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4.2 Fourier Series, Fourier Transforms & Frequency-Domain Analysis
PRC Registered Electrical Engineer REE
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4.2 Fourier Series, Fourier Transforms & Frequency-Domain Analysis
Key Takeaways Explain This Topic AI
Any periodic non-sinusoidal waveform satisfying Dirichlet conditions can be decomposed into an infinite sum of harmonically related sines and cosines via Trigonometric Fourier Series.
Waveform symmetry properties (even, odd, half-wave symmetry) significantly simplify Fourier coefficient calculations by reducing integrals to zero.
The RMS value of a complex periodic voltage or current is determined by the square root of the sum of the DC component squared and half the sum of squared peak harmonic amplitudes: V rms = V 0 2 + ∑ V n 2 2 V_{\text{rms}} = \sqrt{V_0^2 + \sum \frac{V_n^2}{2}} V rms = V 0 2 + ∑ 2 V n 2 .
Total Harmonic Distortion (THD) quantifies waveform distortion by measuring the ratio of the RMS voltage of all higher harmonics combined to the RMS voltage of the fundamental frequency component.
The Continuous Fourier Transform F { f ( t ) } = F ( ω ) = ∫ − ∞ ∞ f ( t ) e − j ω t d t \mathcal{F}\{f(t)\} = F(\omega) = \int_{-\infty}^\infty f(t)e^{-j\omega t} dt F { f ( t )} = F ( ω ) = ∫ − ∞ ∞ f ( t ) e − jω t d t maps non-periodic continuous-time signals into continuous frequency spectra.
Last updated: August 2026
4.2 Fourier Series, Fourier Transforms & Frequency-Domain Analysis
Frequency-domain spectral analysis is indispensable in electrical power systems, telecommunications, and signal processing. While Laplace transforms are optimized for transient IVP analysis with initial conditions, Fourier analysis decomposes arbitrary periodic and non-periodic signals into discrete spectral harmonics or continuous frequency distributions. On the PRC Electrical Engineering Board Exam, Fourier series expansion, RMS evaluation of non-sinusoidal currents, and Total Harmonic Distortion (THD) calculations are primary recurring topics.
1. Trigonometric Fourier Series
Any periodic function f ( t ) f(t) f ( t ) with fundamental period T T T and fundamental angular frequency ω 0 = 2 π T \omega_0 = \frac{2\pi}{T} ω 0 = T 2 π that satisfies the Dirichlet conditions (single-valued, bounded, finite number of discontinuities and extrema within period T T T ) can be expressed as a Trigonometric Fourier Series: f ( t ) = a 0 + ∑ n = 1 ∞ [ a n cos ( n ω 0 t ) + b n sin ( n ω 0 t ) ] f(t) = a_0 + \sum_{n=1}^\infty \left[ a_n \cos(n \omega_0 t) + b_n \sin(n \omega_0 t) \right] f ( t ) = a 0 + n = 1 ∑ ∞ [ a n cos ( n ω 0 t ) + b n sin ( n ω 0 t ) ]
Euler-Fourier Coefficient Formulas
DC Component a 0 a_0 a 0 (Average Value): a 0 = 1 T ∫ 0 T f ( t ) d t a_0 = \frac{1}{T} \int_0^T f(t) \, dt a 0 = T 1 ∫ 0 T f ( t ) d t
Cosine Coefficients a n a_n a n : a n = 2 T ∫ 0 T f ( t ) cos ( n ω 0 t ) d t for n = 1 , 2 , 3 , … a_n = \frac{2}{T} \int_0^T f(t) \cos(n \omega_0 t) \, dt \quad \text{for } n = 1, 2, 3, \dots a n = T 2 ∫ 0 T f ( t ) cos ( n ω 0 t ) d t for n = 1 , 2 , 3 , …
Sine Coefficients b n b_n b n : b n = 2 T ∫ 0 T f ( t ) sin ( n ω 0 t ) d t for n = 1 , 2 , 3 , … b_n = \frac{2}{T} \int_0^T f(t) \sin(n \omega_0 t) \, dt \quad \text{for } n = 1, 2, 3, \dots b n = T 2 ∫ 0 T f ( t ) sin ( n ω 0 t ) d t for n = 1 , 2 , 3 , …
Amplitude-Phase Form
Alternatively, combining sine and cosine terms at each harmonic frequency yields: f ( t ) = A 0 + ∑ n = 1 ∞ A n cos ( n ω 0 t − θ n ) f(t) = A_0 + \sum_{n=1}^\infty A_n \cos(n \omega_0 t - \theta_n) f ( t ) = A 0 + n = 1 ∑ ∞ A n cos ( n ω 0 t − θ n )
where peak harmonic amplitude A n = a n 2 + b n 2 A_n = \sqrt{a_n^2 + b_n^2} A n = a n 2 + b n 2 and phase angle θ n = arctan ( b n a n ) \theta_n = \arctan\left(\frac{b_n}{a_n}\right) θ n = arctan ( a n b n ) .
