Fun With The Fourier Series

Transcription

1 oo = a Σ 0 + a n cos(2nπt/t) + b n sin(2nπt/t) n=1 Presented at: Santa Clara Valley Chapter IEEE EMC Society March 10, 1998 Franz Gisin Silicon Graphics Inc 2011 North Shoreline M/S 43L 946 Phone: (415) Fax: (415) gisinf@engr.sgi.com Fun With The Fourier Series

2 Time Frequency Domain Time Many of us are used to thinking in the time domain. Unfortunately, many EMC specification limits such as conducted and radiated emissions are in the frequency domain. f(ω) Limit Freq The Fourier Series is a method that allows us to travel back and forth between the time and frequency domains. oo = a Σ 0 + a n cos(2nπt/t) + b n sin(2nπt/t) n=1

3 o o o o Teacher Secret policeman Political prisoner Govenor of southern Egypt Baron Jean Babtiste Joseph Fourier o o Prefect of Isère and Rhône Friend of Napoleon o o Author of "Théorie Analytique de la Chaleur" Inventor of the definite integral symbol T/2 o Secretary of the Académie des Sciences T/2 C

4 Fourier s "Claim to Fame" oo π π oo In 1807, Fourier proposed that any bounded periodic function,, defined in the interval [ π, π] could be expressed as the trigonometric series: oo = Σ a n cos(nt) + b n sin(nt) n=0 where a n and b n are real numbers that represent the magnitudes of the corresponding sine and cosine terms.

5 Well Known Spectral Responses o Square Wave Only odd harmonics What kind of waveform produces only even harmonics? o Rectangular Pulse Train Sometimes envelope of harmonics looks like sinc function, sin(x)/x. o Trapezoidal with Equal Rise/Fall Times 20 db/decade What Happens in Here 40 db/decade Envelope of Maximums

6 Not So Well Known Spectral Responses o Trapezoidal with non equal rise/fall times A t r t f T/2 W/2 W/2 T/2 t o Waveforms with over/under shoots A T/2 W/2 W/2 T/2 t o Digital Differential Drivers Do they really cancel each other out? o Most spectrum analyzer displays Can a connection even be made between a Fourier series and a typical spectrum analyzer display?

7 "Fitting" an Equation to a Periodic Waveform oo oo 2T T 0 T 2T o o o Non periodic functions such as polynomials, log, and exp do not fit well since they all (with the exception of a polynomial constant) go to to infinity Periodic functions such as the tangent and cotangent functions also do not fit well because they periodically go to infinity Only the sine and cosine functions fit well since they are periodic and do not go to infinity

8 "Fitting" an Equation to a Periodic Waveform oo oo 2T T 0 T 2T A good start for a general (all encompassing) equation is the following: oo = a Σ 0 + a n cos(2nπt/t) + b n sin(2nπt/t) n=1 The only problem with this equation is it does not tell you how to go about finding the optimum values for a 0, a n and b n.

9 Math Products o Scalar (Times) Product scalar scalar = scalar 2 2 = 4 scalar vector = vector = o Outer (Cross or Vector) Product Only valid for vectors in 3 D space vector x vector = vector x 2 1 = o Inner (Dot) Product vector vector = scalar = = 20

10 Perpendicularity o If the inner product of two vectors is zero then the vectors are perpendicular to each other = ( 1) = 0 ( 1,2) (4,2) o The inner product can be used as a "filter" to "extract" the components of a vector that align with the desired axis = (1) 4 + (0) 2 = 4 2 (4,2) = (0) 4 + (1) 2 = 2 4

11 Orthogonality o Perpendicular vectors exhibit the property of orthogonality A = [a 0, a 1,..., a n ] B = [b 0, b 1,..., b n ] A B = <A,B> = a 0 b 0 +a 1 b a n b n = 0 o The concept of orthogonality can be extended to other mathematical entities such as the definite integral of algebraic expressions T/2 cos(2mπt/t) dt = 0 T/2 n Σ i=0 a i b i = 0

