Paper Title: Practical Application of Wavelet to Power Quality Analysis Author and Presenter: Norman Tse 1
Harmonics Frequency Estimation by Wavelet Transform (WT) Any harmonic signal can be described by its frequency, amplitude and phase Wavelet Transform is used to estimate harmonic frequencies amplitudes of the harmonic frequencies initial phase difference 2
Harmonic Frequencies Estimation Algorithm Step 1 use Continuous Wavelet Transform (CWT) to extract signal time and scale (frequency) information Step 2 - Normalised scalogram and wavelet ridges are used to find out the scale(s) ) at which the absolute value of the CWT coefficients are maximum Step 3 - the scales at which the wavelet ridges are maximum reveal the frequencies of the harmonics 3
Amplitude Estimation Algorithm Step 1 CWT coefficients of scales of the maximum wavelet ridges are extracted (these coefficients contain the time information of the harmonic frequencies) Step 2 the amplitudes of the harmonic frequencies are calculated from these CWT coefficients 4
Phase difference Estimation Algorithm Step 1 estimate the instantaneous phase angles from the CWT coefficients of the harmonic frequencies Step 2 initial phase angles (phase difference) of any two harmonic frequencies are calculated accordingly 5
What is Continuous Wavelet Transform (CWT)? able to extract signal time and scale (frequency) information simultaneously able to preserve phase information 6
Graphical Illustrations of CWT 50Hz Signal Increasing scales (dilating) Shifting 7
The wavelet coefficients will be largest when the wavelet oscillation frequency at a particular scale is the same as the signal harmonic frequencies The scale and the signal frequency is related by f 1 = fs fc S1 where f 1 is the signal frequency f c is the wavelet centre frequency f s is the sampling frequency s 1 is the corresponding scale 8
Wavelet at scale 5a Signal with mixed frequencies of 50Hz + 250Hz Wavelet at scale a 9
Basic Properties Selection of the Wavelet must have finite energy (i.e. finite support) of zero mean, i.e. has no zero frequency component Simplified Complex Morlet Wavelet (CMW) is chosen smooth and harmonic like waveform suitable for harmonic analysis contain phase information 10
Complex Morlet Wavelet (CMW) 11
Characteristics of the CMW with the centre frequency fc and the bandwidth parameter fb large enough,, the mean of the CMW is practically equal to zero the frequency support of the CMW is not a compact support but the entire frequency axis the time support of the CMW is taken from -88 to 8 the waveform is symmetrical 12
Harmonic Frequencies Estimation Algorithm Given a signal f (t) = a(t)cos φ (t) Its wavelet transform is Wf(u,s) = s 2 a(u)e jφ(u) ( ĝ( s[ ξ φ (u)]) + ε(u, ξ )) The corrective term will be appreciable if the waveform f(t) has large variations 13
The normalised scalogram is defined as ξ P w f ( u, ξ ) = η Wf ( u,s ) s 2 which is calculated by ξ Pw f η ( u, ξ ) = 1 4 a 2 ( u ) ĝ( η[1 φ ( u ) ξ ]) + ε( u, ξ ) 2 This term is maximum at ω=0 neglect this corrective term 14
The normalised scalogram is maximum at η s( u ) = ξ( u ) = φ ( u ) The corresponding points ( u, ξ(u) ) calculated by the above equation are called wavelet ridges. 15
Illustrative examples A signal is constructed as x = 10cos(2π50t) + 5cos(2π100t 100t-20 o ) 16
Wavelet Ridges Plot Detected frequency = 50Hz Accuracy = 100% Detected frequency = 100Hz Accuracy = 100% 17
The analytic amplitude is given by Absolute value of the wavelet coefficient ξ 2 Wf( u,s ) 2 P w f ( u, ξ ) 2 η s a ( u ) = = = ĝ(0 ) 1 2Wf( u,s ) s The scale representing a particular harmonic frequency of the signal 18
Absolute Coefficients Plot at 50Hz Detected Amplitude = 9.9995 Accuracy = 99.995% 19
Absolute Coefficients Plot at 100Hz Detected Amplitude = 5.0002 Accuracy = 99.996% 20
Detection of Adjacent Frequencies If a signal f is a sum of two sinusoids f (t ) = acos( ω t ) + acos( ω t ) 1 2 The analytic part fa of the signal f is f a ( t ) = ae iω1 t + ae iω2 t = a cos( ω1 + ω2 t 2 ω1 + ω ( i 2 t 2 The instantaneous frequency is the average of the two frequencies ' ω1 + ω2 φ ( t ) = 2 ) e ) 21
Illustrative Examples Signal = mix frequencies of 400Hz + 500Hz Detected frequency = 456.62Hz (400Hz + 500Hz)/2 = 450Hz 22