2. Waveform Symmetry Rules
Recognizing geometric waveform symmetry eliminates unnecessary integration during board exam computations:
Symmetry Type Mathematical Condition Fourier Coefficients Result Harmonic Content
Even Symmetry f ( − t ) = f ( t ) f(-t) = f(t) f ( − t ) = f ( t ) b n = 0 for all n b_n = 0 \quad \text{for all } n b n = 0 for all n Contains DC and cosine terms only
Odd Symmetry f ( − t ) = − f ( t ) f(-t) = -f(t) f ( − t ) = − f ( t ) a 0 = 0 , a n = 0 for all n a_0 = 0, \quad a_n = 0 \quad \text{for all } n a 0 = 0 , a n = 0 for all n Contains sine terms only
Half-Wave Symmetry f ( t + T 2 ) = − f ( t ) f\left(t + \frac{T}{2}\right) = -f(t) f ( t + 2 T ) = − f ( t ) a 0 = 0 , a n = 0 , b n = 0 for even n a_0 = 0, \quad a_n = 0, b_n = 0 \quad \text{for even } n a 0 = 0 , a n = 0 , b n = 0 for even n Contains odd harmonics only ( n = 1 , 3 , 5 , … n = 1, 3, 5, \dots n = 1 , 3 , 5 , … )
Even Quarter-Wave Even + Half-Wave a 0 = 0 , b n = 0 , a n = 4 T ∫ 0 T / 4 f ( t ) cos ( n ω 0 t ) d t a_0 = 0, \quad b_n = 0, \quad a_n = \frac{4}{T} \int_0^{T/4} f(t) \cos(n\omega_0 t) dt a 0 = 0 , b n = 0 , a n = T 4 ∫ 0 T /4 f ( t ) cos ( n ω 0 t ) d t Odd cosine harmonics only
Odd Quarter-Wave Odd + Half-Wave a 0 = 0 , a n = 0 , b n = 4 T ∫ 0 T / 4 f ( t ) sin ( n ω 0 t ) d t a_0 = 0, \quad a_n = 0, \quad b_n = \frac{4}{T} \int_0^{T/4} f(t) \sin(n\omega_0 t) dt a 0 = 0 , a n = 0 , b n = T 4 ∫ 0 T /4 f ( t ) sin ( n ω 0 t ) d t Odd sine harmonics only
3. Exponential (Complex) Fourier Series & Parseval's Power Theorem
Using Euler's identity e j θ = cos θ + j sin θ e^{j\theta} = \cos\theta + j\sin\theta e j θ = cos θ + j sin θ , the trigonometric series transforms into a compact complex exponential representation: f ( t ) = ∑ n = − ∞ ∞ c n e j n ω 0 t f(t) = \sum_{n=-\infty}^\infty c_n e^{j n \omega_0 t} f ( t ) = n = − ∞ ∑ ∞ c n e jn ω 0 t
where complex coefficients c n c_n c n are given by: c n = 1 T ∫ 0 T f ( t ) e − j n ω 0 t d t c_n = \frac{1}{T} \int_0^T f(t) e^{-j n \omega_0 t} \, dt c n = T 1 ∫ 0 T f ( t ) e − jn ω 0 t d t
Relationships to trigonometric coefficients: c 0 = a 0 , c n = a n − j b n 2 , c − n = c n ∗ = a n + j b n 2 c_0 = a_0, \quad c_n = \frac{a_n - j b_n}{2}, \quad c_{-n} = c_n^* = \frac{a_n + j b_n}{2} c 0 = a 0 , c n = 2 a n − j b n , c − n = c n ∗ = 2 a n + j b n