12 Finding a n Using an "Orthogonal Filter" oo = a Σ 0 + a n cos(2nπt/t) + b n sin(2nπt/t) n=1 T/2 cos(2mπt/t) dt = T/2 + Σ oo n=1 oo + Σ n=1 T/2 a 0 cos(2mπt/t) dt T/2 T/2 a n cos(2nπt/t) cos(2mπt/t) dt T/2 T/2 b n sin(2nπt/t) cos(2mπt/t) dt T/2 = 0 for all m = 0 for n <>m = a n T/2 for n=m = 0 for all n, m T/2 cos(2nπt/t) dt = a n T/2 T/2 Note: n = m

13 The Fourier Series and Associated Euler Formulas oo = a Σ 0 + a n cos(2nπt/t) + b n sin(2nπt/t) n=1 T/2 a 0 = 1 T a n = 2 T dt T/2 T/2 cos(2nπt/t) dt T/2 b n = 2 T T/2 sin(2nπt/t) dt T/2 Mag n = a n 2 + b n 2 θ n = Tan 1 (b n /a n ) hint: cos(nωt) = cos(n2πft) = (2nπt/T)

14 Other Forms of the Fourier Series oo = a Σ 0 + a n cos(2nπt/t) + b n sin(2nπt/t) n=1 = oo Σ a n cos(2nπt/t) + b n sin(2nπt/t) n=0 oo = a Σ 0 + c n cos(2nπt/t + φ n ) n=1 oo = a Σ 0 + c n sin(2nπt/t + θ n ) n=1 = n=+oo Σ g n ei2n πt/t n= oo

15 Rectangular Pulse A T/2 W/2 W/2 T/2 t oo = AW/T + Σ 2A/nπ sin(nπw/t) cos(2nπt/t) n=1 DC Term Harmonic Amplitude a n Frequency of Harmonic Note that there are no b n terms, and for most EMC applications, the DC term, a 0, is of no interest and so is usually ignored. Mag n = a n 2 + b n 2 = a n = a n θ n = Tan 1 (b n /a n ) = Tan 1 (0/a n ) = 0, π Cos(θ) = Cos(θ+π)

16 Rectangular Pulse A t T/2 W/2 W/2 T/2 a n = 2A/nπ sin(nπw/t) = 2AT/n²π²W sinc(nπw/t) 1 If a rectangular pulse produces a sinc shaped spectrum how come the spectrum of a 50% duty cycle pulse does not look like one???

17 1 Rectangular Pulse Only valid for integer n SA only displays absolute value Display magnitude in Log mode

18 Rectangular Pulse A T/2 W/2 W/2 T/2 a n = 2A/nπ sin(nπw/t) t a n is at a local "maximum" when sin(nπw/t) = 1 This occurs when nπw/t = (2m 1)(π/2), m = 1, 2,... or n = (m 1/2) (T/W) For a 50% duty cycle, W/T = 0.5 n = 1, 3, 5,... For a 25% duty cycle, W/T = 0.25 n = 2, 6, 10,... a n is at a local "minimum" when sin(nπw/t) = 0 This occurs when nπw/t = 2mπ, m = 1, 2,... or n = mt/w For a 50% duty cycle, W/T = 0.5 n = 2, 4, 6,... For a 25% duty cycle, W/T = 0.25 n = 4, 8, 12,...

