Objective

• To demonstrate, in the time domain, how a Linear Time-Invariant (LTI) system responds to a sinusoidal input.
• We will start from the convolution integral and use Euler’s identity to show that the steady-state output remains sinusoidal.

Introduction

Euler’s formula shows the deep connection between trigonometry and complex exponential functions:

\[e^{i\alpha} = \cos(\alpha) + i \sin(\alpha)\]

To prove this, we use the Taylor Series method at \(a = 0\). This allows us to write complex functions as infinite sums of simple terms.

Defining the Taylor Series

First, we define the standard power series for the three functions involved:

Exponential Function: \[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots\]

Trigonometric Functions: \[\cos(\alpha) = 1 - \frac{\alpha^2}{2!} + \frac{\alpha^4}{4!} - \frac{\alpha^6}{6!} + \dots\] \[\sin(\alpha) = \alpha - \frac{\alpha^3}{3!} + \frac{\alpha^5}{5!} - \frac{\alpha^7}{7!} + \dots\]

Complex Substitution

Next, we replace the variable \(x\) in the exponential series with the imaginary term \(i\alpha\):

\[e^{i\alpha} = 1 + (i\alpha) + \frac{(i\alpha)^2}{2!} + \frac{(i\alpha)^3}{3!} + \frac{(i\alpha)^4}{4!} + \dots\]

We must apply the powers of the imaginary unit \(i\): * \(i^2 = -1\) * \(i^3 = -i\) * \(i^4 = 1\) * \(i^5 = i\)

Expanding the Series

After applying the powers of \(i\), the series looks like this:

\[e^{i\alpha} = 1 + i\alpha - \frac{\alpha^2}{2!} - i\frac{\alpha^3}{3!} + \frac{\alpha^4}{4!} + i\frac{\alpha^5}{5!} - \dots\]

Now, we group the Real parts and the Imaginary parts:

\[e^{i\alpha} = \underbrace{\left( 1 - \frac{\alpha^2}{2!} + \frac{\alpha^4}{4!} - \dots \right)}_{\text{Real Part}} + i \underbrace{\left( \alpha - \frac{\alpha^3}{3!} + \frac{\alpha^5}{5!} - \dots \right)}_{\text{Imaginary Part}}\]

The Rectangular Form

A complex number \(z\) is typically written in Cartesian coordinates as:

\[z = a + bi\]

Where:

• \(a\): The Real part (\(\text{Re}\{z\}\))
• \(b\): The Imaginary part (\(\text{Im}\{z\}\))
• \(i\): The imaginary unit (\(\sqrt{-1}\))

From Trigonometry to Euler

From the geometry of a right triangle, we know: * \(a = r \cos(\alpha)\) * \(b = r \sin(\alpha)\)

Substituting these into \(z = a + bi\): \[z = r \cos(\alpha) + i(r \sin(\alpha))\] \[z = r (\cos(\alpha) + i \sin(\alpha))\]

Using Euler’s Formula (\(e^{i\alpha} = \cos\alpha + i\sin\alpha\)), we simplify this to: \[z = r e^{i\alpha}\]

Why Use Exponential Form?

In biomedical engineering, the form \(z = r e^{i\alpha}\) is superior for calculations:

1. Multiplication: \[z_1 z_2 = (r_1 e^{i\alpha_1})(r_2 e^{i\alpha_2}) = r_1 r_2 e^{i(\alpha_1 + \alpha_2)}\]

2. Division: \[\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\alpha_1 - \alpha_2)}\]

3. Signal Analysis: It allows us to represent a rotating phasor, which is the basis for the Fourier Transform.

Summary

\[e^{i\alpha} = \cos(\alpha) + i \sin(\alpha)\]

1. System Definition

A continuous-time LTI system is defined by:

\[ y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t - \tau)\,d\tau \]

where

• \(x(t)\) is the input,
• \(h(t)\) is the impulse response,
• \(y(t)\) is the output.