Parseval's Power Theorem
Parseval's identity states that the total average power of a periodic signal f ( t ) f(t) f ( t ) equals the sum of the average powers of its individual harmonic components: P avg = 1 T ∫ 0 T ∣ f ( t ) ∣ 2 d t = ∑ n = − ∞ ∞ ∣ c n ∣ 2 = a 0 2 + ∑ n = 1 ∞ a n 2 + b n 2 2 = A 0 2 + ∑ n = 1 ∞ A n 2 2 P_{\text{avg}} = \frac{1}{T} \int_0^T |f(t)|^2 \, dt = \sum_{n=-\infty}^\infty |c_n|^2 = a_0^2 + \sum_{n=1}^\infty \frac{a_n^2 + b_n^2}{2} = A_0^2 + \sum_{n=1}^\infty \frac{A_n^2}{2} P avg = T 1 ∫ 0 T ∣ f ( t ) ∣ 2 d t = n = − ∞ ∑ ∞ ∣ c n ∣ 2 = a 0 2 + n = 1 ∑ ∞ 2 a n 2 + b n 2 = A 0 2 + n = 1 ∑ ∞ 2 A n 2
- 4. RMS Values of Non-Sinusoidal Signals & Total Harmonic Distortion (THD)
- RMS Value of Complex Non-Sinusoidal Waveforms
When a voltage or current contains multiple orthogonal harmonic frequencies: v ( t ) = V 0 + V 1 , m cos ( ω 0 t + θ 1 ) + V 2 , m cos ( 2 ω 0 t + θ 2 ) + … v(t) = V_0 + V_{1,m} \cos(\omega_0 t + \theta_1) + V_{2,m} \cos(2\omega_0 t + \theta_2) + \dots v ( t ) = V 0 + V 1 , m cos ( ω 0 t + θ 1 ) + V 2 , m cos ( 2 ω 0 t + θ 2 ) + …
The overall Effective (RMS) Voltage V rms V_{\text{rms}} V rms is: V rms = V 0 2 + V 1 , m 2 2 + V 2 , m 2 2 + ⋯ + V n , m 2 2 = V 0 2 + V 1 , rms 2 + V 2 , rms 2 + ⋯ + V n , rms 2 V_{\text{rms}} = \sqrt{V_0^2 + \frac{V_{1,m}^2}{2} + \frac{V_{2,m}^2}{2} + \dots + \frac{V_{n,m}^2}{2}} = \sqrt{V_0^2 + V_{1,\text{rms}}^2 + V_{2,\text{rms}}^2 + \dots + V_{n,\text{rms}}^2} V rms = V 0 2 + 2 V 1 , m 2 + 2 V 2 , m 2 + ⋯ + 2 V n , m 2 = V 0 2 + V 1 , rms 2 + V 2 , rms 2 + ⋯ + V n , rms 2
Total Harmonic Distortion (THD)
Total Harmonic Distortion measures waveform degradation caused by non-linear loads (such as variable frequency drives, rectifiers, and switching power supplies): THD V = ∑ n = 2 ∞ V n , rms 2 V 1 , rms = V 2 , rms 2 + V 3 , rms 2 + V 4 , rms 2 + … V 1 , rms \text{THD}_V = \frac{\sqrt{\sum_{n=2}^\infty V_{n,\text{rms}}^2}}{V_{1,\text{rms}}} = \frac{\sqrt{V_{2,\text{rms}}^2 + V_{3,\text{rms}}^2 + V_{4,\text{rms}}^2 + \dots}}{V_{1,\text{rms}}} THD V = V 1 , rms ∑ n = 2 ∞ V n , rms 2 = V 1 , rms V 2 , rms 2 + V 3 , rms 2 + V 4 , rms 2 + …
Expressed as a percentage, THD % = THD V × 100 % \text{THD}\% = \text{THD}_V \times 100\% THD % = THD V × 100% .
