Lab Part 2

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3101L

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Electrical Engineering

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Feb 20, 2024

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Lab #3-4 Fourier Series pt. 2 ECE 3101-02 February 14, 2023
Objectives Be able to calculate the complex exponential Fourier Series coefficient Xn of periodic filtered and unfiltered sinusoid, rectangular and triangle signals and from them obtain Cn’s Dn’s and θn’s before and after the filters Be able to calculate how the Fourier Coefficients of periodic signals like pulse trains are affected by filters Table 2 – Lab 2 Experiment Waveforms
Filter Transfer Function Given a time domain periodic pulse trains x(t) applied to the RC filters below i. Find the filters’ transfer function H(jω) = Y(jω)/X(jω) as a function of the circuit’s impedances a. Low Pass RC filter Transfer Function VoltageGain Magnitude = | H () | = ¿ H ( jω )∨ ¿ 1 / √ ( 1 + ( ωRC ) 2 ) The absolute Value of H(w) is the magnitude of the transfer function and w is the angular frequency in radians per second. This shows the ratio of output voltage to input voltage (i.e Vout Vin ) b. High Pass RC Filter Transfer Function VoltageGain Magnitude = | H () | = H ( jω )= jωRC /( 1 + jωRC ) This function shows the ratio of the output voltage across the capacitor to the input voltage across the resistor. ii. Find the 3dB cutoff frequency ω3dB as a function of R1 and C1 We first set the magnitude of the transfer function to 1/√2 or -3db and we proceed to solve for w |H(jω)|= |jωR1C1/((1 + jωR1C1) )|=1/√2 Take the magnitude of both sides and simplify. ωR1C1 = 1/√2 Solve for w.
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3dB corner or cutoff frequency ωc=1/((R1C1√2) ) iii. Calculate R1 for ω3dB = 2π3000 rad/s and C1 = 0.01μF If we were to use the formula from before we would get: R1 = 530.84 Ohms
Fourier Coefficient expressions of filtered sine, square & sawtooth waves Derive the complex exponential and trigonometric Fourier coefficients magnitude and phase expressions Y n (Complex exponential Fourier coefficient of the filtered signal) & D n (Trigonometric Fourier coefficient of the filtered signal) of y(t) (Time domain filtered signal) for the rectangular pulse train duty cycle=dc a. Complex Exponential Fourier Coefficient Yn of a low pass RC filtered signal y(t) Yn = 2 π ∗ ∫ ¿ y ( t )∗ e (− jnωt ) dt / T The low-pass RC filter can be represented by a transfer function H ( jω )= 1 /( 1 + jωR 1 C 1 ) only if we assume that the input signal is a periodic function with period T . The output signal y(t) can be obtained if we convolve the input signal x(t) with the impulse response of the filter h(t) = (1/RC) * e^(-t/RC)u(t), u(t) = unit step function. By convolution theorem: y ( t )= x ( t )∗ h ( t )= ∑ [ n =− ∞ ,∞ ] Xn H ( jnω ) e ( jnωt ) Substituting the expression for H(jω) and h(t): y ( t )= ∑ [ n =− ∞ ,∞ ] Xn ( 1 /( 1 + jnωR 1 C 1 ))( ( 1 RC ) ∗ ∫ [ 0 ,∞ ) e − t − τ RC ∗ u ( τ ) e (− jnωτ ) dτ ) Simplifying using Laplace transform of the unit step function & multiplying the and dividing the numerator by jn, we get: y ( t )= ∑ [ n =− ∞ ,∞ ] Xn ( 1 /( 1 + jnωR 1 C 1 ))( 1 /( RC ∗( 1 + jnωRC )))( jn ∗ e (− jnωt ) ) Now take the imaginary part of both sides using Euler's formula, we get: ℑ{ Yn }=− π ∗ Xn ∗( ωRC )/( 1 +( nωRC ) 2 ) where Im{Yn} is the imaginary part of the complex exponential Fourier coefficient Yn of the output signal y(t).
