8.06 Frequency Spectra Phase Spectra and Modulation

8.05 Time-Domain Convolution & Multiplication Properties | 8.07 CTFT System Analysis of Continuous Networks

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8.06 Frequency Spectra, Phase Spectra & Modulation

Core Idea

For real-world engineering signals, continuous Fourier transforms ($X(j\Omega )$) are typically complex-valued, meaning they carry both magnitude and phase information. The Amplitude (Magnitude) Spectrum $|X(j\Omega )|$ and the Phase Spectrum $\angle X(j\Omega )$ provide a complete, visually intuitive frequency-domain portrait of the signal’s harmonics. This note covers the plotting conventions for both single-sided (physical) and double-sided (mathematical) spectra, and details the Modulation Theorem—the physical basis for translating low-frequency message signals onto high-frequency carriers for long-distance propagation.

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1. Polar Representation of Fourier Spectra

Because the continuous-time Fourier transform is a complex function of the angular frequency $\Omega$, it is conventionally analyzed and plotted in polar form:

$X(j\Omega )=|X(j\Omega )|e^{j\angle X(j\Omega )}=|X(j\Omega )|e^{j\theta (\Omega )}$

where:

• $|X(j\Omega )|$ represents the Continuous Amplitude (Magnitude) Spectrum {representing the signal strength per unit frequency bandwidth}.
• $\angle X(j\Omega )$ (or $\theta (\Omega )$) represents the Continuous Phase Spectrum {representing the harmonically related time-alignment offsets of the phasor components}.

1.1 Symmetries of Spectra for Real-Valued Signals

If the time-domain signal $x(t)$ is real-valued ($x^{\ast }(t)=x(t)$), its Fourier transform satisfies the conjugate-symmetric property: $X^{\ast }(j\Omega )=X(-j\Omega )$. This forces strict symmetry bounds on the spectra:

1. Even Magnitude Symmetry: The magnitude spectrum is an even function of frequency. $|X(j\Omega )|=|X(-j\Omega )|$
2. Odd Phase Symmetry: The phase spectrum is an odd function of frequency. $\angle X(-j\Omega )=-\angle X(j\Omega )$

      EVEN MAGNITUDE SPECTRUM             ODD PHASE SPECTRUM
             |X(jΩ)|                           ∠X(jΩ)
                ^                                 ^
             *  |  *                              |   *
           *    |    *                            |  *
         *      |      *             -------------+-------------> Ω
       *        |        *               *        |
   ----+--------+--------+----> Ω         *       |
      -Ω0       0       Ω0                 *      |
                                                 -Ω0

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2. Single-Sided vs. Double-Sided Spectra

When visualizing discrete spectral lines of periodic sinusoids, we use two different plotting conventions:

2.1 Single-Sided Spectra

• Concept: Plots only physical, non-negative frequencies ($\Omega \ge 0$).
• Basis: Originates directly from the trigonometric Fourier Series expansion of a real signal: $x(t)=A_0+{\sum }_{n=1}^{\infty }{A}_n\cos (n{\Omega }_{0}t+{\theta }_n)$
• Plotting Rules:
  • Plot an impulse at each positive harmonic frequency $n{\Omega }_0$ with height equal to the full peak amplitude $A_n$.
  • Plot the corresponding phase ${\theta }_n$ at each positive harmonic frequency $n{\Omega }_0$.