5. Continuous Fourier Transform
For non-periodic signals ( T → ∞ T \to \infty T → ∞ ), the discrete frequency spectrum becomes continuous. The continuous Fourier transform pairs continuous time signal f ( t ) f(t) f ( t ) with continuous frequency spectrum F ( ω ) F(\omega) F ( ω ) :
Direct Continuous Fourier Transform F { f ( t ) } = F ( ω ) = ∫ − ∞ ∞ f ( t ) e − j ω t d t \mathcal{F}\{f(t)\} = F(\omega) = \int_{-\infty}^\infty f(t) e^{-j \omega t} \, dt F { f ( t )} = F ( ω ) = ∫ − ∞ ∞ f ( t ) e − jω t d t
Inverse Continuous Fourier Transform F − 1 { F ( ω ) } = f ( t ) = 1 2 π ∫ − ∞ ∞ F ( ω ) e j ω t d ω \mathcal{F}^{-1}\{F(\omega)\} = f(t) = \frac{1}{2\pi} \int_{-\infty}^\infty F(\omega) e^{j \omega t} \, d\omega F − 1 { F ( ω )} = f ( t ) = 2 π 1 ∫ − ∞ ∞ F ( ω ) e jω t d ω
Operational Properties of Fourier Transform
Property Time Domain f ( t ) f(t) f ( t ) Frequency Domain F ( ω ) F(\omega) F ( ω )
Linearity a f 1 ( t ) + b f 2 ( t ) a f_1(t) + b f_2(t) a f 1 ( t ) + b f 2 ( t ) a F 1 ( ω ) + b F 2 ( ω ) a F_1(\omega) + b F_2(\omega) a F 1 ( ω ) + b F 2 ( ω )
Time Scaling f ( a t ) f(a t) f ( a t ) 1 ∣ a ∣ F ( ω a ) \frac{1}{\vert a \vert} F\left(\frac{\omega}{a}\right) ∣ a ∣ 1 F ( a ω )
Time Shifting f ( t − t 0 ) f(t - t_0) f ( t − t 0 ) e − j ω t 0 F ( ω ) e^{-j \omega t_0} F(\omega) e − jω t 0 F ( ω )
Frequency Shifting e j ω 0 t f ( t ) e^{j \omega_0 t} f(t) e j ω 0 t f ( t ) F ( ω − ω 0 ) F(\omega - \omega_0) F ( ω − ω 0 )
Time Differentiation d n f ( t ) d t n \frac{d^n f(t)}{dt^n} d t n d n f ( t ) ( j ω ) n F ( ω ) (j \omega)^n F(\omega) ( jω ) n F ( ω )
- Duality F ( t ) F(t) F ( t ) 2 π f ( − ω ) 2\pi f(-\omega) 2 π f ( − ω )
- Step-by-Step Worked Numerical Examples
- Example 4.2.1: Fourier Series Coefficients of a Symmetric Square Wave
Problem: A symmetric square wave voltage v ( t ) v(t) v ( t ) alternates between + 100 V +100\text{ V} + 100 V and − 100 V -100\text{ V} − 100 V with period T = 2 π s T = 2\pi\text{ s} T = 2 π s (fundamental frequency ω 0 = 1 rad/s \omega_0 = 1\text{ rad/s} ω 0 = 1 rad/s ). Specifically, v ( t ) = 100 V v(t) = 100\text{ V} v ( t ) = 100 V for 0 < t < π 0 < t < \pi 0 < t < π and v ( t ) = − 100 V v(t) = -100\text{ V} v ( t ) = − 100 V for π < t < 2 π \pi < t < 2\pi π < t < 2 π . Calculate the Fourier series coefficients a 0 a_0 a 0 , a n a_n a n , and b n b_n b n , and write the trigonometric series expansion.