In frequency domain, the complex morlet wavelet is a bandpass filter centred at the centre frequency fc with bandwidth parameter fb To discriminate adjacent frequencies the bandwidth must be narrow It is estimated that to detect adjacent frequencies, fb and fc must satisfy the inequality f c f b 0.87 x f 2 f 2 + f 1 f 1 23
Denoising with Discrete Stationary Wavelet Transform (DSWT) apply to wavelet ridges for frequency detection apply to absolute coefficients for amplitude detection Symlet2 wavelet is used for both wavelet ridges and absolute coefficients Haar wavelet can be used for absolute coefficients 24
Without denoising Illustrative Examples (Wavelet Ridges) signal = 10cos(2π50t) + 5cos(2π100t-20 o ) Detected frequency = 50.1Hz Wavelet ridges Detected frequency = 100Hz 25
With denoising Illustrative Examples signal = 10cos(2π50t) + 5cos(2π100t-20 o ) Detected frequency = 50Hz Accuracy = 100% Wavelet Ridges Detected frequency = 100Hz Accuracy = 100% 26
Without denoising Illustrative Examples (Wavelet Coefficient) signal = 10cos(2π50t) + 5cos(2π100t-20 o ) Detected Amplitude = 10.0023 Accuracy = 99.977% Absolute Coefficient Plot for 50Hz frequency 27
Illustrative Examples (Wavelet Coefficient) signal = 10cos(2π50t) + 5cos(2π100t-20 o ) With denoising Detected Amplitude = 9.9995 Accuracy = 99.995% Absolute Coefficient Plot for 50Hz frequency 28
Illustrative Examples (Wavelet Coefficient) signal = 10cos(2π50t) + 5cos(2π100t-20 o ) Without denoising Detected Amplitude = 5.0067 Accuracy = 99.866% Absolute Coefficient Plot for 100Hz frequency 29
With denoising Illustrative Examples (Wavelet coefficient) signal = 10cos(2π50t) + 5cos(2π100t-20 o ) Detected Amplitude = 5.0002 Accuracy = 99.996% Absolute Coefficient Plot for 100Hz frequency 30
Conclusions The paper has presented a computational algorithm to estimate frequency contents of a signal together with the respective amplitudes The calculation results show that accuracy in frequency detection is 100% accuracy in amplitude detection is practically 100% 31
Further Works Initial phase difference detection Transient detection Increase the computational speed 32
Initial Phase (Phase Difference) Detection Given a signal f = a 1 cos( 2πf 1 t+θ 1 ) + a 2 cos( 2πf 2 t+θ 2 ) The instantaneous phase of the signal components are at time t is ph 1 = 2πf 1 t+θ 1 ph 2 = 2πf 2 t+θ 2 The instantaneous phase difference is given by ph 1 ph 2 = 2π(f 1 f 2 )t + (θ 1 -θ 2 ) changes over time t remains constant 33
Phase information by CWT with Complex Morlet Wavelet The phase angles represented by the wavelet coefficients are instantaneous phase angles which are dependent on the frequency of the signal component and change over time The initial phase angles (at time = 0) of the signal components are preserved by the instantaneous phase information of the wavelet coefficients 34
Accuracy in Initial Phase Angle Detection The accuracy in initial phase angle estimation is dependent on the sampling frequency A 50Hz signal sampled at 4000Hz would have an angular step of o 360 Sampling Frequency x Signal frequency = o 360 4000 x 50 = 4.5 o The higher the sampling frequency, the better the initial phase angle estimation 35
Illustrative Examples signal = 10cos(2π50t) + 5cos(2π100t-20 o ) Initial phase difference = 20 o sampling frequency = 4000Hz angular step for 50Hz = 4.5 o angular step for 100Hz = 9 o 36
Data Point Estimation Results Instantaneous Phase 50Hz Signal 100Hz Signal Initial Phase Difference 1501 1.61 rad -2.7133 rad 22.2932 deg 1502 1.5315 rad -2.8703 rad 22.2955 deg Error 2.2933 deg (11.47% ) 2.2955 deg (11.47% ) The error of 2.29 o is approx. half of the angular step size of the 50Hz signal, i.e. 4.5 o /2 = 2.25 o Take this 2.25 o into account, the error in initial phase difference detection can be reduced to 0.216% This estimation approach needs to be verified 37
Preliminary Study on Transient Detection Given a 50Hz signal containing a transient as shown 38
The 50Hz signal amplitude = 10 Signal Characteristics sampled at 2000Hz No. of data = 2000 The transient duration = 0.5msec magnitude = 14.1221 located at data no. 487 39
Wavelet Ridges Plot Detected frequency = 50Hz Detected Frequency = 1000Hz 40
Absolute Coefficients Plot of the Transient 41
THE END QUESTIONS ARE WELCOME 42