2.2 Double-Sided Spectra

• Concept: Plots both positive and negative frequencies ($\acute{\Omega }\in (-\infty ,\infty )$).
• Basis: Originates from the complex exponential Fourier representation: $x(t)={\sum }_{n=-\infty }^{\infty }{C}_n{e}^{jn{\Omega }_{0}t}$
• Euler Conversion: To represent a real-valued cosine wave $A\cos ({\Omega }_{0}t+\theta )$ exponentially, we expand it using Euler’s formula: $A\cos ({\Omega }_{0}t+\theta )=\frac{A}{2}{e}^{j({\Omega }_{0}t+\theta )}+\frac{A}{2}{e}^{-j({\Omega }_{0}t+\theta )}=\left(\frac{A}{2}{e}^{j\theta }\right)e^{j{\Omega }_{0}t}+\left(\_\frac{A}{2}{e}^{-j\theta }\right)e^{-j{\Omega }_{0}t}$
• Plotting Rules:
  • Halved Amplitude Split: The peak amplitude $A$ splits equally into two rotating phasors of length $A/2$ located at $\pm {\Omega }_0$.
  • Conjugate Phase Split: The phase angle $\theta$ is plotted at the positive frequency $+{\Omega }_0$, while its negative conjugate $-\theta$ is plotted at the negative frequency $-{\Omega }_0$.

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3. The Modulation (Frequency Shifting) Theorem

The Modulation Theorem

Let $m(t)$ be a low-frequency message signal {with bandwidth bounded such that $M(j\Omega )=0$ for $|\Omega |>{\Omega }_m$}. Multiplying this message signal by a high-frequency sinusoidal carrier wave $c(t)=\cos ({\Omega }_{c}t)$ translates the entire baseband spectrum continuously in frequency, centering it at the carrier frequencies $\pm {\Omega }_c$:

$y(t)=m(t)\cos ({\Omega }_{c}t)\leftrightarrow Y(j\Omega )=\frac{1}{2}\left[M(j(\Omega -{\Omega }_c))+M(j(\Omega +{\Omega }_c))\right]$

3.1 Mathematical Derivation

We prove the theorem using the fundamental Frequency Shifting Property of the continuous-time Fourier transform:

$\text{If }m(t)\leftrightarrow M(j\Omega ),\text{then }{e}^{j{\Omega }_{c}t}m(t)\leftrightarrow M(j(\Omega -{\Omega }_c))$

1. Express the carrier wave $\cos ({\Omega }_{c}t)$ in complex exponential form using Euler’s identity: $\cos ({\Omega }_{c}t)=\frac{e^{j{\Omega }_{c}t}+e^{-j{\Omega }_{c}t}}{2}$
2. Substitute this expression into the modulated signal equation: $y(t)=m(t)\cos ({\Omega }_{c}t)=m(t)\left[\frac{e^{j{\Omega }_{c}t}+e^{-j{\Omega }_{c}t}}{2}\right]=\frac{1}{2}{e}^{j{\Omega }_{c}t}m(t)+\frac{1}{2}{e}^{-j{\Omega }_{c}t}m(t)$
3. Take the Fourier transform of both sides using the Linearity and Frequency Shifting properties: $Y(j\Omega )=\mathcal{F}\{y(t)\}=\frac{1}{2}\mathcal{F}\{e^{j{\Omega }_{c}t}m(t)\}+\frac{1}{2}\mathcal{F}\{e^{-j{\Omega }_{c}t}m(t)\}$ $Y(j\Omega )=\frac{1}{2}M(j(\Omega -{\Omega }_c))+\frac{1}{2}M(j(\Omega +{\Omega }_c))■$

   MESSAGE SPECTRUM M(jΩ)                 MODULATED DSB-SC SPECTRUM Y(jΩ)
          |M(jΩ)|                                    |Y(jΩ)|
             ^                                          ^
          ---|---                                  ---|---   ---|---
         -Ωm 0  Ωm  --> Ω                         -Ωc     0     Ωc   --> Ω
                                                    |<-2Ωm->|

Physical Context: Double-Sideband Suppressed-Carrier (DSB-SC)

This spectral translation is called Double-Sideband Suppressed-Carrier (DSB-SC) amplitude modulation. The original baseband signal requires a physical antenna size proportional to the wavelength $\lambda =c/f$. By shifting the message spectrum up to a high-frequency carrier $f_c$ (where $f_c\gg f_m$), we decrease the wavelength drastically, allowing highly efficient transmission over compact, practical antennas.

