \overline{X_3}`. When computing :math:`x_n`, :math:`X_1` becomes:
+
+.. math::
+    X_1 \cdot e^{j2\pi\frac{1}{N}n} &= ({\color{blue}-2}+{\color{red}2j}) \cdot (cos(2\pi\frac{1}{4}n)+j\cdot sin(2\pi\frac{1}{4}n))
+
+    &= {\color{blue}-2 \cdot cos(2\pi\frac{1}{4}n) -2j \cdot sin(2\pi\frac{1}{4}n)} + {\color{red}2j \cdot cos(2\pi\frac{1}{4}n) -2 \cdot sin(2\pi\frac{1}{4}n)}
+
+In turn,  :math:`X_3` becomes:
+
+
+.. math::
+    X_3 \cdot e^{j2\pi\frac{3}{N}n} &= ({\color{blue}-2}{\color{red}-2j}) \cdot (cos(2\pi\frac{3}{4}n)+j\cdot sin(2\pi\frac{3}{4}n))
+
+    &= {\color{blue}-2 \cdot cos(2\pi\frac{3}{4}n) -2j \cdot sin(2\pi\frac{3}{4}n)} {\color{red}- 2j \cdot cos(2\pi\frac{3}{4}n) + 2 \cdot sin(2\pi\frac{3}{4}n)}
+
+Notice that:
+
+.. math::
+   \forall n:\, & cos(2\pi\frac{1}{4}n) = 0
+
+   & cos(2\pi\frac{3}{4}n) = 0
+
+   & sin(2\pi\frac{1}{4}n) = j
+
+   & sin(2\pi\frac{3}{4}n) = -j
+
+In particular, see how the symmetry of the DFT leads to opposite values.
+Plugin in these new values whenever :math:`j` shows up gives the following:
+
+.. math::
+   X_1 \cdot e^{j2\pi\frac{1}{N}n} &= -2 \cdot cos(2\pi\frac{1}{4}n) + (-2j \cdot j) + (2j \cdot 0) -2 \cdot sin(2\pi\frac{1}{4}n)
+
+   &=  -2 \cdot cos(2\pi\frac{1}{4}n) + 2 -2 \cdot sin(2\pi\frac{1}{4}n)
+
+   X_3 \cdot e^{j2\pi\frac{3}{N}n} &= -2 \cdot cos(2\pi\frac{3}{4}n) +(-2j \cdot -j) +(- 2j \cdot 0) + 2 \cdot sin(2\pi\frac{3}{4}n)
+
+   &= -2 \cdot cos(2\pi\frac{3}{4}n) - 2 + 2 \cdot sin(2\pi\frac{3}{4}n)
+
+As these two terms get summed during the inverse DFT gives:
+
+.. math::
+   X_1 \cdot e^{j2\pi\frac{1}{N}n} +  X_3 \cdot e^{j2\pi\frac{3}{N}n} = -2 \cdot cos(2\pi\frac{1}{4}n) \cancel{+ 2} -2 \cdot sin(2\pi\frac{1}{4}n)\\
+
+   -2 \cdot cos(2\pi\frac{3}{4}n) \cancel{- 2} + 2 \cdot sin(2\pi\frac{3}{4}n)
+
+An you are left with sum of real signals.