CT vs. DT Fourier Series: A Comparative Recap
For a periodic CT signal \(x(t)\) with fundamental period \(T_0\) and fundamental frequency \(\omega_0 = 2\pi/T_0\).
Synthesis Equation:
(Reconstruction from coefficients)
\[ x(t) = \sum_{k=-\infty}^{\infty} a_k e^{j k \omega_0 t} \]
Analysis Equation:
(Calculation of coefficients)
\[ a_k = \frac{1}{T_0} \int_{T_0} x(t) e^{-j k \omega_0 t} dt \]
Note
Key Characteristics:
For a periodic DT signal \(x[n]\) with fundamental period \(N\) and fundamental frequency \(\omega_0 = 2\pi/N\).
Synthesis Equation:
(Reconstruction from coefficients)
\[ x[n] = \sum_{k=\langle N\rangle} a_k e^{j k \omega_0 n} = \sum_{k=\langle N\rangle} a_k e^{j k(2\pi/N)n} \]
Analysis Equation:
(Calculation of coefficients)
\[ a_k = \frac{1}{N} \sum_{n=\langle N\rangle} x[n] e^{-j k \omega_0 n} = \frac{1}{N} \sum_{n=\langle N\rangle} x[n] e^{-j k(2\pi/N)n} \]
Note
Key Characteristics:
Consider a CT periodic square wave \(x(t)\) with period \(T_0\), amplitude \(A=1\), and pulse width \(T_1\).
Let \(T_0=1\) and \(T_1=0.5\).
The Fourier series coefficients are given by:
\[ a_k = \frac{A T_1}{T_0} \text{sinc}\left(k \frac{T_1}{T_0}\right) = \frac{A T_1}{T_0} \frac{\sin(\pi k T_1/T_0)}{\pi k T_1/T_0} \]
For \(k=0\), \(a_0 = \frac{A T_1}{T_0}\).
Here we visualize the CT square wave and its Fourier coefficients.
Consider a DT periodic square wave \(x[n]\) with period \(N\), amplitude \(A=1\), and pulse width \(P\).
Let \(N=10\) and \(P=5\) (i.e., \(2N_1+1=5 \implies N_1=2\)).
The Fourier series coefficients are given by:
\[ a_k = \frac{A}{N} \frac{\sin(\pi k P/N)}{\sin(\pi k/N)}, \quad k \neq 0, \pm N, \pm 2N, \ldots \]
For \(k=0, \pm N, \ldots\):
\[ a_k = \frac{A P}{N} \]
Here we visualize the DT square wave and its Fourier coefficients.
k continuous envelope)k sampled points)Using only a finite number of terms to reconstruct a CT square wave will show the Gibbs phenomenon.
viewof A_ct_rec = Inputs.number([0, 5], {
label: "A (Amplitude):",
step: 0.1,
value: 1
})
viewof T0_ct_rec = Inputs.number([0.1, 5], {
label: "T0 (Fundamental Period):",
step: 0.1,
value: 1
})
viewof T1_ct_rec = Inputs.number([0.1, 2], {
label: "T1 (Pulse Width):",
step: 0.1,
value: 0.5
})
viewof K_rec_ct = Inputs.range([1, 20], {
label: "Number of Harmonics (K):",
step: 1,
value: 1
})Using all \(N\) terms perfectly reconstructs a DT periodic square wave.
viewof A_dt_rec = Inputs.number([0, 5], {
label: "A (Amplitude):",
step: 0.1,
value: 1
})
viewof N_dt_rec = Inputs.number([1, 20], {
label: "N (Period):",
step: 1,
value: 10
})
viewof P_dt_rec = Inputs.number([1, 15], {
label: "P (Pulse Width, 2N1+1):",
step: 1,
value: 5
})
viewof M_rec_dt = Inputs.range([0, 4], {
label: "Number of Terms (M):",
step: 1,
value: 0
})Tip
Analogy:
Think of CTFS as trying to describe a smooth painting with an infinite palette of colors, while DTFS is like describing a pixelated image with a finite, repeating set of colors.