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4. High-Yield Worked Examples (The Exam Classics)

4.1 Example 1: Double-Sided Spectrum of a Simple Cosine [PYQ 2019 Q5b/c - 1 Mark]

Question: Sketch the double-sided frequency spectrum of the signal: $x(t)=10\cos \left(20\pi t-\frac{\pi }{6}\right),-\infty <t<\infty$

Step-by-Step Mathematical Analysis:

1. Identify Parameters:
   • Peak Amplitude $A=10$.
   • Fundamental Cyclic Frequency $f_0=10$ Hz $⟹$ Angular Frequency ${\Omega }_0=20\pi$ rad/s.
   • Phase Angle $\theta =-\frac{\pi }{6}$ rad.
2. Apply Euler’s Expansion: $x(t)=5e^{-j\pi /6}{e}^{j20\pi t}+5e^{j\pi /6}{e}^{-j20\pi t}$
3. Determine Double-Sided Coefficients:
   • At $\Omega =+20\pi$ rad/s: Amplitude $=5$, Phase $=-\frac{\pi }{6}$ rad (or $-30^{\circ }$).
   • At $\Omega =-20\pi$ rad/s: Amplitude $=5$, Phase $=+\frac{\pi }{6}$ rad (or $+30^{\circ }$).
   • All other frequency components are exactly zero.

Plotting the Spectra:

       DOUBLE-SIDED AMPLITUDE SPECTRUM            DOUBLE-SIDED PHASE SPECTRUM
                  |X(jΩ)|                                    ∠X(jΩ) (rad)
                     ^                                          ^
                     |                                          |   pi/6
             5       |       5                                  |    o
             |       |       |                                  |    |
       ------+-------+-------+------> Ω                  -------+----+-------> Ω
          -20pi      0     20pi                             -20pi |   20pi
                                                                  |  o
                                                                  | -pi/6

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4.2 Example 2: Complete Single & Double-Sided Spectra [PYQ 2017 Question 5b - 5 Marks]

Question: Sketch the single and double-sided frequency spectra of the following signal: $x(t)=25\cos \left(5\pi t-\frac{\pi }{2}\right)$

Step-by-Step Mathematical Analysis:

1. Identify Parameters:

   • Peak Amplitude $A=25$.
   • Angular Frequency ${\Omega }_0=5\pi$ rad/s.
   • Phase Angle $\theta =-\frac{\pi }{2}$ rad (or $-90^{\circ }$).

2. Single-Sided Representation:

   • Plots only the positive physical frequency $\Omega =5\pi$ rad/s.
   • Amplitude Line: Height $=A=25$ at $\Omega =5\pi$.
   • Phase Line: Angle $=\theta =-\frac{\pi }{2}$ rad at $\Omega =5\pi$.

3. Double-Sided Representation:

   • Phasor amplitude is halved: $A/2=12.5$.
   • At $\Omega =+5\pi$ rad/s: Amplitude $=12.5$, Phase $=-\frac{\pi }{2}$ rad.
   • At $\Omega =-5\pi$ rad/s: Amplitude $=12.5$, Phase $=+\frac{\pi }{2}$ rad.

Plotting both spectrum sets:

       SINGLE-SIDED SPECTRA (Ω >= 0)             DOUBLE-SIDED SPECTRA (All Ω)
  
       |X(jΩ)| single                            |X(jΩ)| double
          ^                                         ^
       25 |       |                              12.5 |     |     |
          |       |                                   |     |     |
       ---+-------+---------> Ω                  -----+-----+-----+-----> Ω
          0      5pi                                 -5pi   0    5pi
  
       ∠X(jΩ) single                             ∠X(jΩ) double (rad)
          ^                                         ^
          |                                         |  pi/2
       ---+-------+---------> Ω                     |   o
          0 |    5pi                            ----+---+---+-----> Ω
            | o                                    -5pi |  5pi
            |-pi/2                                      | o
                                                        |-pi/2