2. Sinusoidal Input

Let the input be a sinusoid:

\[ x(t) = A \cos(\omega_0 t + \phi) \]

Using Euler’s identity:

\[ x(t) = \frac{A}{2} \left[e^{j(\omega_0 t + \phi)} + e^{-j(\omega_0 t + \phi)}\right] \]

3. Linearity of the Convolution Integral

Because the system is linear, we can treat each exponential term separately:

\[ y(t) = \frac{A}{2}\left[ y_1(t) + y_2(t) \right] \] where \[ y_1(t) = \frac{A}{2} e^{j\phi} \int_{-\infty}^{\infty} h(\tau)\, e^{j\omega_0(t-\tau)}\, d\tau \] \[ y_2(t) = \frac{A}{2} e^{-j\phi} \int_{-\infty}^{\infty} h(\tau)\, e^{-j\omega_0(t-\tau)}\, d\tau \]

4. Simplifying Each Integral

We can factor out the term \(e^{j\omega_0 t}\):

\[ y_1(t) = e^{j\phi} e^{j\omega_0 t} \int_{-\infty}^{\infty} h(\tau)\, e^{-j\omega_0 \tau}\, d\tau \]

Let \[ H_1 = \int_{-\infty}^{\infty} h(\tau)\, e^{-j\omega_0 \tau}\, d\tau \]

Then, \[ y_1(t) = e^{j\phi} e^{j\omega_0 t} H_1 \]

Similarly, \[ y_2(t) = e^{-j\phi} e^{-j\omega_0 t} H_1 \]

5. Combine the Two Parts

The total output is:

\[ y(t) = \frac{A}{2}\left[e^{j\phi} H_1 e^{j\omega_0 t} + e^{-j\phi} H_1^* e^{-j\omega_0 t}\right] \]

If we express \(H_1\) in polar form: \[ H_1 = |H_1| e^{j\theta} \]

Then, \[ y(t) = A |H_1| \cos(\omega_0 t + \phi + \theta) \]

6. Interpretation in Time Domain

• The output is a cosine at the same frequency \(\omega_0\) as the input.
• Its amplitude is scaled by \(|H_1|\), which depends on \(h(t)\).
• Its phase is shifted by\(\theta\), the argument of \(H_1\).

\[ \boxed{y(t) = A |H_1| \cos(\omega_0 t + \phi + \theta)} \]

1. System Definition

A discrete-time LTI system is defined by the convolution sum:

\[ y[n] = \sum_{k=-\infty}^{\infty} h[k]\,x[n - k] \]

where

• \(x[n]\) is the input sequence,
• \(h[k]\) is the impulse response,
• \(y[n]\) is the output sequence.

2. Sinusoidal Input

Let the input be a discrete sinusoid:

\[ x[n] = A \cos(\omega_0 n + \phi) \]

Using Euler’s identity:

\[ x[n] = \frac{A}{2}\left[e^{j(\omega_0 n + \phi)} + e^{-j(\omega_0 n + \phi)}\right] \]

3. Linearity of the Convolution Sum

Because the system is linear, each exponential term can be treated independently:

\[ y[n] = \frac{A}{2}\left[y_1[n] + y_2[n]\right] \]

where \[ y_1[n] = e^{j\phi} \sum_{k=-\infty}^{\infty} h[k]\, e^{j\omega_0 (n - k)} \] \[ y_2[n] = e^{-j\phi} \sum_{k=-\infty}^{\infty} h[k]\, e^{-j\omega_0 (n - k)} \]

4. Simplifying Each Sum

We can factor out \(e^{j\omega_0 n}\):

\[ y_1[n] = e^{j\phi} e^{j\omega_0 n} \sum_{k=-\infty}^{\infty} h[k]\, e^{-j\omega_0 k} \]

Let \[ H_1 = \sum_{k=-\infty}^{\infty} h[k]\, e^{-j\omega_0 k} \]

Then, \[ y_1[n] = e^{j\phi} H_1 e^{j\omega_0 n} \]

Similarly, \[ y_2[n] = e^{-j\phi} H_1^* e^{-j\omega_0 n} \]

5. Combine the Two Parts

The total output becomes:

\[ y[n] = \frac{A}{2}\left[e^{j\phi} H_1 e^{j\omega_0 n} + e^{-j\phi} H_1^* e^{-j\omega_0 n}\right] \]

If we write \(H_1\) in polar form: \[ H_1 = |H_1| e^{j\theta} \]

Then, \[ y[n] = A |H_1| \cos(\omega_0 n + \phi + \theta) \]

6. Interpretation in Time Domain

• The output remains sinusoidal with the same frequency \(\omega_0\).
• Its amplitude is scaled by \(|H_1|\).
• Its phase is shifted by \(\theta\).
• Complex exponential are proper functions (eigen functions) of LTI systems.

\[ \boxed{y[n] = A |H_1| \cos(\omega_0 n + \phi + \theta)} \]

Introduction

• Signals can be analyzed in both time domain and frequency domain.
• The frequency content of a signal describes how different frequency components contribute to the overall signal.
• Applications in biomedical signals, audio processing, communications, and image processing.