19 Rectangular Pulses Freq (MHz) a n (dbµv) 50% a n (dbµv) 25% a n (dbµv) 10% Every 2nd One Every 4th One Every 10th One

20 Some Examples of Rectangular Pulses % Duty Cycle (every 5th harmonic =0) % Duty Cycle (every 10th harmonic =0) % Duty Cycle (every 100th harmonic =0)

21 What Goes Down Must Come Back Up For narrow rectangular pulses... a n = 2A/nπ sin(nπw/t) Lim sin(x) = x x 0 Lim [ a n ] = Lim [(2A/nπ) (nπw/t)] = (2AW/T) W 0 W 0 Only for very narrow pulse width do all harmonics decrease equally with duty cycle For not so narrow rectangular pulses... a n W [2A/nπ = sin(nπw/t)] = 2A/T W cos(nπw/t)] For not so narrow rectangular pulses the rate of increase of low level harmonics is greater than the rate of decrease of high level harmonics

22 What Goes Down Must Come Back Up % Duty Cycle Pulse % Duty Cycle Pulse

23 What Goes Down Must Come Back Up Freq (MHz) a n (dbµv) 50% a n (dbµv) 25% a n (dbµv) 10% Very Little Change Very Big Change

24 Symmetrical 1 MHz Trapezoidal Pulses A t r = tr f = t f T/2 W/2 W/2 T/2 t 2AT nπtrf nπw a n = sin( ) sin( ) n 2 π2t rf T T Freq (MHz) a n (dbµv) t rf = 0 a n (dbµv) t rf = 1 nsec a n (dbµv) t rf = 5 nsec These go up while these go down

25 Symmetrical 1 MHz Trapezoidal Pulses t r /t f = 0/0 nsec t r /t f = 1/1 nsec t r /t f = 5/5 nsec

26 Asymmetrical 1 MHz Trapezoidal Pulses A t r t f T/2 W/2 W/2 T/2 t Freq (MHz) Mag (dbµv) t r /t f = 1/1 ns Mag (dbµv) t r /t f = 1/10 ns Mag (dbµv) t r /t f = 10/10 ns Only very high frequencies are substantially affected

27 Asymmetrical 1 MHz Trapezoidal Pulses t r /t f = 1/1 nsec t r /t f = 1/10 nsec t r /t f = 10/10 nsec

28 Trapezoidal Pulses With Over/Under Shoots 140 A t r t f T/2 W/2 W/2 T/2 t A 120 t 100 T/2 W/2 W/2 T/

29 Rectangular Pulse A T/2 W/2 W/2 T/2 t oo = AW/T + Σ 2A/nπ sin(nπw/t) cos(2nπt/t) n=1 DC Term Harmonic Amplitude a n Frequency of Harmonic A What if the waveform was not centered about t = 0?? A T/2 W/2 W/2 T/2 t T/2 W/2 W/2 T/2 t

30 A Little Black Magic!! dbµ_ 114 dbµ_ 0 dbµ_ 114 dbµ_ 120 dbµ_ Mag of b n in direction θ = π/2 Mag of a n,b n in direction θ = 3π/2 Throw away anything less than 120 db 20Log(b n ): θ = π/2 20Log(a n ): θ = 0 20Log( a n ): θ = π 20Log( b n ): θ = 3π/2 n

31 Offset Rectangular Pulse oo = a Σ 0 + a n cos(2nπt/t) + b n sin(2nπt/t) n=1 sin(x) = cos(x+π/2) sin(x) = sin(x+π) 20Log( a n ) 20Log(b n ) 20Log(a n ) 20Log( b n ) n A T/2 W/2 W/2 T/2 t A T/2 W/2 W/2 T/2 t A T/2 W/2 W/2 T/2 t

32 Harmonic Analysis of Differential Signals Consider a differential signal pair [Sig +, Sig ] represented by the Fourier Series [, ]. Sig + = Sig = n = = Σ c n e jnω 0t n = n = = Σ c n e jnω 0t n = If we ignore for the moment the fact that the two conductors are not physically located in the same space, one can intuitively see that the two signals should "cancel each other out". n = = Σ c n e jnω 0 t c n e jnω 0t = 0 n =