Solution:
Check DC component a 0 a_0 a 0 : a 0 = 1 2 π [ ∫ 0 π 100 d t + ∫ π 2 π ( − 100 ) d t ] = 1 2 π [ 100 π − 100 π ] = 0 a_0 = \frac{1}{2\pi} \left[ \int_0^\pi 100 \, dt + \int_\pi^{2\pi} (-100) \, dt \right] = \frac{1}{2\pi} [100\pi - 100\pi] = 0 a 0 = 2 π 1 [ ∫ 0 π 100 d t + ∫ π 2 π ( − 100 ) d t ] = 2 π 1 [ 100 π − 100 π ] = 0
Check symmetry: The wave exhibits odd symmetry v ( − t ) = − v ( t ) v(-t) = -v(t) v ( − t ) = − v ( t ) and half-wave symmetry v ( t + π ) = − v ( t ) v(t + \pi) = -v(t) v ( t + π ) = − v ( t ) .
Therefore, a n = 0 a_n = 0 a n = 0 for all n n n , and b n = 0 b_n = 0 b n = 0 for even n n n .
Compute b n b_n b n for odd n n n using quarter-wave formula: b n = 4 T ∫ 0 T / 2 v ( t ) sin ( n t ) d t = 4 2 π ∫ 0 π 100 sin ( n t ) d t = 200 π [ − cos ( n t ) n ] 0 π b_n = \frac{4}{T} \int_0^{T/2} v(t) \sin(n t) \, dt = \frac{4}{2\pi} \int_0^\pi 100 \sin(n t) \, dt = \frac{200}{\pi} \left[ -\frac{\cos(n t)}{n} \right]_0^\pi b n = T 4 ∫ 0 T /2 v ( t ) sin ( n t ) d t = 2 π 4 ∫ 0 π 100 sin ( n t ) d t = π 200 [ − n cos ( n t ) ] 0 π b n = 200 n π [ 1 − cos ( n π ) ] b_n = \frac{200}{n \pi} [1 - \cos(n \pi)] b n = nπ 200 [ 1 − cos ( nπ )]
For even n n n , cos ( n π ) = 1 ⟹ b n = 0 \cos(n\pi) = 1 \implies b_n = 0 cos ( nπ ) = 1 ⟹ b n = 0 .
For odd n n n , cos ( n π ) = − 1 ⟹ b n = 200 n π [ 1 − ( − 1 ) ] = 400 n π \cos(n\pi) = -1 \implies b_n = \frac{200}{n \pi} [1 - (-1)] = \frac{400}{n \pi} cos ( nπ ) = − 1 ⟹ b n = nπ 200 [ 1 − ( − 1 )] = nπ 400 .