Convolution in Time Domain

• Convolution is a fundamental operation in signal processing.
• Given two signals \(x(t)\) and \(h(t)\), their convolution is defined as:

\[y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\tau\]

• In discrete-time, convolution is:

\[y[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k]\]

Convolution Theorem

• Convolution in time domain corresponds to multiplication in frequency domain:

\[X(f) H(f) = Y(f)\]

• This property is crucial in filter design and system analysis.

Introduction to Fourier Series

(-1.0, 4.0)

(-1.0, 4.0)

Introduction to Fourier Series

• Convolution requiere the representation of the signal in a sum of impulse functions.
• Fourier series represents periodic signals as a sum of sinusoids:

\[x(t) = \sum_{n=-\infty}^{\infty} C_n e^{jn\omega_0 t}\]

where \(C_n\) are the Fourier coefficients.

• Decomposing a signal into sinusoidal components allows frequency analysis.

Fourier Coefficients

• The Fourier coefficients\(C_n\)are computed as:

\[C_n = \frac{1}{T} \int_{0}^{T} x(t) e^{-jn\omega_0 t} dt\]

• Determines how much of each frequency is present in the signal.

Example of Fourier Series Expansion

1. Definition of the Complex Fourier Series

For a periodic signal \(x(t)\) with period \(T\):

\[ x(t) = \sum_{k=-\infty}^{\infty} c_k e^{j k \Omega_0 t}, \]

where:

\[ c_k = \frac{1}{T} \int_{T_0}^{T_0 + T} x(t)e^{-jk\Omega_0 t}\,dt, \quad \Omega_0 = \frac{2\pi}{T}. \]

2. Example Signal

We define a periodic signal \(x(t)\) with period \(2\pi\):

\[ x(t) = \begin{cases} 0, & -\pi \le t < 0,\\ 1, & 0 \le t < \pi. \end{cases} \]

This corresponds to a 50% duty cycle pulse.

3. Symbolic Derivation using SymPy

Expected result:

\[ c_k = \begin{cases} \dfrac{1}{2}, & k=0,\\[4pt] \dfrac{j\left((-1)^k - 1\right)}{2\pi k}, & k \ne 0. \end{cases} \]

4. Numerical Evaluation (NumPy)

import numpy as np
import pandas as pd

def ck_numeric(k):
    k = np.asarray(k, dtype=float)
    return np.where(
        k != 0,
        1j*(np.exp(1j*np.pi*k)-1)/(2*np.pi*k),
        0.5
    )

K = 10
ks = np.arange(-K, K+1)
cks = ck_numeric(ks)

pd.DataFrame({
    "k": ks,
    "Re{c_k}": np.real(cks),
    "Im{c_k}": np.imag(cks),
    "|c_k|": np.abs(cks)
})

     k       Re{c_k}   Im{c_k}         |c_k|
0  -10  1.949086e-17 -0.000000  1.949086e-17
1   -9 -1.949086e-17  0.035368  3.536777e-02
2   -8  1.949086e-17 -0.000000  1.949086e-17
3   -7 -1.949086e-17  0.045473  4.547284e-02
4   -6  1.949086e-17 -0.000000  1.949086e-17
5   -5 -1.949086e-17  0.063662  6.366198e-02
6   -4  1.949086e-17 -0.000000  1.949086e-17
7   -3 -1.949086e-17  0.106103  1.061033e-01
8   -2  1.949086e-17 -0.000000  1.949086e-17
9   -1 -1.949086e-17  0.318310  3.183099e-01
10   0  5.000000e-01  0.000000  5.000000e-01
11   1 -1.949086e-17 -0.318310  3.183099e-01
12   2  1.949086e-17  0.000000  1.949086e-17
13   3 -1.949086e-17 -0.106103  1.061033e-01
14   4  1.949086e-17  0.000000  1.949086e-17
15   5 -1.949086e-17 -0.063662  6.366198e-02
16   6  1.949086e-17  0.000000  1.949086e-17
17   7 -1.949086e-17 -0.045473  4.547284e-02
18   8  1.949086e-17  0.000000  1.949086e-17
19   9 -1.949086e-17 -0.035368  3.536777e-02
20  10  1.949086e-17  0.000000  1.949086e-17