b. Complex Exponential Fourier Coefficient Yn of a high pass RC filtered signal y(t) The complex exponential Fourier coefficient Yn of a high-pass RC filtered signal y(t) can be calculated using the following formula: Yn = 2 π ∗ ∫ ¿ y ( t )∗ e (− jnωt ) dt / T If we assume the input signal is a periodic function with period T, the high-pass RC filter can be represented by a transfer function H(jω) = jωR1C1 / (1 + jωR1C1) Y(t) (output) can be obtained by convolving the input signal x(t) with the impulse response of the filter h(t) = (1/RC) * e^(-t/RC)u(t), where u(t) is the unit step function. Using the convolution theorem, we can write the Fourier series of the output signal y(t) as: y ( t )= x ( t )∗ h ( t )= ∑ [ n =− ∞ ,∞ ] Xn H ( jnω ) e ( jnωt ) where Xn is the complex exponential Fourier coefficient of the input signal x(t). Substituting the expression for H(jω) and h(t) in the above equation, we get: y ( t )= ∑ [ n =− ∞ ,∞ ] Xn ( jnωR 1 C 1 /( 1 + jnωR 1 C 1 ))(( 1 / RC )∗ ∫ ¿ e (−( t − τ )/ RC ) u ( τ ) e (− jnωτ ) dτ ) Now if we simplify using Laplace transform of the unit step function & multiplying the and dividing the numerator by jn, we get: y ( t )= ∑ [ n =− ∞ ,∞ ] Xn ( jnωR 1 C 1 /( 1 + jnωR 1 C 1 ))( 1 /( RC ∗( 1 + jnωRC )))( jn ∗ e (− jnωt ) ) Lastly, we take the imaginary part of both sides and using Euler's formula: ℑ{ Yn }= π ∗ Xn ∗( 1 − e (− 2 πnωRC ) )/( 1 +( nωRC ) 2 )
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c. Trigonometric Fourier Coefficient Dn of a hig nh pass RC filtered signal y(t) The Trigonometric Fourier Coefficient Dn of a high-pass RC filtered signal y(t) can be calculated from its complex exponential Fourier coefficient Yn using the following formula: Dn = 2 ∨ Yn ∨ ¿ π This is the imaginary part of the complex exponential Fourier coefficient Yn of a high-pass RC filtered signal y(t): ℑ{ Yn }= π ∗ Xn ∗( 1 − e (− 2 πnωRC ) )/( 1 +( nωRC ) 2 ) The magnitude |Yn| of Yn can be calculated using the following formula: ¿ Yn ∨ ¿ √ (ℜ{ Yn } 2 +ℑ{ Yn } 2 ) where Re{Yn} is the real part of Yn. Since the input signal is assumed to be a square wave with a 50% duty cycle, its Trigonometric Fourier Coefficients are given by: Dn =( 4 / π )∗( 1 / n )∗ sin ( nπ / 2 ) *Note: odd values of n and Dn = 0 for even values of n.
Simulation of unfiltered sine, square & sawtooth waves using Simulink i. Low pass filtered sine wave Sinewave wave Vpp=V, Vp=V, fo=Hz , 1 st order LPF fc=KHz, scope time domain (time const 53.05e-6) Sinewave wave Vpp=V, Vp=V fo=Hz , 1 st order LPF fc=KHz Spectrum Analyzer frequency domain (time const 53.05e-6)
ii. 1 st order low pass filtered square wave Square wave Vpp=V, Vp=V, fo=Hz, dc=% 1 st order LPF fc=KHz, scope time domain (time const 79.58e-6) Square wave Vpp=V, Vp=V fo=Hz, dc=% 1 st order LPF fc=KHz Spectrum Analyzer frequency domain (time const 79.58e-6)
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iii. 3 rd order low pass filtered square wave Square wave Vpp=V, Vp=V, fo=Hz , 3rd order LPF fc=KHz, scope time domain (time const 79.58e-6)
Square wave Vpp=V, Vp=V fo=Hz , 3rd order LPF fc=KHz Spectrum Analyzer frequency domain (time const 79.58e-6)
iv. 1 st and 3 rd order low pass filtered sawtooth wave Sawtooth Vpp=V, Vp=V, fo=Hz, sym=%, 1st order LPF fc=KHz, scope time domain Sawtooth Vpp=V, Vp=V, fo=Hz, sym=%, 1st order, LPF fc=KHz Spectrum Analyzer frequency domain
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Sawtooth Vpp=V, Vp=V, fo=Hz, sym=%, 3rd order LPF fc=KHz, scope time domain Sawtooth Vpp=V, Vp=V, fo=Hz, sym=%, 3rd order, LPF fc=KHz Spectrum Analyzer frequency domain
v. 1 st order high pass filtered square wave with different cutoff frequencies. Square wave Vpp, Vp=V, fo, dc=%, 1st order HPF fc=KHz, scope time domain Square wave Vpp, Vp fo, dc=%, 1st order, HPF fc=KHz Spectrum Analyzer frequency domain
vi. 3rd order high pass filtered square wave Square wave Vpp, Vp, fo, dc=%, 3rd order HPF fc=KHz, scope time domain
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Square wave Vpp, Vp fo, dc=%, 3rd order, HPF fc=KHz Spectrum Analyzer frequency domain
vii. 1 st and 3rd order high pass filtered sawtooth wave Sawtooth Vpp, Vp, fo, sym=%, 1st order HPF fc=KHz, scope time domain
Sawtooth Vpp, Vp, fo, sym=%, 1st order, HPF fc=KHz Spectrum Analyzer freq domain
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Sawtooth Vpp, Vp, fo, sym=%, 3rd order HPF fc=KHz, scope time domain Sawtooth Vpp, Vp, fo, sym=%, 3rd order, HPF fc=KHz Spectrum Analyzer frequency domain

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