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4.3 Example 3: The Sine Phase-Conversion Trap [PYQ 2018 Question 5b/c - 2 Marks]

Question: Sketch the single and double-sided frequency spectra of the signal: $x(t)=12\sin \left(5\pi t-\frac{\pi }{2}\right)$

Step-by-Step Mathematical Analysis:

Critical Exam Trap: Converting Sine to Cosine

Spectral plotting conventions are strictly derived from the cosine basis wave: $A\cos ({\Omega }_{0}t+\theta )$. Reading the phase directly from a sine representation will result in a zero-mark penalty. You must always convert the sine function to cosine first using the identity: $\sin (\varphi )=\cos \left(\varphi -\frac{\pi }{2}\right)$.

1. Apply Phase Conversion: $x(t)=12\sin \left(5\pi t-\frac{\pi }{2}\right)=12\cos \left(5\pi t-\frac{\pi }{2}-\frac{\pi }{2}\right)=12\cos \left(5\pi t-\pi \right)$
2. Identify Converted Parameters:
   • Peak Amplitude $A=12$.
   • Angular Frequency ${\Omega }_0=5\pi$ rad/s.
   • Phase Angle $\theta =-\pi$ rad (or $+\pi$ rad, since $\pm \pi$ represent the same physical point).
3. Single-Sided Representation:
   • At $\Omega =5\pi$ rad/s: Amplitude $=12$, Phase $=-\pi$ rad.
4. Double-Sided Representation:
   • Halved amplitude: $A/2=6$.
   • At $\Omega =+5\pi$ rad/s: Amplitude $=6$, Phase $=-\pi$ rad.
   • At $\Omega =-5\pi$ rad/s: Amplitude $=6$, Phase $=+\pi$ rad (due to odd symmetry $\angle X(-j\Omega )=-\angle X(j\Omega )$).

Plotting the Spectra:

       SINGLE-SIDED AMPLITUDE & PHASE            DOUBLE-SIDED AMPLITUDE & PHASE
  
         |X| single                               |X| double
            ^                                        ^
         12 |     |                                6 |   |     |
         ---+-----+-----> Ω                          +---+-----+-----> Ω
            0    5pi                               -5pi  0    5pi
  
         ∠X single                                ∠X double (rad)
            ^                                        ^
            |                                    pi  |   o
         ---+-----+-----> Ω                         +---+-----+-----> Ω
          0 |    5pi                               -5pi |    5pi
            | o                                         |   o
            |-pi                                     -pi|

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4.4 Example 4: Large-Mark Symmetrical Double-Sided Spectrum [PYQ 2016 Q5b - 8 Marks]

Question: Draw the double-sided frequency spectrum of: $x(t)=8\sin \left(20\pi t-\frac{\pi }{4}\right),-\infty <t<\infty$

Step-by-Step Mathematical Analysis:

1. Convert Sine to Cosine first: $x(t)=8\cos \left(20\pi t-\frac{\pi }{4}-\frac{\pi }{2}\right)=8\cos \left(20\pi t-\frac{3\pi }{4}\right)$
2. Identify Parameters:
   • Peak Amplitude $A=8$.
   • Angular Frequency ${\Omega }_0=20\pi$ rad/s.
   • Phase Angle $\theta =-\frac{3\pi }{4}$ rad (or $-135^{\circ }$).
3. Determine Double-Sided Components:
   • Halved Amplitude: $A/2=4$.
   • At $\Omega =+20\pi$ rad/s: Amplitude $=4$, Phase $=-\frac{3\pi }{4}$ rad.
   • At $\Omega =-20\pi$ rad/s: Amplitude $=4$, Phase $=+\frac{3\pi }{4}$ rad (odd symmetry).