5. Partial Sum Reconstruction

import numpy as np
import matplotlib.pyplot as plt

def x_numeric(t):
    # Original pulse over [-π, π): 0 on [-π,0), 1 on [0,π).
    tw = ((t + np.pi) % (2*np.pi)) - np.pi
    return np.where((tw >= 0) & (tw < np.pi), 1.0, 0.0)

def partial_sum(t, N):
    ks = np.arange(-N, N+1)
    cks = ck_numeric(ks)
    return np.sum(cks[:, None] * np.exp(1j*np.outer(ks, t)), axis=0)

tgrid = np.linspace(-2*np.pi, 2*np.pi, 4000, endpoint=False)
xg = x_numeric(tgrid)

6. Reconstruction for N = 5

N1 = 5
sN1 = partial_sum(tgrid, N1).real

plt.figure(figsize=(8,3))
plt.plot(tgrid, xg, label='x(t) original')
plt.plot(tgrid, sN1, label=f'N={N1}')
plt.xlim([-2*np.pi, 2*np.pi])

(-6.283185307179586, 6.283185307179586)

plt.legend()
plt.title("Partial Sum Reconstruction (N=5)")
plt.show()

7. Reconstruction for N = 20

(-6.283185307179586, 6.283185307179586)

7. Reconstruction for N = 100

(-6.283185307179586, 6.283185307179586)

8. Discussion

• The average value \(c_0 = 0.5\) matches the mean level of the pulse.
• As \(N\) increases, the reconstruction converges except near discontinuities.
• The Gibbs phenomenon appears at the jump discontinuities.
• The error \(\lVert x(t)-S_N(t)\rVert_2\) decreases with \(N\).

9. Summary

Example 2 of Fourier Series

(array([-5.,  0.,  5., 10., 15., 20., 25., 30., 35.]), [Text(-5.0, 0, '−5'), Text(0.0, 0, '0'), Text(5.0, 0, '5'), Text(10.0, 0, '10'), Text(15.0, 0, '15'), Text(20.0, 0, '20'), Text(25.0, 0, '25'), Text(30.0, 0, '30'), Text(35.0, 0, '35')])

(array([4.5, 5. , 5.5, 6. , 6.5, 7. , 7.5, 8. , 8.5]), [Text(0, 4.5, '4.5'), Text(0, 5.0, '5.0'), Text(0, 5.5, '5.5'), Text(0, 6.0, '6.0'), Text(0, 6.5, '6.5'), Text(0, 7.0, '7.0'), Text(0, 7.5, '7.5'), Text(0, 8.0, '8.0'), Text(0, 8.5, '8.5')])

Example 2 of Fourier Series

(array([-40., -30., -20., -10.,   0.,  10.,  20.,  30.,  40.]), [Text(-40.0, 0, '−40'), Text(-30.0, 0, '−30'), Text(-20.0, 0, '−20'), Text(-10.0, 0, '−10'), Text(0.0, 0, '0'), Text(10.0, 0, '10'), Text(20.0, 0, '20'), Text(30.0, 0, '30'), Text(40.0, 0, '40')])

(array([4.5, 5. , 5.5, 6. , 6.5, 7. , 7.5, 8. , 8.5]), [Text(0, 4.5, '4.5'), Text(0, 5.0, '5.0'), Text(0, 5.5, '5.5'), Text(0, 6.0, '6.0'), Text(0, 6.5, '6.5'), Text(0, 7.0, '7.0'), Text(0, 7.5, '7.5'), Text(0, 8.0, '8.0'), Text(0, 8.5, '8.5')])

Linearity

• If\(f_1(x)\)and\(f_2(x)\)have Fourier series,
• Then for any constants\(a, b\),
• \(a f_1(x) + b f_2(x)\) has a Fourier series,
• With coefficients scaled accordingly.

Time Shifting

• If \(f(x)\) has Fourier coefficients \(a_n, b_n\),
• Then \(f(x - x_0)\) has coefficients:
• \(a_n \cos(n\omega x_0) + b_n \sin(n\omega x_0)\),
• And \(b_n \cos(n\omega x_0) - a_n \sin(n\omega x_0)\).

Frequency Scaling

• If \(g(x) = f(cx)\),
• Then the period scales by \(c\),
• The fundamental frequency changes to \(c\omega\),
• Fourier coefficients adjust accordingly.