Evaluate numerical values for first three odd harmonics:
- b 1 = 400 π ≈ 127.32 V b_1 = \frac{400}{\pi} \approx 127.32\text{ V} b 1 = π 400 ≈ 127.32 V
- b 3 = 400 3 π ≈ 42.44 V b_3 = \frac{400}{3\pi} \approx 42.44\text{ V} b 3 = 3 π 400 ≈ 42.44 V
- b 5 = 400 5 π ≈ 25.46 V b_5 = \frac{400}{5\pi} \approx 25.46\text{ V} b 5 = 5 π 400 ≈ 25.46 V
Write Fourier series expansion: v ( t ) = 400 π ( sin ( t ) + 1 3 sin ( 3 t ) + 1 5 sin ( 5 t ) + … ) V v(t) = \frac{400}{\pi} \left( \sin(t) + \frac{1}{3}\sin(3t) + \frac{1}{5}\sin(5t) + \dots \right) \text{ V} v ( t ) = π 400 ( sin ( t ) + 3 1 sin ( 3 t ) + 5 1 sin ( 5 t ) + … ) V
Example 4.2.2: Total RMS Current and THD of Industrial Load Current
Problem: An industrial load current contains a fundamental component and two higher harmonics: ∋ ( t ) = 10 cos ( ω t ) + 3 cos ( 3 ω t + 30 ∘ ) + 1.5 cos ( 5 ω t − 45 ∘ ) A \ni(t) = 10 \cos(\omega t) + 3 \cos(3\omega t + 30^\circ) + 1.5 \cos(5\omega t - 45^\circ) \text{ A} ∋ ( t ) = 10 cos ( ω t ) + 3 cos ( 3 ω t + 3 0 ∘ ) + 1.5 cos ( 5 ω t − 4 5 ∘ ) A
Calculate: (a) RMS value of fundamental current I 1 , rms I_{1,\text{rms}} I 1 , rms , (b) overall total RMS current I rms I_{\text{rms}} I rms , and (c) Current Total Harmonic Distortion ( THD I \text{THD}_I THD I ).
Solution:
Determine peak amplitudes: I 1 , m = 10 A , I 3 , m = 3 A , I 5 , m = 1.5 A I_{1,m} = 10\text{ A}, \quad I_{3,m} = 3\text{ A}, \quad I_{5,m} = 1.5\text{ A} I 1 , m = 10 A , I 3 , m = 3 A , I 5 , m = 1.5 A
Calculate individual component RMS values: I 1 , rms = 10 2 ≈ 7.0711 A I_{1,\text{rms}} = \frac{10}{\sqrt{2}} \approx 7.0711\text{ A} I 1 , rms = 2 10 ≈ 7.0711 A I 3 , rms = 3 2 ≈ 2.1213 A I_{3,\text{rms}} = \frac{3}{\sqrt{2}} \approx 2.1213\text{ A} I 3 , rms = 2 3 ≈ 2.1213 A I 5 , rms = 1.5 2 ≈ 1.0607 A I_{5,\text{rms}} = \frac{1.5}{\sqrt{2}} \approx 1.0607\text{ A} I 5 , rms = 2 1.5 ≈ 1.0607 A
Calculate total RMS current I rms I_{\text{rms}} I rms : I rms = I 1 , rms 2 + I 3 , rms 2 + I 5 , rms 2 = ( 7.0711 ) 2 + ( 2.1213 ) 2 + ( 1.0607 ) 2 I_{\text{rms}} = \sqrt{I_{1,\text{rms}}^2 + I_{3,\text{rms}}^2 + I_{5,\text{rms}}^2} = \sqrt{(7.0711)^2 + (2.1213)^2 + (1.0607)^2} I rms = I 1 , rms 2 + I 3 , rms 2 + I 5 , rms 2 = ( 7.0711 ) 2 + ( 2.1213 ) 2 + ( 1.0607 ) 2 I rms = 50 + 4.5 + 1.125 = 55.625 ≈ 7.4582 A I_{\text{rms}} = \sqrt{50 + 4.5 + 1.125} = \sqrt{55.625} \approx 7.4582\text{ A} I rms = 50 + 4.5 + 1.125 = 55.625 ≈ 7.4582 A
Compute Total Harmonic Distortion ( THD I \text{THD}_I THD I ): THD I = I 3 , rms 2 + I 5 , rms 2 I 1 , rms = 4.5 + 1.125 7.0711 = 5.625 7.0711 = 2.3717 7.0711 ≈ 0.3354 \text{THD}_I = \frac{\sqrt{I_{3,\text{rms}}^2 + I_{5,\text{rms}}^2}}{I_{1,\text{rms}}} = \frac{\sqrt{4.5 + 1.125}}{7.0711} = \frac{\sqrt{5.625}}{7.0711} = \frac{2.3717}{7.0711} \approx 0.3354 THD I = I 1 , rms I 3 , rms 2 + I 5 , rms 2 = 7.0711 4.5 + 1.125 = 7.0711 5.625 = 7.0711 2.3717 ≈ 0.3354 Expressed as percentage: THD I = 33.54 % \text{THD}_I = 33.54\% THD I = 33.54% .