Plotting the Spectra:

       DOUBLE-SIDED AMPLITUDE SPECTRUM            DOUBLE-SIDED PHASE SPECTRUM
                  |X(jΩ)|                                    ∠X(jΩ) (rad)
                     ^                                          ^
                     |                                    3pi/4 |   o
             4       |       4                                  |   |
             |       |       |                                  |   |
       ------+-------+-------+------> Ω                  -------+---+-------> Ω
          -20pi      0     20pi                             -20pi |  20pi
                                                                  |  o
                                                                  | -3pi/4

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5. Common Mistakes That Cost Marks

The Sine-Basis Phase Reading Slip

This is the #1 reason students lose marks on spectral sketching questions. If the given equation is written in terms of a sine wave (e.g., $\sin ({\Omega }_{0}t+heta)$), you must convert it to a cosine wave using $\cos ({\Omega }_{0}t+heta-\pi /2)$ before reading the phase angle. Reading phase directly from a sine wave yields a phase error of exactly $-90^{\circ }$, causing an automatic zero-mark evaluation on the phase spectrum plot.

Double-Sided Amplitude Halving Omission

When converting a time-domain sinusoid to a double-sided exponential spectrum, you must split the peak amplitude $A$ in half ($A/2$). Forgetting to divide by 2 (e.g., plotting lines of height 10 instead of 5 for a $10\cos ({\Omega }_t)$ wave) is a standard math-syntax error that costs up to 50% of the question’s total marks.

Even Phase Plotting Violation

The phase spectrum of any real-valued signal must have odd symmetry (ngle X(-j\Omega) = -ngle X(j\Omega)). If you draw positive phase angles at both positive and negative frequencies, your plot violates basic complex number theory, leading to a major point reduction.

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6. PYQ Bank — Verbatim Questions & Answer Plans

6.1 PYQ 2019 Question 5c [1 Mark]

Question: Sketch the double-sided frequency spectrum of the signal $x(t)=10\cos (20\pi t-\pi /6),-\infty <t<\infty$.

• Answer Plan:
  1. Identify $A=10,{\Omega }_0=20\pi ,$ and $heta=-\pi /6$ rad.
  2. Divide peak amplitude to get $A/2=5$.
  3. State double-sided values: amplitude 5 at $\pm 20\pi$, phase $-\pi /6$ at $+20\pi$, and phase $+\pi /6$ at $-20\pi$.
  4. Plot the amplitude and phase spectrum as shown in Section 4.1.

6.2 PYQ 2017 Question 5b [5 Marks]

Question: Sketch the single and double sided frequency spectra of the following signal: $x(t)=25\cos (5\pi t-\pi /2)$.

• Answer Plan:
  1. Identify base parameters: $A=25,{\Omega }_0=5\pi ,heta=-\pi /2$ rad.
  2. Draft single-sided plots at $+5\pi$ with amplitude 25 and phase $-\pi /2$.
  3. Draft double-sided plots at $\pm 5\pi$ with amplitude 12.5 and phase $\mp \pi /2$.
  4. Render both sets clearly labeled as shown in Section 4.2.

6.3 PYQ 2018 Question 5c [2 Marks]

Question: Sketch single and double-sided frequency spectra of the signal $x(t)=12\sin (5\pi t-\pi /2)$.

• Answer Plan:
  1. Write down the sine-to-cosine conversion: $12\sin (5\pi t-\pi /2)=12\cos (5\pi t-\pi )$.
  2. Extract parameters: $A=12,{\Omega }_0=5\pi ,heta=-\pi$ rad.
  3. Calculate single-sided parameters and double-sided parameters ($A/2=6,het{a}_{+}=-\pi ,het{a}_{-}=+\pi$).
  4. Draw both plots with accurate axis ticks as shown in Section 4.3.

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7. Self-Check Before Moving On

• Can you explain why real-valued signals must have even amplitude and odd phase spectra? [1.1]
• Do you know how to convert a sine signal to cosine before extracting phase parameters? [4.3]
• Have you memorized the amplitude halving rule ($AoA/2$) for double-sided spectra? [2.2]
• Can you state and derive the Modulation Theorem in under 2 minutes? [3.1]

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Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 02 Fourier Transform.pdf, Rabiul sir class note.pdf.