Differentiation Property

• If \(f(x)\) is differentiable,
• Then \(f'(x)\) has Fourier series,
• With coefficients scaled as \(n a_n, n b_n\),
• Higher frequencies get amplified.

Integration Property

• If \(f(x)\) has a Fourier series,
• Then \(\int f(x) dx\) has a Fourier series,
• With coefficients scaled as \(\frac{a_n}{n}, \frac{b_n}{n}\),
• Lower frequencies get emphasized.

Parseval’s Theorem

• The total signal energy is conserved,
• Energy in time domain equals energy in frequency domain,
• Given by:
• \(\sum (a_n^2 + b_n^2) = \frac{1}{T} \int |f(x)|^2 dx\).

Convolution Property

• Convolution in time domain,
• Is multiplication in Fourier series coefficients,
• If \(f_1\) and \(f_2\) are convoluted,
• Their Fourier coefficients multiply component-wise.

Discrete Time Fourier Series

• Represents periodic discrete signals using harmonics.
• Extends Fourier series to discrete-time domain.
• Fundamental in digital signal processing.
• Basis for the Discrete Fourier Transform (DFT).

Mathematical Expression

• A periodic sequence \(x[n]\) can be expressed as:
• \[x[n] = \sum_{k=0}^{N-1} C_k e^{j(2\pi k n / N)}\].
• The coefficients \(C_k\) are computed as:
• \[C_k = \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-j(2\pi k n / N)}\].

Periodicity and Symmetry

• The coefficients \(C_k\) repeat every \(N\).
• Ensures correct reconstruction of signals.
• Explains frequency domain representation.
• Basis for spectral analysis.

Key Properties

• Linearity: Superposition holds.
• Time Shift: Causes phase shift in coefficients.
• Parseval’s Theorem: Energy conservation.
• Convolution: Time convolution → Frequency multiplication.

Frequency Domain Interpretation

• \(C_k\) represents discrete frequency content.
• The spectrum consists of \(N\) harmonics.
• Resolution improves with larger \(N\).
• Essential for analyzing periodic discrete signals.

Comparison with Continuous Case

• DTFS applies to discrete periodic signals.
• Continuous Fourier series applies to continuous functions.
• Both represent signals as sums of sinusoids.
• DTFS is used in digital communications and audio processing.

Example of th DTFS

DTFS Coefficients:

C[0] = 2.0000+0.0000j
C[1] = -0.6036-0.6036j
C[2] = 0.0000+0.0000j
C[3] = 0.1036-0.1036j
C[4] = 0.0000+0.0000j
C[5] = 0.1036+0.1036j
C[6] = 0.0000+0.0000j
C[7] = -0.6036+0.6036j

Example 02

Conceptual Foundation

• Fourier Series represents periodic signals in terms of sinusoids.
• As period \(T \to \infty\), the signal becomes aperiodic.
• The Fourier Transform generalizes Fourier Series to aperiodic signals.
• Transforms signals from time to frequency domain.

Mathematical Transition

• Fourier Series of a periodic signal:
• \[f(x) = \sum_{n=-\infty}^{\infty} C_n e^{j(2\pi n x / T)}\].
• As \(T \to \infty\), frequency spacing \(\frac{1}{T}\) → differential.
• Leads to the Fourier Transform:
• \[F(\omega) = \int_{-\infty}^{\infty} f(x) e^{-j\omega x} dx\].

Frequency Spectrum Interpretation

• Fourier Series: discrete frequency spectrum.
• Fourier Transform: continuous frequency spectrum.
• Coefficients \(C_n\) become the function \(F(\omega)\).
• Allows analysis of arbitrary signals in frequency domain.

Inverse Fourier Transform

• Recovers time-domain signal from \(F(\omega)\).
• Defined as:
• \[f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega x} d\omega\].
• Ensures complete information preservation.
• Basis for signal reconstruction in DSP.

Energy and Parseval’s Theorem

• Energy conservation in time and frequency domains.
• Parseval’s theorem states:
• \[\int |f(x)|^2 dx = \frac{1}{2\pi} \int |F(\omega)|^2 d\omega\]
• Ensures no energy loss between domains.