Example 4.2.3: Continuous Fourier Transform of an Exponential Pulse
Problem: Find the continuous Fourier transform F ( ω ) F(\omega) F ( ω ) of the causal single-sided exponential decay signal f ( t ) = e − a t u ( t ) f(t) = e^{-a t} u(t) f ( t ) = e − a t u ( t ) with attenuation a = 5 s − 1 a = 5\text{ s}^{-1} a = 5 s − 1 . Calculate the magnitude spectrum ∣ F ( ω ) ∣ \vert F(\omega) \vert ∣ F ( ω ) ∣ at angular frequency ω = 5 rad/s \omega = 5\text{ rad/s} ω = 5 rad/s .
Solution:
Apply continuous Fourier transform integral: F ( ω ) = ∫ 0 ∞ e − 5 t e − j ω t d t = ∫ 0 ∞ e − ( 5 + j ω ) t d t F(\omega) = \int_0^\infty e^{-5 t} e^{-j \omega t} \, dt = \int_0^\infty e^{-(5 + j \omega) t} \, dt F ( ω ) = ∫ 0 ∞ e − 5 t e − jω t d t = ∫ 0 ∞ e − ( 5 + jω ) t d t
Evaluate the definite integral: F ( ω ) = [ e − ( 5 + j ω ) t − ( 5 + j ω ) ] 0 ∞ = 0 − ( 1 − ( 5 + j ω ) ) = 1 5 + j ω F(\omega) = \left[ \frac{e^{-(5 + j \omega) t}}{-(5 + j \omega)} \right]_0^\infty = 0 - \left( \frac{1}{-(5 + j \omega)} \right) = \frac{1}{5 + j \omega} F ( ω ) = [ − ( 5 + jω ) e − ( 5 + jω ) t ] 0 ∞ = 0 − ( − ( 5 + jω ) 1 ) = 5 + jω 1
Determine magnitude formula ∣ F ( ω ) ∣ \vert F(\omega) \vert ∣ F ( ω ) ∣ : ∣ F ( ω ) ∣ = 1 5 2 + ω 2 = 1 25 + ω 2 \vert F(\omega) \vert = \frac{1}{\sqrt{5^2 + \omega^2}} = \frac{1}{\sqrt{25 + \omega^2}} ∣ F ( ω ) ∣ = 5 2 + ω 2 1 = 25 + ω 2 1
Evaluate at ω = 5 rad/s \omega = 5\text{ rad/s} ω = 5 rad/s : ∣ F ( 5 ) ∣ = 1 25 + 5 2 = 1 50 = 1 7.0711 ≈ 0.1414 \vert F(5) \vert = \frac{1}{\sqrt{25 + 5^2}} = \frac{1}{\sqrt{50}} = \frac{1}{7.0711} \approx 0.1414 ∣ F ( 5 ) ∣ = 25 + 5 2 1 = 50 1 = 7.0711 1 ≈ 0.1414
Conclusion: F ( ω ) = 1 5 + j ω F(\omega) = \frac{1}{5 + j \omega} F ( ω ) = 5 + jω 1 and ∣ F ( 5 ) ∣ = 0.1414 \vert F(5) \vert = 0.1414 ∣ F ( 5 ) ∣ = 0.1414 .