33 Harmonic Analysis of Differential Signals If, however, the Sig signal is delayed in time by a small value, τ, (for example if Sig is derived from Sig+ via an inverter, or the propagation time for Sig+ and Sig is not the same) there will not be complete cancellation. n = = Σ c n e jnω 0t n = n = f(t τ) = Σ c n e jnω 0(t τ) n = n = f(t τ) = Σ c n e jnω 0 t c n e jnω 0(t τ) n = n = = Σ c n e jnω 0 t c n e jnω 0t e jnω 0 τ n = n = = Σ c n e jnω 0t (1 e jnω 0 τ ) n = When e jnω 0 τ = 1, Sig + /Sig will cancel. When e jnω 0 τ = 1, Sig + /Sig will be additive. For other values of e jnω 0 τ, Sig + /Sig will partially cancel.

34 Harmonic Analysis of Differential Signals 1 e jnω 0 τ = 1 cos(nω 0 τ) + j sin(nω 0 τ) = 1 cos(2nπf 0 τ) + j sin(2nπf 0 τ) [ω 0 = 2πf 0 ] = 1 cos([nτ/t]2π) + j sin([nτ/t]2π) [f 0 = 1/T] For nτ/t = 0, 1, 2, 3,... cos([nτ/t]2π) = 1, sin([nτ/t]2π) = 0, 1 e jnω 0 τ = 0 For nτ/t = 1/2, 3/2, 5/2,... cos([nτ/t]2π) = 1, sin([nτ/t]2π) = 0, 1 e jnω 0 τ = j Magnitute of 1 e jnω 0 τ [nτ/t]2π 0 + 0j 2 + 0j 0 j

35 Harmonic Analysis of Differential Signals n = f(t τ) = Σ c n e jnω 0 t c n e jnω 0(t τ) n = n = = Σ c n e jnω 0t (1 e jnω 0 τ ) n = nω 0 τ Mag[1 e jnω 0 τ ] 20log 10 (Mag[1 e jnω 0 τ ]) db db db db db db db db If Sig shifts by 180 then Sig is in phase with Sig +

36 Harmonic Analysis of Differential Signals Freq T 5 [ 20 db] 30 [ 6 db] 60 [0 db] (MHz) (nsec) (psec) (psec) (psec) Highest FCC measurement frequency if 108 MHz < F 0 < 500 MHz If the harmonic at 2000 MHz is delayed by more than 83.3 psec, then Sig + and Sig no longer cancel each other out. For typical PCB prop delay = 175 psec/in 6.9 psec 0.04 in (routing s around connectors) 88.3 psec 0.5 in 5 nsec (inverter) 29 in

37 Harmonic Analysis of Differential Signals P 1 t + pd = t pd Sig + and Sig arrive at the same time (cancelling each other out) Sig + Trace Sig + Trace P 2 t + pd t pd Sig + and Sig do not arrive at the same time (complete cancellation does not take place) Propagation velocity in air 113 psec/in. (0.10 in connector pin spacing = 11.3 psec delay.)

38 Asymmetrical 1 MHz Trapezoidal Pulses A t ra t fa T/2 W/2 W/2 T/2 t t ra = t fb t fa = t rb A t fb t rb T/2 W/2 W/2 T/2 t Result f = 1 MHz /T = 1 usec, tr = 2 nsec, tf = 8 nsec, A = 1,W = 0.5 nsec

39 Asymmetrical 1 MHz Trapezoidal Pulses A t ra t fa T/2 W/2 W/2 T/2 t t ra = t rb t fa = t fb A t fb t rb T/2 W/2 W/2 T/2 t Viewed on end

40 Asymmetrical 1 MHz Trapezoidal Pulses A t ra t fa T/2 W/2 W/2 T/2 t t ra = t rb t fa = t fb A t fb t rb T/2 W/2 W/2 T/2 t Result

41 Asymmetrical 1 MHz Trapezoidal Pulses Viewed on End (500 Harmonics)...

42 Asymmetrical 1 MHz Trapezoidal Pulses Resultant Levels (500 Harmonics)...