Proof of symmetry for a real signal

Assume \(x(t)\in\mathbb{R}\). Then

\[ X(\omega)=\int_{-\infty}^{\infty} x(t)e^{-j\omega t}\,dt \]

Take complex conjugate:

\[ X^*(\omega) = \left(\int_{-\infty}^{\infty} x(t)e^{-j\omega t}\,dt\right)^* \]

Because \(x(t)\) is real, \(x^*(t)=x(t)\). Therefore

\[ X^*(\omega) = \int_{-\infty}^{\infty} x(t)e^{j\omega t}\,dt \]

But

\[ X(-\omega) = \int_{-\infty}^{\infty} x(t)e^{-j(-\omega)t}\,dt = \int_{-\infty}^{\infty} x(t)e^{j\omega t}\,dt \]

Hence

\[ \boxed{X(-\omega)=X^*(\omega)} \]

This is called conjugate symmetry.

Consequences of symmetry

From

\[ X(-\omega)=X^*(\omega) \]

we get two useful results:

1. Magnitude is even

\[ |X(-\omega)|=|X(\omega)| \]

2. Phase is odd

If \(X(\omega)=|X(\omega)|e^{j\phi(\omega)}\), then

\[ \phi(-\omega)=-\phi(\omega) \]

So, for a real signal:

• the magnitude spectrum is mirrored around \(0\),
• the phase spectrum changes sign.

The same proof works for the DTFT if \(x[n]\) is real.

Proof of periodicity for the DTFT

Start from the DTFT:

\[ X(e^{j\omega})=\sum_{n=-\infty}^{\infty} x[n]e^{-j\omega n} \]

Now evaluate it at \(\omega+2\pi\):

\[ X(e^{j(\omega+2\pi)}) = \sum_{n=-\infty}^{\infty} x[n]e^{-j(\omega+2\pi)n} \]

Separate the exponential:

\[ X(e^{j(\omega+2\pi)}) = \sum_{n=-\infty}^{\infty} x[n]e^{-j\omega n}e^{-j2\pi n} \]

Because \(n\) is an integer,

\[ e^{-j2\pi n}=1 \]

for every integer \(n\). Then

\[ X(e^{j(\omega+2\pi)}) = \sum_{n=-\infty}^{\infty} x[n]e^{-j\omega n} = X(e^{j\omega}) \]

So,

\[ \boxed{X(e^{j(\omega+2\pi)})=X(e^{j\omega})} \]

More generally,

\[ \boxed{X(e^{j(\omega+2\pi k)})=X(e^{j\omega}),\quad k\in\mathbb{Z}} \]

Why the CTFT is not periodic in general

For the continuous-time Fourier transform,

\[ X(\omega)=\int_{-\infty}^{\infty} x(t)e^{-j\omega t}\,dt \]

Now test \(\omega+2\pi\):

\[ X(\omega+2\pi) = \int_{-\infty}^{\infty} x(t)e^{-j(\omega+2\pi)t}\,dt = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}e^{-j2\pi t}\,dt \]

Here is the key point:

• in DTFT, \(n\) is integer, so \(e^{-j2\pi n}=1\),
• in CTFT, \(t\) is continuous, so \(e^{-j2\pi t}\neq 1\) in general.

Therefore, in general,

\[ \boxed{X(\omega+2\pi)\neq X(\omega)} \]

So the CTFT is not periodic in general.

One simple discrete-time example

Let

\[ x[n]=\delta[n]+\delta[n-1] \]

Its DTFT is

\[ X(e^{j\omega})=1+e^{-j\omega} \]

Factor it:

\[ X(e^{j\omega}) = e^{-j\omega/2}\left(e^{j\omega/2}+e^{-j\omega/2}\right) = 2\cos\left(\frac{\omega}{2}\right)e^{-j\omega/2} \]

Magnitude:

\[ |X(e^{j\omega})|=2\left|\cos\left(\frac{\omega}{2}\right)\right| \]

This magnitude is:

• even in \(\omega\),
• \(2\pi\)-periodic.

So this example shows both properties at the same time.

Visual summary

Final conclusion

Continuous-time Fourier transform (CTFT)

If \(x(t)\) is real:

\[ \boxed{X(-\omega)=X^*(\omega)} \]

So the spectrum is symmetric in the conjugate sense, but not periodic in general.

Discrete-time Fourier transform (DTFT)

For any \(x[n]\):

\[ \boxed{X(e^{j(\omega+2\pi)})=X(e^{j\omega})} \]

So the spectrum is periodic.

If \(x[n]\) is also real, then:

\[ \boxed{X(e^{-j\omega})=X^*(e^{j\omega})} \]

So the DTFT spectrum is both periodic and symmetric.